Convergent Sequence in Metric Space has Unique Limit

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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 1

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

Let $\epsilon > 0$.

Then, provided $n$ is sufficiently large:

\(\displaystyle \map d {l, m}\) \(\le\) \(\displaystyle \map d {l, x_n} + \map d {x_n, m}\) Triangle Inequality
\(\displaystyle \) \(<\) \(\displaystyle \epsilon + \epsilon\) Definition of Limit of Sequence (Metric Space)
\(\displaystyle \) \(=\) \(\displaystyle 2 \epsilon\)

So $0 \le \dfrac {\map d {l, m} } 2 < \epsilon$.

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

Thus from Real Plus Epsilon it follows that $\dfrac {\map d {l, m} } 2 = 0$, that is, that $l = m$.


Proof 2

We have that a Metric Space is Hausdorff.

The result then follows from Convergent Sequence in Hausdorff Space has Unique Limit‎.