129 lines
4.5 KiB
Markdown
129 lines
4.5 KiB
Markdown
# Math4201 Topology I (Lecture 30)
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## Compactness
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### Compactness in Metric Spaces
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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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#### Theorem of equivalence of compactness in metrizable spaces
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If $(X,d)$ is a metric 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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<details>
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<summary>Proof</summary>
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(1) $\implies$ (2):
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We proceed by contradiction,
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Let $X$ be compact and $A\subseteq X$ be an infinite subset of $X$ that doesn't have any limit points.
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Then $X-A$ is open because any $x\in X-A$ isn't in the closure of $A$ otherwise it would be a limit point for $A$, and hence $x$ has an open neighborhood contained in the complement of $A$.
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Next, let $x\in A$. Since $x$ isn't a limit point of $A$, there is an open neighborhood $U_x$ of $x$ in $X$ that $U_x\cap A=\{x\}$. Now consider the open covering of $X$ given as
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$$
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\{X-A\}\cup \{U_x:x\in A\}
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$$
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This is an open cover because either $x\in X-A$ or $x\in A$ and in the latter case, $x\in U_x$ since $X$ is compact, this should have a finite subcover. Any such subcover should contain $U_x$ for any $x$ because $U_x$ is the element in the subcover for $x$.
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This implies that our finite cover contains infinite open sets, which is a contradiction.
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---
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(2) $\implies$ (3):
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Let $\{x_n\}_{n\in\mathbb{N}}$ be an arbitrary sequence in $X$.
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Since $d(z,x_{n_k})\leq \frac{1}{k}$ the subsequence $(x_{n_k})$ converges to $z$.
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This completes the proof.
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except possibly $z$.
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Now we consider
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$$
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B_{r_k}(z)\text { with } r_k=\min \left(\frac{1}{k}, d_k\right)
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$$
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This ball has a point $x_{n_k}$ from $\{x_n\}$ which isn't equal to $z$.
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$r_k\leq d_k\implies n_k\geq n_{k-1}$.
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Since $z$ is a limit point of $\{x_n\}$, there exists $x_{n_k}$ such that $d(z,x_{n_k})<\frac{1}{k}$. So $x_{n_k}\in B_{r_k}(z)$.
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So, we have a convergent subsequence $(x_{n_k})$.
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---
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(3) $\implies$ (1):
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First wee prove the analogue of Lebesgue number lemma for a sequentially compact space $(X,d)$.
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Let $\{U_\alpha\}_{\alpha\in I}$ be an open covering of $X$. **By contradiction**, assume that for any $\delta>0$, there are two points $x,x'$ with $d(x,x')<\delta$ don't belong to the same open set in the covering.
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Take $\delta=\frac{1}{n}$, and let $x_n,x_n'$ be the points as above, then $d(x_n,x_n')<\frac{1}{n}$.
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$x_n,x_n'$ don't belong to the same open set in $\{U_\alpha\}_{\alpha\in I}$.
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By assumption $\{x_n\}$ is convergent after passing to a subsequence
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$$
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\{x_{n_k}\}_i
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$$
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Let $y$ be the limit of this subsequence and $U_\alpha$ be an element of the open covering containing $y$. There is $\epsilon>0$ such that $B_\epsilon(y)\subseteq U_\alpha$.
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If $k$ is large enough, then $x_{n_k}\in B_{\epsilon/2}(y)$ and $d(x_{n_k},x_{n_k}')<\epsilon/2$. (take $k$ such that $\frac{1}{n_k}<\epsilon/2$)
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Then $d(x_{n_k}',y)<\epsilon/2$ this implies that $x_{n_k}'\in U_\alpha$.
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Thus, $d(x_{n_k}',y)\leq d(x_{n_k}',x_{n_k})+d(x_{n_k},y)<\epsilon/2+\epsilon/2=\epsilon$.
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So, $x_{n_k}'\in B_\epsilon(y)\subseteq U_\alpha$.
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This is a contradiction.
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Next we show that for any $\epsilon$, there are
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$$
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y_1,y_2,\cdots,y_k
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$$
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such that $X=\bigcup_{i=1}^k B_{\epsilon}(y_i)$.
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Let's assume that it's not true and construct a sequence of points inductively in the following way:
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- Pick $y_1$ be arbitrary point in $X$.
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- In the $k$-th step, if $X\neq B_{\epsilon}(y_1)\cup \cdots \cup B_{\epsilon}(y_k)$, then pick $y_{k+1}\notin B_{\epsilon}(y_1)\cup \cdots \cup B_{\epsilon}(y_k)$.
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- In particular, $d(y_{k+1},y_j)\geq \epsilon$ for all $j<k$.
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By iteration, this process we obtain a sequence such such that the distance between any two elements is at most $\epsilon$.
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This sequence cannot have a converging subsequence which is a contradiction.
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To prove the compactness of $X$, take an open covering $\{U_\alpha\}_{\alpha\in I}$ of $X$ and let be $\delta>0$ such that any set with diameter at least $\delta$ is one of the $U_\alpha$'s. Let also $y_1,y_2,\cdots,y_k\in X$ be chosen such that $B_{\frac{\delta}{2}}(y_1)\cup \cdots \cup B_{\frac{\delta}{2}}(y_k)=X$.
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Since diameter of $B_{\frac{\delta}{2}}(y_i)$ is less than $\delta$, it belongs to $U_\alpha$ for some $\alpha\in I$.
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Then $\{U_{\alpha_i}\}_{i=1}^k$ is a finite subcover of $X$.
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This completes the proof.
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</details>
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