Convergent Sequence is Cauchy Sequence

From ProofWiki
Jump to: navigation, search

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:

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

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

$\blacksquare$


Also see


Sources