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Professors Road Map to Glory.

  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.

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2 Answers 2

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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.

The second one is your definition you've provided.

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  • $\begingroup$ Great! now I see the connection. $\endgroup$ Commented Nov 3, 2019 at 1:02
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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.

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