# Characterization of pretty compact spaces

This is a cross post from MSE.

I believe that the following problem have already been considered by some sophisticated topologist.

Definition 1. A non-compact Hausdorff topological space $$X$$ is called almost compact if its Stone–Čech compactification coincides with its one point compactification.

An example of almost compact space is $$[0,\omega_1)$$ for the first uncountable ordinal $$\omega_1$$. All almost compact spaces are locally compact and pseudocompact.

Definition 2. A compact Hausdorff space $$X$$ is called pretty compact if $$X\setminus\{p\}$$ is almost compact for all non-isolated points $$p\in X$$.

I know that Stonean spaces are pretty compact. By a result of van Douwen, Kunen and van Mill (There can be $$C^*$$-embedded dense proper subspaces in $$\beta\omega - \omega$$) $$\beta\mathbb{N}\setminus \mathbb{N}$$ is consistently pretty compact. What are other examples of pretty compact spaces? Does there exist any characterization of pretty compact spaces or at least a strong necessary condition?

• From the discussion at MSE, I think I'm not alone in finding the terminology very distracting. For example (let's see if I've got this straight) $\mathbb R$ is not almost compact, so $S^1$ is not pretty compact, so compact does not imply pretty compact. But I think the terminology strongly suggests it should. Already, the fact that $\mathbb R$ is not almost compact makes the terminology feel wrong to me. Feb 2, 2021 at 21:56
• @TimCampion It's nevertheless pretty standard and already used by Gillman and Jerrison (almost compact I mean). Pretty compact is indeed a bit weird. That $\Bbb R$ feels almost compact is due to its completeness/connectedness and linear structure, I think: in a connected ordered space we can always compactfity with at most two points. Feb 3, 2021 at 12:07

A partial answer: other examples of pretty compact spaces are uncountable powers of $$\{0,1\}$$ and $$[0,1]$$, and in general products of uncountably many non-trivial compact Hausdorff spaces. See Problem 3.12.24(c) in Engelking's General Topology, or Glicksberg, Stone-Čech compactifications of products. If $$a$$ is in the product take $$b$$ in the product that differs everywhere from $$a$$. Then $$\Sigma(b)$$ is a subset of $$X\setminus\{a\}$$. As the product is $$\beta\Sigma(b)$$ it is also $$\beta(X\setminus\{a\})$$ (general result: if $$X\subseteq Y\subseteq\beta X$$ then $$\beta X=\beta Y$$).
• Problem 3.12.24(c) in Engelking's General Topology states that the Cartesian product $X$ of compact Hausdorff spaces $\{X_s:s\in S\}$ is the Stone-Cech compactification of their $\Sigma$ product. By 2.7.14 from the same book the $\Sigma$ product of the family $\{X_s:s\in S\}$ is a set of the form $$\{x\in X: \operatorname{Card}(\{s\in S: x_s\neq a_s\})\leq \aleph_0 \}$$ for some $a\in X$. Clearly $\Sigma$ products are quite different from $X\setminus \{a\}$. So I don't think uncountable products of compact Hausdorff spaces are pretty compact. Am I right? Feb 5, 2021 at 18:56