6
$\begingroup$

A space is realcompact if its a closed subspace of an arbitrary product of real lines, with product topology.

A cardinal $\kappa$ is called measurable if there exists a (countably additive) $\{0, 1\}$-valued measure $\mu:\kappa\to \{0, 1\}$ with $\mu(\kappa) = 1$ and $\mu(\{x\}) = 0$ for $x\in \kappa$. This is not the standard meaning of what set theorists call a measurable cardinal. Apparently the proper terminology is "$\sigma$-measurable cardinal".

Its known that a discrete space is realcompact if and only if it has non-measurable cardinality. The proof of this result basically follows trivially from definitions. Its also known that a metrizable space of non-measurable cardinality must be realcompact.

For discussion on above see Rings of continuous functions by Gillman and Jerison.

In the same book, they never question if realcompactness is equivalent to non-measurable cardinality also for metric spaces. Is that true?

Also see this post for some previous discussion of whetever every metrizable space is realcompact.

$\endgroup$
7
  • $\begingroup$ "a discrete space is realcompact if and only if it has non-measurable cardinality": yes, this is the statement in many books. However, I'm not so happy with that: assume, $\kappa$ is a measurable cardinal. If I recall correctly, $\kappa^+$ is not measurable. Hence, the discrete space of size $\kappa^+$ is realcompact. However, it contains the closed, discrete subspace of size $\kappa$, which is not realcompact. Therefore, I think, the correct statement should be "a discrete space is realcompact if and only if it its cardinality is less than every measurable cardinal $\endgroup$
    – Ulli
    Apr 20 at 15:08
  • $\begingroup$ Oh I think I got it. There's a difference between what set theorists call measurable cardinals and what my book calls measurable cardinals. $\endgroup$
    – Jakobian
    Apr 20 at 16:15
  • $\begingroup$ @Ulli If we're going for a statement using what set theorists call measurable cardinals, then the correct statement would be that $X$ is realcompact iff $|X|$ is greater than the least measurable cardinal. $\endgroup$
    – Jakobian
    Apr 20 at 16:25
  • 1
    $\begingroup$ By the way: in your last comment, it should be "$|X|$ is less than the least measurable cardinal". $\endgroup$
    – Ulli
    Apr 20 at 16:43
  • 1
    $\begingroup$ @Ulli I've included a definition in the body of the question, it seems that perhaps there is some ambiguity on what people call a measurable cardinal in topology. I believe it might go back to Ulam's definition of measurable cardinal. The proper terminology that set theorists use seems to be $\sigma$-measurable $\endgroup$
    – Jakobian
    Apr 20 at 16:45

1 Answer 1

3
$\begingroup$

Yes, its true.

See this great answer by KP Hart


I present a slightly different proof below.

Note the following result from the article The sup = max problem for the extent and the Lindelöf degree of generalized metric spaces, II by Hirata:

Theorem. (corollary 2.3) Let $X$ be a semi-stratifiable space with $e(X) = \kappa$, where $\text{cf}(\kappa) > \omega$. Assume that $\tau^\omega < \kappa$ for each $\tau < \kappa$. Then $X$ has a closed discrete subset of size $\kappa$.

Here $e(X) = \sup\{|D| : D\subseteq X, D\text{ is closed and discrete}\}+\omega$ is the extent of $X$.

Suppose that $X$ is a metrizable space (in particular, semi-stratifiable) with $|X|$ a measurable cardinal. Since $X$ is metrizable, $e(X) = w(X)$ and $|X|\leq w(X)^\omega$ so that $|X|\leq e(X)^\omega$, see Handbook of set-theoretic topology by Kunen and Vaughan. From $|X|\leq e(X)^\omega$ it follows that $e(X)$ is a measurable cardinal. If $e(X) = \kappa$ is the least measurable cardinal then $\kappa$ is strongly inaccessible, so conditions of Hirata's theorem are satisfied and there exists a closed discrete subset $D\subseteq X$ with $|D|$ measurable. Since closed subspace of realcompact space is realcompact, $X$ is not realcompact. If $\kappa$ is not the least measurable cardinal then again there exists a closed discrete set $D\subseteq X$ with $|D|$ measurable, which shows that $X$ is not realcompact.

Thus we established that a metrizable space $X$ is realcompact iff $|X|$ is non-measurable.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .