# Finite Union of Compact Sets Clarification

While studying some basic analysis/topology I have come across the proof regarding that the finite union of compact sets is compact using the definition of compactness.

For each compact set choose a finite subcover. The union of those subcovers will be finite and cover the union of the compact sets.

Alright, not a big deal.

However, I think I may have a misconception regarding my notion of a compact set.

When proving a compact set is indeed compact must I not show that EVERY open cover has a finite subcover?

The aforementioned proof, to me, seams as though it is only considering one possible option for an open cover.

I am not doubting the validity of the proof, instead I am looking for some clarification as to why we do not consider the possibility of other open covers.

• Every open cover of the union is an open cover of any component. – André Nicolas Dec 30 '13 at 1:16
• You start with an arbitrary cover of the union. The conclusion is that there is a finite subcover. (And since the cover was arbitrary, this gives you compactness.) To reach the conclusion, you use that there are finite subcovers for each of the compact sets in the union. – Andrés E. Caicedo Dec 30 '13 at 1:17

To clarify what the proof says: Suppose that you have compact sets $K_1, ..., K_N$ for some $N < \infty$. Choose any open cover $\mathcal{A}$ of $K_1 \cup ... \cup K_N$. Then by compactness, there is a finite subcover $\mathcal{A}_1$ of $K_1$; that is, $\mathcal{A}_1$ is a finite collection of open sets whose unions contains $K_1$. Choose a subcover $\mathcal{A}_2$ for $K_2$ in an identical manner, and continue to $\mathcal{A}_n$.
Then $\mathcal{A}_1 \cup ... \cup \mathcal{A}_N$ is a finite collection of open sets whose union covers $K_1 \cup ... \cup K_N$, so we've constructed a finite subcover.
Since $\mathcal{A}$ was arbitrary to begin with, we've covered all possibilites.
The proof should start as follows. Let $A_i$, $i=1$ to $n$ be compact. Let $\mathcal{C}$ be an open cover of $\bigcup A_i$. Then $\mathcal{C}$ is an open cover of any $A_i$. Let $\mathcal{C_i}$ be a finite subcover of $A_i$. Then $\dots$.