# Same open balls

1. Aug 31, 2012

### bedi

1. The problem statement, all variables and given/known data

Let M be a metric space with metric $d$, and let $d_{1}$ be the metric defined below. Show that the two metric spaces $(M,d)$, $(M,d_{1})$ have the same open sets.

2. Relevant equations

$d_1:\frac{d(x,y)}{1+d(x,y)}$

3. The attempt at a solution

I tried to show that the neighborhoods around their elements are the same.

An open ball in $(M,d): B_{r}(x)=\{y \in R: |x-y|<r\}$
An open ball in $(M,d_{1}): B_{r}(x)=\{y \in R: \frac{|x-y|}{1+|x-y|}<r\}=\{y \in R: (|x-y|)(1-r)<r\}$, let $1-r=n$, hence $n(|x-y|)<r$.

I'm stuck.

2. Aug 31, 2012

### sammycaps

All the open sets in a metric space are arbitrary unions of open balls. So, if you can show that every open ball in (M,d) is the same as some open ball in (M,d1), and vice versa, then you're done.

I think what you could do is take some B(x,r) in (M,d) and find the corresponding B(x,r') in (M,d1), such that they have the same elements.

Last edited: Aug 31, 2012
3. Aug 31, 2012

### LCKurtz

In a metric space, there are generally no absolute values. You have a set $M$, which may not be the reals and two metrics on $M$. One is the metric given by $d(x,y)$ and the other, which I will call $D$ is given by $$D(x,y) = \frac{d(x,y)}{1+d(x,y)}$$Since metrics obey many of the same properties that the absolute value function does on the reals, it may help your intuition to work the problem for $M$ a subset of the reals, but in general, you don't have $x,y\in \mathbb R$ and $d(x,y)\ne |x-y|$.

Last edited: Aug 31, 2012
4. Aug 31, 2012

### bedi

Thank you for clarification, am I on the right way though?

5. Aug 31, 2012

### Bacle2

Can you show that the open balls in each of the metrics is also open in the other

metric?

6. Aug 31, 2012

### bedi

As their metrics differ only by a little number n, we can always find a "subball" which is completely surrounded by those open balls?

That was quite sloppy..

7. Aug 31, 2012

### Bacle2

I think it would be helpful if you review what LCKurtz said first, to clear things up.

Like he said, in most spaces your distance is not given by an absolute values.

8. Aug 31, 2012

### LCKurtz

As far as I can tell, I don't think so. Look at Sammycap's suggestion. That leads to an easy solution.

9. Sep 1, 2012

### bedi

Alright, so let $B_{r}(x)=\{y \in M:d<r\}$ and $B_{r'}(x)=\{y \in M:\frac{d}{1+d}<r\}$ (intuitively $d$ and $1+d$ seem to be positive? Is that correct?)

Then I get the same kind of thing: $B_{r'}(x)=\{y \in M:d\times(1-r')<r'\}$

Again, multiplying the distance by $(1-r')$ doesn't make difference if I choose $r'$such that $0<r'=r<1$

Looks like I don't make any progress :(

10. Sep 1, 2012

### LCKurtz

Intuitively?? Seems to be positive??? Isn't $d$ given to be a metric?

At the risk of repeating sammycap's suggestion: Given a ball of radius $r$ in the metric $d$, what value $r'$ in the metric $D$ corresponds to it? Figure that out. Then show $B(x,r)$ in $(M,d)$ is identical to $B(x,r')$ in $(M,D)$.

 Or maybe not. See my next post.

Last edited: Sep 1, 2012
11. Sep 1, 2012

### LCKurtz

Perhaps I was a little hasty about sammycap's suggestion making the question real easy. Try showing that if d(x,y) is small so is D(x,y), and conversely, and use that to show the metrics are equivalent.

Last edited: Sep 1, 2012
12. Sep 1, 2012

### bedi

Thank you, I can do it now.

By the way I actually feel guilty for saying that "intuitive" thing.

13. Sep 1, 2012

### LCKurtz

Don't feel bad. We all, myself included, sometimes type things in haste without taking the time to think it through.