Let $X $ be an infinite-dimensional normed space. I want to prove that the weak topology on $X$ is not metrizable. This is my solution:

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 (for the strong topology), 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.

Is this the right solution?

Please give me a reference about the other solutions.

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    $\begingroup$ Looks correct to me (even though I'm not fully sure whether the results you cite hold for arbitrary infinite-dimensional normed spaces or just Banach ones). In fact, something stronger is true: the weak topology on an infinite-dimensional normed space can never be first-countable (which is a weaker condition than metrizability). See, for example, Theorem 6.26 in Aliprantis–Border (2006, p. 237). $\endgroup$ – triple_sec Aug 2 '15 at 9:37
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    $\begingroup$ @triple_sec The topology of a locally convex space is metrizable if and only if it is first-countable. So the result you quoted appears to be stronger, but is not really. $\endgroup$ – user357151 Apr 10 '17 at 0:17
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    $\begingroup$ Your argument is correct. I don't think there is a simpler proof. $\endgroup$ – Kavi Rama Murthy Mar 19 '18 at 6:35
  • $\begingroup$ I don't find anything wrong. $\endgroup$ – Lin Xuelei Feb 13 at 10:50

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