bug fixed,
remaining issues in mobilenavbar need pruning, need rewrite the class.
This commit is contained in:
Trance-0
@@ -1,123 +0,0 @@
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# Math4201 Topology I (Lecture 10)
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## Continuity
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### Continuous functions
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Let $X,Y$ be topological spaces and $f:X\to Y$. For any $x\in X$ and any open neighborhood $V$ of $f(x)$ in $Y$, $f^{-1}(V)$ contains an open neighborhood of $x$ in $X$.
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#### Lemma for continuous functions
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Let $f:X\to Y$ be a function, then:
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1. $A\subseteq Y$: $f^{-1}(A^c) = (f^{-1}(A))^c$.
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2. $\{A_\alpha\}_{\alpha\in I}\subseteq Y$: $f^{-1}(\bigcup_{\alpha\in I} A_\alpha) = \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$.
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3. $\{A_\alpha\}_{\alpha\in I}\subseteq Y$: $f^{-1}(\bigcap_{\alpha\in I} A_\alpha) = \bigcap_{\alpha\in I} f^{-1}(A_\alpha)$.
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<details>
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<summary>Proof</summary>
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1. By definition of continuous functions, $\forall V$ open in $Y$, $f^{-1}(V)$ is open in $X$.
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2. It is sufficient to shoa that $x\in f^{-1}(\bigcup_{\alpha\in I} A_\alpha)$ if and only if $x\in \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$.
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This condition holds if and only if $\exists \alpha\in I$ such that $f(x)\in A_\alpha$.
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Which is equivalent to $\exists \alpha\in I$ such that $x\in f^{-1}(A_\alpha)$.
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So $x\in f^{-1}(\bigcup_{\alpha\in I} A_\alpha)$
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In particular, $f^{-1}(\bigcup_{\alpha\in I} A_\alpha) = \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$.
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3. Similar to 2 but use forall.
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</details>
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#### Properties of continuous functions
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A function $f:X\to Y$ is continuous if and only if:
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1. $f^{-1}(V)$ is open in $X$ for any open set $V\subset Y$.
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2. $f$ is continuous at any point $x\in X$.
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3. $f^{-1}(C)$ is closed in $X$ for any closed set $C\subset Y$.
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4. Assume $\mathcal{B}$ is a basis for $Y$, then $f^{-1}(\mathcal{B})$ is open in $X$ for any $B\in \mathcal{B}$.
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5. For any $A\subseteq X$, $f(\overline{A})\subseteq \overline{f(A)}$.
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<details>
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<summary>Proof</summary>
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**Showing $1\iff 3$**:
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> Use the lemma for continuous functions (1)
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**Showing $1\iff 4$**:
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$1 \implies 4$:
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Because any $B\in \mathcal{B}$ is open in $Y$, so $f^{-1}(B)$ is open in $X$.
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$4 \implies 1$:
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Let $V\subset Y$ be an open set. Then there are basis elements $\{B_\alpha\}_{\alpha\in I}$ such that $V=\bigcup_{\alpha\in I} B_\alpha$.
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So $f^{-1}(V) = f^{-1}(\bigcup_{\alpha\in I} B_\alpha) = \bigcup_{\alpha\in I} f^{-1}(B_\alpha)$ (by lemma (2)) is a union of open sets, so $f^{-1}(V)$ is open in $X$.
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**Showing $1\implies 5$**:
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Take $A\subseteq X$ and $x\in \overline{A}$. It suffices to show $f(x)$ is an element of the closure of $f(A)$. This is equivalent to say that any open neighborhood $V$ of $f(x)$ intersects $f(A)$ has a non-trivial intersection with $f(A)$.
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For any such $V$, 1 implies that $f^{-1}(V)$ is open in $X$. Moreover, $x\in f^{-1}(V)$ because $f(x)\in V$.
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This means that $f^{-1}(V)$ is an open neighborhood of $x$. Since $x\in \overline{A}$, we have $f^{-1}(V)\cap A\neq \emptyset$ and contains a point $x'\in X$.
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So $x'\in f^{-1}(V)\cap A$, this implies that $f(x')\in V$ and $f(x')\in f(A)$, so $f(x')\in V\cap f(A)$.
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> [!NOTE]
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>
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> This verifies our claim. Proof of $5\implies 1$ is similar and left as an exercise.
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</details>
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<details>
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<summary>Example of property 5</summary>
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Let $X=(0,1)\cup (1,2)$ and $Y=\mathbb{R}$ equipped with the subspace topology induced by the standard topology on $\mathbb{R}$.
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Let $f:X\to Y$ be the inclusion map, $f(x)=x$ for all $x\in X$. This is continuous.
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Let $A=(0,1)\cup (1,2)$. Then $\overline{A}=A$. So $f(\overline{A})=f(A)=(0,1)\cup (1,2)$.
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However, $\overline{f(A)}=\overline{(0,1)\cup (1,2)}=[0,2]$.
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So $f(\overline{A})\subsetneq \overline{f(A)}$.
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</details>
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#### Definition of homeomorphism
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A **homeomorphism** $f:X\to Y$ is a continuous map of topological spaces that is a bijection and $f^{-1}:Y\to X$ is also continuous.
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<details>
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<summary>Example of homeomorphism</summary>
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Let $X=\mathbb{R}$ and $Y=\mathbb{R}+$ with standard topology.
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$f:\mathbb{R}\to \mathbb{R}^+$ be defined by $f(x)=e^x$ is continuous and bijective.
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$f^{-1}:\mathbb{R}^+\to \mathbb{R}$ be defined by $f^{-1}(y)=\ln(y)$ is continuous and homeomorphism.
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</details>
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### Epsilon delta definition of continuity
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Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function where we use the standard topology on $\mathbb{R}$.
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Then [property 4](#properties-of-continuous-functions) implies that for any open interval $(a,b)\in \mathbb{R}$, $f^{-1}((a,b))$ is open in $\mathbb{R}$.
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Now take an arbitrary $x\in \mathbb{R}$ and $\epsilon > 0$. In particular $f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$ is an open set containing $x$.
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In particular, there is an open interval (by the standard topology on $\mathbb{R}$) $(c,d)$ such that $x\in (c,d)\subseteq f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$.
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Let $\delta = \min\{x-c, d-x\}$. Then $(x-\delta, x+\delta)\subseteq (c,d)\subseteq f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$.
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This says that if $|y-x| < \delta$, then $|f(y)-f(x)| < \epsilon$.
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@@ -1,113 +0,0 @@
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# Math4201 Topology I (Lecture 11)
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> [!NOTE]
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>
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> Q: Let $f:X\to Y$ be a continuous bijection. Is it true that $f^{-1}$ is continuous?
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>
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> A: No. Consider $X=[0,2\pi)$ and $Y=\mathbb{S}^1$ with standard topology in $\mathbb{R}^2$.
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>
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> Let $f\coloneqq \theta\in [0,2\pi)\to (\cos \theta, \sin \theta)\in \mathbb{S}^1$ is a continuous bijection. ($\forall f^{-1}(V)$ is open in $X$)
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>
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> But $f^{-1}$ is not continuous, consider the open set in $X, U=[0,\pi)$. Then $f^{-1}(U)=[0,\pi)$ is not open in $Y$.
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## Continuous functions
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### Constructing continuous functions
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#### Theorem composition of continuous functions is continuous
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Let $X,Y,Z$ be topological spaces, $f:X\to Y$ is continuous, and $g:Y\to Z$ is continuous. Then $f\circ g:X\to Z$ is continuous.
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<details>
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<summary>Proof</summary>
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Let $U\subseteq Z$ be open. Then $g^{-1}(U)$ is open in $Y$. Since $f$ is continuous, $f^{-1}(g^{-1}(U))$ is open in $X$.
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</details>
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#### Pasting lemma
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Let $X$ be a topological space and $X=Z_1\cup Z_2$ with $Z_1,Z_2$ closed in $X$ equipped with the subspace topology. (may be not disjoint)
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Let $g_1:Z_1\to Y$ and $g_2:Z_2\to Y$ be two continuous maps and $\forall x\in Z_1\cap Z_2$, $g_1(x)=g_2(x)$.
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Define $f:X\to Y$ by $f(x)\begin{cases}g_1(x), & x\in Z_1 \\ g_2(x), & x\in Z_2\end{cases}$ is continuous.
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<details>
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<summary>Proof</summary>
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Let $U\subseteq Y$ be open. Then $f^{-1}(U)=g_1^{-1}(U)\cup g_2^{-1}(U)$.
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$g_1^{-1}(U)$ and $g_2^{-1}(U)$ are open in $Z_1$ and $Z_2$ respectively.
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> It's a bit annoying to show that $g_1^{-1}(U)$ and $g_2^{-1}(U)$ are open in $X$.
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Different way. Consider the definition of continuous functions using closed sets.
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If $W\subseteq X$ is closed, then $W=Z_1\cap Z_2$ is closed in $X$.
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So $f^{-1}(W)=g_1^{-1}(W)\cup g_2^{-1}(W)$ is closed in $Z_1$ and $Z_2$ respectively.
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Note that $Z_1$ and $Z_2$ are closed in $X$, so $g_1^{-1}(W)$ and $g_2^{-1}(W)$ are closed in $X$. [closed in closed subspace lemma](https://notenextra.trance-0.com/Math4201/Math4201_L7#lemma-of-closed-in-closed-subspace)
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So $f^{-1}(W)$ is closed in $X$.
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</details>
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Let $X$ be a topological space and $X=U_1\cup U_2$ with $U_1,U_2$ open in $X$ equipped with the subspace topology.
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With $g_1:U_1\to Y$ and $g_2:U_2\to Y$ be two continuous maps and $\forall x\in U_1\cap U_2$, $g_1(x)=g_2(x)$.
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Then $f:X\to Y$ by $f(x)\begin{cases}g_1(x), & x\in U_1 \\ g_2(x), & x\in U_2\end{cases}$ is continuous.
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<details>
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<summary>Proof</summary>
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Let $U\subseteq Y$ be open. Then $f^{-1}(U)=g_1^{-1}(U)\cup g_2^{-1}(U)$.
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$g_1^{-1}(U)$ and $g_2^{-1}(U)$ are open in $U_1$ and $U_2$ respectively.
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Apply the [open in open subspace lemma](https://notenextra.trance-0.com/Math4201/Math4201_L6#lemma-of-open-set-in-subspace-topology)
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So $f^{-1}(U)$ is open in $X$.
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</details>
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The open set version holds more generally.
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Let $X$ be a topological space and $X=\bigcup_{\alpha\in I} U_\alpha$ with $U_\alpha$ open in $X$ equipped with the subspace topology.
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Let $g_\alpha:U_\alpha\to Y$ be two continuous maps and $\forall x\in U_\alpha\cap U_\beta$, $g_\alpha(x)=g_\beta(x)$.
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Then $f:X\to Y$ by $f(x)=g_\alpha(x), \text{if } x\in U_\alpha$ is continuous.
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#### Continuous functions on different codomains
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Let $f:X\to Y$ and $g:X\to Z$ be two continuous maps of topological spaces.
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Let $H:X\to Y\times Z$, where $Y\times Z$ is equipped with the product topology, be defined by $H(x)=(f(x),g(x))$. Then $H$ is continuous.
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> A stronger version of this theorem is that $f:X\to Y$ and $g:X\to Z$ are continuous maps of topological spaces if and only if $H:X\to Y\times Z$ is continuous.
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<details>
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<summary>Proof</summary>
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It is sufficient to check the basis elements of the topology on $Y\times Z$.
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The basis for the topology on $Y\times Z$ is $U\times V\subseteq Y\times Z$, where $U\subseteq Y$ and $V\subseteq Z$ are open. This form a basis for the topology on $Y\times Z$.
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We only need to show that $H^{-1}(U\times V)$ is open in $X$.
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Let $H^{-1}(U\times V)=\{x\in X | (f(x),g(x))\in U\times V\}$.
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So $H^{-1}(U\times V)=f^{-1}(U)\cap g^{-1}(V)$.
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Since $f$ and $g$ are continuous, $f^{-1}(U)$ and $g^{-1}(V)$ are open in $X$.
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So $H^{-1}(U\times V)$ is open in $X$.
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</details>
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Exercise: Prove the stronger version of the theorem,
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If $H:X\to Y\times Z$ is continuous, then $f:X\to Y$ and $g:X\to Z$ are continuous.
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@@ -1,180 +0,0 @@
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# Math4201 Topology I (Lecture 12)
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## Metric spaces
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### Basic properties and definitions
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#### Definition of metric space
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A metric space is a set $X$ with a function $d:X\times X\to \mathbb{R}$ that satisfies the following properties:
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1. $\forall x,y\in X, d(x,y)\geq 0$ and $d(x,y)=0$ if and only if $x=y$. (positivity)
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2. $\forall x,y\in X, d(x,y)=d(y,x)$. (symmetry)
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3. $\forall x,y,z\in X, d(x,z)\leq d(x,y)+d(y,z)$. (triangle inequality)
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<details>
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<summary>Example of metric space</summary>
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Let $X=\mathbb{R}$ and $d(x,y)=|x-y|$.
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Check definition of metric space:
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1. Positivity: $d(x,y)=|x-y|\geq 0$ and $d(x,y)=0$ if and only if $x=y$.
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2. Symmetry: $d(x,y)=|x-y|=|y-x|=d(y,x)$.
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3. Triangle inequality: $d(x,z)=|x-z|\leq |x-y|+|y-z|=d(x,y)+d(y,z)$ since $|a+b|\leq |a|+|b|$ for all $a,b\in \mathbb{R}$.
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---
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Let $X$ be arbitrary. The trivial metric is $d(x,y)=\begin{cases}
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0 & \text{if } x=y \\
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1 & \text{if } x\neq y
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\end{cases}$
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Check definition of metric space:
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1. Positivity: $d(x,y)=\begin{cases}
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0 & \text{if } x=y \\
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1 & \text{if } x\neq y
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\end{cases}\geq 0$ and $d(x,y)=0$ if and only if $x=y$.
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1. Symmetry: $d(x,y)=\begin{cases}
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0 & \text{if } x=y \\
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1 & \text{if } x\neq y
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\end{cases}=d(y,x)$.
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1. Triangle inequality use case by case analysis.
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</details>
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#### Balls of a metric space forms a basis for a topology
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Let $(X,d)$ be a metric space. $x\in X$ and $r>0, r\in \mathbb{R}$. We define the ball of radius $r$ centered at $x$ as $B_r(x)=\{y\in X:d(x,y)<r\}$.
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Goal: Show that the balls of a metric space forms a basis for a topology.
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$$
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\{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}\text{ is a basis for a topology on }X
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$$
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<details>
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<summary>Example of balls of a metric space</summary>
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Let $X=\mathbb{R}$ and $d(x,y)=\begin{cases}
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0 & \text{if } x=y \\
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1 & \text{if } x\neq y
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\end{cases}$
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The balls of this metric space are:
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$$
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B_r(x)=\begin{cases}
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\{x\} & \text{if } r<1 \\
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X & \text{if } r\geq 1
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\end{cases}
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$$
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> [!NOTE]
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>
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> This basis generate the discrete topology of $X$.
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---
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Let $X=\mathbb{R}$ and $d(x,y)=|x-y|$.
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The balls of this metric space are:
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$$
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B_r(x)=\{(x-r,x+r)\}
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$$
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This basis is the set of all open sets in $\mathbb{R}$, which generates the standard topology of $\mathbb{R}$.
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</details>
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<details>
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<summary>Proof</summary>
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Let's check the two properties of basis:
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1. $\forall x\in X$, $\exists B_r(x)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}$ such that $x\in B_r(x)$. (Trivial by definition of non-zero radius ball)
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2. $\forall B_r(x),B_r(y)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}$, $\forall z\in B_r(x)\cap B_r(y)$, $\exists B_r(z)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}$ such that $z\in B_r(z)\subseteq B_r(x)\cap B_r(y)$.
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Observe that for any $z\in B_r(x)$, then there exists $\delta>0$ such that $B_\delta(z)\subseteq B_r(x)$.
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Let $\delta=r-d(x,z)$, then $B_\delta(z)\subseteq B_r(x)$ (by triangle inequality)
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Similarly, there exists $\delta'>0$ such that $B_\delta'(z)\subseteq B_r(y)$.
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Take $\lambda=min\{\delta,\delta'\}$, then $B_\lambda(z)\subseteq B_r(x)\cap B_r(y)$.
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</details>
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#### Definition of Metric topology
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For any metric space $(X,d)$, the topology generated by the balls of the metric space is called metric topology.
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#### Definition of metrizable
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A topological space $(X,\mathcal{T})$ is metrizable if it is the metric topology for some metric $d$ on $X$.
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> Q: When is a topological space metrizable?
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#### Lemma: Every metric topology is Hausdorff
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If a topology isn't Hausdorff, then it isn't metrizable.
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<details>
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<summary>Example of non-metrizable space</summary>
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Trivial topology **with at least two points** is not Hausdorff, so it isn't metrizable.
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---
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Finite complement topology on infinite set is not Hausdorff.
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Suppose there exists $x,y\in X$ such that $x\neq y$ and $x\in U\subseteq X$ and $y\in V\subseteq X$ such that $X-U$ and $X-V$ are finite.
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Since $U\cap V=\emptyset$, we have $V\subseteq X-U$, which is finite. So $X-V$ is infinite. (contradiction that $X-V$ is finite)
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So $X$ with finite complement topology is not Hausdorff, so it isn't metrizable.
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</details>
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<details>
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<summary>Proof</summary>
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Let $x,y\in (X,d)$ and $x\neq y$. To show that $X$ is Hausdorff, it is suffices to show that there exists $r,r'>0$ such that $B_r(x)\cap B_r'(y)=\emptyset$.
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Take $r=r'=\frac{1}{2}d(x,y)$, then $B_r(x)\cap B_r'(y)=\emptyset$. (by triangle inequality)
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We prove this by contradiction.
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Suppose $\exists z\in B_r(x)\cap B_r'(y)$, then $d(x,z)<r$ and $d(y,z)<r'$.
