Proving two particular metrics have same topology.

Question

For the metrics $d:\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined as $d_{E}(x,y)=|x-y|$ and $$d(x,y)= \frac{|x-y|}{1+|x-y|} . $$ I want to prove this two metrics generate te same topology. By doing a little research I have two prove the following two things:

(1) For every $x \in \mathbb{R}$ and each $\epsilon > 0$ there is some $\delta > 0$ such we have the contention $B_{d}(x, \delta) \subseteq B_{d_{E}}(x, \epsilon)$.

(2) For every $x \in \mathbb{R}$ and each $\epsilon>0$ there is some $\delta >0$ such we have the contention $B_{d_{E}}(x, \delta) \subseteq B_{d}(x, \epsilon)$.

What I got so far, goes as follows: As

$$d(x,y)=\frac{|x-y|}{1+|x-y|} \leq |x-y|=d_{E}(x,y)$$.

So in order to prove (2), I take $x \in \mathbb{R}$, $\epsilon>0$ We have by the last inequality thath $d_{E}(x,y)< \epsilon$ implies $d(x,y)< \epsilon$. This way, choosing $\delta= \epsilon$ we got that if $y \in B_{d_{E}}(x, \delta)=B_{d_{E}}(x, \epsilon)$ this means that $d_{E}(x,y)< \delta=\epsilon$, then $d(x,y) < \epsilon$. Proving that $y \in B_{d}(x, \epsilon)$. This is $B_{d_{E}}(x, \delta) \subseteq B_{d}(x, \epsilon)$.

Is this part of my proof correct? Can anyone help me proving the first condition in order two show this two metrics generate same topology? I know this question has been asked and answered before but I didnt understand what they did so Im trying to formulate my own proof. Thanks

Answer 1

Your answer to 2) is correct. For 1) suppose $0<\epsilon<1$. Verify that $B_d(x,\delta) \subset B_E (x, \epsilon)$ if $0<\delta <\frac {\epsilon} {1+\epsilon}$. How did I arrive at this condition on $\delta$?: we want to choose $\delta$ such that $\frac {|y-x|} {1+|y-x|}<\delta$ implies $|y-x| <\epsilon$. In other words, we want $ {|y-x|}<(1+|y-x|)<\delta$ implies $|y-x| <\epsilon$ or $|y-x| < \frac {\delta} {1-\delta}$ implies $|y-x| <\epsilon$ . The obvious thing to do is to choose $\delta$ in such a way that $\frac {\delta} {1-\delta} <\epsilon$. Now you can rewrite this last inequality as $\delta < \frac {\epsilon} {1+\epsilon}$.

Finally, what happens if we started with $\epsilon \geq 1$? Note that $B_E(x,\epsilon)$ contains $B_E(x,\frac 1 2)$. If we find $\delta$ such that $B_d(x,\delta) \subset B_E(x,\frac 1 2)$ we would get $B_d(x,\delta) \subset B_E(x,\epsilon)$. But this has already been done (jusr replace $\epsilon$ by $\frac 1 2$ in above argument).

Answer 2

The new distance is a (strictly increasing) function of the old distance. Apply that function to find the new radius of a ball from the old topology, or the inverse of that function to find the old radius of a ball from the new topology. Done that way, we don't even need inclusions both ways - we can have the same sets, and just look at different radii for the balls based on the two distances.

Your inequality, that the new distance is smaller than the old distance, does work to show that the new balls are larger than the old balls of the same radius. To go the other way, you'll need an inequality going the other direction, bounding the new distance below by something related to the old distance.