Convergent Sequence in Hausdorff Space has Unique Limit

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

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

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

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

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

Let $$\epsilon > 0$$.

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

Then, from the definition of convergence:


 * $$\exists N_U: n > N_U \Longrightarrow x_n \in U$$;
 * $$\exists N_V: n > N_V \Longrightarrow x_n \in V$$.

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

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.