# Weak topology is not metrizable: what's wrong with this proof?

Let $(X,\|\cdot\|)$ be an infinite-dimensional normed vector space. Suppose that the weak topology of $X$ is metrizable by a metric $d$. Denote by $B^d(x,r)$ the open balls with respect to $d$; they are therefore weakly open. We have that for every $n$ the ball $B^d(0,\frac{1}{n})$ contains a non-trivial subspace. We could then argue as follows:

Choose in each $B^d(0,\frac{1}{n})$ an $x_n$ such that $\|x_n\|=n$. We have $x_n\rightharpoonup x$ but $\|x_n\|\to \infty$. Which is an absurd, because we know that the sequence $(\|x_n\|)_{n=1}^\infty$ must be bounded.

My question lies in the fact that in the Brezis book (Exercise 3.8) there is a proof which uses Baire's theorem! I don't know why they use such a complicated demonstration when there is so much simpler proof, if my proof is correct of course.

• I had not looked through this point of view, well observed! Thanks for the reference. – O Empalador de Cabras May 29 '14 at 21:30

## 1 Answer

The proof seems correct. The assumption that $X$ is infinite dimensional is used to show that $B^d(0,1/n)$ contains a non-trivial subspace and the argument should be detailed a little more because it is not straightforward. The set $B^d(0,1/n)$ contains an open subset of the form $O:=\bigcap_{j=1}^N\{x,|f_j(x)|<\delta\}$ for some integer $N$, $f_j\in X^*$ and $\delta >0$. There is $y\neq 0$ such that $f_j(y)= 0$ for each $j\in \{1,\dots,N\}$ because $X$ has a dimension $\geqslant N+1$.