I will fix $E \doteq \ell^2$, an infinite dimensional space, with separable dual.
- In here, there was a discussion that the weak topology of $E$ is not metrizable.
- By a result given in here, the disk $\mathbb{D}_E$ is weakly metrizable.
In short, I am struggling to see this is not a contradiction. Of course, subspaces of non metrizable spaces can be metrizable, I'm just curious about the mechanism. If either my proofs (or interpretation of them) is wrong or if I'm missing some hypothesis here, please correct me and let me know :)
Here is the proof I know of the first fact:
The weak topology can not be metrizable because $0 \in E$ is not (weakly) first enumerable.
- Suppose it is, then we have an enumerable local base of sets of type $V_{\epsilon,\phi_1,...,\phi_n}(0) \doteq \{x \in E \, : \, |\phi_i(x)|<\epsilon\}$.
- Enumerating them, an enumerable set of finite operators $n \in \mathbb{N} \mapsto F_n \doteq \{\phi_1^n,...,\phi^n_{k_n}\}$ will appear in the subscript.
- For each $\phi \in E^*$, $V_{1,\phi}(0)$ must contain some set of the basis, say $V_{\epsilon_n,\phi_1^n,...,\phi^n_{k_n}}(0) \subset V_{1,\phi}(0)$.
- From this, we conclude that $\ker \phi_i^n \subset \phi$, and then it must be generated by them, so $\phi \in \langle F_n \rangle$.
- In that case, we can write the (Banach) space $E^* = \bigcup_{n \in \mathbb{N}} F_n$ as an enumerable union of closed subspaces.
- We must have $E^* = F_m$ for some $m$ and then it is finite dimensional, from wich we derive the contradiction with the supposition that $E$ was infinite dimensional.
The second result also seems sound to me, and tells me that $0 \in \mathbb{D}_E$ is (weakly) locally first enumerable.
I can accept it is not a contradiction, but I wish to intuitively understand why it is not. What is going on (with the neighborhood system of $0$) when we restrict to the disk that make this first enumerable? Does someone have a good intuition?
