Compactness Is Preserved By Continuity

Theorem

The continuous image of a compact set is also compact.

Proof

Let $f : K \to R$ be such that $f$ is continuous for all $c \in K$ , and $K$ is compact.

To show that $f (K)$ is compact, we'll take an arbitrary sequence in the set and show that it converges to a point in the set (see Closed iff Convergent Cauchy Sequences). Let $(y_{n}) \subseteq f (K)$ be such an arbitrary sequence. Since $y_{n} \in f (K)$ then $\exists x_{n} \in K$ where $f (x_{n}) = y_{n}$ . Since $K$ is compact, then $\exists (x_{n_{k}}) \subseteq K$ (a subsequence) such that $x_{n_{k}} \to x$ where $x \in K$ . Note that $f (x) \in f (K)$ automatically

But then since $f$ is continuous at $x \in K$ , then any sequence converging to $x$ like $x_{n_{k}}$ (like it does here) implies that $f (x_{n_{k}}) \to f (x)$ (so then we get the right immediately). But since $y_{n} = f (x_{n_{k}}) = y_{n_{k}}$ then $y_{n_{k}} \to f (x) \in f (K)$ as desired.

It is bounded trivially. Thus $f (K)$ must be compact.

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See Extreme Value Theorem for an immediate use of the theorem.