0

Show that if $d$ is a metric for $X$ , then $d′(x,y)=d(x,y)/(1+d(x,y))$ is a bounded metric that gives the topology of $X$.

I rephrase this exercise to my taste:

Show $\mathcal{T}_d =\mathcal{T}_{d’}$

My attempt: It is easy to check $d’(x,y)\leq d(x,y)$ inequality. $d’(x,y)\leq d(x,y)$ and lemma 20.2, shows that $\mathcal{T}_{d’}\subseteq \mathcal{T}_d$. Now how to show $\mathcal{T}_{d’}\supseteq \mathcal{T}_d$? Can we prove similar inequality $d(x,y)\leq C d’(x,y)$, where $C$ is some constant?

user264745
  • 4,143
  • In that proof/post, before using $d(x,y)\leq 2d’(x,y)$ how to show $d(x,y)\leq 1$? Because $x=d(x,y)$. More explicitly let $\epsilon \gt 0$, take $\delta =\delta^{\prime}/2,$ where $\delta^{\prime}=\min { 1,\epsilon /2 }$ so that $B_{d’}(x,\delta)\subseteq B_d (x,\epsilon)$. Proof: $y\in B_{d’}(x,\delta)$. Then $d’(x,y)\lt \delta$……… – user264745 Jan 31 '22 at 18:40
  • If the radius of the ball is $\epsilon>1$, you may "shrink" the ball to a radius of $1$ before proceeding further. The inclusions will still hold. I.e. $B_{d’}(x,\delta)\subseteq B_d (x,\min{\epsilon,1}) \subseteq B_d (x,\epsilon)$. Does this make sense? – Momo Jan 31 '22 at 22:43
  • 1
    You cannot prove similar inequality, it is clear if $d$ is unbounded. – Arctic Char Jan 31 '22 at 23:05
  • @ArcticChar make sense. – user264745 Feb 01 '22 at 07:30
  • @Momo kind of. Not completely. As I said in edit, there’s not a unique way to solve this problem. In this post we can discuss about different(from your approach) way to solve this problem. If you see my past posts, you’ll find out how much I like alternative solutions to the same problem. – user264745 Feb 01 '22 at 07:41
  • I wouldn't go that far to call Lemma 20.2 "a different approach", it is essentially the same thing. – Momo Feb 01 '22 at 12:24
  • @Momo that’s means you don’t that proof, do you? – user264745 Feb 01 '22 at 12:26
  • @Momo in that proof we don’t need any inequality. Relation between $d$ and $d’$ metric. For each $\epsilon$, take $\delta$……… taking $\delta$ is clever and we don’t use any relation between $d$ and $d’$ metric. – user264745 Feb 01 '22 at 12:28
  • To apply the lemma to your problem you still need to find $\delta$ so that $B_{d’}(x,\delta)\subseteq B_d (x,\epsilon)$, so you still need some sort of inequality like the one I used. – Momo Feb 01 '22 at 12:33
  • @Momo understood. I understand what you mean to say, and that is somewhat true. In proof of $\mathcal{T}{d’}\supseteq \mathcal{T}_d$. For each $\epsilon$, take $\delta =\frac {\epsilon}{\epsilon +1}$, claim:$ B{d’}(x,\delta )\subseteq B_d (x,\epsilon)$. From this fact, form inequality. At least I wasn’t able to form any inequality. – user264745 Feb 01 '22 at 12:43
  • @Momo Can you also please reply to my last comment in the post, you referred as duplicate? https://math.stackexchange.com/a/2022105/861687 – user264745 Feb 01 '22 at 12:47

0 Answers0