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A space is compact provided every open cover admits a finite subcover.

A space is pseduocompact provided every continuous image of the space into the Euclidean line $\mathbb R$ is bounded.

A space is realcompact provided it embeds as a closed subspace of $\mathbb R^\kappa$ for some cardinal $\kappa$.

We can quickly see that every pseudocompact+realcompact space is compact: take the space $H\subseteq \mathbb R^\kappa$ (by realcompactness); its projection $H_\alpha\subseteq\mathbb R$ for each factor $\alpha<\kappa$ must be bounded (by psuedocompactness), and thus $\overline{H_\alpha}$ is compact by the Heine-Borel theorem. This makes $H$ a closed subset of the compact space $\prod_{\alpha<\kappa}\overline{H_\alpha}$, and thus compact.

It's immediate that every compact space is pseudocompact. It's also true that every compact Hausdorff space is realcompact.

So we have a cute characterization for Hausdorff spaces: A Hausdorff space is compact if and only if it is both pseudocompact and realcompact.

My question is this: is there a natural property $R$ that implies neither Hausdorff nor pseudocompact, such that an arbitrary space is compact if and only if it is both pseudocompact and $R$?

Steven Clontz
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    For arbitrary spaces, pseudocompactness tells you very little about compactness, since there might just not be any nonconstant functions to $\mathbb{R}$ at all. – Eric Wofsey Jul 01 '23 at 19:54
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    The projections maps $\Bbb R^{\kappa}\to\Bbb R$ are not closed maps in general. What makes you say $H_\alpha$ must be closed? – FShrike Jul 01 '23 at 21:37
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    @FShrike I agree, instead the proof should be as follows, each $H_\alpha$ is bounded, so $H$ is a closed subspace of compact space $\prod_{\alpha < \kappa} \overline{H_\alpha}$, hence compact. – Jakobian Jul 01 '23 at 21:45
  • Thanks for the catch, will edit appropriately when I can. – Steven Clontz Jul 01 '23 at 22:11

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In the article "Supports of continuous functions" by M. Mandelker, the following definition of realcompactness and results are given:

Definition 1. A family $\mathcal{F}\subseteq\mathcal{P}(X)$ of subsets of topological space $X$ is called stable if every $f\in C(X)$ is bounded on some $A\in\mathcal{F}$.

Definition 2. $X$ is realcompact if every stable family of closed subsets with finite intersection property has non-empty intersection.

This agrees with standard definition of realcompactess (theorem 5.1):

Theorem 1. If $X$ is $T_{3.5}$, then $X$ is realcompact iff there is cardinal $\kappa$ such that $X$ embedds as closed subspace of $\mathbb{R}^\kappa$.

Moreover, it's exactly what we need (theorem 5.2):

Theorem 2. $X$ is compact iff $X$ is pseudocompact and realcompact.

Jakobian
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