Convergent Sequence in Metric Space has Unique Limit

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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 1

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 $l = m$.


Proof 2

We have that a Metric Space is Hausdorff.

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