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

Let $(X,\|\cdot\|)$ be a infinite dimensional normed vector space, and Suppose that the weak topology in $X$ is metrizable by a metric $d$. How the opens of $\tau_d$ should be the same as the weak topology, we have that for every $n$ the ball $B^d(0,\frac{1}{n})$ mus be have a non trivial subspace we have the following construction:

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

My question lies in the fact that in the Brezis book:

http://www.springer.com/mathematics/analysis/book/978-0-387-70913-0

exercice 3.8 there is a proof script, which uses up to Baire's theorem! I can not conceive why to use such a complicated demonstration when there is so much simpler proof, if my proof is correct

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There's a big hammer used in your proof also. Namely the Uniform Boundedness Principle, which is used to show weakly convergent sequences are norm bounded. –  David Mitra May 29 '14 at 21:15
Incidentally, most proofs of the Uniform Boundedness principle use Baire; though, the use of Baire can be avoided, see this paper. –  David Mitra May 29 '14 at 21:24
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

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 $\geqslant N+1$.