Cauchy's Convergence Criterion/Real Numbers/Necessary Condition
Theorem
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.
$\blacksquare$
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.
$\blacksquare$
Also known as
Cauchy's Convergence Criterion is also known as the Cauchy convergence condition.