Let $X$ be an infinite dimensional Banach space. Why is $X^*$ of the first category in itself when given the weak* topology.
Very closely related to $X^*$ with its weak*-topology is of the first category in itself, but I can't follow all the steps.
The first step (hopefully) is to show that $B_n = \{x^* : \lVert x^* \rVert \leq n\}$ is nowhere dense, i.e. its closure is has an empty interior. Then $X = \bigcup_n B_n$, so $X$ is meagre.
The answer claims that "It suffices to prove that $\text{int}_{w∗}B_n=\emptyset$", but don't we have to show that $\text{int}_{w∗}\overline{B_n}=\emptyset$? So I'm getting stuck trying to say anything about the closure of $B_n$.
First note that $B_n \equiv \cap_{||x\|\leq 1}\{x^{*}:|x^{*}(x)|\leq 1\}$ is closed in $w^{*}$ topology. Suppose if possible, $x^{*}$ is an interior point of $B_n$. Then there exist $n,x_1,x_2,\cdots,x_n$ and $r_1,r_2,\cdots,r_n$ such that $|y^{*}(x_i)-x^{*}(x_i)|<r_i,1\leq i\leq n$ implies $\|y^{*}\| \leq n$. Take $y^{*}=x^{*}+z^{*}$ where $z^{*}(x_i)=0, 1\leq i\leq n$ and $\|z^{*}\| >\|x^{*}\|+n$ to get a contradiction. The existence of $z^{*}$ is an easy consequence of Hahn - Banach Theorem.
[Since $X$ is infinite dimensional there exists a vector $x$ which does not belong to $span \{x_1,x_2,\cdots,x_n\}$. Let $M$ be the span of $\{x,x_1,x_2,\cdots,x_n\}$ and define $f$ on $M$ by $f(m+ax)=a(n+\|x^{*}\|)\|x\|$ for $m\in M,a \in \mathbb R$ (or $\mathbb C$ as the case may be). This is a continuous linear functional on $M$ with $\|f\| \geq \frac {|f(x)|} {\|x\|}=n+\|x^{*}\|$. Extend $f$ to a norm preserving functional on $X$ and call it $z^{*}$].