Is it true that if every continuous real-valued function on a metric space is bounded, then that metric space is compact
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Suppose $X$ is not compact, then take a injective sequence $(x_n)_n$ of elements of $X$ such that the values set $X_f := \{ x_n\;|\;n\in\mathbf{N}\}$ is discrete and closed. Then look at the function $f : X_f \to\mathbf{R}$ defined by sending $x_n$ to $n$. This is continuous and not bounded. Now Uryshon's lemma ensures that $f$ can be extended to a continuous function (still noted) $f:X\to\mathbb R$ which won't be bounded.

Olórin
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http://math.stackexchange.com/questions/181367/pseudocompactness-does-not-imply-compactness
– Olórin Feb 22 '15 at 18:46