I have read the following sentence in a proof:
It states that the sets $V_{x;\varepsilon}:= \{f \in l_\infty ^* : |f(x)|<\varepsilon \}$ are a basis of neighbourhoods in the weak-* topology on $l_\infty ^*$. But as far as I am concerned they should be as
$$ V_{x_1,\ldots,x_n;\varepsilon}:= \{f \in l_\infty ^* : |f(x_i)|<\varepsilon \ \forall i=1,\ldots n \} $$
in order to form a basis. Are they really a basis? Why? Is this something special in $l_\infty ^*$?
I think it should mean subbase here. If we take, for example, $x_1, x_2\in l_\infty$ linearly independent, and $\varepsilon>0$ there cannot be $x\in l_\infty$ and $\delta >0$ so that $V_{x,\delta} \subset V_{x_1,x_2, \varepsilon}$.
Why is that? Because for all $\delta>0$, $\ker x \subseteq V_{x,\delta} \subset V_{x_1,x_2, \varepsilon}$. So if we take $f\in \ker x$, $|x_1(f)|<\varepsilon$, that is, $x_1$ is bounded in $\ker x$, so $x_1$ vanishes at $\ker x$. That means $\ker x_1=\ker x$, so $x=\lambda_1 x_1$ for some real number.
If we repeat the reasoning with $x_2$, we get that $x=\lambda_2 x_2$. That contradicts the fact that $x_1$ and $x_2$ are linearly independent (why?).