Question: Prove that if any convergent filter on a space X converges to a unique limit, then X is Hausdorff. I think the solution in my textbook is faulty. It says "Suppose X is not Hausdorff. Let Nx and Ny be the collection of all neighbourhoods of x and y, respectively. Then Nx U Ny generates a filter that converges to both x and y, a contradiction." The proof likes right except we must verify that Nx U Ny is actually a filter base, i.e. the intersection of two members of Nx U Ny must contain a member of Nx U Ny. If A belongs to Nx and B belongs to Ny, then indeed AnB is nonempty since by assumption X is not Hausdorff. But in order for AnB to contain a member of Nx U Ny, it must contain either x or y. But if neither x nor y belongs to AnB, where is the contradiction? What am I missing? I stated that if neither x nor y belongs to AnB, then int(A-intB) and int(B-intA) are disjoint open sets containing x and y, respectively, contradicting the assumption that X is not Hausdorff--but is that really so? What if they exist on the boundary of AnB? By the way, I looked up the author's definition of neighbourhood of a point p. He defines it to be a set whose interior contains p. Ok, so a neighbourhood, by his defintion, does not have to be open, but I still don't see how that ensures that either x or y belongs to AnB.