# Is Hausdorffness characterisable by the uniqueness of the limits?

It is clear that if the points $x$ and $y$ are separated by neighbourhoods, then there is no sequence which converges to $x$ and to $y$ as well. But when I try to prove that if $x$ and $y$ aren't separated by neighbourhoods I have to assume that both $x$ and $y$ have countable local base in the following way: let be $U_1\supset U_2\supset U_3\supset\ldots$ a countable local base of $x$ and $V_1\supset V_2\supset V_3\supset\ldots$ a countable local base of $y$ then by the assumption $U_i\cap V_i$ isn't empty for any $i$, so according to the axiom of choice there is a sequence $t_1,t_2,\ldots$ that $t_i \in U_i \cap V_i$ that is $t_i$ converges both to $x$ and $y$. So my question is, how could one eliminate the assumption of first-countability or how can one construct a topological space in which the uniqueness of limits holds but isn't Hausdorff.

• See this. – David Mitra May 14 '15 at 12:27
• @DavidMitra Thank you. My question is answered there. – Leonhardt von M May 14 '15 at 12:30