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.

  • 1
    $\begingroup$ 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, 2015 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, 2017 at 0:17
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    $\begingroup$ Your argument is correct. I don't think there is a simpler proof. $\endgroup$ Mar 19, 2018 at 6:35
  • $\begingroup$ I don't find anything wrong. $\endgroup$
    – Lin Xuelei
    Feb 13, 2019 at 10:50

1 Answer 1


To get this off the unanswered list: the proof is fine. You might clarify at the end that "bounded" is with respect to the norm $\|\cdot\|$, and that this follows from the uniform boundedness principle.

More details can be found at Why the weak topology is never metrizable?

Other questions about this theorem, stolen from the Related list:


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