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218 lines
11 KiB
Markdown
218 lines
11 KiB
Markdown
# Math4201 Topology I (Lecture 35)
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## Countability axioms
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### Kolmogorov classification
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Consider the topological space $X$.
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$X$ is $T_0$ means for every pair of points $x,y\in X$, $x\neq y$, there is one of $x$ and $y$ is in an open set $U$ containing $x$ but not $y$.
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$X$ is $T_1$ means for every pair of points $x,y\in X$, $x\neq y$, each of them have a open set $U$ and $V$ such that $x\in U$ and $y\in V$ and $x\notin V$ and $y\notin U$. (singleton sets are closed)
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$X$ is $T_2$ means for every pair of points $x,y\in X$, $x\neq y$, there exists disjoint open sets $U$ and $V$ such that $x\in U$ and $y\in V$. (Hausdorff)
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$X$ is $T_3$ means that $X$ is regular: for any $x\in X$ and any close set $A\subseteq X$ such that $x\notin A$, there are **disjoint open sets** $U,V$ such that $x\in U$ and $A\subseteq V$.
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$X$ is $T_4$ means that $X$ is normal: for any disjoint closed sets, $A,B\subseteq X$, there are **disjoint open sets** $U,V$ such that $A\subseteq U$ and $B\subseteq V$.
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<details>
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<summary>Example</summary>
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Let $\mathbb{R}_{\ell}$ with lower limit topology.
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$\mathbb{R}_{\ell}$ is normal since for any disjoint closed sets, $A,B\subseteq \mathbb{R}_{\ell}$, $x\in A$ and $B$ is closed and doesn't contain $x$. Then there exists $\epsilon_x>0$ such that $[x,x+\epsilon_x)\subseteq A$ and does not intersect $B$.
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Therefore, there exists $\delta_y>0$ such that $[y,y+\delta_y)\subseteq B$ and does not intersect $A$.
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Let $U=\bigcup_{x\in A}[x,x+\epsilon_x)$ is open and contains $A$.
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$V=\bigcup_{y\in B}[y,y+\delta_y)$ is open and contains $B$.
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We show that $U$ and $V$ are disjoint.
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If $U\cap V\neq \emptyset$, then there exists $x\in A$ and $Y\in B$ such that $[x,x+\epsilon_x)\cap [y,y+\delta_y)\neq \emptyset$.
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This is a contradiction since $[x,x+\epsilon_x)\subseteq A$ and $[y,y+\delta_y)\subseteq B$.
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</details>
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#### Theorem Every metric space is normal
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Use the similar proof above.
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<details>
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<summary>Proof</summary>
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Let $A,B\subseteq X$ be closed.
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Since $B$ is closed, for any $x\in A$, there exists $\epsilon_x>0$ such that $B_{\epsilon_x}(x)\subseteq B$.
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Since $A$ is closed, for any $y\in B$, there exists $\delta_y>0$ such that $A_{\delta_y}(y)\subseteq A$.
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Let $U=\bigcup_{x\in A}B_{\epsilon_x/2}(x)$ and $V=\bigcup_{y\in B}B_{\delta_y/2}(y)$.
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We show that $U$ and $V$ are disjoint.
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If $U\cap V\neq \emptyset$, then there exists $x\in A$ and $Y\in B$ such that $B_{\epsilon_x/2}(x)\cap B_{\delta_y/2}(y)\neq \emptyset$.
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Consider $z\in B_{\epsilon_x/2}(x)\cap B_{\delta_y/2}(y)$. Then $d(x,z)<\epsilon_x/2$ and $d(y,z)<\delta_y/2$. Therefore $d(x,y)\leq d(x,z)+d(z,y)<\epsilon_x/2+\delta_y/2$.
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If $\delta_y<\epsilon_x$, then $d(x,y)<\delta_y/2+\delta_y/2=\delta_y$. Therefore $x\in B_{\delta_y}(y)\subseteq A$. This is a contradiction since $U\cap B=\emptyset$.
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If $\epsilon_x<\delta_y$, then $d(x,y)<\epsilon_x/2+\epsilon_x/2=\epsilon_x$. Therefore $y\in B_{\epsilon_x}(x)\subseteq B$. This is a contradiction since $V\cap A=\emptyset$.
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Therefore, $U$ and $V$ are disjoint.
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</details>
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#### Lemma fo regular topological space
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$X$ is regular topological space if and only if for any $x\in X$ and any open neighborhood $U$ of $x$, there is open neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$.
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#### Lemma of normal topological space
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$X$ is a normal topological space if and only if for any $A\subseteq X$ closed and any open neighborhood $U$ of $A$, there is open neighborhood $V$ of $A$ such that $\overline{V}\subseteq U$.
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<details>
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<summary>Proof</summary>
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$\implies$
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Let $A$ and $U$ are given as in the statement.
