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Homework Help: Equivalent definitions of Equivalent metrics.

  1. Jul 4, 2011 #1
    Two metrics 'd' and 'f' are said to be equivalent on a metric space X, if they have the same set of open sets. This is equivalent to saying every open ball with respect to d contains an open ball with respect to f (different radius) and vice versa. (As every open set is a union of open balls).

    What I don't quite understand is how these definitions imply that there exist 2 positive constants 'a' and 'b' such that a*d(x,y) <= f(x,y) <= b*d(x,y) for all x,y in X

    Can someone please shed some light on this or point me in the direction of some related article.

    PS can someone show me where the "less than equal to" sign is?

  2. jcsd
  3. Jul 4, 2011 #2


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    If you use LaTeX to make your formulas, you can get [itex]\leq[/itex] with the \leq command.

    Are you sure you're not forgetting a hypothesis (compact metric space?) or reversing the implication (the existence of a and b imply equivalence)?

    Naively, I would expect there to be a notion of "uniformly equivalent" which is equivalent to your condition.
  4. Jul 4, 2011 #3

    Gib Z

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    The definition of equivalent metric actually doesn't imply that inequality, you have the implication reversed. That inequality is a sufficient, but not necessary condition for two metrics to be equivalent. Two exercises you can try are to prove that it is a sufficient condition, and to verify the following example:

    Let [itex]X=\mathbb{R} [/itex] and let [itex]d_1, d_2 [/itex] be the metrics on X defined by [tex] d_1(x,y) = |x-y| [/tex] and [tex] d_2(x,y) = \frac{ |x-y| }{1+ |x-y| } [/tex]. Then [itex] d_1 [/itex] and [itex] d_2[/itex] are equivalent metrics on [itex]\mathbb{R}[/itex] yet do not satisfy the inequality condition.

    PS. There is no "less or equal to" sign on your keyboard, but in LaTeX it's code is just \leq .
  5. Jul 4, 2011 #4
    oo thanks. I need to study more.
    haha, I know it's not on the keyboard. I shall get latex. Thanks.
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