# Open set in the proof of metrization of the weak star topology for separable Banach space

This questions concerns proof of the theorem:

Let $E$ be separable Banach space with $q$ being an enumeration of dense countable subset of unit sphere $\mathbb{S}_E$ of $E$. Then $$\rho(f,g) = \sum^\infty_{i = 1} \frac{|f(q_i) -g(q_i)|}{2^i}$$ is a distance metric for unit ball $\mathbb{B}_{E^*}$ of the dual space $E^*$. Moreover, metric space $(\mathbb{B}_{E^*},\rho)$ is homeaomorphic to topological space $(\mathbb{B}_{E^*},\mathrm{w}^*)$, where $\mathrm{w}^*$ denotes weak-star topology, with identity map $I$ being homeomorphism.

We know by the merit of the previous work that $(\mathbb{B}_{E^*},\mathrm{w}^*)$ is compact.

I have no problem proving that $\rho$ indeed is a metric.

In the book functional analysis and infinite-dimensional geometry by Fabian et al the proof of $I : (\mathbb{B}_{E^*},\mathrm{w}^*) \to (\mathbb{B}_{E^*},\rho)$ being continuous uses an argument which I don't understand:

Let $B = \mathbb{B}_{(\mathbb{B}_{E^*},\rho)}(f,\varepsilon)$ be an $\varepsilon$-ball with center $f$. By archemedean propery there is a natural number $n$ such that $2^{-n} < \frac{\varepsilon}{2}$. Then it is shown that for every functional $g$ such that $|f(q_i) - g(q_i)| < \frac{\varepsilon}{4n}$ for all $i$ in range $1,\ldots,n$ the funcional $g$ is in the $B$. This can be easily checked. Textbook infers from this fact that $B$ is in $\mathrm{w}^*$. However, for me it just implies that weakly-star-open set $A = \bigcap^n_{i = 1} \{ g \in \mathbb{B}_{E^*} : |f(q_i) - g(q_i)| < \frac{\varepsilon}{4n} \} \subseteq B$.

I see two ways to approach this problem:

• Prove that indeed $A = B$. (But I don't know how. Why $g$ such that $\rho(g,f) < \epsilon$ and $|g(q_i) -f(q_i)| > \epsilon/4n$ doesn't exist? )
• There is a lost induction step on $n$ and it is possible to construct sequence $A_n$ such that $A_n \uparrow B$. But as far I can see increasing $n$ only makes $A_n$ smaller and decreasing $n$ may make it oversize $B$

I also think that $A$ can be redifined as $$A = \bigcap^n_{i = 1} \left\{ g \in \mathbb{B}_{E^*} : |f(q_i) - g(q_i)| < \frac{\varepsilon}{n} 2^{i - 2} \right\}$$ and still be contained inside $B$.

Thank you.

You are correct in that this only shows that $A \subset B.$ This is sufficient however, because the open balls generate the topology of $(\mathbb B_{E^*},\rho).$ You can complete the argument as follows:
If $U \subset \mathbb B_{E^*}$ is open in $(\mathbb B_{E^*},\rho),$ then for all $f \in U$ there is $\epsilon > 0$ such that $A = \mathbb B_{(\mathbb B_{E^*},\rho)}(f,\epsilon) \subset U.$ Then we can find a weak*-open subset $B \subset A \subset U$ containing $f.$ Hence $U$ is weak*-open, taking the union of all such $B$ for each $f.$
• I got it. Instead of proving that balls are weakly-star-open construct sets $B_f \in \mathrm{w}^*$ such that $f \in B_f \subset U$ and then $\bigcup_{f \in U} B_f = U$. Thank you! This was in my blind spot. I will accept your answer tomorrow. – Nik Pronko Jun 12 '17 at 21:11
A function $I:A\to B$ is continuous iff whenever $I(x)\in V$ where $V$ is open in B, there exists open $U$ in $A$ with $x\in U$ and $I(U)\subset V.$ (This is the topological generalization of the classical "$\epsilon$-$\delta$" statement.) In this case $I(f)=f$ and $V=B_{\rho}(f,\epsilon)$ and $U=\cap_{i=1}^n\{g:|f(q_i)-g(q_i)|<\frac {\epsilon}{2n}\}.$