If X is separable, then ball $X^*$ is weak-star metrizable.

Here is a proof in Conway's book. But there is something I don't understand.

1. Why choose $\{x_n\}$ in ball $X$? Why not in $X$?
2. I didn't see where the condition $\{x_n\}$ is dense using there. Without $\{x_n\}$ dense I can still get $x^*-y^*=0$.

Thank you!

• For those who don't have "Conway's book" (which isn't really specific), what does "ball" mean in this context? – Cameron Buie Sep 8 '17 at 23:11
• If you aren't satisfied with the answer to your previous question, you should ask the answerer for clarification. – Cameron Buie Sep 8 '17 at 23:13
• You need $x_n$ dense otherwise you wouldnt necessarily have that $<x_n, y^{\ast}-x^{\ast}>=0$ implies $y^{\ast}-x^{\ast}>=0$. You choose $x_n$ in the ball because then $|<x_n,x^{\ast}>|\leq 1$, and so $\prod _n <x_n,x^{\ast}>$ lies in a compact set. – clark Sep 8 '17 at 23:57
• @clark But I don't think we need $\Phi_n<x_n,x^*>$ to lie in a compact set. – Answer Lee Sep 9 '17 at 1:17
• We indeed don't need $X$ to be compact. We show $\tau$ from the weak$^\ast$ ball $B$ to $X$ is 1-1 and continuous. Then $\tau[B] \simeq B$ as $B$ is compact (this is used essentially here). And $\tau[B]$ is metrisable. But we could just have used $X = \mathbb{F}^\mathbb{N}$ for that. But the proof is inspired (I think) by the one for Alaoglu, hence the focus on the ball. (see my answer) – Henno Brandsma Sep 9 '17 at 5:57

It follows from $|x^\ast(x)| \le \|x^\ast\|\cdot|x|$ for all $x^\ast \in \mathscr{X}^\ast, x \in \mathscr{X}$ (by definition of the norm on the dual):

If we take the $(x_n) \in \text{ball } \mathscr{X}$. We want the $\tau$ to be well-defined. And as $x^\ast \in \text{ball }\mathscr{X}^\ast$, $\|x^\ast\| \le 1$ and so for any $x_n \in \mathscr{X}$, $|x^\ast(x_n)| (= |\langle x_n,x^\ast \rangle|) \le \|x_n\|$. But we want them all to be in $\prod_n D_n$ (just because we can..), so we want $|\langle x_n, x^\ast \rangle | \le 1$, and so we choose all $x_n$ in $\text{ball } \mathscr{X}$.

This can be done as $\mathscr{X}$ being separable implies that all its subspaces are separable too (this holds in any metric space, in fact). As a matter of fact, in any Banach space, the fact that the ball of the space is separable implies the whole space is, too: just take all $qx$ where $x \in D$ (a countable dense set of the ball) and $q$ rational which is a countable dense subset of the whole space. This sketches that "$\text{ball }\mathscr{X}$ is separable" iff "$\mathscr{X}$ is separable". So we can use whichever is convenient.

For the second point, we use this classical fact from general topology:

If $f$ and $g$ are continuous functions on a space $S$ to a Hausdorff space $H$, and for some dense set $D$ of $S$ we have $\forall x \in D: f(x) = g(x)$, so $f$ and $g$ agree on a dense subset. Then $f=g$ on $S$.

In the proof we suppose we have $\tau(x^\ast) = \tau(y^\ast)$, so $\langle x_n, x^\ast \rangle = \langle x_n, y^\ast \rangle$ for all $n$, or $x^\ast - y^\ast$ is a function from $\text{ball } \mathscr{X}$ to $\mathbb{R}$ that agrees with the $0$-functional on a dense subset of the ball. So the above fact gives that it equals $0$ on all of $\text{ball } \mathscr{X}$, as claimed. As an alternative: use that for any $x \in \text{ball }\mathscr{X}$ there is sequence $x_m$ from the countable dense set such that $x_m \to x$ and use (sequential) continuity of $x^\ast -y^\ast$ to see that its value at $x$ is also $0$).

We indeed don't need $X$ to be compact. We show $\tau$ from the weak$^\ast$ ball $B$ to $X$ is 1-1 and continuous. Then $\tau[B] \simeq B$ as $B$ is compact and $X$ is metrisable (so Hausdorff). And $\tau[B]$ is metrisable, as a subspace of $X$. But we could just have used $X = \mathbb{F}^\mathbb{N}$ for that as well.

This proof is similar to the proof that $\text{ball }\mathscr{X}^\ast$ is compact: there we use all $x \in \text{ball }\mathscr{X}$ and define $\tau': \text{ball }\mathscr{X}^\ast \to D^{|X|}= \prod_{x \in X} D_x$ as $\tau'(x^\ast) = (\langle x, x^\ast \rangle)_{x \in X}$, where $D_x= D$ is the unit disk in $\mathbb{F}$. Here we need the $\le 1$ condition to get all the components in $D$, and have a product of compact spaces. $1-1$-ness is almost by definition here (if functionals agree on all points of the ball they're equal everywhere). Standard theorems in topology gives us that $\tau'$ is an embedding, and then what remains to be shown is that $\tau'[\text{ball }\mathscr{X}]$ is closed in this compact product..

The idea of the asked about proof is to "get away with" using only countably many points from $\text{ball }\mathscr{X}$ instead of all of them, and reduce the product to a countable (hence metrisable) one, but we can only do that if we can approximate all the points of the ball, i.e. we need a countable dense subset of it.
I think this proof was conceived as an expansion on the above idea, so it makes sense to start with a dense set in $\text{ball }\mathscr{X}$, even as for the metrisability it's not strictly needed.

• Thank you so much for your help! I get it right now. Here is me trying an alternative way to show ball $X^*$ is weak-star metrizable but I am stuck. Can you help me please? Thank you so much!! math.stackexchange.com/questions/2424021/… – Answer Lee Sep 10 '17 at 14:54