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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?

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up vote 6 down vote accepted

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.

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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.

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

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