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|
||||
Then $d(x,y)\leq d(x,z)+d(z,y)<r+r'=\frac{1}{2}d(x,y)+\frac{1}{2}d(x,y)=d(x,y)$. (contradiction $d(x,y)<d(x,y)$)
|
||||
|
||||
Therefore, $X$ is Hausdorff.
|
||||
|
||||
</details>
|
||||
|
||||
### Other metrics on $\mathbb{R}^n$
|
||||
|
||||
Let $\mathbb{R}^n$ be the set of all $n$-tuples of real numbers with standard topology.
|
||||
|
||||
Let $d: \mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$ be defined by (the Euclidean distance)
|
||||
|
||||
$$
|
||||
d(u,v)=\sqrt{\sum_{i=1}^n (u_i-v_i)^2}
|
||||
$$
|
||||
|
||||
In $\mathbb{R}^2$ the ball is a circle.
|
||||
|
||||
Let $\rho(u,v)=\max_{i=1}^n |u_i-v_i|$. (Square metric)
|
||||
|
||||
In $\mathbb{R}^2$ the ball is a square.
|
||||
|
||||
Let $m(u,v)=\sum_{i=1}^n |u_i-v_i|$. (Manhattan metric)
|
||||
|
||||
In $\mathbb{R}^2$ the ball is a diamond.
|
||||
|
||||
#### Lemma: Square metric, Manhattan metric, and Euclidean metric are well defined metrics on $\mathbb{R}^n$
|
||||
|
||||
Proof ignored. Hard part is to show the triangle inequality. May use Cauchy-Schwarz inequality.
|
||||
@@ -1,159 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 13)
|
||||
|
||||
## Metic spaces
|
||||
|
||||
### Three different metrics on $\mathbb{R}^n$
|
||||
|
||||
Euclidean metric:
|
||||
|
||||
$$
|
||||
d(x,y)=\sqrt{\sum_{i=1}^n (x_i-y_i)^2}
|
||||
$$
|
||||
|
||||
Square metric:
|
||||
|
||||
$$
|
||||
d(x,y)=\max_{i=1}^n |x_i-y_i|
|
||||
$$
|
||||
|
||||
Manhattan metric:
|
||||
|
||||
$$
|
||||
d(x,y)=\sum_{i=1}^n |x_i-y_i|
|
||||
$$
|
||||
|
||||
So to prove our proposition, we need to show that any pair of metrics $d$ and $d'$ with basis generated by balls defined
|
||||
|
||||
$$
|
||||
\mathcal{B}=\{B_r^{(d)}(x)|x\in X,r>0,r\in \mathbb{R}\}
|
||||
$$
|
||||
|
||||
and
|
||||
|
||||
$$
|
||||
\mathcal{B}'=\{B_r^{(d')}(x)|x\in X,r>0,r\in \mathbb{R}\}
|
||||
$$
|
||||
|
||||
are equivalent.
|
||||
|
||||
#### Proposition: The metrics induce the same topology on $\mathbb{R}^n$
|
||||
|
||||
The three metrics induce the same topology on $\mathbb{R}^n$, and it's the standard topology.
|
||||
|
||||
#### Lemma of equivalent topologies
|
||||
|
||||
If $\mathcal{T}$ and $\mathcal{T}'$ are two topologies on $X$, we say $\mathcal{T}$ and $\mathcal{T}'$ are equivalent to each other if and only if the following two conditions are satisfied:
|
||||
|
||||
1. $\forall B_1\in \mathcal{T}, \exists B_2\in \mathcal{T}'$ such that $\forall x\in B_1, \exists x\in B_2\subseteq B_1$.
|
||||
2. $\forall B_2\in \mathcal{T}', \exists B_1\in \mathcal{T}$ such that $\forall x\in B_2, \exists x\in B_1\subseteq B_2$.
|
||||
|
||||
#### Lemma of equivalent metrics
|
||||
|
||||
Let $d$ and $d'$ be two metrics on $X$. If the following holds, then the metric topology associated to $d$ and $d'$ are equivalent.
|
||||
|
||||
1. $\forall x\in X, \forall \delta>0, \exists \epsilon>0$ such that $B_\delta(x)\subseteq B_\epsilon(x)$
|
||||
2. $\forall x\in X, \forall \epsilon>0, \exists \delta>0$ such that $B_\epsilon(x)\subseteq B_\delta(x)$
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
To apply the lemma, we try to compute the three metrics on $\mathbb{R}^n$.
|
||||
|
||||
$u=(u_1,u_2,\dots,u_n), v=(v_1,v_2,\dots,v_n)\in \mathbb{R}^n$
|
||||
|
||||
For Euclidean metric:
|
||||
|
||||
$$
|
||||
d(u,v)=\sqrt{\sum_{i=1}^n (u_i-v_i)^2}
|
||||
$$
|
||||
|
||||
For square metric:
|
||||
|
||||
$$
|
||||
\rho(u,v)=\max_{i=1}^n |u_i-v_i|
|
||||
$$
|
||||
|
||||
For Manhattan metric:
|
||||
|
||||
$$
|
||||
m(u,v)=\sum_{i=1}^n |u_i-v_i|
|
||||
$$
|
||||
|
||||
**First** we will show that $d$ and $\rho$ are equivalent.
|
||||
|
||||
Note that
|
||||
|
||||
$$
|
||||
\max_{i=1}^n |u_i-v_i|\leq \sqrt{\sum_{i=1}^n (u_i-v_i)^2}
|
||||
$$
|
||||
|
||||
So $\forall u\in B_r^{(d)}(x), d(u,v)<r\implies \rho(u,v)<r$
|
||||
|
||||
So $u\in B_r^{(\rho)}(x)$.
|
||||
|
||||
$$
|
||||
B_r^{(d)}(u)\subseteq B_r^{(\rho)}(u)
|
||||
$$
|
||||
|
||||
Note that
|
||||
|
||||
$$
|
||||
\sqrt{\sum_{i=1}^n (u_i-v_i)^2}\leq \sqrt{\sum_{i=1}^n max\{|u_i-v_i|\}^2}=\sqrt{n}\times max\{|u_i-v_i|\}
|
||||
$$
|
||||
|
||||
So $\forall u\in B_{r/\sqrt{n}}^{(\rho)}(x), \rho(u,v)<r/\sqrt{n}\implies d(u,v)<r$
|
||||
|
||||
So $u\in B_r^{(d)}(x)$.
|
||||
|
||||
So $B_{r/\sqrt{n}}^{(\rho)}(x)\subseteq B_r^{(d)}(x)$.
|
||||
|
||||
> imagine two square capped circle inside
|
||||
|
||||
**Then we will show** that $\rho$ and $m$ are equivalent.
|
||||
|
||||
Observing that
|
||||
|
||||
$$
|
||||
\max_{i=1}^n |u_i-v_i|\leq \sum_{i=1}^n |u_i-v_i|\leq n\times \max_{i=1}^n |u_i-v_i|
|
||||
$$
|
||||
|
||||
Then we have
|
||||
|
||||
$$
|
||||
B_{r/n}^{(\rho)}(x)\subseteq B_r^{(m)}(x)\subseteq B_n^{(\rho)}(x)
|
||||
$$
|
||||
|
||||
> imagine two square capped a diamonds inside
|
||||
|
||||
**Finally**, we will show that the topology generated by the square metric is the same as the product topology on $\mathbb{R}^n$.
|
||||
|
||||
Recall the basis for the product topology on $\mathbb{R}^n$ with standard topology.
|
||||
|
||||
$$
|
||||
\mathcal{B}=\{(a_1,b_1)\times (a_2,b_2)\times \cdots \times (a_n,b_n)|a_i,b_i\in \mathbb{R},a_i<b_i\}
|
||||
$$
|
||||
|
||||
Let $x=(x_1,x_2,\dots,x_n)\in (a_1,b_1)\times (a_2,b_2)\times \cdots \times (a_n,b_n)$.
|
||||
|
||||
Let $\delta=\min_{i=1}^n \{|x_i-a_i|,|x_i-b_i|\}$.
|
||||
|
||||
Then $x\in B_\delta^{(\rho)}(x)\subseteq (a_1,b_1)\times (a_2,b_2)\times \cdots \times (a_n,b_n)$.
|
||||
|
||||
In the other direction, let $x\in B_r^{(\rho)}(x)$, $x=(x_1,x_2,\dots,x_n)$.
|
||||
|
||||
Then $B_r^{(\rho)}(x)\subseteq (x_1-r,x_1+r)\times (x_2-r,x_2+r)\times \cdots \times (x_n-r,x_n+r)$.
|
||||
|
||||
This is an element of $\mathcal{B}$, by combining the two directions, we have the standard topology is the same as the topology generated by the square metric.
|
||||
|
||||
</details>
|
||||
|
||||
#### Proposition of metric induced product topology
|
||||
|
||||
Let $(X,d),(Y,d')$ be two metric spaces with metric topology $\mathcal{T},\mathcal{T}'$. On $X\times Y$, we can define a metric $\rho$ by $\rho((x,y),(x',y'))\coloneqq \max\{d(x,x'),d'(y,y')\}$, $(x,y),(x',y')\in X\times Y$.
|
||||
|
||||
Then this metric topology on $X\times Y$ is the same as the product topology on $X\times Y$.
|
||||
|
||||
> [!NOTE]
|
||||
>
|
||||
> Product of metrizable topological spaces is metrizable.
|
||||
|
||||
@@ -1,113 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 14)
|
||||
|
||||
## Metric topology
|
||||
|
||||
### Product topology and metric topology
|
||||
|
||||
If $X$ and $Y$ are metrizable spaces, then the product space $X\times Y$ is metrizable.
|
||||
|
||||
If $X$ is metrizable, then the subspace $A\subset X$ equipped with subspace topology is metrizable.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Let $d$ be a metric on $X$. Then define $d'$ be the restriction of $d$ to $A$:
|
||||
|
||||
$$
|
||||
d':A\times A\to \mathbb{R}+
|
||||
$$
|
||||
|
||||
$$
|
||||
d'(x,y)=d(x,y)
|
||||
$$
|
||||
|
||||
$x,y\in A\subseteq X$
|
||||
|
||||
$d'$ is a metric on $A$. Since the metric topology on $A$ associated to $d'$ is the same as the subspace topology.
|
||||
|
||||
Note that for any $x\in A$ and $r>0$
|
||||
|
||||
$$
|
||||
B_r^{d}(x)\cap A=B_r^{d'}(x)\tag{*}
|
||||
$$
|
||||
|
||||
A basis for metric topology on $A$ is given by:
|
||||
|
||||
$$
|
||||
\mathcal{B}=\{B_r^{d'}(x)|x\in A,r>0,r\in \mathbb{R}\}
|
||||
$$
|
||||
|
||||
A basis for the subspace topology on $A$ is given by:
|
||||
|
||||
$$
|
||||
\mathcal{B}'=\{B_r^{d}(x)\cap A|x\in A,r>0,r\in \mathbb{R}\}
|
||||
$$
|
||||
|
||||
Since (*) holds, $\mathcal{B}\subseteq \mathcal{B}'$.
|
||||
|
||||
This shows that subspace topology on $A$ is finer than the metric topology on $A$.
|
||||
|
||||
We need to show that for any $B_r^{d}(x)$ with $x\in X$ and $y\in B_r^{d}(x)\cap A$, we have $r'>0$ such that $y\in B_r^{d'}(y)\subseteq B_r^{d}(x)\cap A$.
|
||||
|
||||
Use triangle inequality, we have $r'=r-d(x,y)>0$ such that $y\in B_r^{d'}(y)\subseteq B_r^{d}(x)\cap A$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Proposition on sequence and closure
|
||||
|
||||
Let $X$ be a topological space and $A\subseteq X$. Then the following holds:
|
||||
|
||||
If there is a sequence $\{x_n\}_{n=1}^\infty$ such that $x_n\in A$ and $x_n\to x$, then $x\in \overline{A}$. ($x$ may not be in $A$)
|
||||
|
||||
The reverse holds if $X$ is a metric space. That is, if $X$ is a metric space and $x\in \overline{A}$, then there is a sequence $\{x_n\}_{n=1}^\infty$ such that $x_n\in A$ and $x_n\to x$.
|
||||
|
||||
<details>
|
||||
<summary>Example of non-metrizable space</summary>
|
||||
|
||||
For the second part of the claim
|
||||
|
||||
Let $X=\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots$ with the product topology over infinite product.
|
||||
|
||||
$X=\text{Map}(\mathbb{N},\mathbb{R})=\{(x_1,x_2,x_3,\cdots)|x_i\in \mathbb{R},i\in \mathbb{N}\}$.
|
||||
|
||||
The box topology on $X$ is the topology generated by:
|
||||
|
||||
$$
|
||||
\mathcal{B}=\{(a_1,b_1)\times (a_2,b_2)\times (a_3,b_3)\times \cdots|a_i,b_i\in \mathbb{R},i\in \mathbb{N}\}
|
||||
$$
|
||||
|
||||
It is easy to check that this is a basis.
|
||||
|
||||
Take $A=\mathbb{R}^{> 0}\times \mathbb{R}^{> 0}\times \mathbb{R}^{> 0}\times \cdots=\{(x_1,x_2,x_3,\cdots)|x_i> 0\}$.
|
||||
|
||||
$\overline{A}=\mathbb{R}^{\geq 0}\times \mathbb{R}^{\geq 0}\times \mathbb{R}^{\geq 0}\times \cdots=\{(x_1,x_2,x_3,\cdots)|x_i\geq 0\}$.
|
||||
|
||||
In particular, $(0,0,0,\cdots)\in \overline{A}$
|
||||
|
||||
Take a basis element $B=(a_1,b_1)\times (a_2,b_2)\times (a_3,b_3)\times \cdots\in \mathcal{B}$ containing $(0,0,0,\cdots)$. This means that $a_i<0<b_i$ Then $(\frac {b_1}{2},\frac {b_2}{2},\frac {b_3}{2},\cdots)\in B\cap A$.
|
||||
|
||||
This shows that $(0,0,0,\cdots)\in \overline{A}$.
|
||||
|
||||
We claim that there is no sequence $\{x_n\}_{n=1}^\infty$ such that $x_n\in A$ and $x_n\to (0,0,0,\cdots)$.
|
||||
|
||||
We proceed by contradiction and suppose that there is such a sequence $\{x_n\}_{n=1}^\infty$ such that $x_n\in A$ and $x_n\to (0,0,0,\cdots)$.
|
||||
|
||||
Since $x_n\in A$, we have $v_n=(a_1^n,a_2^n,a_3^n,\cdots)$ with $a_i^n>0$ for all $i\in \mathbb{N}$.
|
||||
|
||||
Consider the following open set around $(0,0,0,\cdots)$:
|
||||
|
||||
$C=(-\frac{a_1^1}{2},\frac{a_1^1}{2})\times (-\frac{a_2^1}{2},\frac{a_2^1}{2})\times (-\frac{a_3^1}{2},\frac{a_3^1}{2})\times \cdots$
|
||||
|
||||
We claim that $v_n\notin C$. Otherwise, we should have $-\frac{a_i^j}{2}<a_j^n<\frac{a_j^j}{2}$ for all $i\in \mathbb{N}$.
|
||||
|
||||
But this doesn't hold for $j=n$. This shows that $v_n\cancel{\to} (0,0,0,\cdots)$. Which is a contradiction.
|
||||
|
||||
---
|
||||
|
||||
For the first part of the claim. Let $\{x_n\}_{n=1}^\infty\subset A$ converge to $x$. Then for any open set $U$ of $x$, we have $N$ such that $x_n\in U$ for $n\geq N$.
|
||||
|
||||
In particular, $U\cap A$ is non-empty because it has $x_n$ for large enough $n$.
|
||||
|
||||
So $x\in \overline{A}$.
|
||||
|
||||
</details>
|
||||
@@ -1,132 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 15)
|
||||
|
||||
## Continue on convergence of sequences
|
||||
|
||||
### Closure and convergence
|
||||
|
||||
#### Proposition in metric space, every convergent sequence converges to a point in the closure
|
||||
|
||||
If $X$ is a metric space, $A\subset X$, and $x\in \overline{A}$, then there is a sequence $\{x_n\}_{n=1}^\infty\subseteq A$ such that $x_n\to x$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Let $U_n=B_{\frac{1}{n}}(x)$ be a sequence of open neighborhoods of $x$.
|
||||
|
||||
Since $x\in \overline{A}$ and $U_n$ is an open neighborhood of $x$, then $U_n\cap A\neq \emptyset$. Take $x_n\in U_n\cap A$, we claim that $\{x_n\}_{n=1}^\infty\subseteq A$ that $x_n\to x$.
|
||||
|
||||
Take an open neighborhood $U$ of $x$. Then by the definition of metric topology, there is $r>0$ such that $B_r(x)\subseteq U$.
|
||||
|
||||
let $N$ be such that $\frac{1}{N}<r$. Then for an $n\geq N$, we have $\frac{1}{n}\leq \frac{1}{N}<r$. This implies that $B_{\frac{1}{n}}(x)\subseteq B_r(x)\subseteq U$.
|
||||
|
||||
So $x_n\in B_{\frac{1}{n}}(x)\subseteq U$ for all $n\geq N$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Corollary $\mathbb{R}^\omega$ with the box topology is not metrizable
|
||||
|
||||
$\mathbb{R}^\omega$ is $\text{Map}(\mathbb{N},\mathbb{R})=\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots$.
|
||||
|
||||
> Note that $\mathbb{R}^\omega$ is Hausdorff. So not all Hausdorff spaces are metrizable.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Otherwise, the last proposition holds for $\mathbb{R}^\omega$ with the box topology.
|
||||
|
||||
Last time we showed that this is not the case.
|
||||
</details>
|
||||
|
||||
#### Definition of first countability axiom
|
||||
|
||||
A topological space $(X,\mathcal{T})$ satisfies the **first countability axiom** if for any point $x\in X$, there is a sequence of open neighborhoods of $x$, $\{V_n\}_{n=1}^\infty$ such that any open neighborhood $U$ of $x$ contains one of $V_n$.
|
||||
|
||||
Note that if we take $U_n=V_1\cap V_2\cap \cdots \cap V_n$, then any open neighborhood $U$ that contains $V_N$, then it also contains $U_n$ for all $n\geq N$.
|
||||
|
||||
> As the previous prof, for metric space, it is natural to try $V_n=B_{\frac{1}{n}}(x)$ for some $x\in X$.
|
||||
|
||||
#### Proposition on first countability axiom
|
||||
|
||||
Rewrite the [Proposition of metric space, every convergent sequence converges to a point in the closure](#proposition-in-metric-space-every-convergent-sequence-converges-to-a-point-in-the-closure)
|
||||
|
||||
We can have the following:
|
||||
|
||||
If $(X,\mathcal{T})$ satisfies the first countability axiom, then every convergent sequence converges to a point in the closure.
|
||||
|
||||
We can easily prove this by takeing the sequence of open neighborhoods $\{V_n\}_{n=1}^\infty$ instead of $U_n=B_{\frac{1}{n}}(x)$.
|
||||
|
||||
#### Proposition of continuous functions
|
||||
|
||||
Let $f:X\to Y$ be a map between two topological spaces.
|
||||
|
||||
1. If $f$ is continuous, then for any convergent sequence $\{x_n\}_{n=1}^\infty$ in $X$ converging to $x$, the sequence $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.
|
||||
2. If $X$ is equipped with the metric topology and for any convergent sequence $\{x_n\}_{n=1}^\infty\to x$ in $X$, the sequence $\{f(x_n)\}_{n=1}^\infty\to f(x)$ in $Y$, then $f$ is continuous.
|
||||
|
||||
<details>
|
||||
<summary>Exercise</summary>
|
||||
|
||||
Find an example of a function $f:X\to Y$ which is not continuous but for any convergent sequence in $X$, $\{x_n\}_{n=1}^\infty\to x$, the sequence $\{f(x_n)\}_{n=1}^\infty\to f(x)$.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Solution</summary>
|
||||
|
||||
Consider $X=\mathbb{R}$ with complement finite topology and $Y=\mathbb{R}$ with the standard topology.
|
||||
|
||||
Take identity function $f(x)=x$.
|
||||
|
||||
This function is not continuous by trivially taking $(0,1)\subseteq \mathbb{R}$ and the complement of $(0,1)$ is not a finite set, so the function is not continuous.
|
||||
|
||||
However, for every convergent sequence in $X$, $\{x_n\}_{n=1}^\infty\to x$, the sequence $\{f(x_n)\}_{n=1}^\infty\to f(x)$ trivially.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Part 1:
|
||||
|
||||
Let $f:X\to Y$ be a continuous map and
|
||||
|
||||
$$
|
||||
\{x_n\}_{n=1}^\infty\subseteq X
|
||||
$$
|
||||
converges to $x$.
|
||||
|
||||
Want to show that $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.
|
||||
|
||||
i.e. for any open neighborhood $U$ of $f(x)$, we want to show $f(x_n)$ is eventually in $U$. Take $f^{-1}(U)$. This is an open neighborhood of $x$ since $x_n\to x$.
|
||||
|
||||
There is $N$ such that $\forall n\geq N$, we have $x_n\in f^{-1}(U)$, $f(x_n)\in U$.
|
||||
|
||||
This implies that $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.
|
||||
|
||||
Part 2:
|
||||
|
||||
Let $f:X\to Y$ be a map between two topological spaces with $X$ being metric such that for any convergent sequence in $X$, $\{x_n\}_{n=1}^\infty\to x$ in $X$, we have $f(x_n)\to f(x)$ in $Y$.
|
||||
|
||||
Want to show that $f$ is continuous.
|
||||
|
||||
Recall that it suffice to show that for any $A\subseteq X$, $f(\overline{A})\subseteq \overline{f(A)}$.
|
||||
|
||||
Take $y\in f(\overline{A})$. Then $y=f(x)$ with $x\in \overline{A}$. By the previous proposition, there is a sequence $\{x_n\}_{n=1}^\infty\subseteq A$ (since $X$ is a metric space) such that $x_n\to x$.
|
||||
|
||||
By our assumption,
|
||||
|
||||
$$
|
||||
\{f(x_n)\}_{n=1}^\infty\to f(x) \tag{*}
|
||||
$$
|
||||
|
||||
Note that $\{f(x_n)\}_{n=1}^\infty$ is a sequence in $f(A)$ and ($*$) implies that $y=f(x)\in f(A)$ (by the first part of the proposition). This gives us the claim.
|
||||
</details>
|
||||
|
||||
> [!NOTE]
|
||||
>
|
||||
> The second part of the proposition is also true when $X$ is not a metric space but satisfies the first countability axiom.