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So $A$ and $(X-U)$ are disjoint closed.
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Since $X$ is normal and $A\subseteq V\subseteq X$ and $V\cap W=\emptyset$. $X-U\subseteq W\subseteq X$. where $W$ is open in $X$.
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And $\overline{V}\subseteq (X-W)\subseteq U$.
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And $A\subseteq V$.
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The proof of reverse direction is similar.
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Let $A,B$ be disjoint and closed.
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Then $A\subseteq U\coloneqq X-B\subseteq X$ and $X-B$ is open in $X$.
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Apply the assumption to find $A\subseteq V\subseteq X$ and $V$ is open in $X$ and $\overline{V}\subseteq U\coloneqq X-B$.
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</details>
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#### Proposition of regular and Hausdorff on subspaces
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1. If $X$ is a regular topological space, and $Y$ is a subspace. Then $Y$ with induced topology is regular. (same holds for Hausdorff)
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2. If $\{X_\alpha\}$ is a collection of regular topological spaces, then their product with the product topology is regular. (same holds for Hausdorff)
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> [!CAUTION]
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>
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> The above does not hold for normal.
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Recall that $\mathbb{R}_{\ell}$ with lower limit topology is normal. But $\mathbb{R}_{\ell}\times \mathbb{R}_{\ell}$ with product topology is not normal.
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<details>
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<summary>Proof that Sorgenfrey plane is not normal</summary>
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The goal of this problem is to show that $\mathbb{R}^2_\ell$ (the Sorgenfrey plane) is not normal. Recall that $\mathbb{R}_\ell$ is the real line with the lower limit topology, and $\mathbb{R}_\ell^2$ is equipped with the product topology. Consider the subset
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$$
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L = \{\, (x,-x) \mid x\in \mathbb{R}_\ell \,\} \subset \mathbb{R}^2_\ell.
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$$
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Let $A\subset L$ be the set points of the form $(x,-x)$ such that $x$ is rational and $B\subset L$ be the set points of the form $(x,-x)$ such that $x$ is irrational.
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1. Show that the subspace topology on $L$ is the discrete topology. Conclude that $A$ and $B$ are closed subspaces of $\mathbb{R}_\ell^2$
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<details>
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<summary>Proof</summary>
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First we show that $L$ is closed.
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Consider $x=(a,b)\in\mathbb{R}^2_\ell-L$, by definition $a\neq -b$.
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If $a>-b$, then there exists open neighborhood $U_x=[\frac{\min\{a,b\}}{2},a+1)\times[\frac{\min\{a,b\}}{2},b+1)$ that is disjoint from $L$ (no points of form $(x,-x)$ in our rectangle), therefore $x\in U_x$.
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If $a<-b$, then there exists open neighborhood $U_x=[a,a+\frac{\min\{a,b\}}{2})\times[b,b+\frac{\min\{a,b\}}{2})$ that is disjoint from $L$, therefore $x\in U_x$.
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Therefore, $\mathbb{R}^2_\ell-L=\bigcup_{x\in \mathbb{R}_\ell^2-L} U_x$ is open in $\mathbb{R}^2_\ell$.
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So $L$ is closed in $\mathbb{R}^2_\ell$.
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To show $L$ with subspace topology on $\mathbb{R}^2_\ell$ is discrete topology, we need to show that every singleton of $L$ is open in $L$.
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For each $\{(x,-x)\}\in L$, $[x,x+1)\times [-x,-x+1)$ is open in $\mathbb{R}_\ell^2$ and $\{(x,-x)\}=([x,x+1)\times [-x,-x+1))\cap L$, therefore $\{(x,-x)\}$ is open in $L$.
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Since $A,B$ are disjoint and $A\cup B=L$, therefore $A=L-B$ and $B=L-A$, by definition of discrete topology, $A,B$ are both open therefore the complement of $A,B$ are closed. So $A,B$ are closed in $L$.
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since $L$ is closed in $\mathbb{R}^2_\ell$, by \textbf{Lemma \ref{closed_set_close_subspace_close}}, $A,B$ is also closed in $\mathbb{R}_\ell^2$. Therefore $A,B$ are closed subspace of $\mathbb{R}_\ell^2$.
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</details>
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2. Let $V$ be an open set of $\mathbb{R}^2_\ell$ containing $B$. Let $K_n$ consist of all irrational numbers $x\in [0,1]$ such that $[x, x+1/n) \times [-x, -x+1/n)$ is contained in $V$. Show that $[0,1]$ is the union of the sets $K_n$ and countably many one-point sets.
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<details>
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<summary>Proof</summary>
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Since $B$ is open in $L$, for each $b\in B$, by definition of basis in $\mathbb{R}_\ell^2$, and $B$ is open, there exists $b\in ([b,b+\epsilon)\times [-b,-b+\delta))\cap L\subseteq V$ and $0<\epsilon,\delta$, so there exists $n_b$ such that $\frac{1}{n_b}<\min\{\epsilon,\delta\}$ such that $b\in ([b,b+\frac{1}{n_b})\times [-b,-b+\frac{1}{n_b}))\cap L\subseteq V$.
