# Convergent Sequence in Metric Space has Unique Limit/Proof 1

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## Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $\sequence {x_n}$ be a sequence in $M$.

Then $\sequence {x_n}$ can have at most one limit in $M$.

## Proof

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

Let $\epsilon > 0$.

Then, provided $n$ is sufficiently large:

\(\ds \map d {l, m}\) | \(\le\) | \(\ds \map d {l, x_n} + \map d {x_n, m}\) | Triangle Inequality | |||||||||||

\(\ds \) | \(<\) | \(\ds \epsilon + \epsilon\) | Definition of Limit of Sequence (Metric Space) | |||||||||||

\(\ds \) | \(=\) | \(\ds 2 \epsilon\) |

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 $l = m$.

$\blacksquare$