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

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

Then $\left \langle {a_n} \right \rangle$ is convergent.

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

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
 * $ |a_m - a_n| < 1 $

for all $m, n \ge N$.

In particular, by the Triangle Inequality:
 * $ |a_m| = |a_N + a_m - a_N| \le |a_N| + |a_m - a_N| \le |a_N| + 1$

for all $m \ge N$.

So the sequence is bounded as claimed.

By the Bolzano-Weierstrass Theorem: Proof 1, $\left \langle {a_n} \right \rangle$ has a convergent subsequence.

Hence the result.