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

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

Let $\epsilon > 0$.

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

Because $\sequence {x_n}$ converges to $l$, we have:
 * $\exists N: \forall n > N: \size {x_n - l} < \dfrac \epsilon 2$

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

Thus $\sequence {x_n}$ is a Cauchy sequence.