Topology Filters Let $X$ be a set and $\mathcal F$ a filter on $X$ with a countable filter basis $\mathcal B$. Prove that $\mathcal F$ has a countable filter basis $\mathcal B´$ which is nested, i.e. $\mathcal B´$={$B_n$} with $n\in \Bbb N$ such that $\ B_{n+1}\subset B_n$ for all n.
I am thinking on takin $B´_n=B_1\cap B_2....\cap B_n$. Am I on a good road? How do I prove that both basis generate the same filter?

You're quite right: define $C_n = \bigcap_{i=1}^n B_i$ which are all in $\mathcal{F}$ too (as filters are closed under finite intersections) and is clealry nested as $C_{n+1} \subseteq C_n$ for all $n$.
If $A \in \mathcal{F}$ then some $B_k \subseteq A$ (definition of the $\{B_n: n \in \mathbb{N}\}$ being a base) and then $C_k \subseteq B_k$ too. So the $\{C_n: n \in \mathbb{N}\}$ also forms a base for $\mathcal{F}$. 