|
||||
|
||||
#### Equivalent formulation of continuity
|
||||
|
||||
If $(X,d)$ and $(Y,d')$ are metric spaces and $f:X\to Y$ is a map, then $f$ is continuous if and only if for any $x_0\in X$ and any $\epsilon > 0$, there exists $\delta > 0$ such that if $\forall n\in X, d(x_n,x_0)<\delta$, then $d'(f(x_n),f(x_0))<\epsilon$.
|
||||
|
||||
Proof similar to the $X=Y=\mathbb{R}$ case.
|
||||
@@ -1,159 +0,0 @@
|
||||
# Math4201 Lecture 16 (Topology I)
|
||||
|
||||
## Continuous maps
|
||||
|
||||
The following maps are continuous:
|
||||
|
||||
$$
|
||||
F_+:\mathbb{R}\times \mathbb{R}\to \mathbb{R}, (x,y)\to x+y
|
||||
$$
|
||||
|
||||
$$
|
||||
F_-:\mathbb{R}\times \mathbb{R}\to \mathbb{R}, (x,y)\to x-y
|
||||
$$
|
||||
|
||||
$$
|
||||
F_\times:\mathbb{R}\times \mathbb{R}\to \mathbb{R}, (x,y)\to x\times y
|
||||
$$
|
||||
|
||||
$$
|
||||
F_\div:\mathbb{R}\times (\mathbb{R}\setminus \{0\})\to \mathbb{R}, (x,y)\to \frac{x}{y}
|
||||
$$
|
||||
|
||||
### Composition of continuous functions is continuous
|
||||
|
||||
Let $f,g:X\to \mathbb{R}$ be continuous functions. $X$ is topological space.
|
||||
|
||||
Then the following functions are continuous:
|
||||
|
||||
$$
|
||||
H:X\to \mathbb{R}\times \mathbb{R}, x\to (f(x),g(x))
|
||||
$$
|
||||
|
||||
Since the composition of continuous functions is continuous, we have
|
||||
|
||||
$$
|
||||
F_+\circ H:X\to \mathbb{R}, x\to f(x)+g(x)
|
||||
$$
|
||||
|
||||
$$
|
||||
F_-\circ H:X\to \mathbb{R}, x\to f(x)-g(x)
|
||||
$$
|
||||
|
||||
$$
|
||||
F_\times\circ H:X\to \mathbb{R}, x\to f(x)\times g(x)
|
||||
$$
|
||||
|
||||
are all continuous.
|
||||
|
||||
More over, if $g(x)\neq 0$ for all $x\in X$, then
|
||||
|
||||
$$
|
||||
F_\div\circ H:X\to \mathbb{R}, x\to \frac{f(x)}{g(x)}
|
||||
$$
|
||||
|
||||
is continuous following the similar argument.
|
||||
|
||||
### Defining metric for functions
|
||||
|
||||
#### Definition of bounded metric space
|
||||
|
||||
A metric space $(Y,d)$ is **bounded** if there is $M\in\mathbb{R}^{\geq 0}$ such that
|
||||
|
||||
$$
|
||||
\forall y,y'\in Y, d(y,y')<M
|
||||
$$
|
||||
|
||||
<details>
|
||||
<summary>Example of bounded metric space</summary>
|
||||
|
||||
If $(Y,d)$ is a bounded metric space, let $M$ be a positive constant, then $\overline{d}=\min\{M,d\}$ is a bounded metric space.
|
||||
|
||||
In fact, the metric topology by $d$ and $\overline{d}$ are the same. (proved in homeworks)
|
||||
|
||||
</details>
|
||||
|
||||
Let $X$ be a topological space. and $(Y,d)$ be a **bounded** metric space.
|
||||
|
||||
$$
|
||||
\operatorname{Map}(X,Y)\coloneqq \{f:X\to Y|f \text{ is a map}\}
|
||||
$$
|
||||
|
||||
Define $\rho:\operatorname{Map}(X,Y)\times \operatorname{Map}(X,Y)\to \mathbb{R}$ by
|
||||
|
||||
$$
|
||||
\rho(f,g)=\sup_{x\in X} d(f(x),g(x))
|
||||
$$
|
||||
|
||||
#### Lemma space of map with metric defined is a metric space
|
||||
|
||||
$(\operatorname{Map}(X,Y),\rho)$ is a metric space.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Proof is similar to showing that the square metric is a metric on $\mathbb{R}^n$.
|
||||
|
||||
$\rho(f,g)=0\implies \sup_{x\in X}(d(f(x),g(x)))=0$
|
||||
|
||||
Since $d(f(x),g(x))\geq 0$, this implies that $d(f(x),g(x))=0$ for all $x\in X$.
|
||||
|
||||
The triangle inequality of being metric for $\rho$ follows from the similar properties for $d$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Lemma continuous maps form a closed subset of the space of maps
|
||||
|
||||
Let $(\operatorname{Map}(X,Y),\rho)$ be a metric space defined before.
|
||||
|
||||
and
|
||||
|
||||
$$
|
||||
Z=\{f:X\to Y|f \text{ is a continuous map}\}
|
||||
$$
|
||||
|
||||
Then $Z$ is a closed subset of $(\operatorname{Map}(X,Y),\rho)$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
We need to show that $\overline{Z}=Z$.
|
||||
|
||||
Since $\operatorname{Map}(X,Y)$ is a metric space, this is equivalent to showing that: Let $f_n:X\to Y\in Z$ be a sequence of continuous maps,
|
||||
|
||||
Which is to prove the uniform convergence,
|
||||
|
||||
$$
|
||||
f_n \to f \in \operatorname{Map}(X,Y)
|
||||
$$
|
||||
|
||||
Then we want to show that $f$ is also continuous.
|
||||
|
||||
It is to show that for any open subspace $V$ of $Y$, $f^{-1}(V)$ is open in $X$.
|
||||
|
||||
Take $x_0\in f^{-1}(V)$, we'd like to show that there is an open neighborhood $U$ of $x_0$ such that $U\subseteq f^{-1}(V)$.
|
||||
|
||||
Since $x_0\in f^{-1}(V)$, then $f(x_0)\in V$. By metric definition, there is $r>0$ such that $B_r(f(x_0))\subseteq V$.
|
||||
|
||||
Take $N$ to be large enough such $\rho(f_N(x), f(x)) < \frac{r}{3}$
|
||||
|
||||
So $\forall x\in X$, $d(f(x),f_N(x))<\frac{r}{3}$
|
||||
|
||||
Since $f_N$ is continuous, $f_N^{-1}(B_{r/3}(f(x_0)))$ is an open set $U\subseteq X$ containing $x_0$.
|
||||
|
||||
Take $x\in U$, $d(f(x),f(x_0))<d(f(x),f_N(x_0))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))$ using triangle inequality.
|
||||
|
||||
Note that,
|
||||
|
||||
$d(f(x),f_N(x))<\frac{r}{3}$ (using $N$ large enough),
|
||||
|
||||
$d(f_N(x),f_N(x_0))<\frac{r}{3}$ (using $x\in U$, then $f_N(x)\in B_{r/3}(f_N(x_0))$, so $d(f_N(x),f_N(x_0))<\frac{r}{3}$),
|
||||
|
||||
$d(f_N(x_0),f(x_0))<\frac{r}{3}$ (using $N$ large enough),
|
||||
|
||||
So $d(f(x),f(x_0))<\frac{r}{3}+\frac{r}{3}+\frac{r}{3}=r$.
|
||||
|
||||
So $f(x)\in B_r(f(x_0))\implies x\in f^{-1}(B_r(f(x_0)))\implies x\in f^{-1}(V)\implies U\subseteq f^{-1}(V)$.
|
||||
|
||||
So $f^{-1}(V)$ is open in $X$.
|
||||
</details>
|
||||
@@ -1,86 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 17)
|
||||
|
||||
## Quotient topology
|
||||
|
||||
How can we define topologies on the space obtained points in a topological space?
|
||||
|
||||
### Quotient map
|
||||
|
||||
Let $(X,\mathcal{T})$ be a topological space. $X^*$ is a set and $q:X\to X^*$ is a surjective map.
|
||||
|
||||
The quotient topology on $X^*$ is defined as follows:
|
||||
|
||||
$$
|
||||
\mathcal{T}^* = \{U\subseteq X^*\mid q^{-1}(U)\in \mathcal{T}\}
|
||||
$$
|
||||
|
||||
$U\subseteq X^*$ is open if and only if $q^{-1}(U)$ is open in $X$.
|
||||
|
||||
In particular, $q$ is continuous map.
|
||||
|
||||
#### Definition of quotient map
|
||||
|
||||
$q:X\to X^*$ defined above is called a **quotient map**.
|
||||
|
||||
#### Definition of quotient space
|
||||
|
||||
$(X^*,\mathcal{T}^*)$ is called the **quotient space** of $X$ by $q$.
|
||||
|
||||
### Typical way of constructing a surjective map
|
||||
|
||||
#### Equivalence relation
|
||||
|
||||
$\sim$ is a subset of $X\times X$ satisfying:
|
||||
|
||||
- reflexive: $\forall x\in X, x\sim x$
|
||||
- symmetric: $\forall x,y\in X, x\sim y\implies y\sim x$
|
||||
- transitive: $\forall x,y,z\in X, x\sim y\text{ and } y\sim z\implies x\sim z$
|
||||
|
||||
#### Equivalence classes
|
||||
|
||||
Check equivalence relation.
|
||||
|
||||
For $x\in X$, the equivalence class of $x$ is denoted as $[x]\coloneqq \{y\in X\mid y\sim x\}$.
|
||||
|
||||
$X^*$ is the set of all equivalence classes on $X$.
|
||||
|
||||
$q:X\to X^*$ is defined as $q(x)=[x]$ will be a surjective map.
|
||||
|
||||
<details>
|
||||
<summary>Example of surjective maps and their quotient spaces</summary>
|
||||
|
||||
Let $X=\mathbb{R}^2$ and $(s,t)\sim (s',t')$ if and only if $s-s'$ and $t-t'$ are both integers.
|
||||
|
||||
This space as a topological space is homeomorphic to the torus.
|
||||
|
||||
---
|
||||
|
||||
Let $X=\{(s,t)\in \mathbb{R}^2\mid s^2+t^2\leq 1\}$ and $(s,t)\sim (s',t')$ if and only if $s^2+t^2$ and $s'^2+t'^2$. with subspace topology as a subspace of $\mathbb{R}^2$.
|
||||
|
||||
This space as a topological space is homeomorphic to the spherical shell $S^2$.
|
||||
|
||||
</details>
|
||||
|
||||
We will show that the quotient topology is a topology on $X^*$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
We need to show that the quotient topology is a topology on $X^*$.
|
||||
|
||||
1. $\emptyset, X^*$ are open in $X^*$.
|
||||
|
||||
$\emptyset, X^*$ are open in $X^*$ because $q^{-1}(\emptyset)=q^{-1}(X^*)=\emptyset$ and $q^{-1}(X^*)=X$ are open in $X$.
|
||||
|
||||
2. $\mathcal{T}^*$ is closed with respect to arbitrary unions.
|
||||
|
||||
$$
|
||||
q^{-1}(\bigcup_{\alpha \in I} U_\alpha)=\bigcup_{\alpha \in I} q^{-1}(U_\alpha)
|
||||
$$
|
||||
|
||||
3. $\mathcal{T}^*$ is closed with respect to finite intersections.
|
||||
|
||||
$$
|
||||
q^{-1}(\bigcap_{\alpha \in I} U_\alpha)=\bigcap_{\alpha \in I} q^{-1}(U_\alpha)
|
||||
$$
|
||||
</details>
|
||||
@@ -1,115 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 18)
|
||||
|
||||
## Quotient topology
|
||||
|
||||
Let $(X,\mathcal{T})$ be a topological space and $X^*$ be a set, $q:X\to X^*$ is a surjective map. The quotient topology on $X^*$:
|
||||
|
||||
$U\subseteq X^*$ is open $\iff q^{-1}(U)$ is open in $X$.
|
||||
|
||||
Equivalently,
|
||||
|
||||
$Z\subseteq X^*$ is closed $\iff q^{-1}(Z)$ is closed in $X$.
|
||||
|
||||
### Open maps
|
||||
|
||||
Let $(X,\mathcal{T})$ and $(Y,\mathcal{T}')$ be two topological spaces
|
||||
|
||||
Let $f:X\to Y$ is a quotient map if and only if $f$ is surjective and
|
||||
|
||||
$U\subseteq Y$ is open $\iff f^{-1}(U)$ is open
|
||||
|
||||
or equivalently
|
||||
|
||||
$Z\subseteq Y$ is closed $\iff f^{-1}(Z)$ is closed.
|
||||
|
||||
#### Definition of open map
|
||||
|
||||
Let $X\to Y$ be **continuous**. We say $f$ is open if for any $V\subseteq X$ be open, $f(V)$ is open in $Y$.
|
||||
|
||||
Let $X\to Y$ be **continuous**. We say $f$ is closed if for any $V\subseteq X$ be closed, $f(V)$ is closed in $Y$.
|
||||
|
||||
$$
|
||||
ff^{-1}(U)=U\text{ if }f \text{ is surjective}=U\cap f(X)
|
||||
$$
|
||||
|
||||
<details>
|
||||
<summary>Examples of open maps</summary>
|
||||
|
||||
Let $X,Y$ be topological spaces. Define the projection map $\pi_X:X\times Y\to X$, $\pi_X(x,y)=x$.
|
||||
|
||||
This is a surjective continuous map $(Y\neq \phi)$
|
||||
|
||||
This map is open. If $U\subseteq X$ is open and $V\subseteq Y$ is open, then $U\times V$ is open in $X\times Y$ and such open sets form a basis.
|
||||
|
||||
$\pi_X(U\times V)=\begin{cases}
|
||||
U&\text{ if }V\neq \emptyset\\
|
||||
\emptyset &\text{ if }V=\emptyset
|
||||
\end{cases}$
|
||||
|
||||
In particular, image of any such open set is open. Since any open $W\subseteq X\times Y$ is a union of such open sets.
|
||||
|
||||
$W=\bigcup_{\alpha\in I}U_\alpha\times V\alpha$
|
||||
|
||||
$\pi_X(W)=\pi_X(\bigcup_{\alpha\in I}U_\alpha\times V_\alpha)=\bigcup_{\alpha\in I}\pi_X(U_\alpha\times V_\alpha)=\bigcup_{\alpha\in I}U_\alpha$
|
||||
|
||||
is open in $X$.
|
||||
|
||||
However, $\pi_X$ is not necessarily a closed map.
|
||||
|
||||
Let $X=Y=\mathbb{R}$ and $X\times Y=\mathbb{R}^2$
|
||||
|
||||
$Z\subseteq \mathbb{R}^2=\{(x,y)\in\mathbb{R}^2|x\neq 0, y=\frac{1}{x}\}$ is a closed set in $\mathbb{R}^2$
|
||||
|
||||
$\pi_X(Z)=\mathbb{R}\setminus \{0\}$ is not closed.
|
||||
|
||||
---
|
||||
|
||||
Let $X=[0,1]\cup [2,3]$, $Y=[0,2]$ with subspace topology on $\mathbb{R}$
|
||||
|
||||
Let $f:X\to Y$ be defined as:
|
||||
|
||||
$$
|
||||
f(x)=\begin{cases}
|
||||
x& \text{ if } x\in [0,1]\\
|
||||
x-1& \text{ if }x\in [2,3]
|
||||
\end{cases}
|
||||
$$
|
||||
|
||||
$f$ is continuous and surjective, $f$ is closed $Z\subseteq [0,1]\cup [2,3]=Z_1\cup Z_2$, $Z_1\subseteq [0,1],Z_2\subseteq [2,3]$ is closed, $f(Z)=f(Z_1)\cup f(Z_2)$ is closed in $X$.
|
||||
|
||||
But $f$ is not open. Take $U=[0,1]\subseteq X$, $f=[0,1]\subseteq [0,2]$ is not open because of the point $1$.
|
||||
|
||||
> In general, and closed surjective map is a quotient map. In particular, this is an example of a closed surjective quotient map which is not open.
|
||||
|
||||
</details>
|
||||
|
||||
Let $f$ be a surjective open map. Then $f$ is a quotient map:
|
||||
|
||||
$U\subseteq Y$ is open and $f$ is continuous, $\implies f^{-1}(U)\subseteq X$ is open
|
||||
|
||||
$f^{-1}(U)\subseteq X$ is open and $f$ is surjective and open, $\implies f(f^{-1}(U))=U$ is open.
|
||||
|
||||
#### Proposition of continuous and open maps
|
||||
|
||||
If $f$ is a continuous bijection, then $f$ is open. if and only if $f^{-1}$ is continuous.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
To show $f^{-1}$ is continuous, we have to show for $U\subseteq X$ open. $(f^{-1})^{-1}(U)=f(U)\subseteq Y$ is open.
|
||||
|
||||
This is the same thing as saying that $f$ is open.
|
||||
|
||||
</details>
|
||||
|
||||
Let $f$ be a quotient map $f: X \to Y$, and $g$ be a continuous map $g:X\to Z$.
|
||||
|
||||
We want to find $\hat{g}$ such that $g=\hat{g}\circ f$.
|
||||
|
||||
If $x_1,x_2\in X$, such that $f(x_1)=f(x_2)$ and $g(x_1)\neq g(x_2)$, then we cannot find $\hat{g}$.
|
||||
|
||||
#### Proposition for continuous and quotient maps
|
||||
|
||||
Let $f$ and $g$ be as above. Moreover, for any $y\in Y$, all the points in $f^{-1}(y)$ are mapped to a single point by $g$. Then there is a unique continuous map $\hat{g}$ such that $g=\hat{g}\circ f$.
|
||||
|
||||
Continue next week.
|
||||
@@ -1,84 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 19)
|
||||
|
||||
## Quotient topology
|
||||
|
||||
### More propositions
|
||||
|
||||
#### Proposition for continuous and quotient maps
|
||||
|
||||
Let $X,Y,Z$ be topological spaces. $p$ is a quotient map from $X$ to $Y$ and $g$ is a continuous map from $X$ to $Z$.
|
||||
|
||||
Moreover, if for any $y\in Y$, the map $g$ is constant on $p^{-1}(y)$, then there is a continuous map $f: Y\to Z$ satisfying $f\circ p=g$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
For any $y\in Y$, take $x\in X$ such that $p(x)=y$ (since $p$ is surjective).
|
||||
|
||||
Define $f(y)\coloneqq g(x)$.
|
||||
|
||||
Note that this is well-defined and it doesn't depend on the specific choice of $x$ that $p(x)=y$ because $g$ is constant on $p^{-1}(y)$.
|
||||
|
||||
Then we check that $f$ is continuous.
|
||||
|
||||
Let $U\subseteq Z$ be open. Then we want to show that $f^{-1}(U)\subseteq Y$ is open.
|
||||
|
||||
Since $p$ is a quotient map, this is equivalent to showing that $p^{-1}(f^{-1}(U))\subseteq X$ is open. Note that $p^{-1}(f^{-1}(U))=g^{-1}(U)$.
|
||||
|
||||
Since $g$ is continuous, $g^{-1}(U)$ is open in $X$.
|
||||
|
||||
Since $g^{-1}(U)$ is open in $X$, $p^{-1}(g^{-1}(U))$ is open in $Y$.
|
||||
|
||||
</details>
|
||||
|
||||
> In general, $p^{-1}(y)$ is called the **fiber** of $p$ over $y$. The $g$ must be constant on the fiber.
|
||||
>
|
||||
> We may define $p^{-1}(y)$ as the equivalence class of $y$ if $p$ is defined using the equivalence relation. By definition $p^{-1}([x])$ is the element of $x$ that are $\sim x$.
|
||||
|
||||
#### Additional to the proposition
|
||||
|
||||
Note that $f$ is unique.
|
||||
|
||||
It is not hard to see that $f$ is a quotient map if and only if $g$ is a quotient map. (check book for detailed proofs)
|
||||
|
||||
#### Definition of saturated map
|
||||
|
||||
Let $p:X\to Y$ be a quotient map. We say $A\subseteq X$ is **saturated** by $p$ if $A=p^{-1}(B)$ for some $B\subseteq Y$.
|
||||
|
||||
Equivalently, if $x\in A$, then $p^{-1}(p(x))\subseteq A$.
