Exhaustion of open sets by closed sets

Question

Let $X$ be a topological space with topology $\tau$. Let $U\in \tau$. Say that $U$ can be countably exhausted by closed sets if there exists a family of sets $F_n \subset U$ indexed by $n\in\mathbb{N}$ such that

• $F_n \subset V \subset F_{n+1}$ for some $V\in \tau$;
• $X\setminus F_n \in \tau$; and
• $\bigcup\limits_{n\in\mathbb{N}}F_n = U$.

Is there a charaterisation of topologies for which every open set can be countably exhausted by closed sets?

---

Trivially if every open set is also closed (the discrete topology, say) then every open set can be exhausted. Just take $U = V = F_n$.

For an example in which exhaustion is impossible, consider the real line with the co-countable topology. If $U\subsetneq \mathbb{R}$ is open, and if $F_n$ is a non-empty closed set, we have that necessarily $F_{n+1}$ contains a non-empty open set, is closed, and hence must be $\mathbb{R}$. Hence exhaustion is impossible in this topology.

The first condition seems to suggest that the topology being normal may be part of a sufficient condition. (Indeed, normal + separable + Hausdorff seems to be enough; but how much weaker can we go?)

Answer 1

A $T_1$ space has the property that every open set can be countably exhausted by closed sets iff it is perfectly normal.

Assume that $X$ is $T_1$ and that every $U\in\tau$ can be countably exhausted by closed sets. Let $x\in U\in\tau$. There are sequences $\langle F_n:n\in\Bbb N\rangle$ of closed sets and $\langle V_n:n\in\Bbb N\rangle$ of open sets such that

$$F_n\subseteq V_n\subseteq\operatorname{cl}V_n\subseteq F_{n+1}\tag{1}$$

for each $n\in\Bbb N$ and $U=\bigcup_{n\in\Bbb N}F_n$. Lemma 1.5.14 of Engelking’s General Topology ensures that $X$ is $T_4$ (normal and $T_1$); the proof is similar to the proof that regular Lindelöf spaces are normal.

It’s also clear from $(1)$ that every open set in $X$ is a countable union of regular closed sets. In particular, every open set is an $F_\sigma$, so $X$ is even perfectly normal. It’s well-known that in a $T_1$ space perfect normality is equivalent to each of the following properties:

• Open sets are cozero-sets.
• Closed sets are zero-sets.
• If $H$ and $K$ are disjoint closed sets, there is a continuous $f:X\to[0,1]$ such that $H=f^{-1}[\{0\}]$ and $K=f^{-1}[\{1\}]$.

Conversely, suppose that $X$ is a perfectly normal $T_1$ space, and let $U$ be a non-empty open subset of $X$. There is a continuous $f:X\to[0,1]$ such that $U=f^{-1}\big[(0,1]\big]$. For $n\in\Bbb N$ let

$$F_n=f^{-1}\left[\left[\frac1{2^n},1\right]\right]\text{ and }V_n=f^{-1}\left[\left(\frac1{2^{n+1}},1\right]\right]\;;$$

clearly the sequences $\langle F_n:n\in\Bbb N\rangle$ and $\langle V_n:n\in\Bbb N\rangle$ satisfy $(1)$, and $U=\bigcup_{n\in\Bbb N}F_n$.