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

Proof
Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$ that converges to the limit $l \in \R$.

Let $\epsilon > 0$.

Then also $\dfrac \epsilon 2 > 0$.

Because $\left \langle {x_n} \right \rangle$ converges to $l$, we have:
 * $\exists N: \forall n > N: \left \lvert{x_n-l}\right \rvert < \dfrac \epsilon 2$

So if $m > N$ and $n > N$, then:

Thus $\left \langle {x_n} \right \rangle$ is a Cauchy sequence.