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Math Help - Metric spaces

  1. #1
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    Metric spaces

    Let A not be the empty set, and be a finite subset of R^2 with the Euclidean metric, d. Prove that A with the relative metric d|A is equivalent to (A, d') where d' is the discrete metric.

    I am not even sure what this is asking me to do. Any help would be great! Thanks!
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  2. #2
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    I guess that by "equivalent" you meant homeomorphic. in the discrete space every point is an open set, meaning \forall x \in A, \; \{ x \} is an open set.
    now, for  A \subset \mathbb{R}^2 where A is finite (nonempty) subspace, define  m = min \{ d(x,y) | x,y \in A, x \neq y \} , or m=1 if |A|=1. because A is finite, m is well defined. now for every  x \in A the ball of radius m/2 around x contain only x, so {x} is an open set - therefore this topology is homeomorphic to the discrete topology
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  3. #3
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    That all makes sense to me, but the only thing I'm confused about is the "relative metric denoted d|A. When it says prove A with the relative metric d|A is equivalent to (A,d') where d' is the discrete metric. I'm not sure what the relative metric means?
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  4. #4
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    the metric d is defined over all R^2. to make A into a metric space you need a metric on A, so d|A is probably the notation for the metric d reduced to the points in A
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  5. #5
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    A function f is homeomorphic if :
    - f is bijective (and continuous)
    - f^{-1} is continuous

    2 distances d (in A) and d' (in A) are equivalent if there exist M,N > 0 such that :
    \forall x,y \in A ~,~ M d(x,y) \leqslant d'(x,y) \leqslant N d(x,y)

    is it ?
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