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# Math 4201 Final Exam Review
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> [!NOTE]
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>
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> This is a review for definitions we covered in the classes. It may serve as a cheat sheet for the exam if you are allowed to use it.
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## Topological space
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### Basic definitions
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#### Definition for topological space
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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:
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1. $\emptyset \in \mathcal{T}$ and $X \in \mathcal{T}$
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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 $\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}$
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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 $\bigcap_{i=1}^n U_i \in \mathcal{T}$
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#### Definition of open set
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$U\subseteq X$ is an open set if $U\in \mathcal{T}$
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#### Definition of closed set
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$Z\subseteq X$ is a closed set if $X\setminus Z\in \mathcal{T}$
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> [!WARNING]
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>
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> A set is closed is not the same as its not open.
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>
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> In all topologies over non-empty sets, $X, \emptyset$ are both closed and open.
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### Basis
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#### Definition of topological basis
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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:
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1. For any $x \in X$, there exists a $B \in \mathcal{B}$ such that $x \in B$ (basis covers the whole space)
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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 $x \in B_3 \subseteq B_1 \cap B_2$ (every non-empty intersection of basis elements are also covered by a basis element)
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#### Definition of topology generated by basis
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Let $\mathcal{B}$ be a basis for a topology on a set $X$. Then the topology generated by $\mathcal{B}$ is defined by the set as follows:
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$$
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\mathcal{T}_{\mathcal{B}} \coloneqq \{ U \subseteq X \mid \forall x\in U, \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq U \}
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$$
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> This is basically a closure of $\mathcal{B}$ under arbitrary unions and finite intersections
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#### Lemma of topology generated by basis
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$U\in \mathcal{T}_{\mathcal{B}}\iff \exists \{B_\alpha\}_{\alpha \in I}\subseteq \mathcal{B}$ such that $U=\bigcup_{\alpha \in I} B_\alpha$
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#### Definition of basis generated from a topology
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Let $(X, \mathcal{T})$ be a topological space. Then the basis generated from a topology is $\mathcal{C}\subseteq \mathcal{B}$ such that $\forall U\in \mathcal{T}$, $\forall x\in U$, $\exists B\in \mathcal{C}$ such that $x\in B\subseteq U$.
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#### Definition of subbasis of topology
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A subbasis of a topology is a collection $\mathcal{S}\subseteq \mathcal{T}$ such that $\bigcup_{U\in \mathcal{S}} U=X$.
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#### Definition of topology generated by subbasis
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Let $\mathcal{S}\subseteq \mathcal{T}$ be a subbasis of a topology on $X$, then the basis generated by such subbasis is the closure of finite intersection of $\mathcal{S}$
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$$
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\mathcal{B}_{\mathcal{S}} \coloneqq \{B\mid B\text{ is the intersection of a finite number of elements of }\mathcal{S}\}
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$$
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Then the topology generated by $\mathcal{B}_{\mathcal{S}}$ is the subbasis topology denoted by $\mathcal{T}_{\mathcal{S}}$.
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Note that all open set with respect to $\mathcal{T}_{\mathcal{S}}$ can be written as a union of finitely intersections of elements of $\mathcal{S}$
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### Comparing topologies
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#### Definition of finer and coarser topology
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Let $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ be topological spaces. Then $\mathcal{T}$ is finer than $\mathcal{T}'$ if $\mathcal{T}'\subseteq \mathcal{T}$. $\mathcal{T}$ is coarser than $\mathcal{T}'$ if $\mathcal{T}\subseteq \mathcal{T}'$.
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#### Lemma of comparing basis
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Let $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ be topological spaces with basis $\mathcal{B}$ and $\mathcal{B}'$. Then $\mathcal{T}$ is finer than $\mathcal{T}'$ if and only if for any $x\in X$, $x\in B'$, $B'\in \mathcal{B}'$, there exists $B\in \mathcal{B}$, such that $x\in B$ and $x\in B\subseteq B'$.
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### Product space
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#### Definition of cartesian product
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Let $X,Y$ be sets. The cartesian product of $X$ and $Y$ is the set of all ordered pairs $(x,y)$ where $x\in X$ and $y\in Y$, denoted by $X\times Y$.
