Is every metric space realcompact?

Question

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?

Answer 1

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.