Why topological spaces in Baire category theorem are required to be Hausdorff?

Question

In Baire category theorem it says a locally compact Hausdorff space $S$ is second category.

In the proof, it choose $V_1,V_2\cdots$ are dense open subset of $S$. $B_0$ is an arbitrary nonempty open set in $S$. Then we choose $B_n\not = \emptyset$,$\bar B_n \subset V_n \cap B_{n-1}$. We can choose $\bar B_n$ is compact. Put $K=\cap_{n=1}^\infty \bar B_n$. Because in compact space, every family of closed subsets having the finite intersection property has non-empty intersection. So $K$ is nonempty. And $K\subset B_0,K\subset V_n$. Hence $B_0$ intersects $\cap V_n$

So where the Hausdorffness is used. It seems that it should only be regular space. So where it use that it should be regular?

Answer 1

You say: "We can find open $B_n$ with $\overline{B_n}\subseteq V_n\cap B_{n-1}$ such that $\overline{B_n}$ is compact". But why? Strong local compactness tells me that the open neighbourhood $V_n\cap B_{n-1}$ contains a compact neighbourhood $K$. That means, there is an open $B_n\subseteq K$. I would now love to conclude that $\overline{B_n}$ is also compact - but I can't! Not without further assumptions. If I assume the space is Hausdorff, then $K$ is closed, so $\overline{B_n}\subseteq\overline{K}=K$ and I get to conclude $\overline{B_n}$ is compact. It's worse than that, though: I need to also conclude $\overline{B_n}\subseteq V_n\cap B_{n-1}$. This is again easy if $K$ is closed, but it doesn't follow if $K$ is not closed (which is possible if the space is not Hausdorff).

If I'm not allowed to choose the $B_\bullet$ to be precompact, then the proof breaks down at the end. Specifically, I need to be able to conclude at least one $B_n$ has $\overline{B_n}$ compact.

---

Consider the example of John Doe. $X=\Bbb N$ with the cofinite topology and we define $V_n:=\Bbb N\setminus\{n\}$ for every $n\in\Bbb N$. The closed sets are precisely the finite sets; an infinite set's closure is always $\Bbb N$ itself. So if I want $\overline{B_n}\subseteq V_n\cap B_{n-1}$ I must at least have every $B_n$ finite (then compactness would also follow nicely). But for $B_n$ to be open, it must be infinite. So I can actually never have a $B_n$ whose closure is (compact and) contained in $V_n$.

Answer 2

Consider $\Bbb{N}$ with the topology defined by a basis consisting of all co-finite subsets of $\Bbb{N}$. This topology is $T_1$ but not Hausdorff. Every point is closed. Every non-empty open set is co-finite. So the space is a countable union of closed sets with empty interior ($\{n\}$ for each $n\in\Bbb{N})$.

The space is compact. For any open cover - one member already covers all but finitely many points - and you can make a finite choice of sets to cover the remaining points.

Edit: To answer the questions in the last paragraph more thoroughly

1. It is enough to assume that $S$ is locally-compact regular.
2. Every locally-compact Hausdorff space is regular. See here.
3. Every $T_0$ regular space is Hausdorff. See my answer here.
4. Where is regularity used? Consider the induction step. $B_{n-1}$ is a non-empty open set by the induction hypothesis. $V_n$ is a dense open set by assumption. Hence $\Gamma_n=V_n \cap B_{n-1}$ is non-empty and open. Let $y\in \Gamma_n$. By local compactness, there exists some compact set $D$ such that $y\in\operatorname{int}D$. The set $\Delta_n=\Gamma_n\cap\operatorname{int}D$ is an open neighborhood of $y$. By regularity, there is a closed set $C$ such that $y\in\operatorname{int} C \subseteq C\subseteq \Delta_n$. Also $\Delta_n\subseteq D$ by definition. Hence $C$ is a relatively-closed subset of the compact space $D$ - so $C$ is a closed compact neighborhood of $y$ and $C\subseteq\Delta_n\subseteq \Gamma_n$. Now take $B_n=\operatorname{int}C$.

Answer 3

The Baire category theorem does hold for plenty of non-Hausdorff spaces. For example, it holds for all Quasi-Polish spaces (https://doi.org/10.1016/j.apal.2012.11.001).