Jtdylktuy

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 $|y^{*}(x_i)-x^{*}(x_i)|<r_i,1leq ileq n$ implies $|y^{*}| leq n$. Take $y^{*}=x^{*}+z^{*}$ where $z^{*}(x_i)=0, 1leq ileq n$ 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 $min M,a in mathbb R$ (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 $z^{*}$].