Convergent Sequence in Hausdorff Space has Unique Limit

Theorem
Let $T = \struct {S, \tau}$ be a Hausdorff space.

Let $\sequence {x_n}$ be a convergent sequence in $T$.

Then $\sequence {x_n}$ has exactly one limit.

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

$\ds \lim_{n \mathop \to \infty} x_n = l$ and $\ds \lim_{n \mathop \to \infty} x_n = m$ 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 $U \cap V = \O$

Then, from the definition of convergent sequence:

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

But $U \cap V = \O$.

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

Hence the result.