If $X$ is a Banach space, the weak topology on $X$ is the weakest topology in which each functional $f$ in $X^\ast$ is continuous.
I have some difficulties in understanding its neighborhood basis in 0.
It should be made by the sets of the form
$N(f_1,...,f_n;\epsilon)=\{x :|f_i(x)|<\epsilon, \text{for }i=1,\dots,n\}$ for each finite $n$.
I don't understand why is this a neighborhood basis and not simply $N(f;\epsilon)=\{x : |f(x)|<\epsilon\}$.
Can you explain to me this as clearly as you can? Thank you very much!
A neighbourhood basis of a point $x$ in a topological space is a family $\mathscr{B}_x$ of neighbourhoods of $x$ such that every neighbourhood of $x$ contains some member of $\mathscr{B}_x$. In particular, since the intersection of finitely many neighbourhoods of $x$ is again a neighbourhood of $x$, we must for every pair $V,W\in \mathscr{B}_x$ have an $U\in \mathscr{B}_x$ with $U \subset V\cap W$.
The intersection $N(f;\epsilon_1) \cap N(g;\epsilon_2)$ does in general not contain any set of the form $N(h;\epsilon_3)$, since for $f,g$ linearly independent, $N(f,\epsilon_1)\cap N(g;\epsilon_2)$ contains a linear subspace of codimension $2$, but no linear subspace of codimension $1$, while $N(h;\epsilon)$ contains a linear subspace of codimension $\leqslant 1$ (the kernel of $h$).
So we need to take finite intersections in order to have a neighbourhood-basis element contained in the intersection of two neighbourhood-basis elements.
The sets $N(f;\epsilon)$ form a neighbourhood-subbasis, however.