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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?

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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.

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    $\begingroup$ @ForeverMozart: A discrete space of cardinality $\omega_1$ is completely metrizable and realcompact but not separable. $\endgroup$ Commented Nov 3, 2016 at 20:36
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    $\begingroup$ @JonasFrey: Yes: it follows from Theorem $2$ in this paper [PDF]. $\endgroup$ Commented Apr 13, 2023 at 6:29
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    $\begingroup$ 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. $\endgroup$
    – Jakobian
    Commented Jun 17, 2023 at 13:42
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    $\begingroup$ @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.) $\endgroup$ Commented Jun 17, 2023 at 17:25
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    $\begingroup$ @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.) $\endgroup$ Commented Jun 18, 2023 at 2:14

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