Cauchy Sequence Converges on Real Number Line

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.

By Convergent Subsequences of Cauchy Sequences, 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 - N| \le |a_N| + 1$

for all $m \ge N$.

So the sequence is bounded as claimed.

By the Bolzano-Weierstrass Theorem (if it has been proven directly with the Continuum Property and not exhaustion, which would make this proof circular) , $\left \langle {a_n} \right \rangle$ has a convergent subsequence.

Hence the result.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 5.19$