I want to show that a weak* open set is also norm open.
I know that if I consider the dual space $X^*$, then a neighbourhood basis of $f\in X^*$ is $\{g\in X^*:|g(x_i)-f(x_i)|<\epsilon, 1\leq i\leq l\}$ with $\epsilon>0$ and $x_i\in X$ for all $i$.
Similarly a neighbourhood basis in the topology induced by the norm is given by open balls.
Now my first question is, how can I think about the neighbourhood basis of $f\in X^*$?
So I know that by definition $\mathcal{B}$ is a neighbourhood basis of $x$ in a topological space $X$ if $x\in V$ for all $V\in \mathcal{B}$ and for all neighbourhood $U\ni x$ there exists $V\in \mathcal{B}$ such that $x\in V\subset U$.
But in our case I don't see how the neighbourhoods in the weak* topology looks like.
Additionaly I don't see how to show that a set which is weak* open also is norm open.
Could maybe someone answer my two questions?
There is no need to know how exactly a weak* open set looks like.
$\{g\in X^*:|g(x_i)-f(x_i)|<\epsilon, 1\leq i\leq l\}$ is open in the norm topology whenever $l\in \mathbb N$, $\epsilon>0$ and $x_i\in X$ for all $i$. This is because $g \mapsto g(x_i)$ is a continuous function for the norm topology . [$\{g\in X^*:|g(x_i)-f(x_i)|<\epsilon, 1\leq i\leq l\}=\bigcap_{1\leq i \leq l} \{g: g(x_i) \in (B(f(x_i),\epsilon))\}$]. Any open set in a topological space is a union of basic neighborhoods of its points.