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I've read this demonstration:

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This may be silly and confusing but the basic idea seems to be that for any point not in $K$, we can find a certain open ball $B_r (x)$ such that $B_r (x) \subset K^c$. This is reasonable to me. But why can't we:

  • Find an open ball $B_r(k)$ for all $k\in K$?

I believe this would generate some contradiction but I couldn't find it. Perhaps what I am asking is a bit nonsense but I need this to have peace. I know that the theorem is demonstrated and I know that the the complement of an open set is closed but I still feel this piece of information is needed. What would break if we assume that we can always find an open ball $B_r(k)$ for all $k\in K$?

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  • $\begingroup$ What do you mean when you write “Find an open ball $B_r(k)$ for all $k\in K$”? A ball such as what happens? $\endgroup$ Feb 12 '20 at 15:40
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    $\begingroup$ What breaks is that if $x\in K$, then union of the sets $G_m$ never cover $x$. You will need to get an open set $G\ni x$ to cover it. Then, when you get the finite cover, $G$ might be part of it. After the finite cover is chosen there is no way to ensure that points of $K^c$ are not outside of the $G_k$ of the finite cover. Sure, the finite cover covers all of $\Omega$, but that might be only thanks to $G$ being in the finite cover. $\endgroup$
    – user748968
    Feb 12 '20 at 16:20
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I think you are asking if we can go about doing the proof differently. I wrote down an alternate proof (though the idea is the same) that hopefully captures your question.

Let $K$ be a compact set and we show that $K^\complement$ is open. To this end, fix $x\in K^\complement$ and notice that for any $y\in K$, $r_y := d(x, y) > 0$ since $y \neq x$. Observe that $$ K\subseteq \bigcup_{y\in K} B\left(y, \frac{r_y}{2}\right). $$ Therefore, we have an open cover for $K$ and can extract a finite sub-cover. Let $y_1, \dots, y_n$ be such that $$ K\subseteq \bigcup_{j=1}^n B\left(y_j, \frac{r_{y_j}}{2}\right). $$ Define $\delta = \min_{1\leq j \leq n}\frac{r_{y_j}}{2}$ and observe that $B(x, \delta)\subseteq K^\complement$. Indeed, for any $y\in B(x, \delta)$ and any $j = 1, \dots, n$ we have $$ d(z, y_j) \geq d(y_j, x) - d(x, z) > r_{y_j} - \delta \geq r_{y_j} - \frac{r_{y_j}}{2} = \frac{r_{y_j}}{2}. $$ hence $z\not\in B(y_j, r_{y_j}/2)$ for any $j=1, \dots, n$. Since these balls cover $K$, we see that $z\not\in K$ or rather $z\in K^\complement$.

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