Comparing Open Sets in Metric Spaces

[bedi]

Homework Statement

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.

Homework Equations

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

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.

 

[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.

 

[LCKurtz]

Homework Statement

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.

Homework Equations

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

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.

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 $d(x,y)\ne |x-y|$.

 

[bedi]

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

 

[Bacle2]

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

metric?

 

[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..

 

[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.

 

[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 $d(x,y)\ne |x-y|$.

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

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

 

[bedi]

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

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 :(

 

[LCKurtz]

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?)

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

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 :(

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 $(M,D)$.

[Edit] Or maybe not. See my next post.

 

[LCKurtz]

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

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 :(

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.

 

[bedi]

Thank you, I can do it now.

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

 

[LCKurtz]

Thank you, I can do it now.

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

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