150 lines
4.8 KiB
Markdown
150 lines
4.8 KiB
Markdown
# Math4201 Topology I (Lecture 30)
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## Compact and connected spaces
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### Locally compact
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#### Theorem of one point compactification
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$X$ is a locally compact Hausdorff space if and only if there exists 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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> $Y$ is called **one point compactification** of $X$.
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<details>
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<summary>Proof for existence of Y (forward direction)</summary>
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Let's defined the topology of $Y$ as follows:
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Let $U\subseteq Y$ is open if and only if either
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1. $U\subseteq X$ and $U$ is open in $X$. ($\infty\notin U$) (Type 1 open set)
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2. $\infty \in U$ and $Y-U\subseteq X$ with subspace topology from $X$ is compact. (Type 2 open set)
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---
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First, we prove that $X$ is a subspace of $Y$. (That is, every open set $U\subseteq X$ implies that $U\cap X$ is open in $X$.)
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Case 1: $U\subseteq X$ is open in $X$, then $U\cap X=U$ is open in $Y$.
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Case 2: $\infty\in U$, then $Y-U$ is a compact subspace of $X$, since $X$ is Hausdorff. So $Y-U$ is a closed subspace of $X$.
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So $X\cap U=X-(Y-U)$ is open in $X$.
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We also need to show any open $U\subseteq X$ can be written as the intersection of some open in $Y$ and $X$.
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Note that for an open set $U\subseteq X$, $U\cap X$ is open in $X$. So $U\cap X$ is open in $Y$.
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---
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The second part is trivial by observation.
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---
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**First we show that $Y$ is Hausdorff.**
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Let $x_1,x_2\in Y$, such that $x_1\neq x_2$.
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If one of $x$, without loss of generality, $x_1$ is $\infty$, then by the assumption on $X$, there is a compact set $K$ containing an open neighborhood $U$ of $x_2$.
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Note that $Y-K$ is an open subspace of Type 2 in $Y$. In particular, it contains $\infty$.
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This is disjoint from the open neighborhood $U$ of $x_2$.
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If $x_1,x_2$ are both in $X$, then by the assumption on $X$, then by Hausdorff property for $X$, there are disjoint open neightbors $U_1$ and $U_2$ such that $x_1\in U_1$ and $x_2\in U_2$. By Type 1 open sets, these are also open and disjoint in $Y$.
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**Then we show that $Y$ is compact.**
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Take an open cover $\{U_\alpha\}_{\alpha\in I}$ of $Y$.
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In particular, there is $\alpha_0\in I$ such that $\infty\in U_{\alpha_0}$
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Note that $Y-U_{\alpha_0}\subseteq X$ with subspace topology from $X$ is compact (by Type 2 set).
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So there exists a finite subcover $\{U_{\alpha_i}\}_{\alpha_i\in I}$ such that $Y-U_{\alpha_0}\subseteq \bigcup_{i=1}^n U_{\alpha_i}$.
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So $U_{\alpha_0},U_{\alpha_1},\dots,U_{\alpha_n}$ is a finite cover of $Y$.
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So $Y$ is compact.
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</details>
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<details>
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<summary>Proof for properties from $Y$.(backward direction)</summary>
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**Property 1**
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$X$ is Hausdorff because it's a subspace of Hausdorff space.
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**Property 2**
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By definition
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**Property 3**
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$X$ is locally compact.
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Let $x\in X$ since $Y$ is Hausdorff, there are disjoint open sets $U,V\subseteq Y$ such that $x\in U$ and $\infty\in B$.
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Let $K=Y-V$, $K$ is a subset of $X$ since $\infty\notin V$.
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To complete the proof, we need to show that $K$ is compact.
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Since $V$ is open in $Y$, then $K$ is closed in $Y$. Since $Y$ is compact, then $K$ is compact. (any closed subspace of compact space is compact)
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</details>
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## Countability axioms
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### First countability axiom
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#### Definition for first countability axiom
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Let $X$ be a topological space, then $X$ satisfies the first countability axiom if
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For any $x\in X$, there is a countable collection $\{B_n}_n$ of open neighborhoods of $x$ such that any open neighborhood $U$ of $x$ contains one of $B_n$.
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<details>
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<summary>Example for metric space satisfies the first countability axiom</summary>
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Any metric space satisfies the first countability axiom.
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Take $\{B_{\frac{1}{n}}(x)\}_{n=1}^\infty$.
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</details>
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#### Properties for topological spaces that satisfy the first countability axiom
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1. If $A\subseteq X$, then for any $x\in \overline{A}$, there is a sequence $\{x_n\}_{n=1}^\infty\subseteq A$ such that $x_n\to x$.
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2. If $f:X\to Y$ such that for any sequence $\{x_n\}_{n=1}^\infty\subseteq X$ such that $x_n\to x$, we have $f(x_n)\to f(x)$ in $Y$, then $f$ is continuous.
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### Second countability axiom
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#### Definition for second countability axiom
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Let $X$ be a topological space, then $X$ satisfies the second countability axiom if
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it has a countable basis.
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Clearly any second countable space also satisfies the first countability axiom.
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But the converse is not true.
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<details>
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<summary>Example for metric space satisfies the second countability axiom</summary>
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$\mathbb{R}$ satisfies the second countability axiom. Take $\{(a,b)|a,b\in\mathbb{Q}\}$ is a basis for $\mathbb{R}$.
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And $\mathbb{Q}$ is countable.
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More generally, $\mathbb{R}^n$ is also countable and satisfies the second countability axiom.
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</details>
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> [!WARNING]
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>
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> Not all topological spaces satisfy the second countability axiom is metrizable.
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