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Evening, guys! I'm looking for applications of the Urysohn Metrization Theorem. Well, My first thought was prove that the unit ball in the dual, with $w^*$-topology, of a separable Banach space $X$ is metrizable.

So, all we need to do is guarantee $B^*[0,1]$ is regular and second-countable. The first thing is done directly if you assume Alaglou's theorem and then show that compact Hausdorff spaces are in fact regular. No problems until here.

So, for countability I'm trying to exhibit a enumerable basis for $w^*$. I'm almost convinced that $T_{x_n}^{-1}(B(q,r))$, in which $T_x: X^* \rightarrow \mathbb{R}$ is the map such that $T_x(\phi) = \phi(x)$, $\{x_n\}_n$ is the dense set of $X$ and finally, $p,q \in \mathbb{Q}$, does the job.

The first observation is that the collection of all pre images $T_x^{-1}(A)$ in which $A$ is a real open set and $x\in X$ intersected to $B^*[0,1]$ and all its finite intersections constitute a basis for the $w^*$-topology. Now we use the dense set $\{x_n\}_n$ and density of $\mathbb{Q}$ on the line to extract a countable collection.

So, if $\phi \in T_x^{-1}(A)$ for some $x$ and $A$, then there exist rationals $q,r$ such that $\phi(x) \in B(q,r) \subset A$ and $x_n \in X$ such that $|x_n - x| < r/2$.

Finally, if $\psi \in T_{x_n}^{-1}(B(q,r/2))$ then $|\psi(x_n) - q| < r/2$ and $\| \psi \| \le 1$. This leads to $$ |\psi(x) - q| \le |\psi(x) - \psi(x_n)| + |\psi(x_n) - q| < \| \psi \| |x-x_n| + r/2 < r. $$

Thus $T_{x_n}^{-1}(B(q,r/2)) \subset T_{x}^{-1}(A)$.

My questions are:

  1. Is this enough for my purpose? Or Am I missing something?
  2. Is there another way still using Urysohn's Theorem?
  3. Is there a better application of the Urysohn's Theorem?

Thank you in advance!

Rodrigo Ribeiro
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  • Yes, I know. But the bounded sets are when X is separable? – Rodrigo Ribeiro Jul 08 '16 at 11:26
  • You are right, I forgot it. – Ningxin Jul 08 '16 at 18:07
  • But now I have this doubt. If X is separable, there's any chance the weak*-topology be second countable? – Rodrigo Ribeiro Jul 08 '16 at 18:19
  • I should have kept the link http://math.stackexchange.com/questions/623642/the-weak-topology-on-x-is-not-first-countable-if-x-has-uncountable-dim/626599#626599. If $X$ is a Banach space, then it either has uncountable dimension (in which case $X^$ with weak$^$-topology is not metrizable), or it is finite-dimensional (then the answer is affirmative). If it's not metrizable, then it cannot be second-countable, because it's not first countable. – Ningxin Jul 08 '16 at 19:10
  • Suppose a Banach space $X$ has countable dimension. Then it can be written as a countable union of finite-dimensional proper subspaces. Finite-dimensional subspaces are closed, and proper subspaces have empty interior. Hence $X$ if of the first category in itself. But complete metric spaces are of the second category in themselves, which is a contradiction. – Ningxin Jul 08 '16 at 19:14
  • @QiyuWen thank you for the answers. But, look, I'm trying to prove that the unit ball of the dual is metrizable with the additional hypothesis that X is separable. According to Alaoglu, the unit ball is a compact space. Thus we are talking about a nicer space, right? Limited, compact. Would this be enough to prove it is countable? – Rodrigo Ribeiro Jul 08 '16 at 19:19
  • If you are just focusing on the unit ball with weak$^*$-topology, then I think it is second countable, because compact spaces are Lindelöf, which is equivalent to second-countable in metrizable spaces. – Ningxin Jul 08 '16 at 19:23
  • In fact I think it's easier to prove the unit ball $B$ (with weak$^*$-topology) is metrizable using just the embedding lemma (to find a sequence of continuous functions separating points and closed sets). In the case $B$ is compact, it reduces to finding a sequence that separates points. But then a countable dense set in $X$ separates points in $B$. – Ningxin Jul 08 '16 at 19:38
  • I believe it is. But my attempt is to prove it using the Urysohn Metrization Theorem. Actually, you can exhibit explicitly a metric on $B$ (I guess this is the standard way for proving the result.) But I would like to give an application of this Metrization theorem, got it? – Rodrigo Ribeiro Jul 08 '16 at 19:56
  • The metrization theorem says "if my space is such and such, then I can find a sequence of nice functions to apply the embedding lemma". Now you already have a sequence of nice functions, so I see no point going backward. If you want an application of metrization theorems, the metrizability of topological manifolds should be a good example. – Ningxin Jul 08 '16 at 20:12
  • Yes... You are right. Thank you @QiyuWen ! – Rodrigo Ribeiro Jul 08 '16 at 20:15

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