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#### Definition of product topology
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Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. Then the product topology on $X\times Y$ is the topology generated by the basis
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$$
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\mathcal{B}_{X\times Y}=\{U\times V, U\in \mathcal{T}_X, V\in \mathcal{T}_Y\}
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$$
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or equivalently,
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$$
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\mathcal{B}_{X\times Y}'=\{U\times V, U\in \mathcal{B}_X, V\in \mathcal{B}_Y\}
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$$
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> Product topology generated from open sets of $X$ and $Y$ is the same as product topology generated from their corresponding basis
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### Subspace topology
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#### Definition of subspace topology
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Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$. Then the subspace topology on $Y$ is the topology given by
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$$
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\mathcal{T}_Y=\{U\cap Y|U\in \mathcal{T}\}
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$$
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or equivalently, let $\mathcal{B}$ be the basis for $(X,\mathcal{T})$. Then the subspace topology on $Y$ is the topology generated by the basis
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$$
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\mathcal{B}_Y=\{U\cap Y| U\in \mathcal{B}\}
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$$
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#### Lemma of open sets in subspace topology
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Let $(X,\mathcal{T})$ be a topological space and $Y\subseteq X$. Then if $U\subseteq Y$, $U$ is open in $(Y,\mathcal{T}_Y)$, then $U$ is open in $(X,\mathcal{T})$.
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> This also holds for closed set in closed subspace topology
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### Interior and closure
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#### Definition of interior
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The interior of $A$ is the largest open subset of $A$.
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$$
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A^\circ=\bigcup_{U\subseteq A, U\text{ is open in }X} U
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$$
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#### Definition of closure
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The closure of $A$ is the smallest closed superset of $A$.
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$$
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\overline{A}=\bigcap_{U\supseteq A, U\text{ is closed in }X} U
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$$
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#### Definition of neighborhood
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A neighborhood of a point $x\in X$ is an open set $U\in \mathcal{T}$ such that $x\in U$.
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#### Definition of limit points
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A point $x\in X$ is a limit point of $A$ if every neighborhood of $x$ contains a point in $A-\{x\}$.
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We denote the set of all limits points of $A$ by $A'$.
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$\overline{A}=A\cup A'$
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### Sequences and continuous functions
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#### Definition of convergence
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Let $X$ be a topological space. A sequence $(x_n)_{n\in\mathbb{N}_+}$ in $X$ converges to $x\in X$ if for any neighborhood $U$ of $x$, there exists $N\in\mathbb{N}_+$ such that $\forall n\geq N, x_n\in U$.
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#### Definition of Hausdoorff space
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A topological space $(X,\mathcal{T})$ is Hausdorff if for any two distinct points $x,y\in X$, there exist open neighborhoods $U$ and $V$ of $x$ and $y$ respectively such that $U\cap V=\emptyset$.
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#### Uniqueness of convergence in Hausdorff spaces
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In a Hausdorff space, if a sequence $(x_n)_{n\in\mathbb{N}_+}$ converges to $x\in X$ and $y\in X$, then $x=y$.
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#### Closed singleton in Hausdorff spaces
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In a Hausdorff space, if $x\in X$, then $\{x\}$ is a closed set.
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#### Definition of continuous function
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Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. A function $f:X\to Y$ is continuous if for any open set $U\subseteq Y$, $f^{-1}(U)$ is open in $X$.
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#### Definition of point-wise continuity
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Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. A function $f:X\to Y$ is point-wise continuous at $x\in X$ if for every openset $V\subseteq Y$, $f(x)\in V$ then there exists an open set $U\subseteq X$ such that $x\in U$ and $f(U)\subseteq V$.
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#### Lemma of continuous functions
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If $f:X\to Y$ is point-wise continuous for all $x\in X$, then $f$ is continuous.
