Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition/Proof 1

Proof
Let $\left \langle {a_n} \right \rangle$ be a Cauchy sequence in $\R$.

We have the result Real Number Line is Metric Space.

Hence by Convergent Subsequence of Cauchy Sequence in Metric Space, it is sufficient to show that $\left \langle {a_n} \right \rangle$ has a convergent subsequence.

We observe that the fact that $\left \langle {a_n} \right \rangle$ is Cauchy implies that $\left \langle {a_n} \right \rangle$ is bounded, as follows.

There exists $N \in \N$ such that
 * $\left\vert{a_m - a_n}\right\vert < 1 $

for all $m, n \ge N$.

In particular, by the Triangle Inequality:
 * $\left\vert{a_m}\right\vert = \left\vert{a_N + a_m - a_N}\right\vert \le \left\vert{a_N}\right\vert + \left\vert{a_m - a_N}\right\vert \le \left\vert{a_N}\right\vert + 1$

for all $m \ge N$.

So the sequence is bounded as claimed.

By the Bolzano-Weierstrass Theorem, $\left \langle {a_n} \right \rangle$ has a convergent subsequence.

Hence the result.