Weierstrass theorem and compactness

Question

On my book the statement of Weierstrass theorem is: If $f$ is a continuous function $f:A\subseteq X\rightarrow \mathbb{R}$ defined on a compact set $C$, where $A$ is the domain of $f$ and $X$ is a metric space, then argmax$f$ and argmin$f$ are non-empty and compact sets.

I have a couple of proof to demonstrate that argmin and argmax are nonempty but I cannot understand why they are also compact sets.

Have you any hint/proof to see that?

Answer 1

• Argmin$(f)$ and Argmax$(f)$ are closed (continuous preimages of point sets, which are closed).
• Argmin$(f)$ and Argmax$(f)$ are subsets of $C$.
• $C$ is compact
• Closed subsets of compact sets are compact. (x)

Thus, Argmin$(f)$ and Argmax$(f)$ are compact.