Cauchy sequences

Suppose $\{ x_n \}_{n=1}^\infty$ is Cauchy. Pick an $M$ such that for all $n,k \geq M\text{,}$ we have $\sabs{x_n-x_k} < 1\text{.}$ In particular, for all $n \geq M\text{,}$

By the reverse triangle inequality, $\sabs{x_n} - \sabs{x_M} \leq \sabs{x_n - x_M} < 1\text{.}$ Hence, for $n \geq M\text{,}$

Let

Then $\sabs{x_n} \leq B$ for all $n \in \N\text{.}$

Suppose $\{ x_n \}_{n=1}^\infty$ converges to $x\text{,}$ and let $\epsilon > 0$ be given. Then there exists an $M$ such that for $n \geq M\text{,}$

Hence, for $n \geq M$ and $k \geq M\text{,}$

Alright, that direction was easy. Now suppose $\{ x_n \}_{n=1}^\infty$ is Cauchy. We have shown that $\{ x_n \}_{n=1}^\infty$ is bounded. For a bounded sequence, liminf and limsup exist, and this is where we use the least-upper-bound property. If we show that

then $\{ x_n \}_{n=1}^\infty$ must be convergent by Proposition 2.3.5.

Define $a \coloneqq \limsup_{n\to\infty} x_n$ and $b \coloneqq \liminf_{n\to\infty} x_n\text{.}$ By Theorem 2.3.4, there exist subsequences $\{ x_{n_i} \}_{i=1}^\infty$ and $\{ x_{m_i} \}_{i=1}^\infty$ such that

Given an $\epsilon > 0\text{,}$ there exists an $M_1$ such that $\sabs{x_{n_i} - a} < \nicefrac{\epsilon}{3}$ for all $i \geq M_1$ and an $M_2$ such that $\sabs{x_{m_i} - b} < \nicefrac{\epsilon}{3}$ for all $i \geq M_2\text{.}$ There also exists an $M_3$ such that $\sabs{x_n-x_k} < \nicefrac{\epsilon}{3}$ for all $n,k \geq M_3\text{.}$ Let $M \coloneqq \max \{ M_1, M_2, M_3 \}\text{.}$ If $i \geq M\text{,}$ then $n_i \geq M$ and $m_i \geq M\text{.}$ Hence,

As $\sabs{a-b} < \epsilon$ for all $\epsilon > 0\text{,}$ we have $a=b$ and the sequence converges.