Proving convergent sequences are Cauchy sequences

Question

Prove that if $x_n \rightarrow a, n \rightarrow \infty$ then $\{x_n\}$ is a Cauchy sequence.

I believe I have found the proof as follows, wondering if there are any simpler methods or added intuition. For me, it makes sense that if a sequence has a limit, then distances between elements in the sequence must be getting smaller, in order for it to converge.

Given $\epsilon > 0, \exists N_1 \ s.t. \ \forall n \geq N_1:$

$|x_n - a| < \frac{\epsilon}{2} < \epsilon$

and for $m > n \geq N_1$ we also have:

$|x_m -a| < \frac{\epsilon}{2} < \epsilon$

Let $N \geq N_1$, then $\forall n,m \geq N$ we have:

$|x_n-x_m| = |x_n - a -x_m+a| < |x_n - a| + |-(x_m -a)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

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

Also, if a sequence is Cauchy does it always converge? In other words, is it sufficient to check if a sequence is Cauchy to check for convergence.

Answer 1

Your proof is correct.

Secondly, the property of having every Cauchy sequence converge is very important, and is known as completeness.

As an example, $\mathbb R $ with the usual metric is complete.

Another important area of study is Banach spaces, which roughly are complete metric spaces where the metric comes from a norm.

More generally, there are Hilbert spaces, which are equipped with an inner product.

Answer 2

With response to the comments above, it is not always true that a Cauchy sequence is convergent. For example, if we think of metric spaces, yes it is true that any Cauchy sequence in $(R, \mathrm{d}E)$ is convergent and any Cauchy sequence in $(X, \mathrm{d}\delta)$ is eventually constant, so convergent. We have the counter example however: Cauchy does not imply convergent in $(0, \infty)$, $d(x, y) = |x − y|$.