# Convergent Sequence is Cauchy Sequence

## Theorem

Let $M = \left({A, d}\right)$ be a metric space.

Every convergent sequence in $M$ is a Cauchy sequence.

### Normed Division Ring

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring.

Every convergent sequence in $R$ is a Cauchy sequence.

### Normed Vector Space

Let $\struct{X, \norm{\,\cdot\,} }$ be a normed vector space.

Every convergent sequence in $X$ is a Cauchy sequence.

## Proof

Let $\left \langle {x_n} \right \rangle$ be a sequence in $A$ that converges to the limit $l \in A$.

Let $\epsilon > 0$.

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

Because $\left \langle {x_n} \right \rangle$ converges to $l$, we have:

$\exists N: \forall n > N: d \left({x_n, l}\right) < \dfrac \epsilon 2$

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

 $\displaystyle d \left({x_n, x_m}\right)$ $\le$ $\displaystyle d \left({x_n, l}\right) + d \left({l, x_m}\right)$ Triangle Inequality $\displaystyle$ $<$ $\displaystyle \frac \epsilon 2 + \frac \epsilon 2$ (by choice of $N$) $\displaystyle$ $=$ $\displaystyle \epsilon$

Thus $\left \langle {x_n} \right \rangle$ is a Cauchy sequence.

$\blacksquare$