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Therefore $\bigcup_{n=1}^\infty K_n$ covers irrational points in $[0,1]$
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Note that $B=L-A$ where $A$ is rational points therefore countable.
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So $[0,1]$ is the union of the sets $K_n$ and countably many one-point sets.
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</details>
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3. Use Problem 5-3 to show that some set $\overline{K_n}$ contains an open interval $(a,b)$ of $\mathbb{R}$. (You don't need to prove Problem 5.3, if it is not your choice of \#5.)
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#### Lemma
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Let $X$ be a compact Hausdorff space; let $\{A_n\}$ be a countable collection of closed sets of $X$. If each sets $A_n$ has empty interior in $X$, then the union $\bigcup_{n=1}^\infty A_n$ has empty interior in $X$.
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<details>
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<summary>Proof</summary>
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We proceed by contradiction, note that $[0,1]$ is a compact Hausdorff space since it's closed and bounded.
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And $\{\overline{K_n}\}_{n=0}^\infty$ is a countable collection of closed sets of $[0,1]$.
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Suppose for the sake of contradiction, $\overline{K_n}$ has empty interior in $X$ for all $n\in \mathbb{N}$, by \textbf{Lemma \ref{countable_closed_sets_empty_interior}}, then $\bigcup_{n=1}^\infty \overline{K_n}$ has empty interior in $[0,1]$, where $\Q\cap[0,1]$ are countably union of singletons, therefore has empty interior in $[0,1]$.
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Therefore $\bigcup_{n=1}^\infty \overline{K_n}$ has empty interior in $[0,1]$, since $\bigcup_{n=1}^\infty K_n\subseteq \bigcup_{n=1}^\infty \overline{K_n}$, $\bigcup_{n=1}^\infty K_n$ also has empty interior in $[0,1]$ by definition of subspace of $[0,1]$, therefore $\bigcup_{n=1}^\infty K_n\cup (\Q\cap[0,1])$ has empty interior in $[0,1]$. This contradicts that $\bigcup_{n=1}^\infty K_n\cup (\Q\cap[0,1])$ covers $[0,1]$ and should at least have interior $(0.1,0.9)$.
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</details>
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4. Show that $V$ contains the open parallelogram consisting of all points of the form
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$$
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x\times (-x+\epsilon)\quad\text{ for which }\quad a<x<b\text{ and }0<\epsilon<\frac{1}{n}.
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$$
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<details>
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<summary>Proof</summary>
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Since $V$ is open, by previous problem we know that there exists $n$ such that $\overline{K_n}$ contains the open interval $(a,b)$.
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If $x\in K_n$, $\forall a<x<b$, by definition of $K_n$ $[x,x+\frac{1}{n})\times [-x,-x+\frac{1}{n})\subseteq V$.
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If $x$ is a limit point of $K_n$, since $V$ is open, there exists $0<\epsilon<\frac{1}{n}$ such that $[x,x+\epsilon)\times [-x,-x+\epsilon)\subseteq V$.
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This gives our desired open parallelogram.
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</details>
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5. Show that if $q$ is a rational number with $a<q<b$, then the point $q\times (-q)$ of $\mathbb{R}_\ell^2$ is a limit point of $V$. Conclude that there are no disjoint open neighborhoods $U$ of $A$ and $V$ of $B$.
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<details>
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<summary>Proof</summary>
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Consider all the open neighborhood of $q\times (-q)$ in $\mathbb{R}_\ell^2$, for all $\delta>0$, $[q,q+\delta)\times (-q,-q+\delta)$ will intersect with some $x\times [-x,-x+\epsilon)\subseteq V$ such that $0<\epsilon<\frac{1}{n}<\delta$.
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Therefore, any open set containing $q\times (-q)\in A$ will intersect with $V$, it is impossible to build disjoint open neighborhoods $U$ of $A$ and $V$ of $B$.
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</details>
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</details>
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This shows that $\mathbb{R}_{\ell}$ is not metrizable. Otherwise $\mathbb{R}_{\ell}\times \mathbb{R}_{\ell}$ would be metrizable. Which could implies that $\mathbb{R}_{\ell}$ is normal.
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#### Theorem of metrizability (Urysohn metirzation theorem)
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If $X$ is normal and second countable, then $X$ is metrizable.
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> [!NOTE]
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>
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> - Every metrizable topological space is normal.
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> - Every metrizable space is first countable.
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> - But there are some metrizable space that is not second countable.
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>
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> Note that if $X$ is normal and first countable, then it is not necessarily metrizable. (Example $\mathbb{R}_{\ell}$) |