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#### Properties of continuous functions
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If $f:X\to Y$ is continuous, then
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1. $\forall A\subseteq Y$, $f^{-1}(A^c)=X\setminus f^{-1}(A)$ (complements maps to complements)
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2. $\forall A_\alpha\subseteq Y, \alpha\in I$, $f^{-1}(\bigcup_{\alpha\in I} A_\alpha)=\bigcup_{\alpha\in I} f^{-1}(A_\alpha)$
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3. $\forall A_\alpha\subseteq Y, \alpha\in I$, $f^{-1}(\bigcap_{\alpha\in I} A_\alpha)=\bigcap_{\alpha\in I} f^{-1}(A_\alpha)$
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4. $f^{-1}(U)$ is open in $X$ for any open set $U\subseteq Y$.
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5. $f$ is continuous at $x\in X$.
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6. $f^{-1}(V)$ is closed in $X$ for any closed set $V\subseteq Y$.
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7. 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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8. $\forall A\subseteq X$, $\overline{f(A)}=f(\overline{A})$
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#### Definition of homeomorphism
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Let $(X,\mathcal{T}_X)$ and $(Y,\mathcal{T}_Y)$ be topological spaces. A function $f:X\to Y$ is a homeomorphism if $f$ is continuous, bijective and $f^{-1}:Y\to X$ is continuous.
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#### Ways to construct continuous functions
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1. If $f:X\to Y$ is constant function, $f(x)=y_0$ for all $x\in X$, then $f$ is continuous. (constant functions are continuous)
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2. If $A$ is a subspace of $X$, $f:A\to X$ is the inclusion map $f(x)=x$ for all $x\in A$, then $f$ is continuous. (inclusion maps are continuous)
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3. If $f:X\to Y$ is continuous, $g:Y\to Z$ is continuous, then $g\circ f:X\to Z$ is continuous. (composition of continuous functions is continuous)
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4. If $f:X\to Y$ is continuous, $A$ is a subspace of $X$, then $f|_A:X\to Y$ is continuous. (domain restriction is continuous)
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5. If $f:X\to Y$ is continuous, $Z$ is a subspace of $Y$, then $f:X\to Z$, $g(x)=f(x)\cap Z$ is continuous. If $Y$ is a subspace of $Z$, then $h:X\to Z$, $h(x)=f(x)$ is continuous (composition of $f$ and inclusion map).
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6. If $f:X\to Y$ is continuous, $X$ can be written as a union of open sets $\{U_\alpha\}_{\alpha\in I}$, then $f|_{U_\alpha}:X\to Y$ is continuous.
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7. If $X=Z_1\cup Z_2$, and $Z_1,Z_2$ are closed equipped with subspace topology, let $g_1:Z_1\to Y$ and $g_2:Z_2\to Y$ be continuous, and for all $x\in Z_1\cap Z_2$, $g_1(x)=g_2(x)$, then $f:X\to Y$ by $f(x)\begin{cases}g_1(x), & x\in Z_1 \\ g_2(x), & x\in Z_2\end{cases}$ is continuous. (pasting lemma)
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8. $f:X\to Y$ is continuous, $g:X\to Z$ is continuous if and only if $H:X\to Y\times Z$, where $Y\times Z$ is equipped with the product topology, $H(x)=(f(x),g(x))$ is continuous. (proved in homework)
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### Metric spaces
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#### Definition of metric
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A metric on $X$ is a function $d:X\times X\to \mathbb{R}$ such that $\forall x,y\in X$,
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1. $d(x,x)=0$
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2. $d(x,y)\geq 0$
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3. $d(x,y)=d(y,x)$
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4. $d(x,y)+d(y,z)\geq d(x,z)$
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#### Definition of metric ball
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The metric ball $B_r^{d}(x)$ is the set of all points $y\in X$ such that $d(x,y)\leq r$.
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#### Definition of metric topology
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Let $X$ be a metric space with metric $d$. Then $X$ is equipped with the metric topology generated by the metric balls $B_r^{d}(x)$ for $r>0$.
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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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#### Hausdorff axiom for metric spaces
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Every metric space is Hausdorff (take metric balls $B_r(x)$ and $B_r(y)$, $r=\frac{d(x,y)}{2}$).
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If a topology isn't Hausdorff, then it isn't metrizable.
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Prove by triangle inequality and contradiction.
