4

Is there a non-compact $T_{3\frac12}$ space $X$ (with at least two points) such that all continuous functions $X\to \mathbf R$ are bounded?

This is true for $T_3$ spaces as described in this answer: in this case, it is possible that all functions are simply constant. This is obviously not the case for $T_{3\frac{1}{2}}$ spaces (with at least two points).

Note that this property is equivalent to the property that the natural restriction embedding $C(\beta X)\to C(X)$ is an isomorphism. As a side question, does this property have a name?

If the answer is yes, then there is a further question: is it true if we replace $\mathbf R$ with an arbitrary (but fixed) $T_1$ space $Y$ (where by bounded we mean, as usual, that the range is contained in a compact subset of $Y$).

The question is motivated by this other question about whether one can characterise compactness of a space $X$ by the first order theory of the ring $C(X)$. A positive answer would imply that even the isomorphism type of $C(X)$ does not characterise compact spaces among $T_{3\frac12}$ spaces.

tomasz
  • 35,474
  • Replacing $\Bbb R$ by a $T_1$ space is pointless: we could take the codomain $X$ itself and the identity would not be bounded (for a non-compact space). – Henno Brandsma Aug 07 '21 at 13:35
  • @HennoBrandsma: I mean choosing a space $X$ for a given $Y$, not having one $X$ for all $Y$. I made an edit to reflect that. – tomasz Aug 07 '21 at 14:15
  • Given some fixed $T_1$ space $Y$ which is not compact (to avoid trivialities), you want to find some $X$ that is Tychonoff so that $C(X,Y) = C_b(X,Y)$? and preferably so that they contain more than just the constant maps? – Henno Brandsma Aug 07 '21 at 22:17
  • @HennoBrandsma: I think the ideal would be if $C(X,Y)$ separated points in $X$. And $X$ should also be non-compact. Simply having any non-constant continuous maps does not seem like a big improvement to me (I might be wrong, though). – tomasz Aug 08 '21 at 11:52

1 Answers1

8

Yes, such spaces ( all continuous real functions bounded) are called pseudocompact and a well-known non-compact example is $\omega_1$, the first uncountable ordinal in the order topology.

Indeed the ring $C(\omega_1+1)$ ( that one point compactification equals the Čech Stone one) is isomorphic to $C(\omega_1)$. See Gillman and Jerrison’s book rings of continuous functions for more examples.

Henno Brandsma
  • 242,131