Let $X$ be an infinite-dimensional normed space. Why the weak topology on $X$ is never metrizable ? I saw a proof here but I don't really understand the argument. Here is his argument
Assume that there is a metric $d$ on $X$ inducing the weak topology, and consider $U_n:=\{x\in X: d(x,0)<\frac{1}{n}\}$. We know each $U_n$ is weakly open and so will be unbounded, and thus $$\forall (n)\exists (x_n\in U_n) \:\text{s.t.}\: \|x_n\|\geq n$$ But $x_n\to 0$ in $(X, d)$, so that $x_n \stackrel{w}{\to}0$, and hence $(x_n)$ is bounded. Contradiction.
Q1) Why $U_n$ is weakly open ? Is it because a basis of the weak topology is of the form $$\{B_d(x,\varepsilon)\mid x\in X, \varepsilon>0 \}$$
where $$B_d(x,\varepsilon)=\{y\in X\mid d(x,y)<\varepsilon\} \ ?$$
Q2) Why are $U_n$ unbounded ? Since $U_n\subset B(0,1)$ for all $n\geq 1$, then $U_n$ is bounded (for me, in a metric space $(X,d)$, a set $A$ is bounded, if it's included in a ball $B_d(a,\varepsilon)$… since here $U_n\subset B_d(0,1)$ for all $n\geq 1$, it should be bounded). I really don't understand the argument here.
Best Answer
Note that we require $X$ to be infinite-dimensional. For if $X$ is finite-dimensional, there is only one vector topology on $X$, which is metrizable.
Q1: That's correct. If the metric $d$ induces the weak topology, then the balls in that metric are weakly open.
Q2: This is part of a more general result that every weakly open subset of $X$ is unbounded in the norm-topology of $X$. The idea is that if $U$ is weakly open and $x_0\in U$, then $U$ there is some $\varepsilon>0$ and $\varphi_1,\ldots,\varphi_n\in X^*$ such that $$x_0\in\bigcap_{k=1}^n\{x\in X:|\varphi_k(x-x_0)|<\varepsilon\}\subset U.$$ and the set $\bigcap_{k=1}^n\{x\in X:|\varphi_k(x-x_0)|<\varepsilon\}$ contains the hyperplane $x_0+\bigcap_{k=1}^n\ker(\varphi_k)$, which is unbounded in the norm-topology of $X$.