Exercise 2.5.1 (Limits in a Hausdorff space)

Proof. Suppose for the sake of contradiction that a sequence ${\{x_n\}}_{n=1}^{\infty }$ in a Hausdorff space $X$ converges to both $x\in X$ and $y\in X$ such that $x\ne y$. Since $X$ is $T_2$, we can find two separating distinct open sets $U,V\subseteq X$ such that $x\in U$, $y\in V$ and $U\cap V=\varnothing$. By definition of convergence, we can find a $N_U$ large enough such that for all $n\ge N$ we have $x_n\in U$. Similarly, there is a $N_V$ large enough such that for all $n\ge N$ we have $x_n\in V$. Setting $N:=\max <!--\; FUNCTION\; APPLICATION\; -->\{N_U,N_V\}$, we see that $x_n$ is eventually contained in both $U$ and $V$ - a contradiction to the fact that both sets are disjoint. □