Prove the following statement about Cauchy sequences...

Question

...without using the fact that a Cauchy sequence of real numbers converges:

If a subsequence of a Cauchy sequence converges, then the Cauchy sequence converges.

Answer 1

Let $(x_{n})$ be Cauchy; let $(x_{n_{k}})$ be a subsequence of $(x_{n})$ that converges to some $l \in \mathbb{R}$; if $n,k \geq 1$, then $|x_{n} - l| \leq |x_{n}-x_{n_{k}}| + |x_{n_{k}}-l|$. Let $\varepsilon > 0$; then by Cauchy assumption there is some $N_{1} \geq 1$ such that $|x_{n} - x_{n_{k}}| < \varepsilon/2$ for all $k \geq N_{1}$ and all $n,n_{k} \geq N_{1}$, and by convergence assumption there is some $N_{2} \geq 1$ such that $|x_{n_{k}} - l| < \varepsilon/2$ for all $k \geq N_{2}$; so for all $n \geq \max \{N_{1}, n_{N_{2}} \}$ we have $|x_{n}-l| < \varepsilon$.