3

Let $X$ be a space and $\beta X$ denote the Stone-Cech compactification of $X$.

$X$ is realcompact if for each $p\in \beta X\setminus X$ there is a continuous $f:X\to[0,\infty)$ such that $\beta f (p)=\infty$, where $\beta f:\beta X\to[0,\infty]$ is the extension of $f$.

It is fairly easy to see that every separable metric space $X$ is realcompact. Embed $X$ into $[0,1]^\omega$. If $p\in \overline X \setminus X$ then there is a sequence of points $(x_n)$ in $X$ converging to $p$. Set $f(x_n)=n$ and extend $f$ to a function from $X$ into $[0,\infty)$ using Tietze's theorem.

But what if the space is not separable?

Are all metric spaces realcompact? How about complete metric spaces?

Jakobian
  • 15,280
Forever Mozart
  • 9,534

1 Answers1

4

It’s a consistency result. If there are no measurable cardinals, every metric space is realcompact; this is $\mathbf{15.24}$ in Gillman and Jerison, Rings of Continuous Functions. If $\kappa$ is a measurable cardinal, the discrete topology on $\kappa$ is metrizable but not realcompact.

Brian M. Scott
  • 631,399
  • Do you know if every completely metrizable realcompact space is separable? – Forever Mozart Nov 03 '16 at 20:30
  • 1
    @ForeverMozart: A discrete space of cardinality $\omega_1$ is completely metrizable and realcompact but not separable. – Brian M. Scott Nov 03 '16 at 20:36
  • If measurable cardinals exist, is it true that every metrisable space whose cardinality is smaller than every measurable cardinal is realcompact? – Jonas Frey Apr 12 '23 at 20:43
  • 1
    @JonasFrey: Yes: it follows from Theorem $2$ in this paper [PDF]. – Brian M. Scott Apr 13 '23 at 06:29
  • 1
    Isn't the precise statement of theorem 15.24 that the metric space $X$ is realcompact if $|X|$ is non-measurable? So it should answer @JonasFrey question directly. – Jakobian Jun 17 '23 at 13:42
  • 1
    @Jakobian: Yes, but I didn’t have G&J at hand when I answered that question (and in any case I liked the argument in the Rice paper.) – Brian M. Scott Jun 17 '23 at 17:25
  • hmm, but the property of being measurable is not "upward closed" in the cardinal hierarchy, is it? so does that mean that there are realcompact spaces with non-realcompact subspaces? – Jonas Frey Jun 17 '23 at 22:42
  • 1
    @JonasFrey: There are. Let $X$ be any Tikhonov space that is not realcompact; then $\beta X$ is a compact (hence realcompact) space having $X$ as a non-realcompact subspace. (And you don’t actually need all of $\beta X$: the Hewitt realcompactification $\nu X$ of $X$, which is a subspace of $\beta X$, is enough.) – Brian M. Scott Jun 18 '23 at 02:14
  • ok that makes sense. – Jonas Frey Jun 18 '23 at 02:27
  • @BrianM.Scott Is the converse of this theorem known? I've proved it and I'm wondering if I should write a Q&A on math.stack about it. That is, a metrizable space $X$ is realcompact iff $|X|$ is non-measurable. My proof isn't too long, it only uses one result from a paper I found – Jakobian Apr 19 '24 at 23:05
  • @Jakobian: I’m afraid I don’t know. It’s not an area close to my main interests, and I fear that after 13 years of retirement I’m rather rusty besides. – Brian M. Scott Apr 20 '24 at 02:21