Sequence in Metric Space has One Limit at Most

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

Let $\epsilon > 0$.

Then, provided $n$ is sufficiently large:

	\(\displaystyle \)	\(\displaystyle d \left({l, m}\right)\)	\(\le\)	\(\displaystyle d \left({l, x_n}\right) + d \left({x_n, m}\right)\)	\(\displaystyle \)	Triangle Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle \epsilon + \epsilon\)	\(\displaystyle \)	by the definition of the limit
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 2 \epsilon\)	\(\displaystyle \)

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

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

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

$\blacksquare$