|
||||
|
||||
#### Proposition for quotient maps from saturated sets
|
||||
|
||||
Let $p:X\to Y$ be a quotient map and $q$ be given by restriction of $p$ to $A\subseteq X$. $q:A\to p(A)$, $q(x)=p(x),x\in A$.
|
||||
|
||||
Assume that $A$ is saturated by $p$.
|
||||
|
||||
1. If $A$ is closed or open, then $q$ is a quotient map.
|
||||
2. If $p$ is closed or open, then $q$ is a quotient map.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
We prove 1 and assume that $A$ is open, (the closed case is similar).
|
||||
|
||||
clearly, $q:A\to p(A)$ is surjective.
|
||||
|
||||
In general, restricting the domain and the range of a continuous map is continuous.
|
||||
|
||||
Since $A$ is saturated by $p$, then $p^{-1}(p(A))=A$ is open, so $p(A)$ is open because $p$ is a quotient map. Let $V\subseteq p(A)$ and $q^{-1}(V)\subseteq A$ is open. Then $q^{-1}(V)=p^{-1}(V)$.
|
||||
|
||||
(i) $q^{-1}(V)\subseteq p^{-1}(V)$: $x\in q^{-1}(V)\implies q(x)\in V$. Then $p(x)=q(x)\in V$
|
||||
|
||||
(ii) $p^{-1}(V)\subseteq q^{-1}(V)$: $x\in p^{-1}(V)\implies p(x)\in V\subseteq p(A)$. This implies that $x\in p^{-1}(p(A))=A$ since $A$ is saturated by $p$. Therefore $x\in q^{-1}(V)$.
|
||||
|
||||
Since $A$ is open in $X$, any open subspace of $A$ is open in $X$. In particular, $q^{-1}(V)=p^{-1}(V)$ is open in $X$.
|
||||
|
||||
Since $p$ is a quotient map, and $p^{-1}(V)$ is open in $X$, $V$ is open in $Y$. So $V\subseteq p(A)$ is open in $Y$.
|
||||
|
||||
This shows $q$ is a quotient map.
|
||||
|
||||
---
|
||||
|
||||
We prove 2 next time...
|
||||
|
||||
</details>
|
||||
@@ -1,110 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 2)
|
||||
|
||||
## Topology
|
||||
|
||||
Distance in $\mathbb{R}$, or more generally in $\mathbb{R}^n$. ([metric space](https://notenextra.trance-0.com/Math4111/Math4111_L9#metric-space))
|
||||
|
||||
Intervals in $\mathbb{R}$, or more generally open balls in $\mathbb{R}^n$. (topological space)
|
||||
|
||||
### Topological Spaces
|
||||
|
||||
#### Definition of Topological Space
|
||||
|
||||
A topological space is a pair of set $X$ and a collection of subsets of $X$, denoted by $\mathcal{T}$ (imitates the set of "open sets" in $X$), satisfying the following axioms:
|
||||
|
||||
1. $\emptyset \in \mathcal{T}$ and $X \in \mathcal{T}$
|
||||
2. $\mathcal{T}$ is closed with respect to arbitrary unions. This means, for any collection of open sets $\{U_\alpha\}_{\alpha \in I}$, we have $\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}$
|
||||
3. $\mathcal{T}$ is closed with respect to finite intersections. This means, for any finite collection of open sets $\{U_1, U_2, \ldots, U_n\}$, we have $\bigcap_{i=1}^n U_i \in \mathcal{T}$
|
||||
|
||||
The elements of $\mathcal{T}$ are called **open sets**.
|
||||
|
||||
The topological space is denoted by $(X, \mathcal{T})$.
|
||||
|
||||
<details>
|
||||
|
||||
<summary>Examples of topological spaces</summary>
|
||||
|
||||
Trivial topology: Let $X$ be arbitrary. The trivial topology is $\mathcal{T}_0 = \{\emptyset, X\}$
|
||||
|
||||
Discrete topology: Let $X$ be arbitrary. The discrete topology is $\mathcal{T}_1 = \mathcal{P}(X)=\{U \subseteq X\}$
|
||||
|
||||
Understanding all possible topologies on a set.
|
||||
|
||||
Let's say $X=\{a,b\}$
|
||||
|
||||
The trivial topology is $\mathcal{T}_0 = \{\emptyset, \{a,b\}\}$
|
||||
|
||||
The discrete topology is $\mathcal{T}_1 = \{\emptyset, \{a\}, \{b\}, \{a,b\}\}$
|
||||
|
||||
Other topologies:
|
||||
|
||||
$\mathcal{T}_2 = \{\emptyset, \{a\}, \{a,b\}\}$
|
||||
|
||||
$\mathcal{T}_3 = \{\emptyset, \{b\}, \{a,b\}\}$
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
|
||||
<summary>Non-example of topological space</summary>
|
||||
|
||||
Let $X=\{a,b,c\}$
|
||||
|
||||
The set $\mathcal{T}_1=\{\emptyset, \{a\}, \{b\}, \{a,b,c\}\}$ is not a topology because it is not closed under union $\{a\} \cup \{b\} = \{a,b\} \notin \mathcal{T}_1$
|
||||
|
||||
</details>
|
||||
|
||||
#### Definition of Complement finite topology
|
||||
|
||||
Let $X$ be arbitrary. The complement finite topology is $\mathcal{T}\coloneqq \{U\subseteq X|X\setminus U \text{ is finite}\}\cup \{\emptyset\}$
|
||||
|
||||
The topology is valid because:
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
1. $\emptyset \in \mathcal{T}$ because $X\setminus \emptyset = X$ is finite.
|
||||
2. Let $\{U_\alpha\}_{\alpha \in I}$ be an arbitrary collection such that $X\setminus U_\alpha$ is finite for each $\alpha \in I$.
|
||||
|
||||
Without loss of generality, we can assume that $U_\alpha \neq \emptyset$ for each $\alpha \in I$, since the union of arbitrary set with $\emptyset$ is the set itself.
|
||||
|
||||
If all of them are empty, then the union is empty, which complement is $X\subset \mathcal{T}$.
|
||||
|
||||
Otherwise,
|
||||
|
||||
$$
|
||||
X\setminus \bigcup_{\alpha \in I} U_\alpha = \bigcap_{\alpha \in I} (X\setminus U_\alpha)
|
||||
$$
|
||||
is finite because each $X\setminus U_\alpha$ is finite. Therefore, $\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}$.
|
||||
3. Let $\{U_1, U_2, \ldots, U_n\}$ be a finite collection such that $X\setminus U_i$ is finite for each $i=1,2,\ldots,n$.
|
||||
|
||||
Without loss of generality, we can assume that $U_i \neq \emptyset$ for each $i=1,2,\ldots,n$, since the intersection of arbitrary set with $\emptyset$ is $\emptyset$.
|
||||
|
||||
If all of them are empty, then the intersection is $X\subset \mathcal{T}$.
|
||||
|
||||
Otherwise,
|
||||
|
||||
$$
|
||||
X\setminus \bigcap_{i=1}^n U_i = \bigcup_{i=1}^n (X\setminus U_i)
|
||||
$$
|
||||
is finite because each $X\setminus U_i$ is finite. Therefore, $\bigcap_{i=1}^n U_i \in \mathcal{T}$.
|
||||
</details>
|
||||
|
||||
Another non-example is $\mathcal{T} = \{U\subseteq X|U \text{ is finite}\}\cup \{X\}$
|
||||
|
||||
The topology is invalid because an arbitrary union of points is not a finite set.
|
||||
|
||||
Consider $X=\mathbb{Z}$ but take $U_1=\{1\}, U_2=\{2\}, U_3=\{3\}, \ldots$
|
||||
|
||||
Then $\bigcup_{i=1}^\infty U_i = \mathbb{Z}^+$ is not a finite set and is not $\{X\}$.
|
||||
|
||||
> [!NOTE]
|
||||
>
|
||||
> If $X$ is finite, then finite complement topology is the same as the discrete topology.
|
||||
|
||||
#### Definition of Topology basis
|
||||
|
||||
For a set $X$, a topology basis, denoted by $\mathcal{B}$, is a collection of subsets of $X$, such that the following properties are satisfied:
|
||||
|
||||
1. For any $x \in X$, there exists a $B \in \mathcal{B}$ such that $x \in B$
|
||||
2. If $B_1, B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$, then there exists a $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$
|
||||
@@ -1,121 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 20)
|
||||
|
||||
## Quotient topology
|
||||
|
||||
### More propositions
|
||||
|
||||
#### Proposition for quotient maps in restrictions
|
||||
|
||||
Let $X,Y$ be topological spaces and $p:X\to Y$ is surjective and open/closed. Let $A\subseteq X$ be saturated by $p$, ($p^{-1}(p(A))=A$).
|
||||
|
||||
Then $q: A\to p(A)$ given by the restriction of $p$ is open/closed surjective map (In particular, it's a quotient map).
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
$q$ is surjective and continuous. Now assume $p$ is open and we will show that $q$ is also open. Any open subspace of $A$ is given as $U\cap A$ where $U$ is open in $X$. By definition, $q(U\cap A)=p(U\cap A)=p(U)\cap p(A)$
|
||||
|
||||
To see the second identity:
|
||||
|
||||
1. $p(U\cap A)\subseteq p(U)\cap p(A)$
|
||||
|
||||
$\forall y\in p(U\cap A)$, $y=p(x)$ with $x\in U\cap A$, since $x\in A$ and $x\in U$, $y=p(x)\in p(U)\cap p(A)$
|
||||
|
||||
2. $p(U)\cap p(A)\subseteq p(U\cap A)$
|
||||
|
||||
$\forall y\in p(U)\cap p(A)$, $y=p(x_1)$ with $x_1\in U$ and $y=p(x_2)$ with $x_2\in A$, since $x_1\in U$ and $x_2\in A$, $y=p(x_1)=p(x_2)\in p(U\cap A)$
|
||||
|
||||
So $x_1=x_2\in U\cap A$, $y=p(x_1)=p(x_2)\in p(U)\cap p(A)$, $y\in p(U\cap A)$.
|
||||
|
||||
Note that $p(U)\subseteq X$ is open by $p$ is an open map.
|
||||
|
||||
So $p(U)\cap p(A)$ is open in $p(A)$.
|
||||
|
||||
$q(U\cap A)=p(U\cap A)=p(U)\cap p(A)$ is open.
|
||||
|
||||
So $q$ is open in $p(A)$.
|
||||
|
||||
</details>
|
||||
|
||||
### Simplicial complexes (extra chapter)
|
||||
|
||||
#### Definition for simplicial complexes
|
||||
|
||||
Simplicial complexes are topological space with simplices ($n$ dimensional triangles) as their building blocks.
|
||||
|
||||
#### Definition for n dimensional simplex
|
||||
|
||||
Let $v_0,\dots,v_n$ be points in $\mathbb{R}^m$ such that $v_n-v_0$, $v_{n-1}-v_0$, $\cdots$, and $v_1-v_0$ are linearly independent in $\mathbb{R}^m$. (in particular $n\leq m$).
|
||||
|
||||
The $n$-dimensional simplex determined by $\{v_0,\dots,v_n\}$ is given as:
|
||||
|
||||
$$
|
||||
\Delta^n\coloneqq [v_0,\dots,v_n]=\{t_0v_0+t_1v_1+\cdots+t_nv_n\vert t_i\geq 0, \sum_{i=0}^n t_i=1\}
|
||||
$$
|
||||
|
||||
The coefficients $t_0,\dots,t_n$ are called barycentric coordinates.
|
||||
|
||||
<details>
|
||||
<summary>Example of simplicial complex</summary>
|
||||
|
||||
$n=0$,
|
||||
|
||||
$\Delta^0=\{v_0\}$
|
||||
|
||||
$n=1$,
|
||||
|
||||
$\Delta^1=\{t_0v_0+t_1v_1\vert t_0+t_1=1\}$, this is the line segment between $v_0$ and $v_1$.
|
||||
|
||||
$n=2$,
|
||||
|
||||
$\Delta^2=\{t_0v_0+t_1v_1+t_2v_2\vert t_0+t_1+t_2=1\}$, this is the triangle with vertices $v_0,v_1,v_2$.
|
||||
|
||||
</details>
|
||||
|
||||
> [!NOTE]
|
||||
>
|
||||
> Every non-empty subset $\{v_{i_0},\dots,v_{i_k}\}$ of $\{v_0,\dots,v_n\}$ determines a $k$ dimensional simplex $[v_{i_0},\dots,v_{i_k}]\subseteq \Delta^n=[v_0,\dots,v_n]$. Inside the $n$ dimensional simplex $t_{i_0}v_{i_0}+\cdots+t_{i_n}v_{i_k}\in \Delta^n$. Where the coefficient $t_j$ of $v_j\notin \{v_{i_0},\dots,v_{i_n}\}$ is $0$.
|
||||
|
||||
Any such $k$ dimensional simplex is called a face of the simplex $[v_{i_0},\dots,v_{i_n}]$.
|
||||
|
||||
<details>
|
||||
<summary>Example of faces for simplicial complex</summary>\
|
||||
|
||||
For a triangle $[v_0,v_1,v_2]$, the faces are $[v_0,v_1]$, $[v_0,v_2]$, and $[v_1,v_2]$ (the edges of the triangle).
|
||||
|
||||
</details>
|
||||
|
||||
#### Definition for abstract simplicial complex
|
||||
|
||||
Let $V$ be a finite or countable set, an abstract simplicial complex on $V$ is a collection of **finite non-empty subset** of $V$, denoted by $K$. And the two conditions are satisfied:
|
||||
|
||||
1. If $\sigma\in K$ and $\tau\subseteq \sigma$, then $\tau\in K$.
|
||||
|
||||
2. For any $v\in V$, $\{v\}\in K$.
|
||||
|
||||
<details>
|
||||
<summary>Example of abstract simplicial complex</summary>
|
||||
|
||||
Let $V=\{a,b,c,d\}$.
|
||||
|
||||
If we want to include $\{a,b,c\}$, then we need to include $\{a,b\}$ and $\{b,c\}$, so we have $K=\{\{a,b,c\},\{a,b\},\{b,c\},\{a\},\{b\},\{c\},\{d\}\}$ is an abstract simplicial complex.
|
||||
|
||||
</details>
|
||||
|
||||
#### Topological realization of abstract simplicial complex
|
||||
|
||||
Let $\bigsqcup_{\sigma\in K}\Delta^{|\sigma|-1}$ be the disjoint union of all $|\sigma|-1$ dimensional simplices in $K$.
|
||||
|
||||
$$
|
||||
\tilde{X_k}=\bigsqcup_{\sigma\in K}\Delta^{|\sigma|-1}
|
||||
$$
|
||||
|
||||
We use subspace topology to define a topology on $\Delta^n$ and the union of such topology for each $\Delta^{|\sigma|-1}$ defines a topology on $\tilde{X_k}$.
|
||||
|
||||
We define the equivalence relation $x\in \Delta_{\sigma}^{|\sigma|-1}\sim x'\in \Delta_{\sigma'}^{|\sigma'|-1}$ if $x\in \Delta_{\sigma'\cap \sigma}^{|\sigma'\cap \sigma|-1}\subseteq \Delta_{\sigma}^{|\sigma|-1}$. and $x'\in \Delta_{\sigma'\cap \sigma}^{|\sigma'\cap \sigma|-1}\subseteq \Delta_{\sigma'}^{|\sigma'|-1}$.
|
||||
|
||||
are the sample points of $\Delta_{\sigma\cap \sigma'}^{|\sigma\cap \sigma'|-1}$.
|
||||
|
||||
$X_K$ is the quotient space of $\tilde{X_k}$ by the equivalence relation.
|
||||
|
||||
Continue next time.
|
||||
@@ -1,98 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 21)
|
||||
|
||||
## Simplicial complexes
|
||||
|
||||
### Recall from last lecture
|
||||
|
||||
Let $\sigma=\{a_0,a_1,\dots,a_n\}$ be a finite set. The $n$-dimensional simplex determined by $\tau$ is given as:
|
||||
|
||||
$$
|
||||
\Delta^n(a_0,a_1,\dots,a_n)=\left\{t_0a_0+t_1a_1+\cdots+t_na_n\mid t_i\geq 0, \sum_{i=0}^n t_i=1\right\}
|
||||
$$
|
||||
|
||||
If we have vertices $\tau=\{a_0,a_1,\dots,a_k\}$, $\tau\subseteq \sigma$, the face of $\Delta^n$ is determined by $\tau$ with dimension $|\tau|-1$.
|
||||
|
||||
$\Delta^n$ is the topologized by the subspace topology inherited by the standard topology on Euclidean space $\mathbb{R}^n$.
|
||||
|
||||
Note that there are different ways to of embedding and all give the same topological space.
|
||||
|
||||
### Abstract simplicial complexes
|
||||
|
||||
#### Definition for abstract simplicial complex
|
||||
|
||||
Let $V=\{v_0,v_1,\dots,v_n\}$ be a finite set (set of vertices of a simplicial complex). $K$ be the collection of subspaces of $V$.
|
||||
|
||||
1. $\sigma\in K$ and $\tau\subseteq \sigma$, then $\tau\in K$.
|
||||
2. For any $v\in V$, $\{v\}\in K$.
|
||||
|
||||
Then $\tilde{X_k}=\bigsqcup_{\sigma\in K}\Delta_\sigma$.
|
||||
|
||||
$\Delta_\sigma$ is a simplex of dimension $|\sigma|-1$.
|
||||
|
||||
$X_K$ is the topological realization of $K$.
|
||||
|
||||
Define an equivalence relation on $\tilde{X_k}$ as follows:
|
||||
|
||||
$x\in \Delta_\sigma\sim x'\in \Delta_{\sigma'}$ if and only if $x\in \Delta_{\sigma'\cap \sigma}^{|\sigma'\cap \sigma|-1}\subseteq \Delta_\sigma$ and $x'\in \Delta_{\sigma'\cap \sigma}^{|\sigma'\cap \sigma|-1}\subseteq \Delta_{\sigma'}$.
|
||||
|
||||
This just means that the two points have the same barycentric coordinates in the simplex.
|
||||
|
||||
#### Definition of barycentric coordinates
|
||||
|
||||
Let $\sigma=\{a_0,a_1,\dots,a_n\}$ be a simplex. The barycentric coordinates of a point $x\in \Delta_\sigma$ are the coefficients $t_0,t_1,\dots,t_n$ such that:
|
||||
|
||||
$$
|
||||
x=t_0a_0+t_1a_1+\cdots+t_na_n
|
||||
$$
|
||||
|
||||
and $t_i\geq 0$ and $\sum_{i=0}^n t_i=1$.
|
||||
|
||||
The point $x$ is in the simplex $\Delta_\sigma$ if and only if $t_i\geq 0$ for all $i$.
|
||||
|
||||
<details>
|
||||
<summary>Example of abstract simplicial complex</summary>
|
||||
|
||||
Let $V=\{v_1,v_2,v_3,v_4,v_5\}$.
|
||||
|
||||
If we want to enclose $K=\{\{v_1,v_2,v_3,v_4\},\{v_3,v_4,v_5\}\}$, we need to fill all the singletons $\{v_1\},\{v_2\},\{v_3\},\{v_4\},\{v_5\}$, all the pairs in $K$, $\{v_1,v_2\},\{v_1,v_3\},\{v_1,v_4\},\{v_2,v_3\},\{v_2,v_4\},\{v_3,v_4\},\{v_3,v_5\},\{v_4,v_5\}$, and the triangle $\{v_1,v_2,v_3\}, \{v_1,v_2,v_4\}, \{v_1,v_3,v_4\}, \{v_2,v_3,v_5\}$.
|
||||
|
||||
The final simplicial complex is $\tilde{X_k}=\bigsqcup_{\sigma\in K}\Delta(v_1,v_2,v_3,v_4)\sqcup \Delta(v_3,v_4,v_5)\sqcup \{v_1,v_2,v_3,v_4,v_5\}$.
|
||||
|
||||
We use $\Delta(v_1,v_2,v_3,v_4)$ to denote the simplex with vertices $v_1,v_2,v_3,v_4$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Defining maps on abstract simplicial complexes
|
||||
|
||||
Let $K$ be an abstract simplicial complex. $V=\{v_1,v_2,\dots,v_m\}$
|
||||
|
||||
A map $\pi:\tilde{X_k}\to X_K$ is a quotient map
|
||||
|
||||
$X_K$ is equipped with the quotient topology.
