Baire space and increasing union of closed subspaces

Question

Let $X$ be a Baire space. Suppose there is an increasing sequence $C_1\subset C_2\subset \cdots $ of closed subspaces of $X$, whose set-theoretical union is $X$. Since $X$ is Baire, we know that some $C_n$ has nonempty interior. I'm curious whether the interiors of $C_n$ covers $X$.

I believe that the claim is false, because of the following reason: If the interiors of $C_n$ covers $X$, then $X$ carries the final topology with respect to the family $C_n$. So if we can take $X$ and $C_n$ so that the $X$ carries a strictly coarser topology than the final topology, then we win. (Actually, I arrived at this problem precisely because I had a Baire space $X$ and its closed subsets whose union is $X$, and I wanted to show that $X$ carries the topology of union of those subsets.) There should be plenty of such examples. However, I'm having trouble coming up with one.

Any help is appreciated. Thanks in advance.

Answer 1

Let us consider $X$ as the complex plane:

$C_n=\{z\in\mathbb{C}\colon z=r e^{i\theta}, r\geq 0, 0\leq \theta \leq 2\pi-1/n\}$

This way $\bigcup C_n=X$, on the other hand the positive real axis is not covered by the interiors of $C_n s$.

Answer 2

The simplest example I think: let $X$ be any Baire Hausdorff space with a non-isolated point $p$ with a countable local base $(U_n)_n$, that we can assume to be decreasing WLOG.

Define $C_n= \{p\} \cup U_n^\complement$ which is an increasing union of closed sets with union $X$, but whose interiors do not cover $p$.