The Wikipedia entry on the extreme value theorem
says that if $f$ is a real-valued continuous function
on a closed and bounded interval $[a,b]$,
then $f$ must attain a maximum value,
i.e. there exists an $x \in [a,b]$
such that $f(x) \geq f(y)$ for all $y \in [a,b]$.
I think that there is a more general version of the extreme value theorem
which states a similar result for a closed and bounded subset of $\mathbb{R}^n$.
(At least I think I remember hearing about this in a class on metric spaces.)
Is there a statement of this more general version of the theorem,
hopefully with a reference as well?
Best Answer
Yes, in fact they both come from two standard results in analysis:
Theorem 1: A subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded.
Theorem 2: Continuous image of a compact set is a compact set.
Corollary: Given $f:\mathbb{R}^n\to \mathbb{R}$ is continuous and $A\subset \mathbb{R}^n$ is closed and bounded. By Theorem 1, $A$ is compact. Hence $f(A)$ is a compact subset of $\mathbb{R}$. So $f(A)$ is closed and bounded. That proves the Extreme Value Theorem.