Nested open cover and compactness

I don't understand how a compact set can have a finite open subcover given certain infinite open covers.

For example, say my set $K\subset \mathbb{R}^n$ is the closure of some ball around a point: $K=\overline{B_r(p)}$. Since we're in Euclidean space and $K$ is closed and bounded, $K$ is compact. Consider the sequence of open sets $\{\overline{B_{r/n}(p)}^c\}_{n=1}^\infty$. Their infinite union is an open cover of $K$ but any finite union isn't; namely, any finite union doesn't contain $p$.

Solution 1:

$p$ isn't in $\cup_{n=1}^\infty \overline{B_{r/n}(p)}^c,$ so it's not an open cover:

$p=\cap_{n=1}^\infty \overline{B_{r/n}(p)}=\left(\cup_{n=1}^\infty \overline{B_{r/n}(p)}^c\right)^c$

Solution 2:

I see your point but note that $\{\overline{(B_{r/n}(p))}^{c}\}_{n=1}^\infty$ is not covering the center $p$. Their infinite union is not an open cover of $K$, since the point $p$ would always be left out from the union.

To see why, consider the unidimensional example of this sequence of intervals $(1/n, 1)$. If you take the countable union among all naturals you are not going to cover the interval $[0,1)$ since in the union you will get $(0,1)$. Formally this is explained by saying that a countable union of open set is open, hence it must be $(0,1)$ and not $[0,1)$ since the latter is not open.