Neighborhood base of weak topology 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: 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.
A: We're considering finite subsets of the dual space, which are less general than bounded subsets. The basis by the latter contain the former and hence this gives the weakest topology. Since we're interested in the weakest topology where the functionals are continuous (hence bounded), it makes sense to consider finite subsets of the dual space.
