Is the $\ell^2(\mathbb N_0)$ weak topology on the unit ball minus origin completely metrizable?

Question

$\newcommand\el{\ell\kern.45mm}$Put more precisely, letting $\mathcal T$ be the $\sigma(\el^2,\el^2)$ topology restricted to the set $\el^2(\mathbb N_0)\cap\{\,x:0<\|\,x\,\|_{\el^2}\le 1\,\}\,$, is there a complete metric metrizing $\mathcal T\,$? I have a "strong feeling" that the answer is yes, and have been trying to construct a metric verifying this "feeling", but with no success so far.

Actually, what I ultimately need is the following: The topology $\sigma(\el^1,c_0)$ restricted to $\el^1(\mathbb N_0)\cap\{\,x:\frac 12<\|\,x\,\|_{\el^1}\le 1\,\}$ is completely metrizable. However, to keep matters as simple as possible, I formulated the above question since I think that the same idea that possibly gives a positive answer to that question can also be applied here.

A related matter is the following: If $\mathcal T$ is any compact metrizable topology on some $\Omega$ with $x_0\in\Omega\,$, and if $\mathcal T_0$ is $\mathcal T$ restricted to $\Omega\setminus\{\,x_0\}\,$, then $\mathcal T_0$ is locally compact, and I "feel" that it should be completely metrizable. A metric verifying this should somehow become "large" near $x_0$ so that sequences converging to $x_0$ in $\mathcal T$ cannot be Cauchy.

Answer 1

The answer to both questions is yes. We have the following result.

Proposition. We have the following.

(a) Let $X$ be a separable normed space. Let $r,s\in [0, \infty )$ with $r < s$. Let $A = \{x\in X^{*} : r < \|x\|_{X^{*}} \leq s \}$. Then $(A, \sigma (X^{*}, X))$ is completely metrisable.

(b) Let $X$ be a separable reflexive normed space. Let $r,s\in [0, \infty )$ with $r < s$. Let $B = \{x\in X : r < \|x\|_{X} \leq s\}$. Then $(B, \sigma (X, X^{*}))$ is completely metrisable.

Proof. We denote the closed unit ball of $X$ by $B_{X}$. We first show (a). The set $sB_{X^{*}}$ is compact in the $\sigma (X^{*}, X)$ topology by the Banach-Alaoglu theorem and is metrisable in the $\sigma (X^{*}, X)$ topology because $X$ is separable. Hence there is a metric on $s B_{X^{*}}$ which makes $(s B_{X^{*}}, \sigma (X^{*}, X))$ a compact metric space. In particular, $(s B_{X^{*}}, \sigma (X^{*}, X))$ is completely metrisable. Since $A = s B_{X^{*}} \setminus r B_{X^{*}}$ is open in $(s B_{X^{*}}, \sigma (X^{*}, X))$ and as open subsets of completely metrisable spaces are completely metrisable by this result we obtain that $(A, \sigma (X^{*}, X))$ is completely metrisable. This shows (a).

We now show (b). This can be done with an analogous argument to the proof of (a). On the other hand, we can also argue as follows. Note that $(B, \sigma (X, X^{*}))$ is homeomorphic to $(J(B), \sigma (X^{**}, X^{*}))$ where $J\colon X \to X^{**}$ is the isometric canonical embedding. As $X$ is separable and reflexive, so is $X^{*}$. Hence the result follows from (a) by this result. This completes the proof.

You can also use the proofs of the results mentioned within the above proof to explicitly construct such a metric. For your case, $(B_{\ell^{1}}, \sigma (\ell^{1}, c_{0}))$ has a complete metric $d$ inducing its topology given by \begin{equation} d(x,y) = \sum_{n\in\mathbb{N}} \min\{|x(n) - y(n)|, 2^{-n} \} . \end{equation} Using the metric from here we see that the metric $\rho$ on $\{x\in X^{*} : \tfrac{1}{2} < \|x\|_{X^{*}} \leq 1 \}$ given by \begin{equation} \rho (x,y) = d(x,y) + \Big| \frac{1}{d(x, \tfrac{1}{2}B_{\ell^{1}})} - \frac{1}{d(y, \tfrac{1}{2}B_{\ell^{1}})} \Big| \end{equation} is a complete metric that induces the $\sigma (\ell^{1}, c_{0})$ topology on $\{x\in X^{*} : \tfrac{1}{2} < \|x\|_{X^{*}} \leq 1 \}$.

Answer 2

Proposition. Let $X$ be a completely metrizable space. Then $Y\subseteq X$ is completely metrizable iff $Y$ is a $G_\delta$-subset of $X$.

Theorem. (Banach-Alaoglu) If $X$ is a normed space then $\{x\in X^\ast : \|x\|\leq 1\}$ is compact in the $\sigma(X^\ast, X)$ topology on $X^\ast$.

Theorem. If $X$ is a normed space then $\{x\in X^\ast : \|x\|\leq 1\}$ is metrizable in the $\sigma(X^\ast, X)$ topology on $X^\ast$ iff $X$ is separable.

It follows that for separable $X$, the ball $\{x\in X^\ast : \|x\|\leq 1\}$ is compact metrizable in the $\sigma(X^\ast, X)$ topology, so in particular its completely metrizable. So any $G_\delta$-subset of it is also completely metrizable. In your examples, both $c_0$ and $\ell^2$ are separable, so the open (and in particular, $G_\delta$) subset $$\{x\in X^\ast : s < \|x\|\leq r\} = \{x\in X^\ast : \|x\|\leq r\}\setminus \{x\in X^\ast : \|x\|\leq s\}$$ of $\{x\in X^\ast : \|x\|\leq r\}$ is completely metrizable for any $0\leq s < r$.