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

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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:

\(\ds \size {x_n - x_m}\) \(=\) \(\ds \size {x_n - l + l - x_m}\)
\(\ds \) \(\le\) \(\ds \size {x_n - l} + \size {l - x_m}\) Triangle Inequality
\(\ds \) \(<\) \(\ds \frac \epsilon 2 + \frac \epsilon 2\) by choice of $N$
\(\ds \) \(=\) \(\ds \epsilon\)

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