Convergent Sequence is Cauchy Sequence/Metric Space

Theorem

Let $M = \struct {A, d}$ be a metric space.

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

Proof

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

Let $\epsilon > 0$.

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

Because $\sequence {x_n}$ converges to $l$ in $A$, we have:

$\exists N_1 \in \R_{>0}: \forall n > N_1: \map d {x_n, l} < \dfrac \epsilon 2$

Because $\sequence {x_n}$ converges to $l$, there must also exist at least one sequence $\sequence {x_m}$ in $A$ that converges to the limit $l \in A$, for some $m \in \R_{>0}$.

We have:

$\exists N_2 \in \R_{>0}: \forall m > N_2: \map d {x_m, l} < \dfrac \epsilon 2$

So if $N = \map \max {N_1, N_2}$, such that $m > N$ and $n > N$, then:

$\ds \map d {x_n, x_m}$

$\le$

$\ds \map d {x_n, l} + \map d {l, x_m}$

Metric Space Axiom $(\text M 2)$: Triangle Inequality

$\ds $

$=$

$\ds \map d {x_n, l} + \map d {x_m, l}$

Metric Space Axiom $(\text M 3)$

$\ds $

$<$

$\ds \frac \epsilon 2 + \frac \epsilon 2$

by choice of $N$

$\ds $

$=$

$\ds \epsilon$

Thus $\sequence {x_n}$ is a Cauchy sequence.

$\blacksquare$

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Also see

Definition:Complete Metric Space, where the converse is true.

Sources

1953: Walter Rudin: Principles of Mathematical Analysis ... (next): $3.11a$
1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $7.2$: Sequential compactness: Proposition $7.2.4$
1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $4$: Complete Normed Spaces