Question about the bounded metric

[AxiomOfChoice]

Question about the "bounded" metric

Homework Statement

I'm trying to show that the topology induced on $\mathbb R$ by the bounded metric
$$\sigma(x,y) = \left| \frac{x}{1+|x|} - \frac{y}{1+|y|} \right|$$
is equivalent to that induced by the standard Euclidean metric $d(x,y) = |x-y|$.

Homework Equations

The Attempt at a Solution

Well, it seems to me that, for all $x,y \in \mathbb R$, we have $\sigma(x,y) \leq d(x,y)$. (This implies (I think) that any open set in $d$ is open in $\sigma$.) But I'm having the darndest time showing $\sigma(x,y) \leq d(x,y)$. I know it works when $|x| = |y|$, but when $|x| \neq |y|$, I run into problems. Am I going to have to break this down by cases? That seems awfully...awful.

 

[lanedance]

haven't tried it, but could you just try and show an open ball in each space is open in the other, setting up a on to one corespondance between open sets... though note the "radius" will be different...

 

[AxiomOfChoice]

haven't tried it, but could you just try and show an open ball in each space is open in the other, setting up a on to one corespondance between open sets... though note the "radius" will be different...

Yes, this is what I've tried to do, but the calculations one needs to go through to confirm this are gruesome. For example, can you show that, for any $x,y\in \mathbb R$, we have

$$\left| \frac{x}{|x|+1} - \frac{y}{|y|+1} \right| \leq |x - y|$$

I've tried in a number of ways to show this, and have done it for a few special cases, but I don't have the general case.

Maybe you are talking about something different, but I'm trying to show that, given an open set $\Omega \in \mathcal O_d$, $\Omega \in \mathcal O_\sigma$ and the converse.

 

[Office_Shredder]

I don't think the cases is really that bad. If x is negative and y is positive, or if x is positive and y is negative, you can easily get rid of the absolute value sign (since you know both are positive or both are negative) and simply observe that you're adding two things of the same sign on both sides, one side's added numbers are smaller.

For the case where x and y are the same sign, you can determine whether $$\frac{x}{|x|+1}$$ is bigger or smaller than $$\frac{y}{|y|+1}$$ based solely on whether x<y or x>y, so you really only have four cases left, not eight.

 

[AxiomOfChoice]

I don't think the cases is really that bad. If x is negative and y is positive, or if x is positive and y is negative, you can easily get rid of the absolute value sign (since you know both are positive or both are negative) and simply observe that you're adding two things of the same sign on both sides, one side's added numbers are smaller.

For the case where x and y are the same sign, you can determine whether $$\frac{x}{|x|+1}$$ is bigger or smaller than $$\frac{y}{|y|+1}$$ based solely on whether x<y or x>y, so you really only have four cases left, not eight.

Thanks very much for your help, but I'm not quite sure what you mean when you say "four cases left, not eight." What were the eight, and what are the four? The four cases I'd originally tried to consider (without success) were

1. $x>y$ and $|x|>|y|$,
2. $x>y$ and $|x|<|y|$,
3. $x<y$ and $|x|>|y|$, and
4. $x<y$ and $|x|<|y|$.

As best I can tell, though, you think I should look instead at:

1. $x > y > 0$,
2. $y > x > 0$,
3. $0 > y > x$, and
4. $0 > x > y$.

Is that correct?

 

[AxiomOfChoice]

Never mind. I actually got it to work out, using the cases you suggested. You were right; it's a LOT easier to do it that way. Thanks again!

 

[AxiomOfChoice]

Ok; I'm actually stuck on something different here. If I'm not mistaken, since $\sigma(x,y) \leq d(x,y)$ for all $x,y\in \mathbb R$, then we have $B_d(x,\rho) \subset B_\sigma(x,r)$ whenever $\rho < r$. But I need to show that, given $r > 0$, there exists $\omega = \omega(r)$ such that $B_\sigma(x,\omega) \subset B_d(x,r)$. I'm having a hard time doing this; i.e., figuring out what form $\omega(r)$ should take...can anyone offer hints or suggestions? (BTW, $B_d(x,r) = \{ y\in \mathbb R : d(x,y) < r\}$.)

 

[Office_Shredder]

Clear the denominator

$$| \frac{x}{1+|x|} - \frac{y}{1+|y|}| = (1+|x|)(1+|y|)|x(1+|y|)-y(1+|x|)| \leq (1+|x|)(1+|y|)(2|x||y|)|x-y|$$.

Do you see what to do from there? (hint: $$\omega$$ is going to depend on x here)

 

[AxiomOfChoice]

Clear the denominator

$$| \frac{x}{1+|x|} - \frac{y}{1+|y|}| = (1+|x|)(1+|y|)|x(1+|y|)-y(1+|x|)| \leq (1+|x|)(1+|y|)(2|x||y|)|x-y|$$.

Do you see what to do from there? (hint: $$\omega$$ is going to depend on x here)

I confess I don't see how you get

$$|x(1+|y|)-y(1+|x|)| \leq 2|x||y||x-y|.$$

I tried to finagle the LHS and apply the triangle inequality, but didn't get the RHS.

 

[AxiomOfChoice]

Ok, I'm now to the point where I actually think $|x(1+|y|)-y(1+|x|)| \leq 2|x||y||x-y|$ is wrong. I think it fails for $x = 1/2, y = 1/4$, though I think the statement is true if $|x|,|y| \geq 1$.

 

[AxiomOfChoice]

Also, I thought the idea at this point of the proof was to show that saying

$$\left| \frac{x}{|x|+1} - \frac{y}{|y|+1} \right| < \omega(x,r)$$

implies

$$|x - y| < r.$$

I don't see how we can accomplish that with the kind of thing you have written above, which (as far as I can tell) would be more helpful in showing the converse of this.

 

[AxiomOfChoice]

I've done a bit more work and discovered that

$$\left| \frac{x}{|x|+1} - \frac{y}{|y|+1} \right| < \omega$$

implies

$$|x - y| < \omega (1+|x|)(1+|y|) + 2|x||y|$$

So it kind of makes sense that $\omega$ should have a factor of $1+|x|$ in the denominator. But I can't seem to pick a POSITIVE $\omega(r,x)$ here that will cancel out all the $y$s. Am I close?