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

Theorem

Let $\sequence {x_n}$ be a sequence in $\R$.

Let $\sequence {x_n}$ be convergent.

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

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:

$\ds \size {x_n - x_m}$

$=$

$\ds \size {x_n - l + l - x_m}$

$\ds $

$\le$

$\ds \size {x_n - l} + \size {l - x_m}$

$\ds $

$<$

$\ds \frac \epsilon 2 + \frac \epsilon 2$

by choice of $N$

$\ds $

$=$

$\ds \epsilon$

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

$\blacksquare$