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?

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    $\begingroup$ 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. $\endgroup$ – Tim Campion Feb 2 at 21:56
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    $\begingroup$ @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. $\endgroup$ – Henno Brandsma Feb 3 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 $\beta X=\beta Y$).

As noted in the question extremally disconnected compact spaces have the same property; as these are quite different from product spaces finding a characterization that is not a direct translation seems difficult.

  • $\begingroup$ 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? $\endgroup$ – Norbert Feb 5 at 18:56
  • $\begingroup$ No, I added further explanation above. $\endgroup$ – KP Hart Feb 5 at 19:23
  • $\begingroup$ Now it is clear. Thank you! $\endgroup$ – Norbert Feb 5 at 21:29

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