The question is: let $F_1, … F_n$ be compact subsets of X. Show that $\cup^{N}_{n=1} F_n$ is compact.
know that a set $ F \subset X$ is compact if every open cover $\mathcal {G}$ of F contains a finite subcover $\mathcal {H}$. Intuitively, I think that since $F_1, .. F_n$ are compact, they contain finite subcovers and so $\cup^{N}_{n=1} F_n$ at most has finite subcovers. I'm just stuck at trying to formally prove this out.
Help would be much appreciated!
Best Answer
Let $G$ be an open cover of $F$. then $G$ is an open cover of each $F_i$ with $i=1,2...n$.
since $F_i$ is compact, we can extract from $G$, a finite open subcover $G_i$ of $F_i$.
put now
$G_0=G_1 \cup G_2 \cup ...G_n$
$G_0$ is then a finite open subcover of $F$ .
Qed.