Convergent Sequence is Cauchy Sequence

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

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


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:

\(\displaystyle d \left({x_n, x_m}\right)\) \(\le\) \(\displaystyle d \left({x_n, l}\right) + d \left({l, x_m}\right)\)          (by the 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.


Also see