About Baire spaces I'm having difficult to solve this:
Determine whether or not $\mathbb{R}_l$ is a Baire space.
I tried to aplly the following lemma: "X is a Baire space iff given any countable collection $\mathbb{U}_n$ of open sets in X, each of which is dense in X, their intersection $\bigcap{U}_n$ is also dense in X."
Where $\mathbb{R}_l$ is the topology of lower limit, generated by the intervals [a,b).
Could you help to solve this?
Thank you.
 A: Suppose topologies $\sigma$ and $\tau$ on a set $X$ are such that every nonempty set from $\sigma$ contains a nonempty set from $\tau$, and the other way around. This is the case for the standard and lower-limit topology on the real line. You should be able to show that, for $Y \subset X$,


*

*the $\sigma$-interior of $Y$ is empty if and only if the $\tau$-interior of $Y$ is empty.

*$Y$ is $\sigma$-nowhere dense if and only if $Y$ is $\tau$-nowhere dense.


It follows that $\sigma$ is Baire if and only if $\tau$ is Baire.
A: Here is a more direct proof.
Let $U_n$ be open and dense in $\mathbb{R}_l$. Consider $\text{Int}(U_n)$ where the interior is taken with respect to $\mathbb{R}$. Then $\text{Int}(U_n)$ is open in $\mathbb{R}$. To show that $\text{Int}(U_n)$ is dense in $\mathbb{R}$, let $V$ be any open set of $\mathbb{R}$. Then $V$ is also open in $\mathbb{R}_l$ so $V\cap U_n$ is non-empty and open in $\mathbb{R}_l$.This means $V\cap U_n$ contains a basis element of the form $[a,b)$ (where $a<b$), so it contains an interval $(a,b)$. In the topology of $\mathbb{R}$, $(a,b)$ is open so $(a,b)\subseteq \text{Int}(U_n)\cap V$. This proves that $\text{Int}(U_n)$ is dense in $\mathbb{R}$.
Since $\mathbb{R}$ is a Baire space, $\bigcap \text{Int}(U_n)$ is dense in $\mathbb{R}$. Therefore, $\bigcap U_n$ is dense in $\mathbb{R}$. To show that $\bigcap U_n$ is dense in $\mathbb{R}_l$, it suffices to show that any given basis element $[a,b)$ intersects $\bigcap U_n$. But $[a,b)$ contains $(a,b)$ which intersects $\bigcap U_n$ since the latter is dense in $\mathbb{R}$.
