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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.

Aweygan
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Peter
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2 Answers2

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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 $X$.

Aweygan
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  • Let $\mathcal T_W$ the weak topology, $d$ the metric s.t. $\mathcal T_W=\mathcal T_d$ and $\mathcal T$ the strong topology. When they say that $U_n$ in unbounded, it's wrt $\mathcal T_d$ or $\mathcal T$ ? Because as I'm thinking, it's wrt $\mathcal T_d$, but maybe it's wrt $\mathcal T$ ? – Peter Aug 24 '18 at 20:25
  • It's with respect to $\mathcal T$. That's how you get that $|x_n|\geq n$. – Aweygan Aug 24 '18 at 20:26
  • It make sense now. Thanks a lot for your great answer, that helped me a lot. (Why didn't the OP made this precision in his post ? ) – Peter Aug 24 '18 at 20:30
  • Glad to help! I'm guessing they knew all this, and took it for granted. – Aweygan Aug 24 '18 at 20:43
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$U_n$ is weakly open since the topology defined by the metric is the weak topology, so the open subsets of the two topologies coincide.

$U_n$ is open since $U_n$ is open for the weak topology: i.e, there exists a bounded linear function $f$, and an interval $I$ containing the origin $f^{-1}(I)$ is not empty and contained in $U_n$. Let $x$ such that $f(x)\in U_n$, for every $y\in Ker(f),f(x+y)\in I$, I assume that that the dimension of the space is superior or equal to $2$, then we can assume that $y\neq 0$ and the sequence $x+ny\in f^{-1}(I)$ is unbounded.

Tsemo Aristide
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  • Your proof of the second fact is not quite true. The collection of sets $f^{-1}(I)$ form a subbasis for the weak topology, not a basis. And for this reason, we require not only that $\dim X\geq2$, but that $\dim X=\infty$. – Aweygan Aug 24 '18 at 20:19
  • Sorry but I don't understand your answer. How do you know that there exists a bounded linear function f, and an interval I containing the origin $f^{−1}(I)$ is not empty and contained in $U_n$ ? I guess $I$ is an open interval no ? – Peter Aug 24 '18 at 20:31