Convergent Sequence in Metric Space has Unique Limit/Proof 1

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 $l = m$.

$\blacksquare$