ANA1_L18

Proof of Rudin, 2.34 using open cover compactness

Alternate proof of compact sets being closed using open cover compactness (we proved boundedness using OCC in L17).

Rudin, 2.34

Theorem 1(Theorem).

Let $C$ be a compact subset of $X$. Then $C$ is closed in $X$.

Proof
We will show $X\setminus C$ is open in $X$. Let $x\in X\setminus C$. We have to show that a neighborhood of $x$ in $X$ is contained in $X\setminus C$. For every $y\in C$ find disjoint open sets $U_y$ containing $x$ and $V_y$ containing $y$. Then, ${\bigcup }_{y\in C}{V}_y\supset C$. Now, observe that

$$\begin{array}{r}\left(\bigcap _{y\in C}{U}_y\right)\cap \left(\bigcup _{y\in C}{V}_y\right)=\varnothing .\end{array}$$

Recall that finite intersections of open sets are open. Since $C$ is compact, there exists a finite subcover $\{V_y|y\in K\subset C\}$ of $C$, i.e, ${\bigcup }_{y\in K}{V}_y\supset C$. Since $K$ is finite, ${\bigcap }_{y\in K}{U}_y$ is open in $X$. So,

$$\begin{array}{rl}& \left(\bigcap _{y\in K}{U}_y\right)\cap \left(\bigcup _{y\in K}{V}_y\right)=\varnothing \\ ⟹& \left(\bigcap _{y\in K}{U}_y\right)\cap C=\varnothing \\ ⟹& \left(\bigcap _{y\in K}{U}_y\right)\subset X\setminus C.\end{array}$$

❏

Alternate proof (Abbot)
We have to show that every point in $X$ which is a limit point of $C$ is in $C$. Let $(y_n)$ be a sequence in $C$, with $(y_n)\to y\in X$. FTSOC, assume $y\notin C$. So, every $x\in C$ is a positive distance away from $y$. Let $U_x\equiv B_{d(x,y)/2}(x,X)$. Clearly, $\{U_x|x\in C\}$ is an open cover of $C$, and $y\notin \bigcup U_x$. Let $\{U_x|x\in K\subset C\}$ be a finite subcover. Let $ϵ$ be the minimum of the radii of the balls in this finite subcover. Since $(y_n)\to y$, we can find terms of the sequence inside an $ϵ$ ball around $y$, not covered by our supposed finite subcover. $\Rightarrow \Leftarrow$ ❏

Note that we actually ended up proving a stronger result: For every $x\in X\setminus C$, there exist disjoint open sets $U\ni x$ and $V\supset C$, where $U={\bigcap }_{y\in K}{U}_y$ and $V={\bigcup }_{y\in K}{V}_y$.

Exercise: Show that for two disjoint compact sets $C_1$ and $C_2$, there exist disjoint open sets $U_1$ and $U_2$ such that $U_1\supset C_1$ and $U_2\supset C_2$.

Proof
It follows from the previous theorem that for every $x\in C_1$, there exist disjoint open sets $U_x$ and $V_x$ such that $U_x\ni x$ and $V_x\supset C_2$. Clearly, $\{U_x|x\in C_1\}$ is an open cover of $C_1$. Since $C_1$ is compact, there exists a finite subcover $\{U_x|x\in K\subset C_1\}$. Now, observe that

$$\begin{array}{r}\left(\bigcup _{x\in K}{U}_x\right)\cap \left(\bigcap _{x\in K}{V}_x\right)=\varnothing .\end{array}$$

Setting $U_1={\bigcup }_{x\in K}{U}_x$ and $U_2={\bigcap }_{x\in K}{V}_x$ works, because

• $U_1\cap U_2=\varnothing$.
• $U_1$ is open, since any union of open sets is open.
• $U_2$ is open, since a finite intersection of open sets is open and $K$ is finite.
• ${\bigcup }_{x\in K}{U}_x\supset C_1$ and ${\bigcap }_{x\in K}{V}_x\supset C_2$.

❏

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Proof of compactness of closed intervals in R using LUB property

Proof
Consider an open cover $B=\{V_{\alpha }{\}}_{\alpha \in I}$ of $[a,b]$. Define $S$ to be the set of all $x\in [a,b]$ such that $[a,x]$ can be covered by a finite subcover of $B$. This set is non empty because $a\in S$. Let $s=\sup S$. Assume $s<b$. There exists a $V_s\in B$ containing $s$. Since $V_s$ is open, it must also contain an open interval around $s$, say $(s-ϵ,s+ϵ)$. Now, $[a,s-ϵ]$ must have a finite subcover $\{V_{\beta }{\}}_{\beta \in J\subset I}$. Thus, $\{V_{\beta }{\}}_{\beta \in J\subset I}\cup \{V_s\}$ covers $\left[a,s+\frac{ϵ}{2}\right]$. So, $s+\frac{ϵ}{2}\in S$. This contradicts the hypothesis that $s$ is the supremum of $S$. Hence, $s=b$. ❏

