The problem is from Rudin's Functional Analysis, first part of exercise 7, chapter 1.
Here is what it says:
Let $X$ be the vector space of all complex functions $f:[0,1]\rightarrow \mathbb{C}$ topologized by a family of seminorms $p_x(x)=|f(x)|$, i.e. X has a topology of pointwise convergence. Show that there is a sequence $\{f_n\}$ in $X$ such that $f_n\rightarrow 0$.
I started with $f_n\rightarrow 0$ iff for every $x\in [0,1]$, $|f_n(x)|\rightarrow 0$. So I constructed $f_n$ such that for every $k\in\mathbb{N}$ there is $n_k$ such that $\forall x\in [0,1](|f_n(x)|<\frac{1}{k})$, whenever $n>n_k$, which satisfies the statement earlier.
But then I tried to fit this in the context of the topology defined on $X$. I recall that a collection of all finite intersections of $V(p_x,n)=\{f\in X: p_x(f)=|f(x)|<\frac{1}{n}\}$ forms a local base for the topology. The ${f_n}$ constructed above implies that $f_n \in V(p_x,k), n>n_k$ for all $p_x$ which means $f_n\in \cap_{x\in[0,1]}V(p_x,k)$; this is definitely not a finite intersection therefore is not a member of the local base and this confuses me a lot. So where did I get it wrong?
Thank you in advance.
$f_n (x)=\frac 1 n$ is such a sequence.