Convergent Sequence in Hausdorff Space has Unique Limit

Theorem
Let $T = \left({S, \tau}\right)$ be a Hausdorff space.

Let $\left \langle {x_n} \right \rangle$ be a convergent sequence in $T$.

Then $\left \langle {x_n} \right \rangle$ has exactly one limit.

Proof
From the definition of convergent sequence, we have that $\left \langle {x_n} \right \rangle$ converges to at least one limit.

Suppose $\displaystyle \lim_{n \to \infty} x_n = l$ and $\displaystyle \lim_{n \to \infty} x_n = m$ such that $l \ne m$.

As $T$ is Hausdorff, $\exists U \in \tau: l \in U$ and $\exists V \in \tau: m \in V$ such that $U \cap V = \varnothing$.

Then, from the definition of convergent sequence:

Taking $N = \max \left\{{N_U, N_V}\right\}$ we then have:
 * $\exists N \in \R: n > N \implies x_n \in U, x_n \in V$

But $U \cap V = \varnothing$.

From that contradiction we can see that there can be no such two distinct $l$ and $m$.

Hence the result.

Souces

 * : $4.2$: Separation axioms: Proposition $4.2.2$