Let $X$ be an infinite dimensional normed space. I'm trying to proof that $(X^\star, w^\star)$ is Baire first cathegory, i.e. $X^\star$ is a countable union of nowhere dense sets. Here $X^\star$ denotes the topological dual of $X$.
My attempt: Write $X^\star = \bigcup_{n \in \mathbb N} nB_{X^\star}$, where $B_{X^\star}$ denotes the unit closed ball of $X^\star$ in norm, which is $w^\star-$compact by Banach-Alaogulu-Bourbaki theorem. So, it remains us to proof that $B = B_{X^\star}$ has empty interior and I got stucked in this part.
Let $\varphi_0$ in the $w^\star$-interior of $B$. Then we can take an neighborhood $$ V = \left \{ \varphi \in B : |\varphi_0 (x_i) - \varphi(x)| < \epsilon, \, i=1,\cdots, m \right \}, $$ where $x_1, \cdots, x_m \in X$ and $\epsilon > 0$. How can I get a contradiction?
Picking up where you left off, without loss of generality, $\varphi_0=0.$ There is a $0\neq\varphi \in X^*$ such that $\varphi\in \text{ker}\ x_i;\ 1\le i\le n,$ for if not, then the map $\varphi\mapsto (x_1(\varphi),\cdots,x_n(\varphi))$ injects $X^*$ into a finite dmensional space. But then, $t\varphi\in V\subseteq B$ for arbitrarily large $t\in \mathbb R,$ which is a contradiction.