Convergent Sequence in Metric Space has Unique Limit

Theorem
Let $$\left({X, d}\right)$$ be a metric space.

Let $$\left \langle {x_n} \right \rangle$$ be a sequence in $$\left({X, d}\right)$$.

Then $$\left \langle {x_n} \right \rangle$$ can have at most one limit.

Proof
Suppose $$\lim_{n \to \infty} x_n = l$$ and $$\lim_{n \to \infty} x_n = m$$.

Let $$\epsilon > 0$$.

Then, provided $$n$$ is sufficiently large:

$$ $$ $$

So $$0 \le \frac {d \left({l, m}\right)} 2 < \epsilon$$.

This holds for any value of $$\epsilon > 0$$.

Thus from Real Plus Epsilon it follows that $$\frac {d \left({l, m}\right)} 2 = 0$$, that is, that $$l = m$$.

Alternative Proof
From the fact that a Metric Space is Hausdorff, we can directly use Convergent Sequence in Hausdorff Space has Unique Limit‎.