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Is this proof fine or does it lack rigor? Please help me spot mistakes or improve on it. Thanks!

Proof Attempt: Let $C$ be compact, so it must be closed and bounded due to the Heine-Borel Theorem. Consider the open interval $(a,b)$ and, w.l.o.g., suppose that $C\cap(a,b)=[a,x]$. Then $(a,x]\nsubseteq C\setminus (a,b)$, but $a\in C\setminus (a,b)$. For the sake of contradiction, suppose $C\setminus (a,b)$ is not closed, then $C\setminus (a,b)$ does not contain all its points of closure; that is $\exists$ a point $p$ of $C\setminus (a,b)$ such that $\in(a,x]$, then $a<p\leq x$ which implies $\exists\delta>0$ such that $(p-\delta, p+\delta)\cap C\setminus (a,b)=\emptyset$. So $\forall y\in(a,x]$, $y$ cannot be a point of closure of $C\setminus (a,b)$ ~ a contradiction.

EDIT

Attempt(Direct Approach): Let $U$ be an open cover of $C\setminus (a,b)$ such that $C$ is closed and bounded. Since $U\cup\{(a,b)\}$ is an open cover of $C$. But, $C$ is compact so $U\cup\{(a,b)\}$ has a finite subcover. Knowing that $\{(a,b)\}$ does not cover $C\setminus (a,b)$ then $U$ must be finite.

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    $\begingroup$ Why can we suppose that $C\cap(a,b)$ is an interval? What if, say, $C=\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}$? $\endgroup$ Commented Jul 15, 2018 at 15:36
  • $\begingroup$ I was thinking of considering a compact set $C$ such that $C\setminus (a,b)$ is not empty $\endgroup$ Commented Jul 15, 2018 at 15:38
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    $\begingroup$ Don't use Heine-Borel. The definition of $C\setminus(a,b)$ makes it perfectly suited for a direct argument with open covers $\endgroup$ Commented Jul 15, 2018 at 15:49
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    $\begingroup$ $C$ is closed (because it's compact). $C\setminus (a, b)$ is closed, it's intersection of $(a, b)^c$ and $C$, two closed sets. Now, we have a closed subset $C\setminus (a, b)$ of $C$, a compact set. So $C\setminus (a, b)$ is compact $\endgroup$ Commented Jul 15, 2018 at 16:10
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    $\begingroup$ @TheLastCipher As Adam's argument shows, in general removing an open set $U$ from a closed set $F$ gives a closed set, as $F \setminus U = F \cap U^\mathsf{c}$ is an intersection of closed sets. $\endgroup$ Commented Jul 15, 2018 at 16:57

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You seem to be assuming that $C\cap (a,b)$ is a closed interval. That is not at all the case. For instance, what if $C=\{1,2,3,\ldots, n\}$ for some $n$? Or $C=\{0,\frac11, \frac12,\frac13,\ldots\}$? Or the Cantor set? As you can see, it doesn't even have to be a union of closed intervals in any nice way. As a general intuition, open sets look "nice", but closed sets can be quite "ugly".

However, if $C$ is closed and bounded, what can you say about $C\setminus (a,b)$? Is it closed? Is it bounded?

You can do it a lot more directly from the definition of compact as well. Let $\mathcal U$ be an open cover of $C\setminus (a,b)$. Then $\mathcal U\cup \{(a,b)\}$ is an open cover of $C$. Can you do the rest?

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  • $\begingroup$ +1 Started typing then saw your answer. Lol. $\endgroup$ Commented Jul 15, 2018 at 15:38
  • $\begingroup$ I want to say that $C\setminus (a,b)$ is closed before I've written this proof attempt. But was unsure how to show. $\endgroup$ Commented Jul 15, 2018 at 15:45
  • $\begingroup$ @TheLastCipher It's easier to show that $\Bbb R\setminus (C\setminus(a,b))$ is open, using that $(a,b)$ and $\Bbb R\setminus C$ are both open. $\endgroup$ Commented Jul 15, 2018 at 15:48
  • $\begingroup$ I'll also give the direct approach a try. I'll comment if I have any more problems. Thank you. $\endgroup$ Commented Jul 15, 2018 at 15:49
  • $\begingroup$ To the proposer: The point of the last paragraph of this Answer is that if $ C$ is a compact subset of any space $X$ and if $U$ is an open subset of $X$ then $C$ \ $U$ is compact. $\endgroup$ Commented Jul 16, 2018 at 0:42

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