Definition of weak-star open.

Question

Definition:

Let $X$ be a norm space. The weak-star topology on $X^*$, is the topology on $X^*$ defined by the family $\{p_x:x\in X\}$ of seminorms on $X^*$, where $p_x(x^*)=|\langle x,x^*\rangle|$ $(x^*\in X^*)$. And we have a subset U of $X^*$ is weak-star open if and only if every $x_0^*$ in $U$, there are $x_1,\cdots$, $x_n$ in $X$ and $\varepsilon_1,\cdots$, $\varepsilon_n>0$ such that $$\bigcap\limits_{j=1}^{n}\{x^*\in X^*: |\langle x_j,x^*-x_0^*\rangle|<\varepsilon_j\}\subseteq U.$$

And my question is how to get that there exists a finite set F in X such that $\{x^*\in X^*: |\langle x,x^*-x_0^*\rangle|<1\text{ for all } x\in F\}\subseteq U$ from $\bigcap\limits_{j=1}^{n}\{x^*\in X^*: |\langle x_j,x^*-x_0^*\rangle|<\varepsilon_j\}\subseteq U$. And how is $F$ like?

Thank you in advance!

Answer 1

Given a basic open set $$ \bigcap\limits_{j=1}^{n}\{x^*\in X^*: |\langle x_j,x^*-x_0^*\rangle|<\varepsilon_j\}\subseteq U, $$ take $\varepsilon = \min \varepsilon_j$ and set $F = \{y_1,\dots,y_n\}$, where $y_j = x_j/\varepsilon$. Then

$$ |\langle y_j, x^* - x_0^* \rangle| = \frac{1}{\varepsilon} |\langle x_j, x^* - x_0^* \rangle| < 1 $$

implies

$$ |\langle x_j, x^* - x_0^* \rangle| < \varepsilon \le \varepsilon_j, $$

for all $j$.

Therefore,

$$ \{x^*\in X^*: |\langle x,x^*-x_0^*\rangle|<1\text{ for all } x\in F\}\subseteq \bigcap\limits_{j=1}^{n}\{x^*\in X^*: |\langle x_j,x^*-x_0^*\rangle|<\varepsilon_j\}. $$