Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition/Proof 2

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

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

By the Bolzano-Weierstrass Theorem, $\sequence {a_n}$ has a convergent subsequence $\sequence {a_{n_r} }$.

Let $a_{n_r} \to l$ as $r \to \infty$.

It is to be shown that $a_n \to l$ as $n \to \infty$.

Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.

Then $\dfrac \epsilon 2 > 0$.

Hence:
 * $(1): \quad \exists R \in \R: \forall r > R: \size {a_{n_r} - 1} < \dfrac \epsilon 2$

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

Hence:
 * $(2): \quad \exists N \in \R: \forall m > N, n > N: \size {x_m - x_n} \le \dfrac \epsilon 2$

Let $n > N$.

Let $r \in \N$ be sufficiently large that:
 * $n_r > N$

and:
 * $r > R$

Then $(1)$ is satisfied, and $(2)$ is satisfied with $m = n_r$.

So:

So, given $\epsilon > 0$, we have found $n \in \R$ such that:
 * $\forall n > N: \size {a_n - l} < \epsilon$

Thus:
 * $x_n \to l$ as $n \to \infty$.