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#### Common metrics in $\mathbb{R}^n$
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Euclidean metric
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$$
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d(x,y)=\sqrt{\sum_{i=1}^n (x_i-y_i)^2}
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$$
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Square metric
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$$
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\rho(x,y)=\max_{i=1}^n |x_i-y_i|
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$$
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Manhattan metric
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$$
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m(x,y)=\sum_{i=1}^n |x_i-y_i|
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$$
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These metrics are equivalent.
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#### Product topology and metric
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If $(X,d),(Y,d')$ are metric spaces, then $X\times Y$ is metric space with metric $d(x,y)=\max\{d(x_1,y_1),d(x_2,y_2)\}$.
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#### Uniform metric
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Let $\mathbb{R}^\omega$ be the set of all infinite sequences of real numbers. Then $\overline{d(x,y)}=\sup_{i=1}^\omega \min\{1,|x_i-y_i|\}$, the uniform metric on $\mathbb{R}^\omega$ is a metric.
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#### Metric space and converging sequences
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Let $X$ be a topological space, $A\subseteq X$, $x_n\to x$ such that $x_n\in A$. Then $x\in \overline{A}$.
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If $X$ is a **metric space**, $A\subseteq X$, $x\in \overline{A}$, then there exists converging sequence $x_n\to x$ such that $x_n\in A$.
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### Metric defined for functions
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#### Definition for bounded metric space
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A metric space $(Y,d)$ is bounded if there is $M\in \mathbb{R}^{\geq 0}$ such that for all $y_1,y_2\in Y$, $d(y_1,y_2)\leq M$.
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#### Definition for metric defined for functions
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Let $X$ be a topological space and $Y$ be a bounded metric space, then the set of all maps, denoted by $\operatorname{Map}(X,Y)$, $f:X\to Y\in \operatorname{Map}(X,Y)$ is a metric space with metric $\rho(f,g)=\sup_{x\in X} d(f(x),g(x))$.
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#### Space of continuous map is closed
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Let $(\operatorname{Map}(X,Y),\rho)$ be a metric space defined above, then every continuous map is a limit point of some sequence of continuous maps.
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$$
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Z=\{f\in \operatorname{Map}(X,Y)|f\text{ is continuous}\}
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$$
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$Z$ is closed in $(\operatorname{Map}(X,Y),\rho)$.
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### Quotient space
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#### Quotient map
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Let $X$ be a topological space and $X^*$ is a set. $q:X\to X^*$ is a surjective map. Then $q$ is a quotient map.
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#### Quotient topology
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Let $(X,\mathcal{T})$ be a topological space and $X^*$ be a set, $q:X\to X^*$ is a surjective map. Then
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$$
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\mathcal{T}^* \coloneqq \{U\subseteq X^*\mid q^{-1}(U)\in \mathcal{T}\}
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$$
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is a topology on $X^*$ called quotient topology.
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**That is equivalent to say that $U\subseteq X^*$ is open in $X^*$ if and only if $p^{-1}(U)\subseteq X$ is open in $X$.**
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This is also called "strong continuity" since compared with the continuous condition, it requires if $p^{-1}(U)$ is open in $X$, then $U$ is open in $X^*$.
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$(X^*,\mathcal{T}^*)$ is called the quotient space of $X$ by $q$.
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#### Closed map and open map
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$f:X\to Y$ is a open map if for each open set $U$ of $X$, $f(U)$ is open in $Y$; it is a closed map if for each closed set $U$ of $X$, $f(U)$ is closed in $Y$.
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> [!WARNING]
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>
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> Not all quotient map are closed or open:
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>
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> 1. Example of quotient map that is not open nor closed:
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>
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> Consider the projection map $f:[0,1]\to S^1$, this map maps open set $[0,0.5)$ in $[0,1]$ to non open map $[0,\pi)$
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>
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> 2. Example of open map that is not closed:
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>
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> Consider projection map $f:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ to first coordinate, this map is open but not closed, consider $C\coloneqq\{x\times y\mid xy=1\}$ This set is closed in $\mathbb{R}\times \mathbb{R}$ but $f(C)=\mathbb{R}-\{0\}$ is not closed in $\mathbb{R}$.