|
||||
|
||||
Let $f:V\to \mathbb{R}^m$, then $u_i=f(v_i)$.
|
||||
|
||||
$\tilde{X_k}=\bigsqcup_{\sigma\in K}\Delta_\sigma$ is the disjoint union of all simplices in $K$.
|
||||
|
||||
For $\sigma=\{v_{i_0},\dots,v_{i_k}\}$, we have a map $\Delta_\sigma\to \mathbb{R}^\ell$ given by $[t_{i_0}u_0+t_{i_1}u_1+\cdots+t_{i_k}u_k\mid t_j\geq 0, \sum_{j=0}^k t_j=1]$.
|
||||
|
||||
This is well-defined because the coefficients $t_j$ are uniquely determined by the vertices $v_{i_0},\dots,v_{i_k}$.
|
||||
|
||||
This induces $F:\tilde{X_k}\to \mathbb{R}^\ell$. This map is continuous because $F\vert_{\Delta_\sigma}$ is continuous for all $\sigma\in K$.
|
||||
|
||||
Recall that if for any $x\in X_K$, the map $F$ restricted to $\pi^{-1}(x)$ is constant, then there is a unique continuous map $g$ satisfying $F=g\circ \pi$.
|
||||
|
||||
In fact, this condition is satisfied and there is such a map $G$.
|
||||
|
||||
<details>
|
||||
<summary>Example of map on abstract simplicial complexes</summary>
|
||||
|
||||
Consider the previous example of abstract simplicial complex.
|
||||
|
||||
Let $f:V\to \mathbb{R}$ by $f(v_i)=i$.
|
||||
|
||||
Then $f(\Delta_{\{v_1,v_2,v_3,v_4\}})=[1,4]$
|
||||
|
||||
Then $f(\Delta_{\{v_1,v_3\}})=[1,3]$
|
||||
|
||||
</details>
|
||||
@@ -1,162 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 22)
|
||||
|
||||
## Connectedness of topological spaces
|
||||
|
||||
### Connectedness
|
||||
|
||||
#### Definition of separation
|
||||
|
||||
A separate of a topological space $X$ is a pair of disjoint nonempty open subsets $U,V\subset X$ such that $X=U\cup V$. A space is connected if there is no separation.
|
||||
|
||||
Otherwise, it is disconnected.
|
||||
|
||||
<details>
|
||||
<summary>Example of separation</summary>
|
||||
|
||||
Let $X$ be an arbitrary set with trivial topology. (The only open sets are $\emptyset$ and $X$.)
|
||||
|
||||
This space is connected.
|
||||
|
||||
---
|
||||
|
||||
Let $X=\{a,b\}$ with discrete topology. Then $X$ is disconnected.
|
||||
|
||||
A separation is given by $U=\{a\}$ and $V=\{b\}$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Theorem of separation and clopen sets
|
||||
|
||||
Note that $U,V$ give a separation of $X$ if and only if $U=V^c$ and $V$ is open, then $U$ is closed and open.
|
||||
|
||||
So if $X$ is connected, then there is a non-empty proper (not the same as $X$) closed and open set.
|
||||
|
||||
The reverse is also true. (If the only clopen sets are $\emptyset$ and $X$, then $X$ is connected.)
|
||||
|
||||
<details>
|
||||
<summary>Example of connected and disconnected space in real numbers</summary>
|
||||
|
||||
Let $X=[a,b]$ with subspace topology inherited from $\mathbb{R}$ is connected.
|
||||
|
||||
Then other connected subspace of $\mathbb{R}$ are $(a,b)$, $[a,b)$, $(a,b]$, $(-\infty,b)$, $(-\infty,b]$, $(a,\infty)$, $[a,\infty)$, and $\mathbb{R}$.
|
||||
|
||||
---
|
||||
|
||||
If $X\subseteq \mathbb{R}$ with the subspace topology such that there are $a<b<c$ with $a,c\in X, b\notin X$, then $X$ is disconnected.
|
||||
|
||||
Note that $X=((-\infy,b)\cap X)\cup ((b,\infty)\cap X)$ are two disjoint open sets whose union is $X$.
|
||||
|
||||
$U$ is not empty because $a\in U$.
|
||||
|
||||
$V$ is not empty because $c\in V$.
|
||||
|
||||
$U\cap V=\phi$ because $b\notin U\cap V$, is a valid separation of $X$.
|
||||
|
||||
So $X$ is disconnected.
|
||||
|
||||
</details>
|
||||
|
||||
#### Definition of totally disconnected space
|
||||
|
||||
Any topological space that any subset of it with at least two elements is disconnected is called a totally disconnected space.
|
||||
|
||||
<details>
|
||||
<summary>Example of totally disconnected space</summary>
|
||||
|
||||
In $\mathbb{R}$, any subset of rational numbers with at least two elements is disconnected.
|
||||
|
||||
Because there is a irrational number between any two rational numbers.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
|
||||
<summary>Example of disconnected space</summary>
|
||||
|
||||
Let $X\subseteq \mathbb{R}^2$ and $X=U\cap V$, where $U=\{(x,y)\in \mathbb{R}^2\mid y=1/x\}$ and $V=\{(x,y)\in \mathbb{R}^2\mid x=0\}$.
|
||||
|
||||
Then $X$ is disconnected since $U, V$ gives a separation of $X$ (In this case, $U$ and $V$ are closed sets in $\mathbb{R}^2$).
|
||||
|
||||
</details>
|
||||
|
||||
#### Lemma of separated subsets
|
||||
|
||||
Let $U,V$ give a separation of a topological space $X$. Let $Y\subseteq X$ with the subspace topology is connected. Then $Y$ is either contained in $U$ or $V$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Consider $U'=U\cap Y$ and $V'=V\cap Y$. Then $U'$ and $V'$ are disjoint nonempty open subsets of $Y$. and $Y=U'\cup V'$.
|
||||
|
||||
Since $Y$ is connected, then $U'$ or $V'$ are not a separation, so $U'$ or $V'$ is empty.
|
||||
|
||||
</details>
|
||||
|
||||
#### Theorem of connectedness of union of connected subsets
|
||||
|
||||
Let $X=\bigcup_{\alpha\in I} X_\alpha$ such that $\bigcap_{\alpha\in I} X_\alpha$ is non-empty. And $X_\alpha$ are connected. Then $X$ is also connected.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Let $x\in \bigcap_{\alpha\in I} X_\alpha$. By contradiction, suppose $U,V$ give a separation of $X$. Assume $x\in U$ and $x\notin V$, Applying the lemma to $Y=X_\alpha$ for each $\alpha\in I$, we have $X_\alpha\subseteq U$ or $X_\alpha\subseteq V$.
|
||||
|
||||
Since $x\in X_\alpha$ is an element of $u$, the fist possibility holds, so $\bigcap_{\alpha\in I} X_\alpha\subseteq U$ implies $X\subseteq U$, then $U=x$, $V=\emptyset$, which is a contradiction.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example</summary>
|
||||
|
||||
Let $X=S^1\subseteq \mathbb{R}^2$ with the subspace topology. Let $X_0=S^1\cap \{(x,y)\mid x\leq 0\}$ and $X_1=S^1\cap \{(x,y)\mid x\geq 0\}$.
|
||||
|
||||
Then $X_0\cap X_1=\{(0,1), (0,-1)\}$.
|
||||
|
||||
Note that both of them are homeomorphic to $[0,1]\subseteq \mathbb{R}$, which are known to be connected.
|
||||
|
||||
</details>
|
||||
|
||||
#### Proposition of connectedness and homeomorphism
|
||||
|
||||
Connectedness is a topological property (preserve under homeomorphism).
|
||||
|
||||
i.e. If $X$ and $Y$ are homeomorphic, then $X$ is connected if and only if $Y$ is connected.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
By contradiction, $U,V$ give a separation of $X$ let $\phi:X\to Y$ be a homeomorphism. Then $\phi(U)$ and $\phi(V)$ are disjoint nonempty open subsets of $Y$ whose union is $Y$.
|
||||
|
||||
So $Y$ is disconnected.
|
||||
|
||||
This contradicts the assumption that $Y$ is connected.
|
||||
|
||||
Therefore, $X$ is connected.
|
||||
|
||||
The reverse direction is similar.
|
||||
|
||||
</details>
|
||||
|
||||
> [!NOTE]
|
||||
>
|
||||
> The connectedness of two topological spaces can be preserved under weaker conditions than homeomorphism.
|
||||
|
||||
#### Proposition of connectedness and continuous map
|
||||
|
||||
If $X$ is connected and $f:X\to Y$ is a continuous map, then $f(X)\subseteq Y$ with subspace topology is connected.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
By contradiction, suppose $f(X)$ is disconnected. Then there are disjoint nonempty open subsets $U,V$ of $f(X)$ such that $f(X)=U\cup V$.
|
||||
|
||||
Since $f$ is continuous, $f^{-1}(U)$ and $f^{-1}(V)$ are open in $X$ and $X=f^{-1}(U)\cup f^{-1}(V)$.
|
||||
|
||||
Since $X$ is connected, then $f^{-1}(U)$ and $f^{-1}(V)$ are not a separation, so $f^{-1}(U)$ or $f^{-1}(V)$ is empty.
|
||||
|
||||
This contradicts the assumption that $X$ is connected.
|
||||
|
||||
Therefore, $f(X)$ is connected.
|
||||
|
||||
</details>
|
||||
|
||||
@@ -1,133 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 23)
|
||||
|
||||
## Connectedness of topological spaces
|
||||
|
||||
### Connected space
|
||||
|
||||
#### Definition of connected space
|
||||
|
||||
Let $X$ be a topological space. $X$ is separated if there exist two disjoint nonempty open subsets $U,V\subset X$ such that $X=U\cup V$.
|
||||
|
||||
If $X$ is not separated, then $X$ is connected.
|
||||
|
||||
#### Any interval in $\mathbb{R}$ with standard topology is connected
|
||||
|
||||
Let $I=[a,b]$ be an interval in $\mathbb{R}$ with standard topology. Then $I$ is connected.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
By contradiction, we assume that $U,V$ give a separation of $\mathbb{R}$. In particular, $\exists a\in U, b\in V$.
|
||||
|
||||
Let $a_0\coloneqq \sup\{x\in U\cap [a,b]\}$. Note that $a\in U\cap [a,b]$, so $a_0\geq a$. Since any element of $U\cap [a,b]$ is less than or equal to $b$, $a_0\leq b$.
|
||||
|
||||
**Case 1**: $a_0=a$
|
||||
|
||||
Since $U$ is open, there is $\epsilon>0$ such that $[a,a+\epsilon)\subset U\cap [a,b]$. So $a_0\geq a+\epsilon>a$, which contradicts the definition of $a_0$.
|
||||
|
||||
**Case 2**: $a_0=b$.
|
||||
|
||||
Since $V$ is open, there is $\epsilon>0$ such that $(b-\epsilon,b]\subseteq V\cap [a,b]$. This implies that $b-\epsilon$ is also an upper bound of $U\cap [a,b]$, so $a_0\leq b-\epsilon<b$, which contradicts the definition of $a_0$.
|
||||
|
||||
**Case 3**: $a<a_0<b$.
|
||||
|
||||
**Subcase I**: $a_0\in U$.
|
||||
|
||||
There is an $\epsilon>0$ such that $(a_0-\epsilon,a_0+\epsilon)\subset U\cap [a,b]$ because $U$ is open.
|
||||
|
||||
In particular, $a_0$ is greater than any element of $(a_0-\epsilon,a_0+\epsilon)$, which contradicts the definition of $a_0$.
|
||||
|
||||
**Subcase II**: $a_0\in V$.
|
||||
|
||||
There is an $\epsilon>0$ such that $(a_0-\epsilon,a_0+\epsilon)\subset V\cap [a,b]$ because $V$ is open.
|
||||
|
||||
In particular, $a_0$ is less than any element of $(a_0-\epsilon,a_0+\epsilon)$. Since $a_0$ is an upper bound of $U\cap [a,b]$, any point $>a_0$ is not in $U\cap [a,b]$.
|
||||
|
||||
So, $U\cap [a,b]\subseteq [a,a_0-\epsilon)$. This shows that $a_0-\epsilon$ is an upper bound of $U\cap [a,b]$, which contradicts the definition of $a_0$.
|
||||
|
||||
</details>
|
||||
|
||||
_Intuitively, since both sets in $\mathbb{R}$ are open, you cannot set a clear boundary between the two sets by least upper bound argument._
|
||||
|
||||
#### Corollary as Intermediate Value Theorem
|
||||
|
||||
If $f:[a,b]\to \mathbb{R}$ is continuous, and $c\in\mathbb{R}$ is such that $f(a)<c<f(b)$, then there exists $x\in [a,b]$ such that $f(x)=c$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Since $[a,b]$ is connected, since $f$ is continuous, $f([a,b])$ is connected.
|
||||
|
||||
By contradiction, if $c\notin f([a,b])$, then $f([a,b])$ has two points $f(a),f(b)$ and $c$ is a point between that isn't in $f([a,b])$. This contradicts the connectedness of $f([a,b])$.
|
||||
|
||||
So $f(a)<c<f(b)$ or $f(b)<c<f(a)$ must hold.
|
||||
|
||||
</details>
|
||||
|
||||
#### Definition of path-connected space
|
||||
|
||||
A topological space $X$ is path-connected if for any two points $x,x'\in X$, there is a continuous map $\gamma:[0,1]\to X$ such that $\gamma(0)=x$ and $\gamma(1)=x'$. Any such continuous map is called a path from $x$ to $x'$.
|
||||
|
||||
> [!NOTE]
|
||||
>
|
||||
> Path-connectedness is a stronger condition than connectedness.
|
||||
|
||||
#### Theorem of path-connectedness and connectedness
|
||||
|
||||
Any path-connected space is connected.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
By contradiction, let $U,V$ be a separation of $X$. In particular, $\exists x\in U, x'\in V$.
|
||||
|
||||
Since $X$ is path-connected, $\exists \gamma:[0,1]\to X$ such that $\gamma(0)=x$ and $\gamma(1)=x'$.
|
||||
|
||||
Then since $\gamma$ is continuous, $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ are open in $[0,1]$ and $[0,1]=\gamma^{-1}(U)\cup \gamma^{-1}(V)$. We want to show that this gives a separation of $[0,1]$.
|
||||
|
||||
Since $U\cap V=\emptyset$, $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ are disjoint.
|
||||
|
||||
$U\cup V=X$ so $\gamma^{-1}(U)\cup \gamma^{-1}(V)=[0,1]$.
|
||||
|
||||
Each of $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ is non-empty because $x\in U\implies 0\in \gamma^{-1}(U)$ and $x'\in V\implies 1\in \gamma^{-1}(V)$.
|
||||
|
||||
This contradicts the assumption that $[0,1]$ is connected.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example of path-connected space</summary>
|
||||
|
||||
A subspace $X$ of $\mathbb{R}^n$ is convex if for any two points $x,x'\in X$, the line segment connecting $x$ and $x'$ is contained in $X$.
|
||||
|
||||
In particular $B_R(x)$ is convex. So $X$ is path-connected.
|
||||
|
||||
---
|
||||
|
||||
Let $X=\mathbb{R}^n\setminus\{0\}$. with $n\geq 2$. Then $X$ is path-connected. (simply walk around the origin)
|
||||
|
||||
</details>
|
||||
|
||||
#### Theorem for invariant property of connectedness
|
||||
|
||||
If $f:X\to Y$ is a continuous and surjective map, and $X$ is connected, then $Y$ is connected.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Take $y,y'\in Y$, since $f$ is surjective, $\exists x,x'\in X$ such that $f(x)=y$ and $f(x')=y'$. Let $\gamma:[0,1]\to X$ be a path from $x$ to $x'$.
|
||||
|
||||
Then $f\circ \gamma:[0,1]\to Y$ is a continuous map. and $f\circ \gamma(0)=y$ and $f\circ \gamma(1)=y'$.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example of connected but not path-connected space</summary>
|
||||
|
||||
Let $A=\{(x,y)\in \mathbb{R}^2\mid y=\sin(1/x), x>0\}$. Then $A$ is connected, and also path-connected.
|
||||
|
||||
However, take $\overline{A}=A\cup \{0\}\times [-1,1]$. Then $\overline{A}$ is not path-connected but connected.
|
||||
|
||||
_Show next time_
|
||||
|
||||
</details>
|
||||
@@ -1,110 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 24)
|
||||
|
||||
## Connected and compact spaces
|
||||
|
||||
### Connectedness
|
||||
|
||||
Recall from example last lecture, there exists a connected space but not path-connected space.
|
||||
|
||||
#### Lemma on connectedness
|
||||
|
||||
Let $X$ be a topological space and $A\subseteq X$ is a connected subspace. If $B\subseteq X$ satisfies $A\subseteq B\subseteq \overline{A}$, then $B$ is connected. In particular, $\overline{A}$ is connected.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Assume that $B$ is not connected. In particular, there are open subspaces $U$ and $V$ of $X$ such that $U\cap B, V\cap B$ is a separation of $B$.
|
||||
|
||||
Take $U\cap A, V\cap A$, we show that this gives a separation of $A$.
|
||||
|
||||
(i) Since $U,V$ are open, $U\cap A, V\cap A$ are open in $A$.
|
||||
|
||||
(ii) Since $(U\cap B)\cap (V\cap B)=\emptyset$, $(U\cap A)\cap (V\cap A)=\emptyset$.
|
||||
|
||||
(iii) Since $(U\cap B)\cup (V\cap B)=B$, any point in $B$ is in either $U\cap B$ or $V\cap B$.
|
||||
|
||||
Since $A\subseteq B$, $(U\cap A)\cup (V\cap A)=A$.
|
||||
|
||||
(iv) $U\cap A$ and $V\cap A$ is nonempty by assumption $U\cap B$ is nonempty and contains $x\in B\cap U\subseteq \overline{A}$. So any open neighborhood of $x$ have non-empty intersection $x'\in A$, so $x'\in U\cap A$ and $U\cap A$ is nonempty. Similarly, $V\cap A$ is nonempty.
|
||||
|
||||
So $U\cap A$ and $V\cap A$ is a separation of $A$, which contradicts the assumption that $A$ is connected.
|
||||
|
||||
Therefore, $B$ is connected.
|
||||
|
||||
</details>
|
||||
|
||||
#### Topologists' sine curve
|
||||
|
||||
Let $A=\{(x,y)\in \mathbb{R}^2\mid y=\sin(1/x), x>0\}$. Then $A$ is connected, and also path-connected.
|
||||
|
||||
$$
|
||||
\gamma(t) = (t, \sin(1/t)) \text{ for } t\in (0,1]
|
||||
$$
|
||||
|
||||
However, take $\overline{A}=A\cup \{0\}\times [-1,1]$. Then $\overline{A}$ is not path-connected but connected.
|
||||
|
||||
<details>
|
||||
<summary>Proof that topologists' sine curve is not path-connected</summary>
|
||||
|
||||
We want to show $X=\overline{A}$ has no continuous path
|
||||
|
||||
$$
|
||||
\gamma:([0,1])\to X
|
||||
$$
|
||||
|
||||
such that $\gamma(0)=(0,0)$ and $\gamma(1)=(1,\sin(1))$.
|
||||
|
||||
If there exists such a path, let $t_0\in [0,1]$ be defined as
|
||||
|
||||
$$
|
||||
t_0=\sup\{t\in [0,1]\mid \gamma(t)=(0,x), x\in [0,1]\}
|
||||
$$
|
||||
|
||||
By the assumption on $t_0$, we can find a sequence $\{t_n\}_{n\in\mathbb{N}_+}\subseteq A$ such that $t_n\to t$.
|
||||
|
||||
By continuity of $\gamma$, we have $\gamma(t_n)\to \gamma(t_0)$, $(0,y_n)\to (0,y_0)$.
|
||||
|
||||
Now focus on the restriction of $\gamma$ to $[t_0,1]$, $\gamma:[t_0,1]\to X$, $\gamma(t_0)=(0,y_0)$, $\gamma(1)=(1,\sin(1))$.
|
||||
|
||||
$t\in (t_0,1]$, then $\gamma(t)\in$ graph of $y=\sin(1/x)$.
