# Convergent Sequence in Metric Space has Unique Limit

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## 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 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 $l = m$.

$\blacksquare$

## Proof 2

We have that a Metric Space is Hausdorff.

The result then follows from Convergent Sequence in Hausdorff Space has Unique Limitâ€Ž.

$\blacksquare$