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?)

  • $\begingroup$ Haus + exhaustion $\implies$ normal it seems, but I'm not sure whether points being closed + exhaustion $\implies$ Haus. $\endgroup$ – Eugene Shvarts Jul 9 '12 at 16:24
  • $\begingroup$ Should your definition include a requirement that the $F_n$ be closed? I know $F$ is usually reserved for closed sets (from the french, but it might be worth mentioning explicitly anyway). $\endgroup$ – Arturo Magidin Jul 9 '12 at 16:51
  • 3
    $\begingroup$ @Arturo: The second bullet point implies $F_n$ is closed. $\endgroup$ – Jason DeVito Jul 9 '12 at 17:15

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 $U=\bigcup_{n\in\Bbb N}F_n$.

  • $\begingroup$ Ah, that's great. I was wondering if there is a condition weaker than your third bullet. At least this means that the proof I am working on is sharp from the point of view of this line of argument. Thanks. $\endgroup$ – Willie Wong Jul 10 '12 at 6:28

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