Comparing Metrics on R: d1 vs. d2

[mrbohn1]

Let <r,θ> be polar coordinates on R².

Define f:R-->R² by f(t)=<e^t,t>.

Let d₁ and d₂ be the metrics on R defined by: d₁(t₁,t₂) = |t₁ - t₂|, d₂(t₁,t₂) = ||f(t₁) - f(t₂)||

Are d₁ and d₂ equivalent? If not, is one finer than the other?

 

[tiny-tim]

Let <r,@> be polar coordinates on R^2.

Define f:R-->R^2 by f(t)=<e^t,t>.

Let d_1 and d_2 be the metrics on R defined by: d_1(t_1,t_2) = |t_1 - t_2|, d_2(t_1,t_2) = ||f(t_1) - f(t_2)||

Are d_1 and d_2 equivalent? If not, is one finer than the other?

Hi mrbohn1!

(have a theta: θ and a squared: ² )

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[mrbohn1]

Hi, and thankyou for the symbols, I see where they are now.

I am not really sure where to go with this...usually to show 2 metrics are equivalent I would find constants k₁ and k₂ such that:

k₁d₁(t₁,t₂) < d₂(t₁,t₂), and:

k₂d₂(t₁,t₂) < d₁(t₁,t₂).

I have found that:

||f(t₁) - f(t₂)|| = e^2t₁ + e^2t₂ - 2e^{t₁ + t₂}cos(t₁ - t₂)

By using the nature of polar coordinates, and trig identities. But am not sure whether this is helpful or not.

Perhaps a better approach would be to find a set that is open in the topology induced by one metric, but not in the topology induced by the other, and thus prove the metrics are not equivalent?

 

[Dick]

Hi, and thankyou for the symbols, I see where they are now.

I am not really sure where to go with this...usually to show 2 metrics are equivalent I would find constants k₁ and k₂ such that:

k₁d₁(t₁,t₂) < d₂(t₁,t₂), and:

k₂d₂(t₁,t₂) < d₁(t₁,t₂).

I have found that:

||f(t₁) - f(t₂)|| = e^2t₁ + e^2t₂ - 2e^{t₁ + t₂}cos(t₁ - t₂)

By using the nature of polar coordinates, and trig identities. But am not sure whether this is helpful or not.

Perhaps a better approach would be to find a set that is open in the topology induced by one metric, but not in the topology induced by the other, and thus prove the metrics are not equivalent?

Drawing a picture is more useful than that formula. You are mapping the real line into a logarithmic spiral. Can you see any way to make an open set in one that's not an open set in the other? The only place that looks like it might be a problem is t -> -infinity. But is that really a problem?

 

[mrbohn1]

Thanks...I hadn't even noticed that f mapped into a logarithmic spiral. Unfortunately I'm still stuck!

I'm thinking it has something to do with the fact that f would map -\infty to the origin, but as we are mapping from the real line, and not the extended real line, the image of f does not contain the origin.

I can't see anything else which might lead to there being a set which is open in one induced topology and closed in the other. Unfortunately that isn't quite enough to show they are equivalent metrics...

 

[Dick]

Locally the metrics are equivalent. But consider the sequence t_n=(-2*pi*n). In the reals that converges to -infinity. In R^2 it converges to (0,0) in cartesian coordinates. That means in the d2 metric it's Cauchy. In the d1 metric, it's not. What does that tell you? And I'm actually just asking. I'm not taking this course and I don't have this stuff at the tip of my tongue anymore.