# Is the metric ${d(x,y)}\over {1+d(x,y)}$ complete where $d$ is the usual Euclidean metric on $\mathbb R^{2}$

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.

• You have $d'=\dfrac1{1+\frac1d}=1-\dfrac1{1+d}$, not sure if that helps. – 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.
• 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 $d$. – user118494 Sep 4 '15 at 16:02
• 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. – Sean Clark Sep 4 '15 at 16:06
• 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 $e(x,y)=| \tan x - \tan y|$. – DanielWainfleet Sep 4 '15 at 17:26