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>
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> 3. Example of closed map that is not open:
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>
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> Consider $f:[0,1]\cup[2,3]\to [0,2]$ by taking -1 to elements in $[2,3]$, this map is closed map but not open, since $f([2,3])=[1,2]$ is not open in $[0,2]$ but $[2,3]$ is open in $[0,1]\cup[2,3]$
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#### Equivalent classes
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$\sim$ is a subset of $X\times X$ with the following properties:
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1. $x\sim x$ for all $x\in X$.
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2. If $(x,y)\in \sim$, then $(y,x)\in \sim$.
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3. If $(x,y)\in \sim$ and $(y,z)\in \sim$, then $(x,z)\in \sim$.
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The equivalence classes of $x\in X$ is denoted by $[x]=\{y\in X|y\sim x\}$.
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We can use equivalent classes to define quotient space.
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#### Theorem 22.2
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Let $p:X\to Y$ be a quotient map. Let $Z$ be a space and let $g:X\to Z$ be a map that is constant on each set $p^{-1}(\{y\})$, for $y\in Y$. Then $g$ induces a map $f:Y\to Z$ such that $f\circ p=g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.
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*Prove by setting $f(p(x))=g(x)$, then $g^{-1}(V)=p^{-1}(f^{-1}(V))$ for $V$ open in $Z$.*
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## Connectedness and compactness of metric spaces
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### Connectedness and separation
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#### Definition of separation
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Let $X=(X,\mathcal{T})$ be a topological space. A separation of $X$ is a pair of open sets $U,V\in \mathcal{T}$ that:
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1. $U\neq \emptyset$ and $V\neq \emptyset$ (that also equivalent to $U\neq X$ and $V\neq X$)
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2. $U\cap V=\emptyset$
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3. $X=U\cup V$ ($\forall x\in X$, $x\in U$ or $x\in V$)
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Some interesting corollary:
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- Any non-trivial (not $\emptyset$ or $X$) clopen set can create a separation.
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- Proof: Let $U$ be a non-trivial clopen set. Then $U$ and $U^c$ are disjoint open sets whose union is $X$.
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- For subspace $Y\subset X$, a separation of $Y$ is a pair of open sets $U,V\in \mathcal{T}_Y$ such that:
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1. $U\neq \emptyset$ and $V\neq \emptyset$ (that also equivalent to $U\neq Y$ and $V\neq Y$)
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2. $U\cap V=\emptyset$
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3. $Y=U\cup V$ ($\forall y\in Y$, $y\in U$ or $y\in V$)
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- If $\overline{A}$ is closure of $A$ in $X$, same for $\overline{B}$, then the closure of $A$ in $Y$ is $\overline{A}\cap Y$ and the closure of $B$ in $Y$ is $\overline{B}\cap Y$. Then for separation $U,V$ of $Y$, $\overline{A}\cap B=A\cap \overline{B}=\emptyset$.
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#### Definition of connectedness
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A topological space $X$ is connected if there is no separation of $X$.
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> [!TIP]
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>
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> Connectedness is a local property. (That is, even the big space is connected, the subspace may not be connected. Consider $\mathbb{R}$ with the usual metric. $\mathbb{R}$ is connected, but $\mathbb{R}\setminus\{0\}$ is not connected.)
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>
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> Connectedness is a topological property. (That is, if $X$ and $Y$ are homeomorphic, then $X$ is connected if and only if $Y$ is connected. Consider if not, then separation of $X$ gives a separation of $Y$.)
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|
#### Lemma of connected subspace
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If $A,B$ is a separation of a topological space $X$, and $Y\subseteq X$ is a **connected** subspace with subspace topology, then $Y$ is either contained in $A$ or $B$.
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*Easy to prove by contradiction. Try to construct a separation of $Y$.*
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|
#### Theorem of connectedness of union of connected subsets
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Let $\{A_\alpha\}_{\alpha\in I}$ be a collection of connected subsets of a topological space $X$ such that $\bigcap_{\alpha\in I} A_\alpha$ is non-empty. Then $\bigcup_{\alpha\in I} A_\alpha$ is connected.