|
||||
|
||||
Consider $\pi$ be the projection map to $x$-axis, $\pi\circ \gamma:[t_0,1]\to \mathbb{R}$, $\pi\circ \gamma(t_0)=0$ and $\pi\circ \gamma(1)=1$.
|
||||
|
||||
In particular, there is a sequence $s_n\in [t_0,1]$ such that $s_n\to t_0$ and $\pi\circ \gamma(s_n)=\frac{1}{n\pi+\frac{\pi}{2}}$. (using intermediate value theorem)
|
||||
|
||||
Then $\gamma(s_n)=(\frac{1}{n\pi},\sin(n\pi+\frac{\pi}{2}))=(\frac{1}{n\pi},(-1)^n)$.
|
||||
|
||||
Since as $s_n\to t_0$, and $\gamma$ is continuous, then we get a contradiction that the sequence $\gamma(s_n)$ should converge to $(0,t_0)$ where it is not.
|
||||
|
||||
</details>
|
||||
|
||||
### Compactness
|
||||
|
||||
Motivation: in real numbers.
|
||||
|
||||
#### Extreme value theorem
|
||||
|
||||
Let $f:[a,b]\to \mathbb{R}$ be continuous. Then there are $x_m,x_M\in [a,b]$ such that $f(x_m)\leq f(x)\leq f(x_M)$ for all $x\in [a,b]$.
|
||||
|
||||
#### Definition of cover
|
||||
|
||||
Let $X$ be a topological space. A covering of $X$ is a collection of subsets of $X$ that covers $X$.
|
||||
|
||||
$$
|
||||
\{U_\alpha\}_{\alpha\in I}
|
||||
$$
|
||||
|
||||
such that $X=\bigcup_{\alpha\in I} U_\alpha$.
|
||||
|
||||
An open cover of $X$ is a covering of $X$ such that each $U_\alpha$ is open.
|
||||
|
||||
#### Definition of compact space
|
||||
|
||||
A topological space $X$ is compact if for any open covering $\{U_\alpha\}_{\alpha\in I}$ of $X$, there exists a finite subcovering $\{U_{\alpha_i}\}_{i=1}^n$ such that $X=\bigcup_{i=1}^n U_{\alpha_i}$.
|
||||
|
||||
<details>
|
||||
<summary>Example of non-compact space</summary>
|
||||
|
||||
Consider the interval $(0,1]$, the open covering $(\frac{1}{n},1]$ open in $(0,1]$, $\{(\frac{1}{n},1]\}_{n\in \mathbb{N}_+}$ has no finite subcovering.
|
||||
|
||||
</details>
|
||||
@@ -1,164 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 3)
|
||||
|
||||
## Recall form last lecture
|
||||
|
||||
### Topological Spaces
|
||||
|
||||
#### Basis for a topology
|
||||
|
||||
Let $X$ be a set. A basis for a topology on $X$ is a collection $\mathcal{B}$ (elements of $\mathcal{B}$ are called basis elements) of subsets of $X$ such that:
|
||||
|
||||
1. $\forall x\in X$, $\exists B\in \mathcal{B}$ such that $x\in B$
|
||||
2. $\forall B_1,B_2\in \mathcal{B}$, $\forall x\in B_1\cap B_2$, $\exists B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$
|
||||
|
||||
<details>
|
||||
<summary>Example of standard basis in real numbers</summary>
|
||||
|
||||
Let $X=\mathbb{R}$ and $\mathcal{B}=\{(a,b)|a,b\in \mathbb{R},a<b\}$ (collection of all open intervals).
|
||||
|
||||
Check properties 1:
|
||||
|
||||
for any $x\in \mathbb{R}$, $\exists (x-1,x+1)\in \mathcal{B}$ such that $x\in (x-1,x+1)$
|
||||
|
||||
Check properties 2:
|
||||
|
||||
let $B_1=(a,b)$ and $B_2=(c,d)$ be two basis elements, and $x\in B_1\cap B_2=(\max(a,c),\min(b,d))\in \mathcal{B}$.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example of lower limit basis in real numbers</summary>
|
||||
|
||||
Let $X=\mathbb{R}$ and $\mathcal{B}_{LL}=\{[a,b)|a,b\in \mathbb{R},a<b\}$ (collection of all open intervals).
|
||||
|
||||
Check properties 1:
|
||||
|
||||
for any $x\in \mathbb{R}$, $\exists [x,x+1)\in \mathcal{B}_{LL}$ such that $x\in [x,x+1)$
|
||||
|
||||
Check properties 2:
|
||||
|
||||
let $B_1=[a,b)$ and $B_2=[c,d)$ be two basis elements, and $x\in B_1\cap B_2=[max(a,c),min(b,d))\in \mathcal{B}_{LL}$.
|
||||
|
||||
</details>
|
||||
|
||||
Extend this to $\mathbb{R}^2$.
|
||||
|
||||
#### Definition for cartesian product
|
||||
|
||||
Let $X$ and $Y$ be sets. The cartesian product of $X$ and $Y$ is the set $X\times Y=\{(x,y)|x\in X,y\in Y\}$.
|
||||
|
||||
<details>
|
||||
<summary>Example of open rectangles basis for real plane</summary>
|
||||
|
||||
Let $X=\mathbb{R}^2$ and $\mathcal{B}$ be the collection of rectangle of the form $(a,b)\times (c,d)$ where $a,b,c,d\in \mathbb{R}$ and $a<b,c<d$. (boundary is not included)
|
||||
|
||||
Check properties 1:
|
||||
|
||||
for any $(x,y)\in \mathbb{R}^2$, $\exists (x,y)\in \mathcal{B}$ such that $(x,y)\in (x,y)$
|
||||
|
||||
Check properties 2:
|
||||
|
||||
let $B_1=(a,b)\times (c,d)$ and $B_2=(e,f)\times (g,h)$ be two basis elements, and $(x,y)\in B_1\cap B_2=(max(a,e),min(b,f))\times (max(c,g),min(d,h))\in \mathcal{B}$.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example of open disks basis for real plane</summary>
|
||||
|
||||
Let $X=\mathbb{R}^2$ and $\mathcal{B}$ be the collection of open disks.
|
||||
|
||||
Check properties 1:
|
||||
|
||||
for any $x\in \mathbb{R}^2$, $\exists B_1(x)\in \mathcal{B}$ such that $x\in B_1(x)$.
|
||||
|
||||
Check properties 2:
|
||||
|
||||
let $B_{r_1}(x)$ and $B_{r_2}(y)$ be two basis elements, for every $z\in B_{r_1}(x)\cap B_{r_2}(y)$, $\exists B_{r_3}(z)\in \mathcal{B}$ such that $z\in B_{r_3}(z)\subseteq B_{r_1}(x)\cap B_{r_2}(y)$.
|
||||
|
||||
(even $B_{r_1}(x)\cap B_{r_2}(y)\notin \mathcal{B}$)
|
||||
</details>
|
||||
|
||||
#### Topology generated by a basis
|
||||
|
||||
Let $\mathcal{B}$ be a basis for a topology on $X$. The topology generated by $\mathcal{B}$, denoted by $\mathcal{T}_{\mathcal{B}}$.
|
||||
|
||||
$U\in \mathcal{T}_{\mathcal{B}}\iff \forall x\in U, \exists B\in \mathcal{B}$ such that $x\in B\subseteq U$
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
$\mathcal{T}_{\mathcal{B}}$ is a topology on $X$ because:
|
||||
|
||||
1. $\emptyset \in \mathcal{T}_{\mathcal{B}}$ because $\emptyset \in \mathcal{B}$. $X\in \mathcal{T}_{\mathcal{B}}$ because $\forall x\in X, \exists B\in \mathcal{B}$ such that $x\in B\subseteq X$ (by definition of basis (property 1)))
|
||||
|
||||
2. $\mathcal{T}_{\mathcal{B}}$ is closed under arbitrary unions.
|
||||
|
||||
Want to show $\{U_\alpha | U_\alpha\in \mathcal{T}_{\mathcal{B}}\}_{\alpha \in I}\implies \bigcup_{\alpha \in I} U_\alpha\in \mathcal{T}_{\mathcal{B}}$.
|
||||
|
||||
Because $\forall x\in \bigcup_{\alpha \in I} U_\alpha$, $\exists \alpha_0$ such that $x\in U_{\alpha_0}$. Since $U_{\alpha_0}\in \mathcal{T}_{\mathcal{B}}$, $\exists B\in \mathcal{B}$ such that $x\in B\subseteq U_{\alpha_0}\subseteq \bigcup_{\alpha \in I} U_\alpha$.
|
||||
|
||||
3. $\mathcal{T}_{\mathcal{B}}$ is closed under finite intersections.
|
||||
|
||||
Want to show $U_1,U_2,\ldots,U_n\in \mathcal{T}_{\mathcal{B}}\implies \bigcap_{i=1}^n U_i\in \mathcal{T}_{\mathcal{B}}$.
|
||||
|
||||
If $n=2$, since $\forall x\in U_1\cap U_2$, $x\in U_1$ and $x\in U_2$, $\exists B_1\in \mathcal{B}$ such that $x\in B_1\subseteq U_1$ and $\exists B_2\in \mathcal{B}$.
|
||||
|
||||
Applying the second property of basis, $\exists B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2\subseteq U_1\cap U_2$.
|
||||
|
||||
By induction, we can show that $\bigcap_{i=1}^n U_i\in \mathcal{T}_{\mathcal{B}}$.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example of topology generated by a basis</summary>
|
||||
|
||||
Let $X$ be arbitrary.
|
||||
|
||||
Let $\mathcal{B}=\{x|x\in X\}$ (collection of all singleton subsets of $X$).
|
||||
|
||||
Then $\mathcal{T}$ is the discrete topology.
|
||||
|
||||
</details>
|
||||
|
||||
#### Properties of basis in generated topology
|
||||
|
||||
**Observation 1**: Any $B\in \mathcal{B}$ is an open set in $\mathcal{T}_{\mathcal{B}}$.
|
||||
|
||||
By the defining property of basis, $\forall x\in B$, $\exists x\in B\subseteq B$.
|
||||
|
||||
**Observation 2**: For any collection $\{B_\alpha\}_{\alpha \in I}$, $\bigcup_{\alpha \in I} B_\alpha\in \mathcal{T}_{\mathcal{B}}$.
|
||||
|
||||
By observation 1, each $B_\alpha\in \mathcal{T}_{\mathcal{B}}$. Since $\mathcal{T}_{\mathcal{B}}$ is a topology, $\bigcup_{\alpha \in I} B_\alpha\in \mathcal{T}_{\mathcal{B}}$.
|
||||
|
||||
#### Lemma
|
||||
|
||||
Let $\mathcal{B}$ and $\mathcal{T}_{\mathcal{B}}$ be a basis and the topology generated by $\mathcal{B}$ on $X$. Then,
|
||||
|
||||
$U\in \mathcal{T}_{\mathcal{B}}\iff$ there are basis elements $\{B_\alpha\}_{\alpha \in I}$ such that $U=\bigcup_{\alpha \in I} B_\alpha$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
$(\Rightarrow)$
|
||||
|
||||
If $U\in \mathcal{T}_{\mathcal{B}}$, we want to show that $U$ is a union of basis elements.
|
||||
|
||||
For any $x\in U$, by the definition of $\mathcal{T}_{\mathcal{B}}$, there is a basis element $B_x$ such that $x\in B_x\subseteq U$.
|
||||
|
||||
So, $U\subseteq \bigcup_{x\in U} B_x$.
|
||||
|
||||
Since $\forall B_x\in \{B_x\}_{\alpha \in I}$, $B_x\subseteq U$, we have $U\supseteq\bigcup_{x\in U} B_x$.
|
||||
|
||||
So, $U=\bigcup_{x\in U} B_x$.
|
||||
|
||||
$(\Leftarrow)$
|
||||
|
||||
Applies observation 2.
|
||||
|
||||
</details>
|
||||
|
||||
> [!NOTE]
|
||||
>
|
||||
> A basis for a topology is like a basis for a vector space in the sense that any open set/vector can be represented in terms of basis elements.
|
||||
>
|
||||
> But unlike linear algebra, it's not true that any open set can be written as a union of basis element in a **unique** way.
|
||||
@@ -1,151 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 4)
|
||||
|
||||
## Recall from last lecture
|
||||
|
||||
Assignment due next Thursday. 10PM
|
||||
|
||||
Let $\mathcal{B}$ be a basis for a topology. Then the topology ($\mathcal{T}_{\mathcal{B}}$) **generated** by $\mathcal{B}$ is $\{U\in \mathcal{T}_{\mathcal{B}} \mid \forall x\in U, \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq U\}$.
|
||||
|
||||
## New materials
|
||||
|
||||
### Topology basis
|
||||
|
||||
Given a topology on a set $X$, When is a given collection of subsets of $X$ a basis for a topology?
|
||||
|
||||
Suppose $U\in\mathcal{T}$ is an open set in $X$. If an arbitrary set $\mathcal{C}$ is a basis for $\mathcal{T}$, then by the definition of a topology generated by a basis, we should have the following:
|
||||
|
||||
$$
|
||||
\exists C\in \mathcal{C} \text{ such that } x\in C\subseteq U
|
||||
$$
|
||||
|
||||
#### Theorem of basis of topology
|
||||
|
||||
> [!CAUTION]
|
||||
>
|
||||
> In this course, we use lowercase letters to denote element of a set, and uppercase letters to denote sets. We use $\mathcal{X}$ to denote set of subsets of $X$.
|
||||
|
||||
Let $(X,\mathcal{T})$ be a topological space. Let $\mathcal{C}\subseteq \mathcal{T}$ be a collection of subsets of $X$ satisfying the following property:
|
||||
|
||||
$$
|
||||
\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } U\subseteq C
|
||||
$$
|
||||
|
||||
Then $\mathcal{C}$ is a basis and the topology generated by $\mathcal{C}$ is $\mathcal{T}$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
We want to show that $\mathcal{C}$ is a basis.
|
||||
|
||||
> Recall the definition of a basis:
|
||||
>
|
||||
> 1. $\forall x\in X$, there is $B\in \mathcal{B}$ such that $x\in B$
|
||||
> 2. $\forall B_1,B_2\in \mathcal{B}$, $\forall x\in B_1\cap B_2$, there is $B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$
|
||||
|
||||
First, we want to show that $\mathcal{C}$ satisfies the first property.
|
||||
|
||||
Take $x\in X$. Since $X\in \mathcal{T}$, we can apply the given condition ($
|
||||
\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } U\subseteq C
|
||||
$) to get $C\in \mathcal{C}$ such that $x\in C\subseteq X$.
|
||||
|
||||
Next, we want to show that $\mathcal{C}$ satisfies the second property.
|
||||
|
||||
Let $C_1,C_2\in \mathcal{C}$ and $x\in C_1\cap C_2$. Since $C_1,C_2\in \mathcal{T}$, by the definition of $\mathcal{T}$, we have $U=C_1\cap C_2\in \mathcal{T}$.
|
||||
|
||||
We can apply the given condition to get $C_3\in \mathcal{C}$ such that $x\in C_3\subseteq U=C_1\cap C_2$.
|
||||
|
||||
---
|
||||
|
||||
Then we want to show that the topology generated by $\mathcal{C}$ is $\mathcal{T}$.
|
||||
|
||||
> Recall the definition of the topology generated by a basis:
|
||||
>
|
||||
> To prove this, we need to show that $\forall U\in \mathcal{T}\implies U\in \mathcal{T}_{\mathcal{C}}$ and $\forall U\in \mathcal{T}_{\mathcal{C}}\implies U\in \mathcal{T}$.
|
||||
>
|
||||
> Moreover, from last lecture, we have $U\in \mathcal{T}_{\mathcal{B}}\iff U=\bigcup_{\alpha \in I} B_\alpha$ for some $\{B_\alpha\}_{\alpha \in I}\subseteq \mathcal{B}$.
|
||||
|
||||
First, we want to show that $\forall U\in \mathcal{T}_{\mathcal{C}}\implies U\in \mathcal{T}$.
|
||||
|
||||
Let $U=\bigcup_{\alpha \in I} C_\alpha$ for some $\{C_\alpha\}_{\alpha \in I}\subseteq \mathcal{C}$. Then since $C_\alpha\in \mathcal{T}$, by the definition of $\mathcal{T}$, we have $U\in \mathcal{T}$.
|
||||
|
||||
Next, we want to show that $\forall U\in \mathcal{T}\implies U\in \mathcal{T}_{\mathcal{C}}$.
|
||||
|
||||
Let $U\in \mathcal{T}$. Then $\forall x\in U$ by the given condition, we have $C\in \mathcal{C}$ such that $x\in C\subseteq U$.
|
||||
|
||||
So, $U=\bigcup_{\alpha \in I} C_\alpha\in \mathcal{T}_{\mathcal{C}}$. (using the [same trick last time](https://notenextra.trance-0.com/Math4201/Math4201_L3#lemma))
|
||||
|
||||
</details>
|
||||
|
||||
Let $\mathcal{T}$ be the topology on $X$. Then $\mathcal{T}$ itself satisfies the basis condition.
|
||||
|
||||
#### Definition of subbasis of topology
|
||||
|
||||
A subbasis of a topology on a set $X$ is a collection $\mathcal{S}\subseteq \mathcal{T}$ of subsets of $X$ such that their union is $X$.
|
||||
|
||||
$$
|
||||
\mathcal{S}=\{S_{\alpha}\mid S_\alpha\subseteq X\}_{\alpha \in I}\text{ and }\bigcup_{\alpha \in I} S_\alpha=X
|
||||
$$
|
||||
|
||||
#### Definition of topology generated by a subbasis
|
||||
|
||||
If we consider the basis generated by the subbasis $\mathcal{S}$ by the following:
|
||||
|
||||
$$
|
||||
\mathcal{B}=\{B\mid B\text{ is the intersection of a finite number of elements of }\mathcal{S}\}
|
||||
$$
|
||||
|
||||
Then $\mathcal{B}$ is a basis.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
First, $\forall x\in X$, there is $S_\alpha\in \mathcal{S}$ such that $x\in S_\alpha$. In particular, $x\in \mathcal{B}$.
|
||||
|
||||
Second, let $B_1,B_2\in \mathcal{B}$. Since $B_1$ is the intersection of a finite number of elements of $\mathcal{S}$, we have $B_1=\bigcap_{i=1}^n S_{i_1}, B_2=\bigcap_{i=1}^n S_{i_2}$ for some $S_{i_1},S_{i_2}\in \mathcal{S}$.
|
||||
|
||||
So $B_1\cap B_2$ is the intersection of finitely many elements of $\mathcal{S}$.
|
||||
|
||||
So $B_1\cap B_2\in \mathcal{B}$.
|
||||
|
||||
</details>
|
||||
|
||||
We call the topology generated by $\mathcal{B}$ the topology generated by the subbasis $\mathcal{S}$. Denote it by $\mathcal{T}_{\mathcal{S}}$.
|
||||
|
||||
An open set with respect to $\mathcal{T}_{\mathcal{S}}$ is a subset of $X$ such that it can be written as a union of finitely intersections of elements of $\mathcal{S}$.
|
||||
|
||||
<details>
|
||||
|
||||
<summary>Example (standard topology on real numbers)</summary>
|
||||
|
||||
Let $X=\mathbb{R}$. Take $\mathcal{S}=\{(-\infty, a)|a\in \mathbb{R}\}\cup \{(a,+\infty)|a\in \mathbb{R}\}$.
|
||||
|
||||
We claim this is a subbasis of the standard topology on $\mathbb{R}$.
|
||||
|
||||
The basis $\mathcal{B}$ associated with $\mathcal{S}$ is the collection of all open intervals.
|
||||
|
||||
$$
|
||||
\mathcal{B}=\{(a,b)=(-\infty, b)\cap (a,+\infty)\}
|
||||
$$
|
||||
|
||||
So, $\mathcal{B}=\mathcal{B}_{st}$ (the standard basis).
|
||||
|
||||
This topology on $\mathbb{R}$ is the same as the standard topology on $\mathbb{R}$.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example (finite complement topology)</summary>
|
||||
|
||||
Let $X$ be an arbitrary set. Let $\mathcal{S}$ defined as follows:
|
||||
|
||||
$$
|
||||
\mathcal{S}=\{S\subseteq X\mid S=X\setminus \{x\} \text{ for some } x\in X\}
|
||||
$$
|
||||
|
||||
Let $x,y\in X$ and $x\neq y$. Then $S_x=X\setminus \{x\}$ and $S_y=X\setminus \{y\}$ are two elements of $\mathcal{S}$. Since $x\neq y$, we have $S_x\cup S_y=X\setminus \{x\}\cup X\setminus \{y\}=X$. So $\mathcal{S}$ is a subbasis of $X$.
