The question arises when I was reading Hopf Algebra written by Eiichi Abe, page 67.
Given a $k$-linear space $V$, let the dual $k$- linear space $\mathrm{Hom}_{k}(V,k)$ be denoted by $V^*$.
For $x\in V$, set $$x^{\perp}=\{f\in V^*| f(x)=0\}$$ Given $f\in V^{*}$, by letting the family of subsets of $V^*$ $$\{f+x^{\perp}|x\in V\}$$ be a base for a system of neighborhoods of $f$, $V^*$ becomes a linear topological space. A $k$- linear space whose underlying abelian group is a topological group and such that a family of $k$- linear subspaces can be chosen for its system of neighborhoods is called a linear topological space.
My question is why are the family of subsets of $V^*$ as above a base for a system of neighborhoods? The context does not refer to some topology of $V^*$ and hence I think that the family of subsets should be a base for the topology: that is to say, the intersection of two sets of the given form is a union of sets of the form. Can anyone explain the idea here?
It's not actually a basis of neighborhoods. Let $V = \mathbb{R}^2$, and consider $f = 0$ for simplicity. Let $e_1 = (1,0)$ and $e_2 = (0,1)$ be a basis of $V$. Then $e_1^\perp \cap e_2^\perp = \{ 0 \}$, which cannot be written as a union of subsets of the form $v^\perp$: for any $v \in V$, $v^\perp$ contains a nonzero element.
(Indeed, if $v = (v_1, v_2) \neq 0$, then you can consider $f \in V^*$ given by $f(x,y) = v_2 x - v_1 y$. Otherwise if $v = 0$, then $v^\perp = V^*$ and you can choose any nonzero element.)
In fact, I'm pretty sure that the author(s) meant that this was a subbase for a system of neighborhoods.