In the completion of a metric space, a distance is defined on the set of equivalence classes of Cauchy sequences:

$$ \begin{align} \tilde d:\tilde X\times \tilde X &\to \mathbb{R^+}\\ ([x_n],[y_n]) &\mapsto \lim_{n\to \infty}(d(x_n,y_n)) \end{align}$$ with $x_n,y_n$ Cauchy sequences in the metric space $(X,d)$.

A detail troubles me. I can see that this is well-defined (w.r.t. various representatives of the equivalence classes), except for the fact that this limit needs not exist? What if $d(x_n,y_n)$ was periodic for instance. Is that clear that it can't be?


By definition, $$d(\bar x,\bar y)=\lim\limits_{n\to\infty}d(x_n,y_n)$$

Now, since $x_n,y_n$ are Cauchy, and $$|d(x_m,y_m)-d(x_n,y_n)|\leq d(x_n,x_m)+d(y_n,y_m)$$ $d_n:=d(x_n,y_n)$ is also Cauchy, but in $\Bbb R$; which is complete!

ADD The inequality

$$|d(x,y)-d(z,w)|\leq d(x,z)+d(y,w)$$ is known as the quadrilateral inequality.


Another argument, conceptually slightly different than Peter's, for $d(x_n,y_n)$ being Cauchy: The distance function $d:X\times X\to \mathbb R$ is uniformly continuous. $x_n$ and $y_n$ are Cauchy in $X$, thus $(x_n,y_n)$ is Cauchy in $X\times X$. A uniformly continuous function maps Cauchy sequences to Cauchy sequences. QED.

  • $\begingroup$ @PeterTamaroff absolutely! Both proofs are essentially identical, but conceptually slightly different. $\endgroup$ – Ittay Weiss Jul 25 '13 at 22:08
  • $\begingroup$ I see. Again, how would you prove that the function is uniformly continuous? $\endgroup$ – Pedro Tamaroff Jul 25 '13 at 22:34
  • $\begingroup$ Essentially, the same argument you give. I'm not claiming anything significantly different here, except for the packaging. $\endgroup$ – Ittay Weiss Jul 25 '13 at 23:35
  • $\begingroup$ @Pedro : $\:$ "it sends Cauchy sequences to Cauchy sequences" is weaker than "it is uniformly continuous". $\hspace{.25 in}$ $\endgroup$ – user57159 Oct 1 '13 at 4:23
  • $\begingroup$ @RickyDemer $x\mapsto x^2$ is the simplest counterexample I can think of now. $\endgroup$ – Pedro Tamaroff Oct 1 '13 at 4:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.