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!

  • 2
    $\begingroup$ It's not arbitrary, it's a finite union. $\endgroup$
    – xyzzyz
    Oct 17, 2016 at 23:09
  • $\begingroup$ Thanks for pointing that out! I gleaned this question off of a website and I blindly followed it without realizing. $\endgroup$
    – Nikitau
    Oct 17, 2016 at 23:10
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    $\begingroup$ Where you wrote "they contain finite subcovers," it seems as though "they" refers to the $F_n$'s. I trust that's not what you intended, since the $F_n$'s are not covers. I think that, once you figure out what "they" should refer to, you'll be well on the way to a proof. $\endgroup$ Oct 17, 2016 at 23:13

3 Answers 3


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 $F$ .



$F = \bigcup^n F_i$ be the union in question. We want to show that $F$ is compact. Take any open cover $F \subset \bigcup U_j$. Clearly $F_i \subset F$, and so each $F_i$ is also covered by $\bigcup U_j$. Thus for each $i$ there exist a finite subcover $U_{i,1}, \ldots U_{i,k_i}$ of $F_i$. Now, clearly $\bigcup_{1 \leq i \leq n} \bigcup_{1 \leq j \leq k_i} U_{i, k_j}$ is finite, covers each $F_i$, and so it also covers $F$.


Just show that the union of two compact sets is compact. The finite union will follow by induction.:)


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