# Convergent Sequence is Cauchy Sequence

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## Theorem

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

Every convergent sequence in $M$ 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:

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

In the same way:

$\forall m > N: d \left({x_m, l}\right) < \dfrac \epsilon 2$

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

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

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

$\blacksquare$