Convergent Sequence in Metric Space has Unique Limit

From ProofWiki
Jump to navigation Jump to search

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

Let $\epsilon > 0$.


Then, provided $n$ is sufficiently large:

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

So $0 \le \dfrac {d \left({l, m}\right)} 2 < \epsilon$.

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

Thus from Real Plus Epsilon it follows that $\dfrac {d \left({l, m}\right)} 2 = 0$, 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 Limit‎.

$\blacksquare$