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.

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:

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

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

$\blacksquare$

Also see

A metric space in which the converse also holds is called complete.

An example of a complete metric space is given by the real number line.

Thus, every Cauchy sequence in $\R$ is convergent.

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$: Complete Metric Spaces
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 5.17$