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Homework Help: Equivalent metrics?

  1. Sep 11, 2008 #1
    Let <r,θ> be polar coordinates on R².

    Define f:R-->R² by f(t)=<et,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?
    Last edited: Sep 11, 2008
  2. jcsd
  3. Sep 11, 2008 #2


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    Hi mrbohn1! :smile:

    (have a theta: θ and a squared: ² :smile:)

    Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
  4. Sep 11, 2008 #3
    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)|| = e2t1 + e2t2 - 2et1 + t2cos(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?
  5. Sep 11, 2008 #4


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    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?
  6. Sep 11, 2008 #5
    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 -[tex]\infty[/tex] 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...
  7. Sep 11, 2008 #6


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    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.
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