Characterization of Compactness in R

Theorem

A set $K \subseteq R$ is compact iff it is closed and bounded.

Proof

( $\Rightarrow$ ): Let $K$ be compact. Let's first prove that $K$ must be bounded. Assume for contradiction that $K$ is not a bounded set. The idea is to show that there is a sequences that marches off to $\infty$ in such a way that it cannot have a convergent subsequence as the definition of compact requires. To do this, notice that because $K$ is not bounded then $\exists x_{1} \in K$ satisfying $| x_{1} | > 1$ . Like wise $\exists x_{2} \in K$ with $| x_{2} | > 2$ , and so on, where for all $k$ then $x_{k} \in K, | x_{k} | > k$ .

Now because $K$ is compact, then $(x_{n})$ should have a convergent subsequence $(x_{n_{k}})$ . But the elements of the sequence satisfy $| x_{n_{k}} | > n_{k}$ and thus $(x_{n_{k}})$ is unbounded. But that contradicts $(x_{n_{k}})$ being a convergent sequence (see Cauchy Sequences Are Bounded). Thus $K$ must be bounded.

Now for the closed part, let $x$ be a limit point of $K$ , so then $\exists (x_{n}) \in K - {x}$ where $x_{n} \to x$ . By the definition of compactness, then $(x_{n})$ has a convergent subsequence $(x_{n_{k}})$ where because subsequences converge to the same value as their convergent parent sequence, then $(x_{n_{k}}) \to x$ . The definition of compactness also requires that $x \in K$ , completing the proof that $K$ is closed.

( $\Leftarrow$ ): Let $K$ be closed and bounded and let $(x_{n}) \subseteq K$ . The Balzano-Weierstrass Theorem says that $\exists (x_{n_{k}})$ such that $x_{n_{k}} \to x$ exists since $K$ is bounded. Since $K$ is closed then $x \in K$ . Thus every sequence in $K$ contains a subsequence converging to a limit in $K$ , showing the definition of compact.

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