Why can't we find an open ball $B_r(k)$ for all $k\in K$?

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 $$k\in K$$?

• 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? Feb 12 '20 at 15:40
• 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.
– user748968
Feb 12 '20 at 16:20

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 $$z\in K^\complement$$.