Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition/Proof 4

Proof
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.

By Real Cauchy Sequence is Bounded, $\sequence {a_n}$ is bounded.

Thus $\sequence {a_n}$ is both bounded above and bounded below.

Let us create a monotone subsequence $\sequence {b_n}$ of $\sequence {a_n}$ using the following construction:

For each $m \in \N$, let $S_m$ denote the set of elements of $\sequence {a_n}$ from $m$ onwards:
 * $S_m = \set {a_n: n \ge m}$

From Real Cauchy Sequence is Bounded, we have that $S_m$ is also bounded.

Hence, by the Continuum Property, $\sup S_m$ exists.

Let $b_m := \sup S_m$.

Because $S_{m + 1} \subseteq S_m$, it follows from Supremum of Set of Real Numbers is at least Supremum of Subset that:
 * $\sup S_{m + 1} \le \sup S_m$

Thus $\sequence {b_m}$ is decreasing.

By definition of $b_m$, we also have that:
 * $b_m \ge a_m$

So, as $\sequence {a_n}$ is bounded below, so is $\sequence {b_m}$.

So, by the Monotone Convergence Theorem (Real Analysis), $\sequence {b_m}$ is convergent.

Let $\sequence {b_m}$ converge to $l$.

Let $\epsilon \in \R_{>0}$ be arbitrary.

We have that $\sequence {a_n}$ is a Cauchy sequence in $\R$.

We also have from Subsequence of Real Cauchy Sequence is Cauchy that $\sequence {b_n}$ is also a Cauchy sequence in $\R$.

Then there exists:
 * $N_1 \in \N$ such that $\size {a_m - a_n} < \epsilon$ for $m, n \ge N_1$
 * $N_2 \in \N$ such that $\size {l - b_n} < \epsilon$ for $m \ge N_2$

Let $N = \max \set {N_1, N_2}$.

By definition of $b_N$, we have that $b_N - \epsilon$ is not an upper bound of $S_N = \set {a_n: n \ge N}$.

Thus there exists $M \ge N$ such that $a_M > b_N - \epsilon$.

We note also that $a_M \le b_N$ because $b_N$ is an upper bound for $S_N$.

Now let $n \ge N$.

We have that:

Hence by Convergence by Multiple of Error Term we have that $\sequence {a_n}$ converges to $l$.