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A continuous real-valued function defined on a compact space is bounded and attains its bounds. Is it so that on a non-compact space it is always possible to define a continuous function that does not have these properties? (In fact, this is the case if the non-compact set lies in a finite-dimensional normed vector space, because in this case the set is either not closed or not bounded.)

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No. A space with your property is called pseudocompact. A pseudocompact space need not be compact: for example, an infinite set with the particular point topology.

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  • $\begingroup$ $X$ is pseudocompact if image of any continuous function $f \colon X \to \mathbb R$ is bounded. But nothing is said about attainability of its bounds. So author of question speaks about some subclass of pseudocompact spaces, right? Is there a name for such subclass? $\endgroup$ – Appliqué Apr 25 '13 at 20:38
  • $\begingroup$ @Nimza: It's easy to see that this subclass is in fact, as I stated in my answer, the whole thing. $\endgroup$ – Chris Eagle Apr 25 '13 at 20:43
  • $\begingroup$ Ah, yes, if $f^\ast$ for example is upper not-atteinable bound of $f$, then $x \mapsto 1/(f^\ast-f(x))$ will be unbounded $\endgroup$ – Appliqué Apr 25 '13 at 21:01
  • $\begingroup$ What can we say if the compact set is a subset of an infinite- dimensional normed vector space or, more generally, is a metric space? $\endgroup$ – Rodney Coleman Apr 26 '13 at 6:59
  • $\begingroup$ @RodneyColeman: We can say lots of things about such sets. Do you have an actual question about them? $\endgroup$ – Chris Eagle Apr 26 '13 at 9:08
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No, it is not: there are $T_3$ spaces on which every real-valued continuous function is constant. E. K. van Douwen, A regular space on which every continuous real-valued function is constant, Nieuw Arch. Wisk. 20 (1972), 143-145, is a particularly nice example.

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