# 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 $$p$$.

• Your union does not cover $p$. Commented Jan 1, 2022 at 21:41
• In fact, none of your sets contain $p$. Commented Jan 1, 2022 at 21:44
• That's not a cover of $K$. The union doesn't contains the whole $K$ Commented Jan 1, 2022 at 21:44
• And a clarification about the way you are using the term "open cover". An open cover of $A$ is a set of open sets whose union contains $A$. You used the term "open cover" to refer to the union, which is not right. Commented Jan 1, 2022 at 21:50
• @jjagmath that's fair - imprecise wording on my part. Though I suppose to be pedantic the union is also an open cover singleton. Commented Jan 1, 2022 at 21:55

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