Let $d(x,y)$ be the usual Euclidean metric on $\mathbb R^{2}.$ $\mathbb R^{2}$ is complete under $d(x,y)$. I have this subspace given $$[0,1]\times [0,\infty )\ \ of\ \ \mathbb R^{2}.$$ I thought this is also complete under $d$ for I could not think of any sequence that is not convergent in this space. Correct me if I am wrong. Now the metric $$d'(x,y)={{d(x,y)}\over {1+d(x,y)}}$$ on the subspace $[0,1]\times [0,\infty )$. Is this complete $?$

I was thinking if I could prove that $d'(x,y)$ and $d(x,y)$ are equivalent then completeness would be readily proved. Am I thinking right $?$ Need help to further the proof.

Thanks for any help.

  • $\begingroup$ You have $d'=\dfrac1{1+\frac1d}=1-\dfrac1{1+d}$, not sure if that helps. $\endgroup$ – Akiva Weinberger Sep 4 '15 at 15:47

Note that $d'(x,y)\leq d(x,y)$ for any $x,y$ in your space, and $d(x,y)\leq 2d'(x,y)$ if $d(x,y)\leq 1$, so a sequence is Cauchy with respect to one metric if and only if it is Cauchy in the other metric.

| cite | improve this answer | |
  • $\begingroup$ Thanks. Another thing here. For the space $[0,1]\times [0,\infty )$ I did not find counter example but how to prove for sure that this space is complete under $d$. $\endgroup$ – user118494 Sep 4 '15 at 16:02
  • 2
    $\begingroup$ Well, as you said $\mathbb R^2$ is complete under $d$, and any Cauchy sequence in the subspace $[0,1]\times [0,\infty)$ (under $d$) is also Cauchy as a sequence in $\mathbb R^2$, so has a limit in $\mathbb R^2$. The last thing you need is that $[0,1]\times [0,\infty)$ is a closed subspace of $\mathbb R^2$, so any limit point of a sequence of points in the subspace lies in the subspace. $\endgroup$ – Sean Clark Sep 4 '15 at 16:06
  • $\begingroup$ Equivalence of metrics means that they generate the same topology. Completeness of a metric does not automatically imply completeness of an equivalent one. For example on the open interval $(0,\pi /2)$ let $d(x,y)= |x-y|$ and $e(x,y)=| \tan x - \tan y|$. $\endgroup$ – DanielWainfleet Sep 4 '15 at 17:26

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.