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*Easy to prove by lemma of connected subspace.*
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|
#### Lemma of compressing connectedness
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|
Let $A\subseteq X$ be a connected subspace of a topological space $X$ and $A\subseteq B\subseteq \overline{A}$. Then $B$ is connected.
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|
*Easy to prove by lemma of connected subspace. Suppose $C,D$ is a separation of $B$, then $A$ lies completely in either $C$ or $D$. Without loss of generality, assume $A\subseteq C$. Then $\overline{A}\subseteq\overline{C}$ and $\overline{A}\cap D=\emptyset$ (from $\overline{C}\cap D=\emptyset$ by closure of $A$). (contradiction that $D$ is nonempty) So $D$ is disjoint from $\overline{A}$, and hence from $B$. Therefore, $B$ is connected.*
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|
#### Theorem of connected product space
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|
|
Any finite cartesian product of connected spaces is connected.
|
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|
*Prove using the union of connected subsets theorem. Using fiber bundle like structure union with non-empty intersection.*
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|
|
|
|
### Application of connectedness in real numbers
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|
Real numbers are connected.
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|
Using the least upper bound and greatest lower bound property, we can prove that any interval in real numbers is connected.
|
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|
|
#### Intermediate Value Theorem
|
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|
|
Let $f:[a,b]\to \mathbb{R}$ be continuous. If $c\in\mathbb{R}$ is such that $f(a)<c<f(b)$, then there exists $x\in [a,b]$ such that $f(x)=c$.
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|
*If false, then we can use the disjoint interval with projective map to create a separation of $[a,b]$.*
|
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|
#### Definition of path-connected space
|
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|
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 $x'$.
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|
- Every connected space is path-connected.
|
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|
|
- The converse may not be true, consider the topologists' sine curve.
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|
### Compactness
|
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|
|
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|
#### Definition of compactness via open cover and finite subcover
|
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|
Let $X=(X,\mathcal{T})$ be a topological space. An open cover of $X$ is $\mathcal{A}\subset \mathcal{T}$ such that $X=\bigcup_{A\in \mathcal{A}} A$. A finite subcover of $\mathcal{A}$ is a finite subset of $\mathcal{A}$ that covers $X$.
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$X$ is compact if every open cover of $X$ has a finite subcover (i.e. $X=\bigcup_{A\in \mathcal{A}} A\implies \exists \mathcal{A}'\subset \mathcal{A}$ finite such that $X=\bigcup_{A\in \mathcal{A}'} A$).
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|
#### Definition of compactness via finite intersection property
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|
A collection $\{C_\alpha\}_{\alpha\in I}$ of subsets of a set $X$ has finite intersection property if for every finite subcollection $\{C_{\alpha_1}, ..., C_{\alpha_n}\}$ of $\{C_\alpha\}_{\alpha\in I}$, we have $\bigcap_{i=1}^n C_{\alpha_i}\neq \emptyset$.
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|
Let $X=(X,\mathcal{T})$ be a topological space. $X$ is compact if every collection $\{Z_\alpha\}_{\alpha\in I}$ of closed subsets of $X$ satisfies the finite intersection property has a non-empty intersection (i.e. $\forall \{Z_{\alpha_1}, ..., Z_{\alpha_n}\}\subset \{Z_\alpha\}_{\alpha\in I}, \bigcap_{i=1}^n Z_{\alpha_i} \neq \emptyset\implies \bigcap_{\alpha\in I} Z_\alpha \neq \emptyset$).
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|
|
#### Compactness is a local property
|
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|
|
|
|
|
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|
Let $X$ be a topological space. A subset $Y\subseteq X$ is compact if and only if every open covering of $Y$ (set open in $X$) has a finite subcovering of $Y$.
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|
- A space $X$ is compact but the subspace may not be compact.
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|
- Consider $X=[0,1]$ and $Y=[0,1/2)$. $Y$ is not compact because the open cover $\{(0,1/n):n\in \mathbb{N}\}$ does not have a finite subcover.
|
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|
- A compact subspace may live in a space that is not compact.
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|
- Consider $X=\mathbb{R}$ and $Y=[0,1]$. $Y$ is compact but $X$ is not compact.
