Convergent Sequence is Cauchy Sequence/Metric Space

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Theorem

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


Every convergent sequence in $M$ 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 $l$, we have:

$\exists N: \forall n > N: \map d {x_n, l} < \dfrac \epsilon 2$

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

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

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

$\blacksquare$


Also see


Sources