general-topology
What is name for those topological spaces that satisfy the following property?
Any space with all singleton sets closed satisfies this property, but the converse need not hold.
Are there any interesting characterization of these spaces?
This is equivalent to local compactness.
If a topological space satisfies your property, then a singleton point set $\{x\}$ is compact, and hence there must be a relatively compact set $T$ whose interior contains $x$ (in fact, it contains $\overline{\{x\}}$, which may be strictly larger). The closure of $T$ is a compact neighbourhood of $x$.
If, on the other hand, if we have locally compact space and a relatively compact set $S$, then $\overline{S}$ is compact. For each point $s \in \overline{S}$, there is an open neighbourhood $\mathcal{N}_s$ of $s$ that is compact. Note that the interiors of these neighbourhoods cover $\overline{S}$, hence there must be a finite-subcover of $\overline{S}$. Unioning this finite subcover yields an open, relatively compact neighbourhood containing $\overline{S}$, establishing equivalence.
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.
Best Answer
Suppose that $X$ has this property, and let $x\in X$. If $y\in\operatorname{cl}\{x\}$, then of course $\operatorname{cl}\{y\}\subseteq\operatorname{cl}\{x\}$, so we must actually have $\operatorname{cl}\{y\}=\operatorname{cl}\{x\}$. Thus, the specialization order on $X$ is symmetric, and it’s clear that the converse also holds: $X$ has the property in question if and only if its specialization order is symmetric.
The specialization order is actually a pre-order: it is reflexive and transitive but not necessarily antisymmetric. Points that are equivalent with respect to this order are topologically indistinguishable: they have the same open nbhds. The order is symmetric precisely when topologically distinguishable points are separated, meaning that each of them has an open nbhd that does not contain the other. Such a space is said to be $R_0$ (or symmetric), and its Kolmogorov quotient is $T_1$.
In other words, it’s basically a $T_1$ space in which some of the points may have been ‘fattened up’ into sets of topologically indistinguishable points.