3

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?

Ben Grossmann
  • 225,327
Junior
  • 33

2 Answers2

8

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.

Pedro
  • 122,002
2

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.

Ittay Weiss
  • 79,840
  • 7
  • 141
  • 236
  • @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 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". $\hspace{.25 in}$ –  Oct 01 '13 at 04:23
  • @RickyDemer $x\mapsto x^2$ is the simplest counterexample I can think of now. – Pedro Oct 01 '13 at 04:36
  • @RickyDemer To redeem myself. Call two sequences cofinal if $\rho(x_n,y_n)\to 0$. Then $f$ is uniformly continuous $\iff$ it maps cofinal sequences to cofinal sequences. – Pedro Oct 01 '13 at 04:40