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Let $f$ be a continuous function on $A$ which is a compact subset of $\mathbb{R}^2$. Due to Weierstrass theorem, I know that there exists $(a,b) \in A$ such that $$f(a,b) = \max_{(x,y)\in A} f(x,y).$$ This implies that the set $$U = \arg \max_{(x,y)\in A} f(x,y)$$ is non-empty. My question is the following: is it always possible to "choose" an element of U when U is non-singleton ?

I thought considering some lexicographical order on U, but I don't see how to prove that a maximal element always exists.

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  • $\begingroup$ $U$ will always be compact (as a closed subset of the compact set $A$), so lexicographical order will have a maximal element. $\endgroup$ – martini Oct 16 '12 at 14:48
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    $\begingroup$ @martini : the topology on $\mathbb{R}^2$ induced by the lexicographical order is not the same as the standard topology. I don't think $U$ has to be compact with respect to the topology induced by the lexicographical order. So it must be proven that $U$ has a maximal element in that order (but I don't that will be difficult). $\endgroup$ – Stefan Smith Oct 16 '12 at 22:02
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    $\begingroup$ @Romain : If you mean the standard lexicographical ordering on $\mathbb{R}^2$, then $U$ is not necessarily compact in that order topology, but it is not difficult to show that $U$ has a maximal element in that ordering. $\endgroup$ – Stefan Smith Oct 17 '12 at 1:03

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