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.

  • 3
    $\begingroup$ "if a sequence is not Cauchy can it converge?" you literally just proved if it converges it is Cauchy.... $\endgroup$ – mathworker21 Jan 29 '17 at 23:17
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    $\begingroup$ Very good proof. Indeed, if a sequence is convergent, then it is Cauchy (it can't be not Cauchy, you have just proved that!). However, the converse is not true: A space where all Cauchy sequences are convergent, is called a complete space. In complete spaces, Cauchy property is equivalent to convergence. However (for example Riemann integrable functions on $\mathbb R$, say) incomplete spaces certainly do exist, but there is an abstract notion of "completion" of a space, which is used time and again to unify these concepts. $\endgroup$ – астон вілла олоф мэллбэрг Jan 29 '17 at 23:18
  • $\begingroup$ Yes haha sorry, what I meant to ask was if a sequence is Cauchy does it always converge, which астон вілла олоф мэллбэрг explained nicely! $\endgroup$ – student_t Jan 29 '17 at 23:24
  • $\begingroup$ It is true that a Cauchy sequence in $\Bbb R$ converges to a member of $\Bbb R.$ This is a consequence of the definition of $\Bbb R.$ A Cauchy sequence in $\Bbb Q$ will converge in $\Bbb R$ but it may or may not converge to a member of $\Bbb Q$. So whether a sequence is said to converge depends on the $space$ (e.g. $\Bbb Q$ or $\Bbb R$ ) that is under consideration. $\endgroup$ – DanielWainfleet Dec 17 '17 at 5:53
  • $\begingroup$ the title is opposed to your question $\endgroup$ – Guy Fsone Jan 16 '18 at 12:51

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 $(0, \infty)$, $d(x, y) = |x − y|$.

  • $\begingroup$ I guess I should have clarified. This is for metric spaces. $\endgroup$ – student_t Dec 16 '17 at 23:00
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    $\begingroup$ It's not true for all metric spaces. $\mathbb Q$ with the normal metric is a metric space where it is not true. $\endgroup$ – fleablood Feb 20 '18 at 17:04
  • $\begingroup$ Or try $\Bbb R-\{0\}$. $\endgroup$ – Ted Shifrin Nov 7 '18 at 18:06

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