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I know that compact sets imply the existence of a maximizer, but is the converse true:

Let $(X,d)$ be a metric space. Suppose that whenever $f$ is a continuous (and real) function on $X$, there exists a point $y \in X$ such that $f(y) \geq f(x)$ for all $x \in X$. Then X is compact.

Cool
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    It seems not, since no bounds from below follow (take $X=(-\infty,0]$). – Milly Oct 18 '14 at 18:30
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    ^That doesn't work, since the function $f(x)=-x$ on $(-\infty, 0]$ has no maximum. – Nishant Oct 18 '14 at 18:30
  • (I don't think that the observation "by applying the statement to $|f|$, any real-valued function on $X$ is bounded" warrants having a whole new question) – Najib Idrissi Oct 18 '14 at 18:38

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If $X$ is metric and not compact, then $X$ has an infinite closed discrete subset $D$. By passing to a subset if necessary, we may assume that $D=\{x_n:n\in\Bbb N\}$ is countably infinite. Now define $f:D\to\Bbb R:x_n\mapsto n$, and apply the Tietze extension theorem to get an unbounded continuous real-valued function on $X$.

Brian M. Scott
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