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Question about relationship between metrics

  1. Jun 15, 2010 #1
    Hello,
    Let X and Y be two spaces equipped respectively with two metrics [itex]d_1[/itex] and [itex]d_2[/itex], and let's consider a mapping [itex]f:(X,d_1)\rightarrow(Y,d_2)[/itex]

    How can I formalize the fact that for any point x in X, all the points that are "very close" to x (respect to the metric [itex]d_1[/itex]) must be mapped into points in Y that are "very close" to f(x) (with respect to the metric [itex]d_2[/itex]).

    Thanks!
     
  2. jcsd
  3. Jun 15, 2010 #2

    radou

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    You're talking about a continuous mapping f between X and Y. You can express this very similar as (for example) in real analysis (epsilon-delta), only you have to invoke the different metrics into your notation. So, for any given ε > 0 there exists some δ > 0 such that d1(p, x) < δ implies d2(f(p), f(x)) < ε - this means that f is continuous at the point p from X. If f is continuous at any point of X, then it is continuous on X.
     
  4. Jun 15, 2010 #3
    Ah...I see.
    and if I add the requirement that f is bijective I obtain a homeomorphism between X and Y? is it correct?

    Thanks a lot.
     
  5. Jun 15, 2010 #4

    radou

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    Yes, along with the condition that the inverse of f is continuous too.
     
  6. Jun 20, 2010 #5
    Now here is a point I am still confused with: when is a homeomorphism between
    metric spaces as isometry.? I know every isometry is an isomorphism, but not
    the other way around ( I guess the homeo. between (0,1) and R is maybe the
    clearest counterexample.). Do we need some sort of Lipschitz condition.?
    Anyone know.?
     
  7. Jun 21, 2010 #6
    Bacle, it's precisely when the pullback metric (if it's a homomorphism, this becomes a metric) agrees with the given metric. It's an interesting result in Riemannian geometry that a homeomorphism which preserves the distance given by the Riemannian metric is actually an isometry (in the Riemannian sense) - in particular, it's smooth.
     
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