## September 2000

Problem Statement: Use the open cover definition of compactness to show that any finite union of compact sets is compact.

Proof: Let $C_1,C_2,\dots,C_n$ be compact sets in $\mathbb{R}$. We wish to show that $C=C_1\cup C_2\cup\dots\cup C_n$ is a compact set.

Let $T=\bigcup\limits_{\alpha}T_{\alpha}$ be an open cover of $C$. Then $T$ is an open cover for $C_j$ for every $j\in[1,n]$. Since each $C_j$ is compact there is a finite subcover of $T$ which covers $C_j$ for each $j\in[1,n]$. So for each $C_j$ we may construct $T_{j}=T_{j_1}\cup T_{j_2}\cup\dots\cup T_{j_m}$, a finite subset of $T$ which covers $C_j$, for each $C_j$.

I claim that $T_C=\bigcup\limits_{j=1}^{n}T_j$ is a finite subcover of $C$.

First note that $T_C$ is a union of finitely many unions of open sets and so it is an open set. Now to verify that $T_C$ covers $C$. Let $x\in C$, then $x\in C_j$ for some $C_j$. Thus, by construction, $x\in T_j$ since $T_j$ is an open cover of $C_j$. Therefore, $x\in T_C$ and we may conclude that $T_C$ is a finite subcover of $T$.

Therefore, every open cover of $C$ has a finite subcover and so $C$ is compact.

$\Box$

Reflection: The key in this proof is that we had a finite union of compact sets. From there all we had to do was union up all of their finite subcovers. A finite union of a finite union is simply a finite union of open sets.