Baire Category Theorem in weak topology

Question

We say that a topological space is Baire if intersection of open dense sets is dense.

Suppose that we know that a certain space is Baire in a given topology $\tau$. Can we determine whether this space will still be Baire in a weaker topology $\sigma \subset \tau$?

What I attempted: Sets which are open in weak topology are also open in the strong topology, however sets that are dense in the weak topology may not be dense in the strong topology. This makes it difficult to make any conclusions about the Baire property.

Answer 1

This is false in general, in fact for any topology $\tau$ on a set $X$ there is a finer topology $\tau'$ which makes $X$ into a Baire space, just take the discrete one, discrete spaces are Baire since they're completely metrizable.

Answer 2

In general, the weaker topology need not be Baire.

Let $H$ be an infinite-dimensional Hilbert space. In the norm topology $\tau$, $H$ is a complete metric space and hence Baire. Let $\sigma$ be the usual weak topology on $H$ (i.e. the weakest topology in which each map $x \mapsto \langle x,y \rangle$ is continuous). It's a standard fact that any open set in $\sigma$ is unbounded with respect to the norm. So every closed norm-ball in $H$ is closed and nowhere dense in $\sigma$. Since $H$ is a countable union of closed balls, $\sigma$ is not Baire.