|
||||
|
||||
So, the basis associated with $\mathcal{S}$, $\mathcal{B}$, is the collection of subsets of $X$ with finite complement.
|
||||
|
||||
This is in fact a topology, which is the **finite complement topology** on $X$.
|
||||
|
||||
</details>
|
||||
@@ -1,106 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 5 Bonus)
|
||||
|
||||
## Comparison of two types of topologies
|
||||
|
||||
Let $X=\mathbb{R}^2$ and the two types of topologies are:
|
||||
|
||||
The "circular topology":
|
||||
|
||||
$$
|
||||
\mathcal{T}_c=\{B_r(p)\mid p\in \mathbb{R}^2,r>0\}
|
||||
$$
|
||||
|
||||
The "rectangle topology":
|
||||
|
||||
$$
|
||||
\mathcal{T}_r=\{(a,b)\times (c,d)\mid a,b,c,d\in \mathbb{R},a<b,c<d\}
|
||||
$$
|
||||
|
||||
> Are these two topologies the same?
|
||||
|
||||
### Comparison of two topologies
|
||||
|
||||
#### Definition of finer and coarser
|
||||
|
||||
Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on $X$. We say $\mathcal{T}$ is finer than $\mathcal{T}'$ if $\mathcal{T}'\subseteq \mathcal{T}$. We say $\mathcal{T}$ is coarser than $\mathcal{T}'$ if $\mathcal{T}\subseteq \mathcal{T}'$. We say $\mathcal{T}$ and $\mathcal{T}'$ are equivalent if $\mathcal{T}=\mathcal{T}'$.
|
||||
|
||||
$\mathcal{T}$ is strictly finer than $\mathcal{T}'$ if $\mathcal{T}'\subsetneq \mathcal{T}$. (that is, $\mathcal{T}'$ is finer and not equivalent to $\mathcal{T}$)
|
||||
$\mathcal{T}$ is strictly coarser than $\mathcal{T}'$ if $\mathcal{T}\subsetneq \mathcal{T}'$. (that is, $\mathcal{T}$ is coarser and not equivalent to $\mathcal{T}'$)
|
||||
|
||||
<details>
|
||||
<summary>Example (discrete topology is finer than the trivial topology)</summary>
|
||||
|
||||
Let $X$ be an arbitrary set. The discrete topology is $\mathcal{T}_1 = \mathcal{P}(X)=\{U \subseteq X\}$
|
||||
|
||||
The trivial topology is $\mathcal{T}_0 = \{\emptyset, X\}$
|
||||
|
||||
Clearly, $\mathcal{T}_1 \subseteq \mathcal{T}_0$.
|
||||
|
||||
So the discrete topology is finer than the trivial topology.
|
||||
|
||||
</details>
|
||||
|
||||
#### Lemma
|
||||
|
||||
> [!TIP]
|
||||
>
|
||||
> Motivating condition:
|
||||
>
|
||||
> We want $U$ be an open set in $\mathcal{T}'$, then $U$ has to be open with respect to $\mathcal{T}$. In other words, $\forall x\in U, \exists$ some $B\in \mathcal{B}$ such that $x\in B\subseteq U$.
|
||||
|
||||
Let $\mathcal{T}$ and $\mathcal{T}'$ be topologies on $X$ associated with bases $\mathcal{B}$ and $\mathcal{B}'$. Then
|
||||
|
||||
$$
|
||||
\mathcal{T}\text{ is finer than } \mathcal{T}'\iff \text{ for any } x\in X, x\in B'\in \mathcal{B}', \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq B'
|
||||
$$
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
$(\Rightarrow)$
|
||||
|
||||
Let $B'\in \mathcal{B}'$. If $x\in B'$, then $B'\in \mathcal{T}'$ and $T$ is finer than $T'$, so $B'\in \mathcal{T}$.
|
||||
|
||||
Take $T=\mathcal{T}_{\mathcal{B}}$. $\exists B\in \mathcal{B}$ such that $x\in B\subseteq B'$.
|
||||
|
||||
$(\Leftarrow)$
|
||||
|
||||
Let $U\in \mathcal{T}$. Then $U=\bigcup_{\alpha \in I} B_\alpha'$ for some $\{B_\alpha'\}_{\alpha \in I}\subseteq \mathcal{B}'$.
|
||||
|
||||
For any $B_\alpha'$ and any $x\in \mathcal{B}_\alpha'$, $\exists B_\alpha\in \mathcal{B}$ such that $x\in B_\alpha\subseteq B_\alpha'$.
|
||||
|
||||
Then $B_\alpha'$ is open set in $\mathcal{T}$.
|
||||
|
||||
So $U$ is open in $\mathcal{T}$.
|
||||
|
||||
$T$ is finer than $T'$.
|
||||
|
||||
</details>
|
||||
|
||||
Back to the example:
|
||||
|
||||
For every point in open circle, we can find a rectangle that contains it.
|
||||
|
||||
For every point in open rectangle, we can find a circle that contains it.
|
||||
|
||||
So these two topologies are equivalent.
|
||||
|
||||
#### Standard topology in $\mathbb{R}^2$
|
||||
|
||||
The standard topology in $\mathbb{R}^2$ is the topology generated by the basis $\mathcal{B}_{st}=\{(a,b)\times (c,d)\mid a,b,c,d\in \mathbb{R},a<b,c<d\}$. This is equivalent to the topology generated by the basis $\mathcal{B}_{disk}=\{(x,y)\in \mathbb{R}^2|d((x,y),(a,b))<r\}$.
|
||||
|
||||
<details>
|
||||
|
||||
<summary>Example (lower limit topology is strictly finer than the standard topology)</summary>
|
||||
|
||||
The lower limit topology is the topology generated by the basis $\mathcal{B}_{ll}=\{[a,b)\mid a,b\in \mathbb{R},a<b\}$.
|
||||
|
||||
This is finer than the standard topology.
|
||||
|
||||
Since $(a,b)\in \mathcal{B}_{st}$, we have $\forall x\in (a,b), \exists B=[x,b)\in \mathcal{B}_{ll}$ such that $x\in B\subsetneq (a,b)$.
|
||||
|
||||
So the lower limit topology is strictly finer than the standard topology.
|
||||
|
||||
$[0,1)$ is not open in the standard topology. but it is open in the lower limit topology.
|
||||
|
||||
</details>
|
||||
@@ -1,144 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 6)
|
||||
|
||||
## Product topology
|
||||
|
||||
### Define topological spaces on cartesian product of two topological spaces
|
||||
|
||||
Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be two topological spaces.
|
||||
|
||||
$$
|
||||
X\times Y=\{(x,y)|x\in X,y\in Y\}
|
||||
$$
|
||||
|
||||
Goal: Define a topology on $X\times Y$.
|
||||
|
||||
One way is to take $\mathcal{B}_{X\times Y}=\{(U\times V)|U\in \mathcal{T}_X,V\in \mathcal{T}_Y\}$ is a basis for the topology on $X\times Y$.
|
||||
|
||||
> There are two ways to define the topology on $\mathbb{R}^2$: By rectangles ($\mathcal{B}_{rect}=\{(a,b)\times (c,d)|a,b,c,d\in \mathbb{R},a<b,c<d\}$) or by open disks ($\mathcal{B}_{disk}=\{(x,y)\in \mathbb{R}^2|d((x,y),(a,b))<r\}$). (check bonus video)
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Take $U=X,V=Y$, then $(x,y)\in (U\times V)=X\times Y$.
|
||||
|
||||
So the first property of basis is satisfied.
|
||||
|
||||
Check the second property of basis:
|
||||
|
||||
Let $B_1=U_1\times V_1,B_2=U_2\times V_2$ be two basis elements, and $(x,y)\in B_1\cap B_2$.
|
||||
|
||||
$$
|
||||
B_1\cap B_2=(U_1\times V_1)\cap (U_2\times V_2)=(U_1\cap U_2)\times (V_1\cap V_2)
|
||||
$$
|
||||
|
||||
Since $U_1\cap U_2\in \mathcal{T}_X$ and $V_1\cap V_2\in \mathcal{T}_Y$, we have $(U_1\cap U_2)\times (V_1\cap V_2)\in \mathcal{B}_{X\times Y}$.
|
||||
|
||||
Take $B_3=(U_1\cap U_2)\times (V_1\cap V_2)$, then $(x,y)\in B_3=B_1\cap B_2$.
|
||||
|
||||
</details>
|
||||
|
||||
> [!CAUTION]
|
||||
>
|
||||
> $\mathcal{B}_{X\times Y}$ is not a topology on $X\times Y$.
|
||||
>
|
||||
> $\mathcal{B}_{X\times Y}$ is not closed with respect to arbitrary unions.
|
||||
|
||||
#### Definition of product topology
|
||||
|
||||
Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be two topological spaces.
|
||||
|
||||
$$
|
||||
\mathcal{B}_{X\times Y}=\{(U\times V)|U\in \mathcal{T}_X,V\in \mathcal{T}_Y\}
|
||||
$$
|
||||
|
||||
The product topology or $X\times Y$ is the topology generated by $\mathcal{B}_{X\times Y}$.
|
||||
|
||||
Let $\mathcal{B}_X$ be a basis for $\mathcal{T}_X$ and $\mathcal{B}_Y$ be a basis for $\mathcal{T}_Y$.
|
||||
|
||||
If we define
|
||||
|
||||
$$
|
||||
\mathcal{B}'_{X\times Y}=\{(U\times V)|U\in \mathcal{B}_X,V\in \mathcal{B}_Y\}
|
||||
$$
|
||||
|
||||
#### Proposition
|
||||
|
||||
$\mathcal{B}'_{X\times Y}$ is a basis for the product topology on $X\times Y$.
|
||||
|
||||
> [!NOTE]
|
||||
>
|
||||
> This basis is smaller than $\mathcal{B}_{X\times Y}$.
|
||||
>
|
||||
> Consider $X=Y=\mathbb{R}$ and $U=(a,b)\cap (c,d)$ and $V=(e,f)\cap (g,h)$. (Assume $a<b,c<d,e<f,g<h$) The union of $U$ and $V$ is four rectangles, which is not in $\mathcal{B}'_{X\times Y}$. but it is in $\mathcal{B}_{X\times Y}$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Using the [lemma](https://notenextra.trance-0.com/Math4201/Math4201_L4#theorem-of-basis-of-topology) from Friday. it suffices to show that:
|
||||
|
||||
Let $W\in X\times Y$ and $(x,y)\in W$, we need to show that there are $B\in \mathcal{B}_X, B'\in \mathcal{B}_Y$ such that $(x,y)\in (B\times B')\subseteq W$.
|
||||
|
||||
Since $W$ is open with respect to the topology generated by $\mathcal{B}_{X\times Y}$, in particular, there is $U\times V$ such that $(x,y)\in U\times V\subseteq W$. And $x\in U$ and $y\in V$.
|
||||
|
||||
Since $U\in \mathcal{B}_X$ and $V\in \mathcal{B}_Y$, by property of basis $\mathcal{B}_X$ and $\mathcal{B}_Y$, $\forall x\in U$, $\exists B\in \mathcal{B}_X$ such that $x\in B\subseteq U$ and $\forall y\in V$, $\exists B'\in \mathcal{B}_Y$ such that $y\in B'\subseteq V$.
|
||||
|
||||
So $(x,y)\in (B\times B')\subseteq U\times V\subseteq W$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Definition of standard topology on $\mathbb{R}^n$
|
||||
|
||||
The standard topology on $\mathbb{R}^n$ is defined as the product topology on $\mathbb{R}^{n-1}\times \mathbb{R}$.
|
||||
|
||||
## Subspace topology
|
||||
|
||||
Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$.
|
||||
|
||||
Want to define a topology on $Y$.
|
||||
|
||||
$$
|
||||
\mathcal{T}_Y=\{U\cap Y|U\in \mathcal{T}\}
|
||||
$$
|
||||
|
||||
We claim that $\mathcal{T}_Y$ is a topology on $Y$, called as the subspace topology on $Y$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
First, $\emptyset \cap Y=\emptyset \in \mathcal{T}_Y$ and $Y=X\cap Y\in \mathcal{T}_Y$.
|
||||
|
||||
Second, let $\{U_\alpha\cap Y\}_{\alpha \in I}$ be collection of open sets in $\mathcal{T}_Y$. Note that $U_\alpha\in \mathcal{T}$ for all $\alpha \in I$.
|
||||
|
||||
So, $\bigcup_{\alpha \in I} U_\alpha\cap Y=\left(\bigcup_{\alpha \in I} U_\alpha\right)\cap Y\in \mathcal{T}_Y$ because $\bigcup_{\alpha \in I} U_\alpha\in \mathcal{T}$.
|
||||
|
||||
Third, let $\{U_i\cap Y\}_{i=1}^n$ be a finite collection of open sets in $\mathcal{T}_Y$. Note that $U_i\in \mathcal{T}$ for all $i=1,2,\ldots,n$.
|
||||
So, $\bigcap_{i=1}^n U_i\cap Y=\left(\bigcap_{i=1}^n U_i\right)\cap Y\in \mathcal{T}_Y$ because $\bigcap_{i=1}^n U_i\in \mathcal{T}$.
|
||||
|
||||
</details>
|
||||
|
||||
### Generate basis for subspace topology
|
||||
|
||||
Let $\mathcal{B}$ is a basis for $(X,\mathcal{T})$. We'd like to use $\mathcal{B}$ to define a basis for $(Y,\mathcal{T}_Y)$.
|
||||
|
||||
$$
|
||||
\mathcal{B}_Y=\{B\cap Y|B\in \mathcal{B}\}
|
||||
$$
|
||||
|
||||
#### Proposition for basis of subspace topology
|
||||
|
||||
$\mathcal{B}_Y$ is a basis for a topology on $Y$ that generates the subspace topology on $Y$.
|
||||
|
||||
Proof as exercise. (same as the proof for the basis of product topology)
|
||||
|
||||
<details>
|
||||
<summary>Example: not every open set in subspace topology is open in the original space</summary>
|
||||
|
||||
Let $X=\mathbb{R}$ with standard topology and $Y=[0,1]\cup [2,3]$. equipped with subspace topology generated by the standard basis for $\mathbb{R}$.
|
||||
|
||||
so $[0,1]=(-1,\frac{3}{2})\cap Y$ In particular, $[0,1]$ is open set in $Y$, but not an open set in $\mathbb{R}$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Lemma of open set in subspace topology
|
||||
|
||||
Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$ is open. $Z\subseteq Y$ is open subset with respect to the subspace topology on $Y$. Then $Z$ is open in $X$.
|
||||
@@ -1,163 +0,0 @@
|
||||
# Math 4201 Topology I (Lecture 7)
|
||||
|
||||
## Review from last lecture
|
||||
|
||||
Not every open set in subspace topology is open in the original space
|
||||
|
||||
Let $X=\mathbb{R}$ with standard topology and $Y=[0,1]\cup [2,3]$. equipped with subspace topology generated by the standard basis for $\mathbb{R}$.
|
||||
|
||||
so $[0,1]=(-1,\frac{3}{2})\cap Y$ In particular, $[0,1]$ is open set in $Y$, but not an open set in $\mathbb{R}$.
|
||||
|
||||
## New materials
|
||||
|
||||
### Closed sets in topological space
|
||||
|
||||
#### Proposition of open set in subspace topology
|
||||
|
||||
If $X$ is a topological space, then $Y\subseteq X$ is open and is with the subspace topology. If $Z\subset Y$ is open subspace of $Y$, then $Z$ is also an open subspace of $X$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Since $Z\subset Y$ is open in the subspace topology, there is an open $U\subset Y$ such that $Z=U\cap Y$.
|
||||
|
||||
SInce $Z$ is the intersection of open sets in $X$, then $Z$ is open in $X$.
|
||||
</details>
|
||||
|
||||
|
||||
#### Definition of closed set
|
||||
|
||||
For any topology $\mathcal{T}$ on a set $X$, a subset $Z\subseteq X$ is said to be **closed** if its complement $Z^c$ is an open set (in $X$).
|
||||
|
||||
> Note the complement is defined $Z=X\setminus Z$.
|
||||
|
||||
<details>
|
||||
<summary>Example of closed set in standard topology of real numbers</summary>
|
||||
|
||||
For example, $[a,b]$ is a closed set in the standard topology of real numbers. since $\mathbb{R}-[a,b]=(-\infty,a)\cup (b,\infty)$ is an open set.
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example of closed set in trivial topology</summary>
|
||||
|
||||
For any set $X$, the trivial topology is $\mathcal{T}_0=\{\emptyset, X\}$.
|
||||
Since $X^c=\emptyset$ is an open set, $X$ is a closed set.
|
||||
Since $\emptyset^c=X$ is an open set, $\emptyset$ is a closed set.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example of closed set in finite complement topology</summary>
|
||||
|
||||
For any set $X$, the finite complement topology is $\mathcal{T}_1=\{U\subseteq X\mid X\setminus U\text{ is finite}\}$.
|
||||
|
||||
Then the set of all finite subsets of $X$ is a closed set.
|
||||
</details>
|
||||
|
||||
For general, if $\mathcal{T}$ is a topology on $X$, then:
|
||||
|
||||
1. $\emptyset, X$ are closed sets.
|
||||
2. $\mathcal{T}$ is closed with respect to arbitrary unions.
|
||||
|
||||
Let $\{Z_\alpha\}_{\alpha \in I}$ be an arbitrary collection of closed sets in $X$. Then $X-Z_\alpha$ is open for each $\alpha \in I$. So $\forall \alpha \in I$. $\bigcup_{\alpha \in I} (X-Z_\alpha)=X-\bigcap_{\alpha \in I} Z_\alpha$ is open. So $\bigcap_{\alpha \in I} Z_\alpha$ is closed.
|
||||
|
||||
So the corollary is: an arbitrary intersection of closed sets is closed.
|
||||
|
||||
3. $\mathcal{T}$ is closed with respect to finite intersections. This also implies that a finite union of closed sets is closed.
|
||||
|
||||
If $\{Z_1, Z_2, \ldots, Z_n\}$ is a closed subset of $X$, then $X-Z_i$ is open for each $i=1,2,\ldots,n$. So $\forall i=1,2,\ldots,n$. $\bigcap_{i=1}^n (X-Z_i)=X-\bigcup_{i=1}^n Z_i$ is open. So $\bigcup_{i=1}^n Z_i$ is closed.
|
||||
|
||||
> [!NOTE]
|
||||
>
|
||||
> We can also define the topology using the closed sets instead of the open sets.
|
||||
>
|
||||
> 1. $\emptyset, X$ are closed sets.
|
||||
> 2. $\mathcal{T}$ is closed with respect to arbitrary intersections.
|
||||
> 3. $\mathcal{T}$ is closed with respect to finite unions.
|
||||
>
|
||||
> This yields the same topology.
|
||||
|
||||
#### Theorem of closed set in subspace topology
|
||||
|
||||
Let $X$ is a topological space and $Y\subseteq X$ equipped with the subspace topology.
|
||||
|
||||
A subset $Z\subseteq Y$ is closed in $Y$ if and only if $Z$ is the intersection of a closed $W\subseteq X$ and $Y$. That is $Z=W\cap Y$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
$\Rightarrow$
|
||||
|
||||
If $Z$ is closed in $Y$, then $Y-Z$ is open in $Y$.
|
||||
|
||||
Then, there is open set $U\subseteq X$ such that $Y-Z=U\cap Y$.
|
||||
|
||||
So $Z=(X-U)\cap Y$, $X-U$ is closed in $X$ because $U$ is open in $X$.
|
||||
|
||||
Take $W=X-U$.
|
||||
|
||||
$\Leftarrow$
|
||||
|
||||
If $Z=W\cap Y$ for some closed $W\subseteq X$, then $Y-Z=Y-(W\cap Y)=(Y-W)\cap Y$ is open in $Y$.
|
||||
So $Z$ is closed in $Y$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Lemma of closed in closed subspace
|
||||
|
||||
Let $X$ is a topological space and $Y\subseteq X$ is closed and is equipped with the subspace topology. Then any closed subset of $Y$ is also closed in $X$.
