Cauchy's Convergence Criterion/Real Numbers/Necessary Condition

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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.

Proof 1

Let $\sequence {x_n}$ be convergent.

Let $\struct {\R, d}$ be the metric space formed from $\R$ and the usual (Euclidean) metric:

$\map d {x_1, x_2} = \size {x_1 - x_2}$

where $\size x$ is the absolute value of $x$.

This is proven to be a metric space in Real Number Line is Metric Space.

From Convergent Sequence in Metric Space is Cauchy Sequence, we have that every convergent sequence in a metric space is a Cauchy sequence.

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


Proof 2

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.


Also known as

Cauchy's Convergence Criterion is also known as the Cauchy Convergence Condition.