# Proving that the Union of Two Compact Sets is Compact

Prove if $S_1,S_2$ are compact, then their union $S_1\cup S_2$ is compact as well.

The attempt at a proof:

Since $S_1$ and $S_2$ are compact, every open cover contains a finite subcover. Let the open cover of $S_1$ and $S_2$ be $\mathscr{F}_1$ and $\mathscr{F}_2$, and let the finite subcover of $\mathscr{F}_1$ and $\mathscr{F}_2$ be $\mathscr{G}_1$ and $\mathscr{G}_2$, respectively. If I can show that $S_1\cup S_2$ contains a finite subcover for every open cover, then I will have showed that the union is indeed compact. I note that $\mathscr{G}_1\subset\mathscr{F}_1$ and that $\mathscr{G}_2\subset\mathscr{F}_2$ (by definition of an open subcover). Then, I note that $\mathscr{G}_1\subset\mathscr{F}_1\cup\mathscr{F}_2$ and that $\mathscr{G}_2\subset\mathscr{F}_1\cup\mathscr{F}_2.$

I’m not sure how to proceed from this point. I think I am on the right track though. Any suggestions to proceed would be appreciated. Thanks in advance.

#### Solutions Collecting From Web of "Proving that the Union of Two Compact Sets is Compact"

HINT: You’re starting in the wrong place. In order to show that $S_1\cup S_2$ is compact, you should start with an arbitrary open cover $\mathscr{U}$ of $S_1\cup S_2$ and show that it has a finite subcover. The hypothesis that $\mathscr{U}$ covers $S_1\cup S_2$ simply means that $S_1\cup S_2\subseteq\bigcup\mathscr{U}$. Clearly this implies that $S_1\subseteq\bigcup\mathscr{U}$ and $S_2\subseteq\bigcup\mathscr{U}$. Thus, $\mathscr{U}$ is an open cover of $S_1$ and also an open cover of $S_2$. Now use the compactness of $S_1$ and $S_2$ to produce a finite subset of $\mathscr{U}$ that covers $S_1\cup S_2$.

Every open cover of the union is an open cover of each set.
So for each set there is a finite subcover.
The union of the finite subcover is still finite and covers the union of the two sets.
So the union is indeed compact.

Suppose you have an open cover of $S_1 \cup S_2$. Since they are separately compact, there is a finite open cover for each. Then combine the finite covers, this will still be finite. Hence the union is compact,

Start with an open cover of the union. For $i=1,2$ it is also an open cover of $S_{i}$. These are compact so there are finite subcovers. The union of these subcovers is a finite subcover of the union.