|
||||
|
||||
> [!WARNING]
|
||||
>
|
||||
> Not any subset of a topological space $X$ is either open or closed.
|
||||
|
||||
<details>
|
||||
<summary>Example of open and closed subset</summary>
|
||||
|
||||
Let $X=\mathbb{R}$ with standard topology.
|
||||
|
||||
$(a,b)$ is open, but not closed.
|
||||
$[a,b]$ is closed, but not open.
|
||||
$[a,b)$ is neither open nor closed.
|
||||
$\emptyset,\mathbb{R}$ is both open and closed.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example of open and closed subset in other topologies</summary>
|
||||
|
||||
Let $X=[0,1]\cup (2,3)$ induced by the standard topology of $\mathbb{R}$.
|
||||
|
||||
$Z=[0,1]$ is an open subset of $X$.
|
||||
|
||||
$Z=[0,1]$ is also closed subset of $X$ since $Z=[0,1]\cap X$ is open in $\mathbb{R}$.
|
||||
|
||||
</details>
|
||||
|
||||
We can associate an open and a closed to any subset $A$ of a topological space $X$.
|
||||
|
||||
#### Interior and closure of a set
|
||||
|
||||
The interior of a set $A$ is defined as follows:
|
||||
|
||||
$$
|
||||
\operatorname{Int}(A)=\bigcup_{U\subseteq A, U\text{ is open in }X} U
|
||||
$$
|
||||
|
||||
Also denoted as $A^\circ$.
|
||||
|
||||
The interior of a set $A$ is the largest open subset of $A$.
|
||||
|
||||
That is $\forall U\subseteq A, U\text{ is open in }X$, then $U\subseteq \operatorname{Int}(A)$. (by definition that $U$ must be in collection of open sets that is a subset of $A$)
|
||||
|
||||
#### Closure of a set
|
||||
|
||||
The closure of a set $A$ is the smallest closed subset of $X$ that contains $A$.
|
||||
|
||||
> Note that if we change the definition as the intersection of all closed subsets of $X$ that **contained in $A$**, we will get the empty set.
|
||||
|
||||
$$
|
||||
\overline{A}=\bigcap_{A\subseteq C, C\text{ is closed in }X} C
|
||||
$$
|
||||
|
||||
The closure of a set $A$ is the smallest closed subset of $X$ that contains $A$. (follows the same logic as the previous definition)
|
||||
@@ -1,142 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 8)
|
||||
|
||||
Recall from real analysis, a set is closed if and only if it has limit points.
|
||||
|
||||
## New materials
|
||||
|
||||
### Limit points
|
||||
|
||||
Let $(X,\mathcal{T})$ be a topological space. $A$ is a subset of $X$, then we say $x\in X$ is a limit point of $A$ if any open set $U\subset X$ containing $x$ has another point $y\in A-\{x\}$.
|
||||
|
||||
<details>
|
||||
<summary>Example of limit points</summary>
|
||||
|
||||
Let $X=\mathbb{R}$ with standard topology.
|
||||
|
||||
Let $A=(0,1)$, then set of limit points of $A$ is $[0,1]$.
|
||||
|
||||
Let $A=\left\{\frac{1}{n}\right\}_{n\in \mathbb{N}}$, then set of limit points of $A$ is $\{0\}$.
|
||||
|
||||
Let $A=\{0\}\cup (1,2)$, then set of limit points of $A$ is $[1,2]$
|
||||
|
||||
Let $A=\mathbb{Z}$, then set of limit points of $A$ is $\emptyset$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Proposition of limit points and closed sets
|
||||
|
||||
A set is close if and only if it has limit points.
|
||||
|
||||
Theorem: For any subset $A$ of a topological space $X$, the closure of $A$ is $\overline{A}=A\cup A'$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
First we want to prove the theorem implies the proposition,
|
||||
|
||||
$\Rightarrow$
|
||||
|
||||
Let $A$ be a close set in $X$, then $\overline{A}=A$ because $A$ in the intersection of all closed subsets $Z\subseteq A$ in $X$ that contains $A$.
|
||||
|
||||
So $Z=A$ is such a closed subset of $X$ that contains $A$.
|
||||
|
||||
By the theorem, $\overline{A}=A\cup A'$. Combining this with the fact that $A$ is closed, we have $A=A\cup A'$.
|
||||
|
||||
So $A'\subseteq A$.
|
||||
|
||||
$\Leftarrow$
|
||||
|
||||
Suppose $A\subseteq X$ is a set that includes all its limit points, then $A'\subseteq A$.Then $A'\cup A=A$.
|
||||
|
||||
By the theorem, $\overline{A}=A\cup A'=A$.
|
||||
|
||||
Since $\overline{A}$ is the smallest closed subset of $X$ that contains $A$, we have $A$ is closed.
|
||||
|
||||
</details>
|
||||
|
||||
#### Definition of neighborhood
|
||||
|
||||
Let $(X,\mathcal{T})$ be a topological space. A neighborhood of a point $x\in X$ is an open set $U\in \mathcal{T}$ such that $x\in U$.
|
||||
|
||||
#### Lemma of intersection of neighborhoods for closure of a set
|
||||
|
||||
$x\in \overline{A}$ if and only if any neighborhood of $U$ of $x$ non-trivial intersects $A$. ($A\cap U\neq \emptyset$)
|
||||
|
||||
<details>
|
||||
<summary>Proof of Lemma</summary>
|
||||
|
||||
$\Leftarrow$
|
||||
|
||||
We proceed by contradiction.
|
||||
|
||||
Suppose $A\notin \overline{A}$, then $x\notin \overline{A}$.
|
||||
|
||||
Then $\overline{A}=\bigcap_{A\subseteq Z, Z\text{ is closed}} Z$
|
||||
|
||||
So, there is $A\subseteq Z\subset X$ and $Z$ is closed.
|
||||
|
||||
So this implies that $x\in X-Z\coloneq U$ and $U$ is open since it a complement of a closed set $Z$.
|
||||
|
||||
Since $A\subseteq Z$, we have $A\cap U= \emptyset$. (disjoint)
|
||||
|
||||
So $U$ and $A$ are disjoint. So $U$ is an open neighborhood of $x$ that is disjoint from $A$.
|
||||
|
||||
This contradicts the assumption that $x\in \overline{A}$.
|
||||
|
||||
$\Rightarrow$
|
||||
|
||||
Let $x\in \overline{A}$, and we want to show that any neighborhood of $U$ of $x$ non-trivial intersects $A$. ($A\cap U\neq \emptyset$)
|
||||
|
||||
By contradiction, suppose that there is an open neighborhood of $x$ that is disjoint from $A$. Then $Z\coloneq X-U$ is closed and $A\subseteq Z$ because $U\cap A= \emptyset$.
|
||||
|
||||
Also $x\notin Z$.
|
||||
|
||||
By the definition of closure, $\overline{A}=\subset Z$.
|
||||
|
||||
Since $x\notin Z$, we have $x\notin \overline{A}$.
|
||||
|
||||
This contradicts the assumption that $x\in \overline{A}$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Proof of theorem
|
||||
|
||||
For any subset $A$ of a topological space $X$, the closure of $A$ is $\overline{A}=A\cup A'$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
First we show $A\subseteq \overline{A}$.
|
||||
|
||||
If $x\in A'$, then any open neighborhood $U$ of $x$ has a non-trivial intersection with $A$ by the lemma.
|
||||
|
||||
So $x\in \overline{A}$.
|
||||
|
||||
We already know $A\subseteq \overline{A}$.
|
||||
|
||||
Therese two inductions implies $A\cup A'\subseteq \overline{A}$.
|
||||
|
||||
Next we show that $\overline{A}\subseteq A\cup A'$.
|
||||
|
||||
If $x\in \overline{A}$, then by the lemma, any open neighborhood $U$ of $x$ has a non-trivial intersection with $A$.
|
||||
|
||||
If $x\in A$, then $x\in A\cup A'$.
|
||||
|
||||
If $x\notin A$, then the intersection of any open neighborhood $U$ of $x$ with $A$ does not contain $x$.
|
||||
|
||||
This implies that this intersection has to include a point $y$ that is not $x$.
|
||||
|
||||
Since this holds for any open neighborhood $U$ of $x$, we have $x\in A'$. ($x$ is a limit point of $A$)
|
||||
|
||||
So $x\in A'$.
|
||||
|
||||
Therese two inductions implies $\overline{A}\subseteq A\cup A'$.
|
||||
|
||||
</details>
|
||||
|
||||
> [!TIP]
|
||||
>
|
||||
> Now the three definition of closure are equivalent.
|
||||
>
|
||||
> 1. The smallest closed subset of $X$ that contains $A$.
|
||||
> 2. $A\cup A'$.
|
||||
@@ -1,172 +0,0 @@
|
||||
# Math4201 Topology I (Lecture 9)
|
||||
|
||||
## Convergence of sequences
|
||||
|
||||
Let $(X,\mathcal{T})$ be a topological space and $\{x_n\}_{n\in\mathbb{N}_+}$ be a sequence of points in $X$. We say $x_n\to x$ as $n\to \infty$ ($x_n$ converges to $x$ as $n\to \infty$)
|
||||
|
||||
if for any open neighborhood $U$ of $x$, there exists $N\in\mathbb{N}_+$ such that $\forall n\geq N, x_n\in U$.
|
||||
|
||||
<details>
|
||||
<summary>Example of convergence of sequences</summary>
|
||||
|
||||
Let $X=\{a,b,c\}$ with the topology $\mathcal{T}=\{\emptyset, \{b\}, \{a,b\}, \{b,c\}, \{a,b,c\}\}$
|
||||
|
||||
Let $x_n=b$ for all $n\in\mathbb{N}_+$. Then $x_n\to b$ as $n\to \infty$.
|
||||
|
||||
Moreover, $x_n\to a$ as $n\to \infty$ since any open neighborhood of $a$ ($\{a,b\}$,$\{a,b,c\}$) contains $b$.
|
||||
|
||||
Without loss of generality, $x_n\to c$ as $n\to \infty$ since any open neighborhood of $c$ ($\{b,c\}$,$\{a,b,c\}$) contains $b$.
|
||||
|
||||
> You may find convergence of sequences in more than one point.
|
||||
|
||||
Let $x_n=a$ for all $n\in\mathbb{N}_+$. Then $x_n\to a$ as $n\to \infty$ (take $U=\{a,b\}$)
|
||||
|
||||
A non-example of convergence of sequences:
|
||||
|
||||
Let $x_n=\begin{cases}a, & n\text{ is even} \\ c, & n\text{ is odd}\end{cases}$. Then $x_n$ does not converge to any point in $X$. So this sequence does not have a limit in $(X,\mathcal{T})$.
|
||||
|
||||
</details>
|
||||
|
||||
### More special topologies
|
||||
|
||||
#### Hausdorff space
|
||||
|
||||
A topological space $(X,\mathcal{T})$ is a Hausdorff space if for any two distinct points $x,y\in X$, there exist open neighborhoods $U$ and $V$ of $x$ and $y$ respectively such that $U\cap V=\emptyset$.
|
||||
|
||||
<details>
|
||||
<summary>Non-example of Hausdorff space</summary>
|
||||
|
||||
Let $X=\{a,b,c\}$ with the topology $\mathcal{T}=\{\emptyset, \{b\}, \{a,b\}, \{b,c\}, \{a,b,c\}\}$
|
||||
|
||||
Let $x=a,y=b$. Then they don't have disjoint open neighborhoods.
|
||||
|
||||
This topology is not a Hausdorff space.
|
||||
|
||||
</details>
|
||||
|
||||
If a topological space is a Hausdorff space, then every sequence has a unique limit point or have no limit point.
|
||||
|
||||
#### Properties of Hausdorff space
|
||||
|
||||
Let $(X,\mathcal{T})$ be a Hausdorff space. Then every sequence $\{x_n\}_{n\in\mathbb{N}_+}$ converges to $x$ and $y$, then $x=y$.
|
||||
|
||||
> [!TIP]
|
||||
>
|
||||
> We want to show if $x\neq y$, then there exists an open neighborhood $U$ of $x$ and $V$ of $y$ such that $U\cap V=\emptyset$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
We proceed by contradiction.
|
||||
|
||||
Suppose $x\neq y$, then by definition of Hausdorff space, there exists an open neighborhood $U$ of $x$ and $V$ of $y$ such that $U\cap V=\emptyset$.
|
||||
|
||||
If $x_n$ converges to $x$, then for any open neighborhood $U_x$ of $x$, there exists $N_x\in\mathbb{N}_+$ such that $\forall n\geq N_x, x_n\in U_x$. Similarly, for any open neighborhood $U_y$ of $y$, there exists $N_y\in\mathbb{N}_+$ such that $\forall n\geq N_y, x_n\in U_y$.
|
||||
|
||||
Then we can find $N=max\{N_x,N_y\}$ such that $x_n\in U_x\cap U_y$ for all $n\geq N$. This contradicts the assumption that $U\cap V=\emptyset$.
|
||||
|
||||
Therefore, $x=y$.
|
||||
|
||||
</details>
|
||||
|
||||
#### Properties of closed singleton
|
||||
|
||||
Let $(X,\mathcal{T})$ be a Hausdorff topological space. Then $\forall x\in X$, $\{x\}$ is a closed set.
|
||||
|
||||
<details>
|
||||
<summary>Non-example of closed singleton</summary>
|
||||
|
||||
Let $X=\{a,b,c\}$ with the topology $\mathcal{T}=\{\emptyset, \{b\}, \{a,b\}, \{b,c\}, \{a,b,c\}\}$
|
||||
|
||||
Then $\{b\}$ is not a closed set, since $X\setminus \{b\}=\{a,c\}$ is not an open set.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
We need to show that $A=X\setminus \{x\}$ is an open set.
|
||||
|
||||
Take $y\in A$, then by the assumption, $x$ and $y$ have disjoint open neighborhoods $U_x$ and $V_y$ respectively. $x\in U_x$ and $y\in V_y$ and $U_x\cap V_y=\emptyset$.
|
||||
|
||||
So $x\notin V_y$, $y\in V_y$. So $A\subseteq\bigcup_{y\in A,y\neq x} V_y$.
|
||||
|
||||
Since $\forall V_y,x\notin V_y$, So $A\subseteq\bigcup_{y\in A,y\neq x} V_y$.
|
||||
|
||||
So $A=\bigcup_{y\in A,y\neq x} V_y$ is an arbitrary union of open sets, so $A$ is an open set.
|
||||
|
||||
Therefore, $\{x\}$ is a closed set.
|
||||
|
||||
</details>
|
||||
|
||||
## Continuous
|
||||
|
||||
### Continuous functions
|
||||
|
||||
#### Definition for continuous functions
|
||||
|
||||
Let $(X,\mathcal{T})$ and $(Y,\mathcal{T}')$ be topological spaces and $f:X\to Y$. We say that $f$ is continuous if for every open set $V\in Y$, $f^{-1}(V)\coloneqq\{x\in X: f(x)\in V\}$ is open in $X$.
|
||||
|
||||
That is, $\forall V\in \mathcal{T}', f^{-1}(V)\in \mathcal{T}$.
|
||||
|
||||
#### Definition for point-wise continuity
|
||||
|
||||
Let $(X,\mathcal{T})$ and $(Y,\mathcal{T}')$ be topological spaces and $f:X\to Y$. We say that $f$ is continuous at $x\in X$ if for every open set $V\in \mathcal{T}'$ such that $f(x)\in V$, there exists an open set $U\in \mathcal{T}$ such that $x\in U$ and $f(U)\subseteq V$. ($f^{-1}(V)$ contains an open neighborhood of $x$)
|
||||
|
||||
#### Lemma for continuous functions
|
||||
|
||||
$f:X\to Y$ is continuous if and only if $\forall x\in X$, $f$ is continuous at $x$.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
$\Rightarrow$:
|
||||
|
||||
Trivial
|
||||
|
||||
$\Leftarrow$:
|
||||
|
||||
Take an open set $V\in \mathcal{T}'$. Then for any point $x\in f^{-1}(V)$, we have $f(x)\in V$.
|
||||
|
||||
In particular, by definition of continuity at $x$, there exists an open set $U_x$ of $x$ such that $U_x\subseteq f^{-1}(V)$.
|
||||
|
||||
Then $f^{-1}(V)=\bigcup_{x\in f^{-1}(V)} U_x$ is an arbitrary union of open sets, so $f^{-1}(V)$ is open in $X$.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example of continuous functions</summary>
|
||||
|
||||
Let $X$ be any set and $\mathcal{T}$ be the discrete topology on $X$, $\mathcal{T}'$ be the trivial topology on $X$.
|
||||
|
||||
Let $f:(X,\mathcal{T})\to (X,\mathcal{T}')$ be the identity function. Then $f$ is continuous.
|
||||
|
||||
Since forall $V\in \mathcal{T}'$, $V$ is open in $X$, we can find $f^{-1}(V)$ is open in $X$. (only neet to test $X,\emptyset$)
|
||||
|
||||
In general, if $T$ is a finer than $T'$, then $f:(X,\mathcal{T})\to (X,\mathcal{T}')$ be the identity map is continuous.
|
||||
|
||||
However, if we let $f:(X,\mathcal{T}')\to (X,\mathcal{T})$ be the identity function, then $f$ is not continuous unless $X$ is a singleton.
|
||||
|
||||
</details>
|
||||
|
||||
#### Definition for constant maps
|
||||
|
||||
Let $X,Y$ be topological spaces and $y_0\in Y$, $f:X\to Y$ is defined by $f(x)=y_0$ for all $x\in X$. Then $f$ is continuous.
|
||||
|
||||
<details>
|
||||
<summary>Proof</summary>
|
||||
|
||||
Take an open set $V\in \mathcal{T}'$. $f^{-1}(V)=\begin{cases}X, & y_0\in V \\ \emptyset, & y_0\notin V\end{cases}$ is open in $X$. (by definition of topology, $X,\emptyset$ are open in $X$)
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>Example of inclusion maps</summary>
|
||||
|
||||
Let $X$ be a topological space and $A\subseteq X$ equipped with the subspace topology.
|
||||
|
||||
Let $f:A\to X$ be the inclusion map $f(x)=x$ for all $x\in A$.
|
||||
|
||||
Then take $V\subseteq X$ be an open set. $f^{-1}(V)=V\cap A\subseteq A$ is open in $A$ (by subspace topology).
|
||||
|
||||
</details>
|
||||
@@ -1,31 +0,0 @@
|
||||
export default {
|
||||
index: "Course Description",
|
||||
"---":{
|
||||
type: 'separator'
|
||||
},
|
||||
Exam_reviews: "Exam reviews",
|
||||
Math4201_L1: "Topology I (Lecture 1)",
|
||||
Math4201_L2: "Topology I (Lecture 2)",
|
||||
Math4201_L3: "Topology I (Lecture 3)",
|
||||
Math4201_L4: "Topology I (Lecture 4)",
|
||||
Math4201_L5: "Topology I (Lecture 5) Bonus",
|
||||
Math4201_L6: "Topology I (Lecture 6)",
|
||||
Math4201_L7: "Topology I (Lecture 7)",
|
||||
Math4201_L8: "Topology I (Lecture 8)",
|
||||
Math4201_L9: "Topology I (Lecture 9)",
|
||||
Math4201_L10: "Topology I (Lecture 10)",
|
||||
Math4201_L11: "Topology I (Lecture 11)",
|
||||
Math4201_L12: "Topology I (Lecture 12)",
|
||||
Math4201_L13: "Topology I (Lecture 13)",
|
||||
Math4201_L14: "Topology I (Lecture 14)",
|
||||
Math4201_L15: "Topology I (Lecture 15)",
|
||||
Math4201_L16: "Topology I (Lecture 16)",
|
||||
Math4201_L17: "Topology I (Lecture 17)",
|
||||
Math4201_L18: "Topology I (Lecture 18)",
|
||||
Math4201_L19: "Topology I (Lecture 19)",
|
||||
Math4201_L20: "Topology I (Lecture 20)",
|
||||
Math4201_L21: "Topology I (Lecture 21)",
|
||||
Math4201_L22: "Topology I (Lecture 22)",
|
||||
Math4201_L23: "Topology I (Lecture 23)",
|
||||
Math4201_L24: "Topology I (Lecture 24)",
|
||||
}
|
||||
Reference in New Issue
Block a user