# 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.

• "$S$ is compact" does not mean "there is an open cover $F$ of $S$ such that $F$ has a finite subcover" but rather it means "for every open cover $F$ of $S$, $F$ has a finite subcover". – Trevor Wilson Oct 24 '13 at 18:04
• So your proof should begin "let $F$ be an open cover of $S_1 \cup S_2$" and proceed to find a finite subset of $F$ that covers $S_1 \cup S_2$. – Trevor Wilson Oct 24 '13 at 18:05

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 $S_1\cup S_2$.
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.
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,