# Suppose that $E_1,E_2,...E_n$ are compact sets, Prove that their union is also compact.

1. Explain why if $$\{a_n\}_{n}^{\infty}$$ is a sequence of numbers, and $$a_n \in E_1\cup E_2,...\cup E_n$$ for all $$n\in \mathbb{N}$$, then their is a subsequence $$\{a_{n(k)}\}_{k}^{\infty}$$ completely contained in one of the compact sets $$E_k$$ for some $$k \in \{1,2,...n\}$$
2. Explain why, for the subsequence $$\{a_{n(k)}\}_{k}^{\infty}$$ contained completely in one compact set $$E_k$$, there is a convergent subsequence of the subsequence with a limit contained in $$E_k$$.

these are current defenitions that I have for compactness, looking at the RTG, I more than likely will use #2

A set $$E$$ is compact iff

1. Every open cover of E permits a finite subcover.
2. Every sequence of values in E contains a convergent subsequence whose limit is in E.
3. E is closed and bounded.

My intuition:

If $$\{a_n\}_{n}^{\infty}$$ is a sequence in a union of compact sets.

I am super hazy in the part that reads, "then their is a subsequence $$\{a_{n(k)}\}_{k}^{\infty}$$ completely contained in one of the compact sets $$E_k$$ for some $$k \in \{1,2,...n\}$$"

however given then there is a subsequence $$\{a_{n(k)}\}_{k}^{\infty}$$ completely contained in one of the compact sets $$E_k$$ for some $$k \in \{1,2,...n\}$$ the subsequence of the subsequence is convergent to a point in $$E_k$$ thus by the second definition of compactness, the union of the sets are compact.

The first one is pigeonhole principle. If there are only finitely many terms in each $$E_i$$ then there's only finitely many terms in their union, contradiction.
Suppose $$A_1,A_2,...$$ cover $$E_1 \cup E_2 \cup ... \cup E_n$$.
Since this sequence of sets covers $$E_1$$, we take a finite subcover. Since this sequence of sets also covers $$E_2$$, we take another finite subcover.
In total, we have $$N$$ finite subcovers, which unioned together still makes a finite subcover. Therefore, $$E_1 \cup E_2 \cup ... \cup E_n$$ is also compact.