Hausdorff spaces and convergence How can I construct a filterbase converging to two points in in a non-Hausdorff space?
 A: HINT: First construct a sequence converging to two different points in some non-Hausdorff space. That sequence is a net. Build a filter base from it the way I did in this answer to one of your earlier questions.
One handy non-Hausdorff space is any infinite set (e.g., $\Bbb N$) with the cofinite topology: every non-trivial sequence then converges to every point of the space!
A: Let $X$ be a non-Hausdorff space. This means that there exist 2 points $p$ and $q$ such that for all neighbourhoods $U$ of $x$ and for all neighbourhoods $V$ of $y$ we have $U \cap V \neq \emptyset$. This is just the logical negation of the Hausdorff condition.
Now define $\mathcal{B} = \left\{U \cap V: U \mbox{ a neighbourhood of } p, V \mbox{ a neighbourhood of } q \right\}$. By the choice of $p$ and $q$, all sets in $\mathcal{B}$ are non-empty, and it's easy to check it is indeed a filterbase (as neighbourhood systems are themselves filterbases), and as we can choose $U = X$ or $V = X$ in particular, e.g., it's clear that $\mathcal{B}$ contains the complete neighbourhood system of both $p$ and $q$, which just means, by definition, that $\mathcal{B} \rightarrow p$ and $\mathcal{B} \rightarrow q$, as required.