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|
#### Closed subspaces of compact spaces
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|
A closed subspace of a compact space is compact.
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A compact subspace of Hausdorff space is closed.
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|
*Each point not in the closed set have disjoint open neighborhoods with the closed set in Hausdorff space.*
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|
|
#### Theorem of compact subspaces with Hausdorff property
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|
If $Y$ is compact subspace of a **Hausdorff space** $X$, $x_0\in X-Y$, then there are disjoint open neighborhoods $U,V\subseteq X$ such that $x_0\in U$ and $Y\subseteq V$.
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|
#### Image of compact space under continuous map is compact
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|
Let $f:X\to Y$ be a continuous map and $X$ is compact. Then $f(X)$ is compact.
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|
|
|
#### Tube lemma
|
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|
|
Let $X,Y$ be topological spaces and $Y$ is compact. Let $N\subseteq X\times Y$ be an open set contains $X\times \{y_0\}$ for $y_0\in Y$. Then there exists an open set $W\subseteq Y$ is open containing $y_0$ such that $N$ contains $X\times W$.
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*Apply the finite intersection property of open sets in $X\times Y$. Projection map is continuous.*
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|
#### Product of compact spaces is compact
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|
Let $X,Y$ be compact spaces, then $X\times Y$ is compact.
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Any finite product of compact spaces is compact.
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|
|
### Compact subspaces of real numbers
|
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|
|
|
|
|
|
|
|
#### Every closed and bounded subset of real numbers is compact
|
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$[a,b]$ is compact in $\mathbb{R}$ with standard topology.
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|
|
#### Good news for real numbers
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|
|
Any of the three properties is equivalent for subsets of real numbers (product of real numbers):
|
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|
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|
|
1. $A\subseteq \mathbb{R}^n$ is closed and bounded (with respect to the standard metric or spherical metric on $\mathbb{R}^n$).
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|
2. $A\subseteq \mathbb{R}^n$ is compact.
|
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|
|
#### Extreme value theorem
|
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|
|
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|
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|
|
If $f:X\to \mathbb{R}$ is continuous map with $X$ being compact. Then $f$ attains its minimum and maximum. (there exists $x_m,x_M\in X$ such that $f(x_m)\leq f(x)\leq f(x_M)$ for all $x\in X$)
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|
#### Lebesgue number lemma
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|
|
For a compact metric space $(X,d)$ and an open covering $\{U_\alpha\}_{\alpha\in I}$ of $X$. Then there is $\delta>0$ such that for every subset $A\subseteq X$ with diameter less than $\delta$, there is $\alpha\in I$ such that $A\subseteq U_\alpha$.
|
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|
*Apply the extreme value theorem over the mapping of the averaging function for distance of points to the $X-U_\alpha$. Find minimum radius of balls that have some $U_\alpha$ containing the ball.*
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|
|
#### Definition for uniform continuous function
|
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|
$f$ is uniformly continuous if for any $\epsilon > 0$, there exists $\delta > 0$ such that for any $x_1,x_2\in X$, if $d(x_1,x_2)<\delta$, then $d(f(x_1),f(x_2))<\epsilon$.
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|
|
#### Uniform continuity theorem
|
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|
|
Let $f:X\to Y$ be a continuous map between two metric spaces. If $X$ is compact, then $f$ is uniformly continuous.
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|
#### Definition of isolated point
|
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|
A point $x\in X$ is an isolated point if $\{x\}$ is an open subset of $X$.
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|
|
#### Theorem of isolated point in compact spaces
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|
|
Let $X$ be a nonempty compact Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.
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|
|
*Proof using infinite nested closed intervals should be nonempty.*
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|
|
### Variation of compactness
|
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|
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|
#### Limit point compactness
|
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|
A topological space $X$ is limit point compact if every infinite subset of $X$ has a limit point in $X$.
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|
- Every compact space is limit point compact.
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|
#### Sequentially compact
|
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|
A topological space $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.
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|
- Every compact space is sequentially compact.
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|
#### Equivalence of three in metrizable spaces
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|
If $X$ is a metrizable space, then the following are equivalent:
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1. $X$ is compact.
