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$K\subset R$ is compact, i.e closed and bounded.

I am almost at the end of the proof but not clear how the contradiction implies the statement.

Assume for contradiction, that there is not finite sub cover.

  1. $I_0$ is a closed interval containing $K$ and $A_1, B_1$ are first and second half od $I_0$ then either $K \cap A_1$ or $K \cap B_1$ has not finite subcover, since $K$ doesn't have a finite subcover then atleast one of these shouldn't have a finite subcover.
  2. We can make a series $(I_n)$ where each $I_n$ is the half of the $I_{n-1}$ whose intersection with $K$ didn't have an open subcover. By construction $I_0 \supseteq I_1 \supseteq I_2 ...$ and $\lim|I_n|=0$.
  3. Since all the $I_n$ are non empty there exist $x\in I_n \quad\forall n\in\mathbb{N}$.
  4. Since $O_\lambda$ is an open cover there must exist some $O_{\lambda_0}$ which contains $x$.

In last step I am further required to show that there is an $n_0$ large enough to guarantee that $I_{n_0}\subseteq O_{\lambda_0}$. And this is the contradiction we need somehow. I am not understanding what argument to give for the last part.

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  • $\begingroup$ You should point out what definition of compactness you are using. The most common definition of compactness is what you have in the title (every open cover admits a finite subcover). $\endgroup$ – parsiad Mar 28 '18 at 5:58
  • $\begingroup$ We are initially using the closed and bounded definition. $\endgroup$ – Sonal_sqrt Mar 28 '18 at 6:03
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Let $\mathcal{O}=\{O_{\lambda}\}_{\lambda}$ be an open cover of $K^{(0)}=[-1,1]$. Assume, in order to arrive at a contradiction, that no finite subset of $\mathcal{O}$ covers $K^{(0)}$. By our assumption, at least one of $[-1,0]$ or $[0,1]$ are also not coverable by a finite subset of $\mathcal{O}$. Call one of these noncoverable subsets $K^{(1)}$. Repeating this bisection inductively, we obtain nested intervals $$ K^{(0)}\supset K^{(1)}\supset K^{(2)}\supset\cdots $$ such that each interval $K^{(n)}$ is of length $2^{-n+1}$ and not coverable by a finite subset of $\mathcal{O}$. Now for each $n$, pick $x_{n}\in K^{(n)}$. Since the sequence $(x_{n})_{n}$ is Cauchy, it has a limit $x$. Moreover, $x\in K^{(n)}$ for all $n$. Note that $x\in O_{\lambda(x)}$ for some $\lambda(x)$, and hence for $n$ sufficiently large, $K^{(n)}$ is in $O_{\lambda(x)}$, a contradiction.

The above establishes the desired property for $[-1,1]$. Now, let $K$ be an arbitrary closed subset of $[-1,1]$. Let $\mathcal{O}=\{O_{\lambda}\}_{\lambda}$ be an open cover of $K$. Since $K^{c}$ is open, note that $\mathcal{O}\cup\{K^{c}\}$ is a cover of $[-1,1]$. Therefore, we can find a finite subcover of $\mathcal{O}\cup\{K^{c}\}$, call it $\{O_{i}\}_{i=1}^{n}\cup\{K^{c}\}$, that covers $[-1,1]$. Therefore, $\{O_{i}\}_{i=1}^{n}$ covers $K$.

What about for arbitrary closed and bounded sets? Note that an arbitrary closed and bounded set $K$ can be contained in some interval $[-k,k]$. Then, the situation is identical to the previous paragraph, and we can conclude that $K$ satisfies the desired property.

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  • $\begingroup$ I got it. Thanks of the nice explanation. Is there any reason you started with $K^{(0)}=[-1,1]$. Can't you prove it using the first two paragraph if you started with arbitrary $ K$ $\endgroup$ – Sonal_sqrt Mar 28 '18 at 7:01
  • $\begingroup$ I used $[-1,1]$ for the bisection argument (this is how it's commonly proved in textbooks). You can probably do it directly, but I suspect this is clearer. $\endgroup$ – parsiad Mar 28 '18 at 7:01

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