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This is a follow-up to my questions here and here. A subset of a topological space is called relatively compact if its closure is compact. Let's call a sequence $(U_n)$ of open sets a bounding sequence if the closure of $U_n$ is a subset of $U_{n+1}$ for all $n$ and every relatively compact set is a subset of some $U_n$. And let's call a set $S$ saturated if for every bounding sequence $(U_n)$, $S$ is a subset of some $U_n$. My question is, for what topological spaces is every saturated set relatively compact?

Is there some category of topological spaces which satisfies this property? And is there an example of a completely regular space which doesn’t satisfy this property?

My reason for asking this question, by the way, is that relatively compact sets form a bornology for $T_1$ spaces, and this property is the conditions for a bornology to be induced by a compatible uniformity, as I discuss here.

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  • $\begingroup$ Have you considered hemicompact spaces? Bornological sets are simply subset ordered ideals, the order dual of filters. $\endgroup$ – William Elliot Dec 21 '18 at 7:15
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    $\begingroup$ Wait, why can't you just take each $U_n = X$ to be the whole topological space? As long as $X$ is not compact, not every saturated set will be relatively compact. $\endgroup$ – mathworker21 Dec 21 '18 at 7:26
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    $\begingroup$ @mathworker21 There's nothing stopping you from considering the bounding sequence $(X,X,X,...)$. But remember, a saturated set must be contained in some element of EVERY bounding sequence, not just one particular bounding sequence. $\endgroup$ – Keshav Srinivasan Dec 21 '18 at 14:16
  • $\begingroup$ @KeshavSrinivasan oh wow, I can't believe I didn't read that. thanks $\endgroup$ – mathworker21 Dec 21 '18 at 14:27
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Clearly, the property is equivalent to that every closed saturated set is compact, so all compact spaces have the property. On the other hand, in a trivial case when the space with the property has no bounding sequences then it is saturated, so compact.

For a space $X$ put $X_c=\{x\in X:\overline{\{x\}}$ is compact$ \}$ and $$X^+_c=\bigcap\{U\subset X: X_c\subset\operatorname{int} U\}=\{x\in X: \overline{\{x\}}\cap X_c\ne\varnothing\}.$$ It is easy to check that each relatively compact subset of $X$ is contained in $X_c$. If $X$ is $T_1$ then $X_c=X$. Recently Paolo Lipparini suggested us to consider spaces $X$ with the property (P) that $X=X_c$ (for other reasons). He observed that:

This property (P) is much weaker than $T_1$,

In general, the closure of a point might be non compact. E.g., $\Bbb N$ with the topology of left intervals, whose open sets are the intervals $[0, n)$, plus the whole of $\Bbb N$. Here the closure of $0$ is the noncompact $\Bbb N$. The example also shows that feeble compactness does not imply (P).

Clearly, each bounding sequence of a space $X$ is a cover of $X^+_c$. Conversely, if $(U_n)$ is a cover of a set $X_c$ by open sets such that $\overline{U_n}\subset U_{n+1}$ for all $n$ then $(U_n)$ is a bounding sequence. Indeed, let $B$ be a relatively compact subset of $X$. Then $\overline{B}\subset X_c$ is compact. Thus $(U_n)$ is a non-decreasing open cover of a compact set $\overline{B}$, so there exists $n$ such that $B\subset \overline{B}\subset U_n$. Also remark that in this case a family $X^+_c\setminus \overline{U_n}$ is locally finite.

As usual, a space $X$ is called feebly compact if every locally finite family of open sets in X is finite. Feebly compact spaces are not assumed to satisfy any separation axiom, while pseudocompact spaces are necessarily Tychonoff. It is clear that feeble compactness and pseudocompactness coincide in Tychonoff spaces, so the former notion is a ’right’ extension of the latter one to non-Tychonoff spaces.

Let us recall that a subset $B$ of a space $X$ is said to be bounded in $X$ if every locally finite family of open sets in $X$ contains only finitely many elements that meet $B$. Hence boundedness is a relative version of feeble compactness. It is clear from the definition that a subset $B$ of a Tychonoff space $X$ is bounded in $X$ if and only if every continuous real-valued function defined on $X$ is bounded on $B$. [TS]

The above implies that each bounded subset of a space with property (P) is saturated. On the other hand, each saturated subset $B$ of $X$ is functionally bounded, that is if every continuous real-valued function defined on $X$ is bounded on $B$. Indeed, if $f:X\to\Bbb R$ is a continuous function unbounded on $B$, then $(f^{-1}(-n,n))$ is a bounding sequence witnessing that $B$ is not saturated. That is, a subset of a completely regular space is saturated iff it is functionally bounded. So a completely regular space has the property iff each its closed functionally bounded subset is compact. In particular, each pseudocompact space satisfying the property is compact. Recall that a subset $B$ of a normal space $X$ is functionally bounded iff $B$ is pseudocompact. Indeed, if $B$ is not pseudocompact then there exists a continuous real-valued unbounded function $f$ on $B$. Since $B$ is a closed subset of a normal space, by Tietze-Urysohn’s Theorem $f$ can be extended to a continuous real-valued (unbounded) function on the whole space, which violates functional boundedness of $B$.

Thus space $X$ has the property under the known condition assuring a pseudocompact space is compact.

  • $X$ is paracompact. Indeed, by [Eng, Theorem 5.1.5] $X$ is normal. Let $B$ be any closed saturated subset of $X$. The set $B$ is paracompact as a closed subset of a paracompact space. Since $B$ is feebly compact, it is compact.

  • $X$ is a normal space with $G_\delta$-diagonal. Indeed, let $B$ be any closed saturated subset of $X$. Since $B$ is feebly compact and normal, by [Eng, Theorem 4.3.28] $B$ is countably compact. By Chaber’s Theorem [Gru, Theorem 2.14], $B$ is compact. See, for instance this my answer for the formulations and proofs of the mentioned theorems.

References

[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

[Gru] Gary Gruenhage Generalized Metric Spaces, in: K.Kunen, J.E.Vaughan (eds.) Handbook of Set-theoretic Topology, Elsevier Science Publishers B.V., 1984.

[TS] Iván Sánchez, Mikhail Tkachenko, Products of bounded subsets of paratopological groups, Topology Appl., 190 (2015), 42-58.

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  • $\begingroup$ Thanks for your answer. By the way, do know of an example of a completely regular space that does not have the property? $\endgroup$ – Keshav Srinivasan Dec 24 '18 at 15:14
  • $\begingroup$ @KeshavSrinivasan Yes, of course. Recall that each pseudocompact space with the property is compact, so any pseudocompact non-compact space is an example. For instance, an ordinal $\omega_1$ endowed with the order topology or any proper space $X$ of $[0,1]^\kappa$, where $\kappa$ is an uncountable cardinal and $$X\supset \Sigma_\kappa I=\{(x_\alpha)_{\alpha<\kappa}\in [0,1]^\kappa: |\{\alpha: x_\alpha>0\}|\le\omega\}.$$ $\endgroup$ – Alex Ravsky Dec 24 '18 at 16:41
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    $\begingroup$ OK thanks for your help! $\endgroup$ – Keshav Srinivasan Dec 24 '18 at 17:02

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