0
$\begingroup$

I have a function $\rho(x,y) = |x^{1/3} - y^{1/3}|$ and I need to prove if the function is metric, and if it is, the next step is to prove if this metric space is complete.

So metric is a function with following conditions:

  1. $\rho(x,y) \iff y=x$
  2. $\rho(x,y) = \rho(y,x)$ because of absolute value's properties
  3. $\rho(x,z) \leq \rho(x,y) + \rho(y,z)$ — I can't prove it myself

But I know that the function is metric indeed and how do I prove that this metric space is complete?

$\endgroup$
1
$\begingroup$

I will assume that your space is $\mathbb R$. Let $d$ be the usual metric on $\mathbb R$. Then$$\begin{array}{ccc}(\mathbb R,\rho)&\longrightarrow&(\mathbb R,d)\\x&\mapsto&\sqrt[3]x\end{array}$$is an isometry. So, since $(\mathbb R,d)$ is complete, so is your space.

$\endgroup$
  • $\begingroup$ This is an isometric homeomorphism, that's why $(\mathbb{R},\rho)$ is complete. $\endgroup$ – Rodrigo Dias May 10 at 12:50
  • $\begingroup$ And there's no need to do anything with Cauchy sequences at all? $\endgroup$ – Alexander May 10 at 12:51
  • $\begingroup$ @RodrigoDias I've edited my answer. Thank you a lot. $\endgroup$ – José Carlos Santos May 10 at 13:02
  • $\begingroup$ @Alexander Only if you really want to. $\endgroup$ – José Carlos Santos May 10 at 13:02

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.