# An extension of Baire's category theorem

In a topological space, a set is said to be rare if its closure has empty interior, and a set is said to be meager if it is a countable union of rare sets. If meager sets all have empty interior, then the topological space is said to be a Baire space. The following fact is known as Baire's category theorem.

Theorem. Complete metric spaces and locally compact Hausdorff spaces are Baire spaces.

[Zălinescu 2002] provides the following extension of this theorem for complete metric spaces, attributing it to C. Ursescu.

Theorem. ([Zălinescu 2002, Theorem 1.4.5]) Let $$X$$ be a complete metric space, and $$\{S_n\}$$ be a sequence of open sets in $$X$$. Then $$\mathrm{cl}(\cap_{n=1}^\infty S_n)$$ and $$\cap_{n=1}^\infty \mathrm{cl}(S_n)$$ have the same interior.

It is easy to check that a topological space is a Baire space if and only if any countable intersection of dense open sets is still dense in this space. Consequently, the theorem above implies that complete metric spaces are Baire spaces. Thus it is considered as an extension.

Question. Does the extension hold for locally compact Hausdorff spaces?

Update: As pointed out by @Alex Kruckman, the property in Ursescu's theorem is indeed equivalent to the fact that $$X$$ is a Baire space.

Theorem. A topological space is a Baire space if and only if $$\mathrm{cl}(\cap_{n=1}^\infty S_n)$$ and $$\cap_{n=1}^\infty \mathrm{cl}(S_n)$$ have the same interior for any sequence of open sets $$\{S_n\}$$.

Recall that a space is a Baire space if and only if any countable intersection of dense open sets is still dense. The theorem is essentially another way to state that a space is a Baire space if and only if all its open subspaces are Baire spaces.

Combined with Baire's category theorem, this observation by Alex proves the desired theorem. My answer below proves it from scratch, the proof being essentially the same as that of Baire's category theorem for locally compact Hausdorff spaces.

I will accept Alex's answer a few days later if nobody else has anything to add.

Thanks.

## 2 Answers

This is not really a strengthening of the Baire category theorem - it is equivalent to it. That is, any Baire space (and in particular any compact Hausdorff space) satisfies Ursescu's theorem.

Let's assume $$X$$ is a Baire space. Suppose $$(S_n)_{n\in \mathbb{N}}$$ is a sequence of open sets in $$X$$. We always have $$\text{int}(\text{cl}(\bigcap_{n=1}^\infty S_n))\subseteq \text{int}(\bigcap_{n=1}^\infty \text{cl}(S_n))$$, so it suffices to prove the reverse inclusion. And to do this, it suffices to prove that for any open $$U\subseteq \bigcap_{n=1}^\infty \text{cl}(S_n)$$, we have $$U\subseteq \text{cl}(\bigcap_{n=1}^\infty S_n)$$.

So fix some $$U$$. Note that an open subspace of a Baire space is Baire. [Why? If $$(V_n)_{n\in\mathbb{N}}$$ is a sequence of dense open subsets of $$U$$, then $$(V_n\cup (X\setminus \text{cl}(U)))_{n\in\mathbb{N}}$$ is a sequence of dense open subsets of $$X$$. The intersection of these sets is $$(\bigcap_{n=1}^\infty V_n)\cup (X\setminus \text{cl}(U))$$, and the fact that this is dense in $$X$$ implies that $$(\bigcap_{n=1}^\infty V_n)$$ is dense in $$U$$.]

So defining $$V_n = U\cap S_n$$, we have that each $$V_n$$ is open in $$U$$, and since $$U\subseteq \text{cl}(S_n)$$ for all $$n$$, each $$V_n$$ is dense in $$U$$. Since $$U$$ is Baire, $$\bigcap_{n=1}^\infty V_n$$ is dense in $$U$$, which implies $$U\subseteq \text{cl}(\bigcap_{n=0}^\infty S_n)$$, as was to be shown.

• Thanks, @Alex Kruckman. I modified the question to highlight your observation. In addition, I modified your answer, changing "Zălinescu's theorem" to "Ursescu's theorem", as suggested by [Zălinescu 2002]. Hope this fine with you. Thanks again. – Nuno Nov 15 '20 at 3:12

Here is my attempt to prove it. It is inspired by the proof of [Zălinescu 2002, Theorem 1.4.5] and the classical proof of Baire's theorem for locally compact Hausdorff spaces. Any comments or criticism will be appreciated. Thanks.

Theorem. Let $$X$$ be a locally compact Hausdorff space, and $$\{S_n\}$$ be a sequence of open sets in X. Then $$\mathrm{cl}(\cap_{n=1}^\infty S_n)$$ and $$\cap_{n=1}^\infty \mathrm{cl}(S_n)$$ have the same interior.

proof. It suffices to show that $$\mathrm{int}(\cap_{n=1}^\infty \mathrm{cl}(S_n))\subset \mathrm{cl}(\cap_{n=1}^\infty S_n)$$. To this end, we only need to prove for any given nonempty open set $$U \subset \cap_{n=1}^\infty \mathrm{cl}(S_n)$$ that $$\begin{equation} \label{eq:usnonempty} U\cap(\cap_{n=1}^\infty S_n) \neq \emptyset. \end{equation}$$ In the sequel, we will define a sequence of compact sets $$\{C_n\}_{n=0}^\infty$$ such that $$C_n$$ has nonempty interior and $$\begin{equation} C_{n+1} \subset C_{n} \subset U\cap(\cap_{k=1}^n S_k) \quad \text{ for each } \quad n \ge 0, \end{equation}$$ where $$\cap_{k=1}^0 S_k=X$$. Once this is done, Cantor's Theorem will yield $$\begin{equation} \label{eq:nestnonempty} \emptyset \;\neq \;\cap_{n=1}^\infty C_n \;\subset\; U\cap (\cap_{n=1}^\infty S_n), \end{equation}$$ which gives us what we want.

We define $$\{C_n\}$$ inductively. As $$U$$ is a nonempty open set and $$X$$ is locally compact, we can take a compact set $$C_0\subset U$$ such that $$\mathrm{int}(C_0)\neq \emptyset$$. Assume that $$C_n$$ is already defined for an $$n\ge 0$$ so that $$C_n\subset U\cap(\cap_{k=1}^n S_k)$$ and $$\mathrm{int}(C_n)\neq \emptyset$$. Recalling that $$U\subset\mathrm{cl}(S_{n+1})$$, we have $$\mathrm{int}(C_n)\subset \mathrm{cl}(S_{n+1})$$, which implies that $$\mathrm{int}(C_n) \cap S_{n+1} \neq\emptyset$$. Since $$\mathrm{int}(C_n)\cap S_{n+1}$$ is open, invoking again the local compactness of $$X$$, we can take a compact set $$C_{n+1}\subset \mathrm{int}(C_n) \cap S_{n+1}$$ such that $$\mathrm{int}(C_{n+1})\neq \emptyset$$. Clearly, $$C_{n+1}\subset C_n$$ and $$C_{n+1}\subset C_{n}\cap S_{n+1} \subset U\cap(\cap_{k=1}^{n+1} S_k)$$. This finishes the induction and completes the proof.