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To prove the Heine-Borel theorem you need to show that a compact set is both closed and bounded. There is a proof of the theorem in the book The Elements of Real Analysis by Bartle. In the proof to show that a compact set K is closed, a specific open cover is used: $$ G_{m} =\left \{ y \, \epsilon \, \mathbb{R}^{n}: ||y-x|| > 1/m \: \forall m \, \epsilon \, \mathbb{N} \right \}$$ where $$ x \, \epsilon \, \complement_{K} $$

I'm curious as to why we are allowed to use a specific open cover to prove a compact set is closed. My thought was that a general open cover should be used because compactness of a set is a property which is related to every open cover on that set.

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    $\begingroup$ We are using the compactness of the given set, not proving the compactness of the given set. If you were proving compactness you would have to start with an arbitrary open cover, but if you are using compactness of some set, then you must start with some specific open cover, and then use its finite subcover in whatever way you want.You should get the difference between proving compactness of a set and using compactness of a set. $\endgroup$ – астон вілла олоф мэллбэрг Dec 11 '17 at 3:09
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The open cover that you mentioned are used to prove that if it is a compact set, then it is closed and bounded. Hence a particular open cover can be used.

The $G_{\alpha}$ in the proof in the converse direction should be viewed as a general open cover.

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Bartle considered these $G_m$'s to show $C(K)$ is open, as he wanted to show every $x\in C(K)$ lies in an open set which doesn't intersect $K$.

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Given a union of open intervals $\bigcup_i (A_i,B_i)$ such that $\bigcup_i (A_i,B_i) \supset [A,B]$ we can take finite number of them $\bigcup_{i\le N} (A_i,B_i)$ such that $\bigcup_{i\le N} (A_i,B_i) \supset [A,B]$

Proof:

Step 1: Cover $A$ with an open interval $(A_i,B_i)$

Step 2: Choose an open interval $(A_{i_1},B_{i_1})$ that intersect with $(A_i,B_i)$ to the left for which

case 1: Length $L=\mu((A_{i_1},B_{i_1})\bigcap (A_i,B_i)^c)$ is maximum if the maximum exists. If $B$ has not been covered repeat step 2 with $(A_{i_1},B_{i_1})$ and $(A_i,B_i)$ replaced with $(A_{i_2},B_{i_2})$ and $(A_{i_1},B_{i_1})$ respectively.

case 2: if the maximum of $L$ doesn't exist that means there is a subsequence of intervals that intersect with $(A_i,B_i)$ for which the subsequence {$(A_{i_m},B_{i_m})$}for which {$B_{i_m}$} is increasing and the limit is $k$, since $k$ is also covered by a certain interval $(A_{i_{2}},B_{i_{2}})$ we can find an interval $(A_{i_1},B_{i_1})$ from the subsequence {$(A_{i_m},B_{i_m})$} that intersects both $(A_{i_{2}},B_{i_{2}})$ and $(A_i,B_i)$. If $B$ has not been covered Repeat step 2 with $(A_{i_1},B_{i_1})$ and $(A_i,B_i)$ replaced with $(A_{i_3},B_{i_3})$ and $(A_{i_2},B_{i_2})$ respectively.

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It must be that $\lim_{n \to\infty} B_{i_{n}} > B$ , if not that would have implied that $B$ is not covered in any open interval. Which means there is an $N$ such that $B_{N}>B $ and $\bigcup_{i\le N} (A_i,B_i) \supset [A,B]$

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