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Rudin, 2.36

Let $B=\{V_{\alpha }{\}}_{\alpha \in I}$, be a collection of open sets in $X$. Then, $K\subset X$ is compact iff

$$\begin{array}{r}\bigcup _{\alpha \in I}{V}_{\alpha }\supset K⟹\exists \text{ finite }J\subset I\text{ such that }\bigcup _{\beta \in J}{V}_{\beta }\supset K.\end{array}$$

Now, we will rephrase the hypothesis in terms of closed sets $V_{\alpha }^c$.

$$\begin{array}{rl}& \bigcup _{\alpha \in I}{V}_{\alpha }={\left(\bigcap _{\alpha \in I}{V}_{\alpha }^c\right)}^c\supset K\\ ⟺& \left(\bigcap _{\alpha \in I}{V}_{\alpha }^c\right)\cap K=\varnothing .\end{array}$$

Ditto for the conclusion.

$$\begin{array}{rl}& \bigcup _{\beta \in J}{V}_{\beta }\supset K\\ ⟺& \left(\bigcap _{\beta \in J}{V}_{\beta }^c\right)\cap K=\varnothing .\end{array}$$

Thus, we say $K$ is compact iff, for any collection of closed sets $B^{\prime }=\{K_{\alpha }{\}}_{\alpha \in I}$,

$$\begin{array}{r}\left(\bigcap _{\alpha \in I}{K}_{\alpha }\right)\cap K=\varnothing ⟹\exists \text{ finite }J\subset I\text{ such that }\left(\bigcap _{\beta \in J}{K}_{\beta }\right)\cap K=\varnothing .\end{array}$$

We can consider all $K_{\alpha }$ to be compact subsets of $X$, since compact sets are in particular closed sets. Now, say $K$ is compact. Augment $B^{\prime }$ such that it now contains $K$ (update $I$ accordingly). This gives us

$$\begin{array}{r}\bigcap _{\alpha \in I}{K}_{\alpha }=\varnothing ⟹\exists \text{ finite }J\subset I\text{ such that }\bigcap _{\beta \in J}{K}_{\beta }=\varnothing .\end{array}$$

Take the contrapositive of this statement.

$$\begin{array}{r}\forall \text{ finite }J\subset I\text{ such that }\bigcap _{\beta \in J}{K}_{\beta }\ne \varnothing ⟹\bigcap _{\alpha \in I}{K}_{\alpha }\ne \varnothing .\end{array}$$

This is essentially Rudin, 2.36.

Theorem 2(Theorem).

If $\{K_{\alpha }\}$ is a set of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_{\alpha }\}$ is nonempty, then $\bigcap K_{\alpha }$ is nonempty.

Analogue of nested interval property in metric spaces

A special case of the above theorem is the analogue of the nested interval property in metric spaces.

Theorem 3(Corollary).

If $\{K_n\}$ is a sequence of non empty compact sets such that $K_n\supset K_{n+1}$, then $\bigcap K_i$ is nonempty.

Alternate proof using sequential compactness
For every $n\in \mathbb{N}$, pick $x_n\in K_n$. It follows that $K_n$ should contain the tails of $(x_n)$ from the $n$th term. Since $K_1$ is compact, $(x_n)$ must have a convergent subsequence $(x_{k_n})\to x\in K_1$. Since tails of $(x_{k_n})$ also converge to $x$ (and all subsequences of a convergent sequence converge to the limit of the sequence), we have $x\in K_n\forall n$. Thus, $x\in \bigcap K_i$. ❏

Rudin, 3.10 b

If we have the additional constraint that ${\lim }_{n\to \infty }\text{diam }{K}_n=0$, then $\bigcap K_i$ contains only one element.

Theorem 4(Theorem).

If $\{K_n\}$ is a sequence of non empty compact sets such that $K_n\supset K_{n+1}$, and if ${\lim }_{n\to \infty }\text{diam }{K}_n=0$, then $\bigcap K_i$ contains only one element.

Proof
$\bigcap K_i$ is non empty, so it has some element $p$. Let it also have another element $q$. There exists $N$ such that for all $n\ge N$, $\text{diam }{K}_n<d(q,p)$. Thus, we are forced to conclude $q=p$. ❏

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Weierstrass theorem, rephrased

Rudin, 2.42

Theorem 5(Theorem).

Every bounded infinite subset of ${\mathbb{R}}^k$ has a limit point in ${\mathbb{R}}^k$.

Proof
Let $S\subset {\mathbb{R}}^k$ be bounded and infinite. Since it is bounded, you can enclose it in a box, which is compact (in particular, limit point compact), which guarantees that $S$ has a limit point in the box. ❏