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.
Best Answer
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 $S_1\cup S_2$.