I am trying to understand exactly why a weakly open subset of a Banach space is also open in the norm topology. The lecture notes defined "weakly open" as follows:
Suppose $X$ is a Banach space. $x∈X$, $φ∈X^*$.
Recall the isometric embedding $X↪X^{**}$, $x$⟼$x^{**}$ where $\hat{x}(f)=f(x)$.
A subset $U$ of $X$ is weakly open if $∀x_0∈U$, $∃ε>0$ and $φ_1,…,φ_n∈X^*$ with
$$ ⋂_{k=1}^n \{x∈X∶|φ_k(x-x_0)|<ε\} \subset U.$$
I know it has something to do with the fact that the intersection of finitely many open sets is open, but somehow I just can't get the picture. Would you mind help me understand this?
$\{x\in X:|\phi_k(x-x_0)|<\epsilon \}$ is the inverse image under $\phi_k$ of $\{a: |a-\phi_k(x_0)| <\epsilon\}$. This last set is an open ball in the scalar field and since $\phi_k$ is continuous on $X$ the inverse image is open. Since finite intersection of open sets is open we are done.