I have a question about the topological space underlying a Banach space.

A topological space $X$ is realcompact iff it is homeomorphic to a closed subset of an infinite product of the form $\mathbb R^\kappa$. Closed subsets of realcompact spaces are realcompact.

A classical result in infinite topology states that every infinite dimensional separable Banach space is hemeomorphic to $\mathbb R ^ \omega$, so in particular, a separable Banach space is realcompact.

What about non-separable Banach spaces? Are they realcompact?

Since it is consistent with ZFC that there are discrete space which are not realcompact and every discrete set $X$ is a closed discrete subset of a space $\ell^p(X)$, it seems that in general the answer will be no. But of course, this counterexample is a huge space and in particular, it is not a ZFC-counter-example.

So, one question would be: Is there a ZFC-example of a Banach space which is not realcompact? Another question would be: Is there an easy way to determine which Banach spaces are realcompact and which are not?

Thank you!

  • $\begingroup$ FYI, closed subsets of Euclidean space $\mathbb R^n$ (with $n$ finite) are also realcompact. $\endgroup$
    – PatrickR
    Jul 7, 2023 at 4:26

1 Answer 1


Every metric space of nonmeasurable cardinality is realcompact. [1] 15.24. Thus, if there are no measurable cardinals, then every metric space is realcompact

As you noted...To get a Banach space that is not realcompact: Let $X$ be a set with measurable cardinal, and then the discrete topology on it is not realcompact. So for example Banach space $l^1(X)$ has a closed subset homeomorphic to that, so is not realcompact.

There is some nice work starting with [2] on when the weak topology of a Banach space is realcompact (or normal, or Lindelof...).

[1] L. Gillmann & M. Jerison, Rings of Continuous Functions (1960)

[2] H. H. Corson, The weak topology of a Banach space, Trans. Amer. Math. Soc. 101 (1961), 1--15.

  • $\begingroup$ Thanks for the fact that without measurable cardinals every metric space is realcompact. I was not aware of that. (and of course, also thanks for the link regarding the weak topology) $\endgroup$
    – Tom
    Nov 24, 2015 at 15:29
  • 1
    $\begingroup$ Perhaps it is more illuminating to see the chain of implications from metrizability to realcompactness. It turns out that every metric space is paracompact. The paracompact spaces are precisely the spaces with a compatible supercomplete uniformity (supercomplete simply means that the hyperspace uniformity is complete). Every supercomplete uniform space is a complete uniform space. And below the first measurable cardinal, the realcompact spaces are precisely the topological spaces with a compatible complete uniformity. Therefore, every metric space of non-measurable cardinality is realcompact. $\endgroup$ Nov 27, 2015 at 16:12

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