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I am looking at a subsets of a particular topological space $\mathbf{X}$, and I care about subsets $A$ of $\mathbf{X}$ which are a subset of a compact subset $B$ of $\mathbf{X}$. Since I am not restricting myself to Hausdorff $\mathbf{X}$ here, this notion is not the same as being relatively compact.

My question is whether the notion I am looking at has a name or some "further reading" attached to it.

Arno
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    Why are you interested in such a concept? It is often the case that people call compact spaces quasi-compact instead and include being Hausdorff in the definition of compactness: the reason for this is that compactness isn't as strong and useful of a property in non Hausdorff spaces. So I kinda doubt that the one you are looking for is a notion much studied, but if you give some context there might be something more to say. –  Jan 02 '23 at 00:53
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    @milore Basically, I want to prove that a set satisfying certain conditions can be extended to a compact set while avoiding adding any particular other point to it. I'm hoping that there is something in the literature that could help, but just searching for individual key words gets me other stuff. The Hausdorff case is trivial (and many people really underestimate how much nice stuff there is outside of the Hausdorff world). – Arno Jan 02 '23 at 09:01
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    Maybe it would be reasonable to call such subsets 'bounded' – FShrike Jan 27 '24 at 12:23
  • @FShrike I've considered this before, but note that there are non-compact metrizable spaces for which every subset is bounded in the metric. – Steven Clontz Jan 27 '24 at 15:22
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    @StevenClontz Of course, every metrisable space is boundedly metrisable. And also bounded is not equivalent to precompact even for complete metric spaces so I agree it’s not the best solution. – FShrike Jan 27 '24 at 15:28
  • I do not think there is a standard name for these. You can call such subsets "subcompact." – Moishe Kohan Jan 27 '24 at 17:06
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    Subcompact is defined differently here: https://arxiv.org/pdf/2102.11243.pdf https://eudml.org/doc/298522 – Steven Clontz Jan 27 '24 at 17:08
  • Other candidates that are already "taken" besides the two in my answer: almost compact, nearly compact, pseudocompact... – Steven Clontz Jan 27 '24 at 22:24

1 Answers1

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I had to do some digging a few years back for similar reasons for this paper: https://cmuc.karlin.mff.cuni.cz/pdf/cmuc1702/clontz.pdf (see section 4)

  • A subset $Y$ of $X$ is relatively compact to $X$ if for every open cover of $X$, there exists a finite subcollection which covers $Y$.

  • A subset $Y$ of $X$ is precompact in $X$ if $\operatorname{cl}_X(Y)$ is compact.

Note that every precompact set is relatively compact, and we get the converse for regular spaces (Prop 4.4).

Without regularity these are distinct. Let $X$ be the real line with the topology generated by sets of the form $(a,b)$ and $(a,b)\setminus\mathbb Q$ (example 5.8). This is a refinement of a Hausdorff space and is thus Hausdorff. Then $[-1,1]\setminus\mathbb Q$ is relatively compact to $X$, but not precompact: its closure is $[-1,1]$, which is not compact as it contains the closed infinite discrete set $[-1,1]\cap\mathbb Q$.

However, you're asking about this:

  • A subset $Y$ of $X$ is $(\star)$ in $X$ if it is contained in some compact $K\subseteq X$.

We see then that every precompact set is $(\star)$, and every $(\star)$ set is relatively compact. It follows that all three are equivalent under regularity.

Note our example $[-1,1]\setminus\mathbb Q$ is not $(\star)$ in $X$ as compacts in $X$ are nowhere dense in $\mathbb R$ and thus cannot contain the set $[-1,1]\setminus\mathbb Q$.

So then, might $(\star)$ and precompact be equivalent? This is true for Hausdorff (more generally, KC (P100)) spaces: given a ($\star$) in $X$ set $Y$, its compact $K\supseteq Y$ is closed, so $\operatorname{cl}_X(Y)\subseteq K$ is compact. So we'd require a counterexample to be both non-Hausdorff and non-regular.

Note that every compact subset is $(\star)$; therefore every finite subset is $(\star)$. Consider the particular point topology on $\mathbb Z$ (S8): the non-empty open sets are exactly the sets that contain $0$. Then $\{0\}$ is $(\star)$ because it is finite. But $\operatorname{cl}(\{0\})=\mathbb Z$, which is not compact as $\mathbb Z\setminus\{0\}$ is a closed infinite discrete subset.

We can improve this to a $T_1$ (but not $T_2$, or even US (P99)) space. Consider the union $X=A\cup B$ of disjoint infinite sets $A,B$ where non-empty sets are open provided they contain a cofinite subset of $A$. Then A is compact and thus $(\star)$ in $X$, but its closure is the whole space $X$ which contains the infinite closed discrete subset $B$. (It'd be nice to obtain a US, if not weak Hausdorff (P143), example.)

Which I guess is a very long way of saying, I don't think I've exactly seen $(\star)$ or an equivalent notion named in the literature, but here's some related ideas at least. :-) If you want to give it a name, perhaps "antecompact" would be a cute choice?

Steven Clontz
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    See also the discussion in https://math.stackexchange.com/questions/4702452/definitions-of-relatively-compact. Def 2 in that question is what the OP is asking about. – PatrickR Jan 29 '24 at 01:10
  • That explains the sense of deja vu I had while writing this answer! – Steven Clontz Jan 31 '24 at 18:23