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2. $X$ is limit point compact.
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3. $X$ is sequentially compact.
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#### Local compactness
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A space $X$ is locally compact if every point $x\in X$, **there is a compact subspace $K$ of $X$ containing a neighborhood $U$ of $x$** $x\in U\subseteq K$ such that $K$ is compact.
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#### Theorem of one point compactification
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Let $X$ be a locally compact Hausdorff space if and only if there exists a topological space $Y$ satisfying the following properties:
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1. $X$ is a subspace of $Y$.
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2. $Y-X$ has one point, usually denoted by $\infty$.
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3. $Y$ is compact and Hausdorff.
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The $Y$ is defined as follows:
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$U\subseteq Y$ is open if and only if one of the following holds.
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1. $U\subseteq X$ and $U$ is open in $X$
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2. $\infty \in U$ and $Y-U\subseteq X$, and $Y-U$ is compact.
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## Countability and Separation Axioms
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### Countability Axioms
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#### First countability axiom
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A topological space $(X,\mathcal{T})$ satisfies the first countability axiom if 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 $V_n$.
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Apply the theorem above, we have if $(X,\mathcal{T})$ satisfies the first countability axiom, then:
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1. Every convergent sequence converges to a point in the closure of the sequence.
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<details>
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<summary>Space that every convergent sequence not converges to a point in the closure of the sequence.</summary>
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Consider $\mathbb{R}^\omega$ with the box topology.
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And $A=(0,1)\times (0,1)\times \cdots$ and $x=(0,0,\cdots)$.
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$x\in \overline{A}$ but no sequence converges to $x$.
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Suppose there exists such sequence, $\{x_n=(x_1^n,x_2^n,\cdots)\}_{n=1}^\infty$.
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Take $B=(-x_1^1,x_1^1)\times(-x_2^2,x_2^2)\times \cdots$, this is basis containing $x$ but none of $x_n$.
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</details>
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2. If $f:X\to Y$ such that for any sequence $\{x_n\}_{n=1}^\infty$ in $X$, $f(x_n)\to f(x)$, then $f$ is continuous.
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#### Second countability axiom
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Let $(X,\mathcal{T})$ be a topological space, then $X$ satisfies the second countability axiom if $X$ has a countable basis.
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If $X$ is second countable, then:
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1. Any discrete subspace $Y$ of $X$ is countable
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2. There exists a countable subset of $X$ that is dense in $X$.
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3. Every open covering of $X$ has a **countable** subcover (That is if $X=\bigcup_{\alpha\in I} U_\alpha$, then there exists a **countable** subcover $\{U_{\alpha_1}, ..., U_{\alpha_\infty}\}$ of $X$) (*also called Lindelof spaces*)
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### Separation Axioms
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#### Hausdorff spaces
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A topological space $(X,\mathcal{T})$ is Hausdorff if for any two distinct points $x,y\in X$, there are **disjoint open sets** $U,V$ such that $x\in U$ and $y\in V$. (note that $U\cup V$ may not be $X$, compared with definition of separation)
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Some corollaries:
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1. A subspace of Hausdorff space is Hausdorff, and a product of Hausdorff spaces is Hausdorff.
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#### Regular spaces
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A topological space $(X,\mathcal{T})$ is regular if for any $x\in X$ and any closed 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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Some corollaries:
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1. $X$ is regular if and only if given a point $x$ and a open neighborhood $U$ of $x$, there is open neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$.
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2. A subspace of regular space is regular, and a product of regular spaces is regular.
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#### Normal spaces
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A topological space $(X,\mathcal{T})$ is normal if 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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Some corollaries:
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1. $X$ is normal if and only if given a closed set $A\subseteq X$, there is open neighborhood $V$ of $A$ such that $\overline{V}\subseteq U$.
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> [!CAUTION]
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>
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> Product of normal spaces may not be normal (consider Sorgenfrey plane)
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#### Regular space with countable basis is normal
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Let $X$ be a regular space with countable basis, then $X$ is normal.
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*Prove by taking disjoint open neighborhoods by countable cover.*
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Trance-0