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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?

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  • $\begingroup$ The general thing is compactness. A continuous function on a [non-empty] compact space attains its maximum (and minimum). That's because a continuous image of a compact space is compact. $\endgroup$ – Daniel Fischer Jul 29 '14 at 14:04
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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: If $f:\mathbb{R}^n\to \mathbb{R}$ 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.

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There is this document that states the theorem even more générally:

Theorem 3.4. (The Extreme Value Theorem) If $f : X \to R$ is real valued [edit: and continuous] function from a compact space to the real numbers, then f attains a greatest value, that is there is an $x \in X$ such that $f(x) \geq f(y)$ for all $y \in X$.

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    $\begingroup$ In proof of Theorem 3.4 the Lemma 3.1 is used (but not referenced with name). In the lemma 3.1 f is correctly required to be continuous. Thus is should be included in Theorem 3.4 that f is required to be continuous. $\endgroup$ – dioid Jul 29 '14 at 14:35
  • $\begingroup$ Thanks, dioid, I've edited my post $\endgroup$ – Igor Jul 29 '14 at 19:59

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