123 lines
4.3 KiB
Markdown
123 lines
4.3 KiB
Markdown
# Math4201 Topology I (Lecture 31)
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## Compactness
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### Local compactness
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$\mathbb{R}$ is not compact but it has a "lot" of compact subspaces.
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An arbitrary point $x\in\mathbb{R}$ then there is a subset $(x-\epsilon,x+\epsilon)U\subseteq \mathbb{R}$ such that $x\in U$ and $U$ is compact.
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#### Definition of 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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<details>
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<summary>Example</summary>
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$\mathbb{R}^\omega=\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots$ with product topology.
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where basis is
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$$
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B=\{(a_1,b_1)\times (a_2,b_2)\times (a_3,b_3)\times \cdots\times \mathbb{R}\times \mathbb{R}\times \cdots \mid a_i,b_i\in \mathbb{R},a_i<b_i\}
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$$
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all except finitely many of these open intervals are $\mathbb{R}$.
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This space isn't locally compact.
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Consider $\underline{0}=(0,0,0,...)\in \mathbb{R}^\omega$. If there is a compact subspace $K$ of $\mathbb{R}^\omega$ containing a neighborhood $U$ of $0$, then it should contain a basis element around $\underline{0}$.
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And $\overline{U}\subseteq K$.
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Since $\overline{U}$ is closed in $K$, it has to be compact.
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But $\overline{U}=[a_1,b_1]\times [a_2,b_2]\times [a_3,b_3]\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots$ which is not compact.
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So we can find a open covering
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$$
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V=\{[a_1,b_1]\times [a_2,b_2]\times [a_3,b_3]\times (-j,j)\times \mathbb{R}\times \mathbb{R}\times \cdots\mid j\in \mathbb{N}\}
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$$
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which doesn't have a finite subcover.
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</details>
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#### Theorem of Homeomorphism over locally compact Hausdorff spaces
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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 unique in the following sense:
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If $Y'$ is another such space, then there is a homeomorphism between $Y$ and $Y'$ $f(x)=x$ for any $x\in X$.
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<details>
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<summary>Proof for existence of Y</summary>
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Let $Y=X\cup \{\infty\}$. as a set.
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Topology on $Y$:
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$U\subseteq Y$ is open if and only if either
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1. $U\subseteq X$ and $U$ is open in $X$. ($\infty\notin U$)
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2. $Y-U\subseteq X$ and $Y-U$ with the subspace topology from $X$ is compact. ($\infty\in U$)
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We need to show that there is a topology on $Y$ that satisfies the definition.
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1. $\emptyset\in \mathcal{T}$ because $\emptyset\subseteq X$, $Y\in \mathcal{T}$ because $Y-Y=\emptyset$ is compact.
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2. This topology is closed with respect to finite intersections.
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Consider $U,U'\in \mathcal{T}$. Then $U\cap U'$ is open.
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- Case 1: $\infty\notin U,U'$, then $U\cap U'$ is open in $X$.
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- Case 2: $\infty\in U,U'$ both, then $Y-U$, $Y-U'$ with subspace topology from $X$ are compact. Note that $Y-(U\cap U')=(Y-U)\cup (Y-U')$ is compact.
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- Case 3: $\infty\in U$ but not $U'$, then $Y-U$ with subspace topology from $X$ is compact. So $Y-U\subseteq X$ is compact, and $U'\subseteq X$ is open. And $Y-U\subseteq X$ is closed because $X$ is Hausdorff. and $Y-U\subseteq X$ is compact. So $U\cap U'$ is open in our topology.
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</details>
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<details>
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<summary>Example for such Y</summary>
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Consider $X=(0,1)$, we can build $Y=(0,1)\cup \{\infty\}=S^1$.
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</details>
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<details>
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<summary>Proof for the theorem</summary>
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First we prove the uniqueness of $f$.
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$Y=X\cup \{\infty\}$.
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$Y'=X\cup \{\infty'\}$.
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the function $f:Y\to Y'$ is defined $f(x)=x$ for $x\in X$ and $f(\infty)=\infty$.
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We show that $f$ is a homeomorphism.
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If $f$ is clearly a bijection, we need to show $U\subseteq Y$ is open if and only if $f(U)\subseteq Y'$ is open.
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1. Suppose $U\subseteq Y$ is open.
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Case 1, $\infty\notin U$, so $U\subseteq X$. (Note $X$ is in $Y$) is open. $\{\infty'\}$ is closed in $Y'$ (since $Y'$ is Hausdorff). $f(U)=U\subseteq X$ (Note $X$ is in $Y'$) is open. So $U\subseteq X'$ is open.
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Case 2, $\infty\in U$.
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Since $U\subseteq Y$ is open, then $Y-U$ is closed. Note that $Y-U$ is closed in $Y$ and $Y$ is Hausdorff. So $Y-U$ is also compact.
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Since $\infty\in U$, then $Y-U\subseteq X$.
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This implies that $f(Y-U)\subseteq X\subset Y'$ is also compact.
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Since $Y-U\subseteq Y'$ and $Y'$ is Hausdorff, then $Y-U\subseteq Y'$ is closed.
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So $f(U)=U\subseteq Y'$ is open.
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2. Suppose $f(U)\subseteq Y'$ is open.
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</details> |