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Metric space proof

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that if [itex] (X,\rho) [/itex]is a metric space then so is [itex](X,\bar\rho)[/itex], where
    \bar\rho:X \times X \Rightarrow R_{0}^{+}, (x,y) \Rightarrow \frac{\rho(x,y)}{1+\rho(x,y)}.

    2. Relevant equations

    I'm trying to prove the axiom that a metric space is positive definate.

    3. The attempt at a solution

    because given [itex] (X,\rho) [/itex] is a metric space is it enough to say that [itex](X,\bar\rho)[/itex], cannot be [itex]< 0 [/itex] because [itex] (X,\rho) [/itex] cannot be [itex]< 0 [/itex] ? ie the limit of [itex](X,\bar\rho)[/itex] is 0 as [itex] (X,\rho) [/itex] tends to zero ?
    Last edited: Feb 26, 2010
  2. jcsd
  3. Feb 26, 2010 #2
    Yes, if x is not y, then [itex]
    [/itex] is greater than 0 since [itex]
    [/itex] is greater than 0 for this case.
    If x=y, then it is clear that [itex]
    [/itex] is 0 as you have [itex]

    [/itex] = 0 in the numerator.
  4. Feb 27, 2010 #3
    Thanks for your help.

    I have also got to prove the following
    Let [itex](Y, \theta)[/itex] be a metric space. Prove the following.
    [itex] f : X \rightarrow Y [/itex] is continuous with respect to [itex]\bar\rho [/itex] if and only if it is continuous with respect to [itex]\rho[/itex]

    I know I have to show that the inverse of Open sets in Y are open in X and visa versa how ever I'm not sure how to represent the open sets in [itex]\bar\rho[/itex]

    ie if I was proving [itex](X,\rho)[/itex] and [itex](Y,\theta)[/itex]

    I would show that for each [itex] G \subseteq Y[/itex] that [itex]f^{-1}(G) [/itex]is an open [itex] \subseteq X [/itex] whenever [itex]G[/itex] isan open subset of [itex]Y[/itex]

    As the metric space [itex]\bar\rho[/itex] uses the metric [itex]\rho[/itex] how to I incorporate that into the proof ?
  5. Feb 27, 2010 #4
    Let f be continuous wrt [itex]
    [/itex] and let G be open in Y. Then, as you pointed out, [itex]
    [/itex] is open in X. Let a be a point in [itex]
    [/itex] = A. Then, A is open in X implies that neighborhoods of a are in A (under the rho metric). Write out what these neighborhoods look like. They look like {y | ||a-y||[itex]
    < e }, and these neighborhoods are in X. Now what do the neighborhoods look like under the [itex]
    [/itex] metric and how do they relate to the usual neighborhoods?
  6. Mar 1, 2010 #5
    Does this look alright ?
    I've tried to use Open balls of radius [itex]\delta[/itex] and [itex]\psi[/itex]

    So I have three metric spaces

    [itex](Y, \theta)[/itex]
    and a function
    [itex] f : X \rightarrow Y [/itex]
    which is continous wrt [itex](X,\rho)[/itex]

    So given [itex]f : X \rightarrow Y [/itex]continous wrt [itex]\rho[/itex].

    Take a neighbourhood [itex]B_{\rho}(a,\delta) = {x \in X \mid \rho(x,a) < \delta } [/itex]

    Then for each [itex]\epsilon > 0
    [/itex] there is an [itex] \delta > 0 [/itex]

    Such that we have [itex]B_{\rho}(a,\delta) \leq \theta(f(x),f(a)) < \epsilon[/itex] As [itex] f [/itex] is continuous

    To show that [itex] f : X \rightarrow Y [/itex] is continuous wrt [itex]\bar\rho[/itex] we need to to find a [itex]B_{\rho}(b,\psi) \subset B_{\rho}(a,\delta)[/itex]

    So we can take [itex]\psi = \delta - \rho(b,a)[/itex]

    [itex]\rho(x,b) < \psi[/itex]
    [itex]\rho(x,a) \leq \rho(x,b)+\rho(x,a)[/itex]
    [itex] \rho(x,a) < \psi+ \delta - \psi = \delta[/itex]

    So we can know show that:
    [itex] f : X \rightarrow Y [/itex] is continuous wrt [itex]\bar\rho[/itex]

    [itex]0 < B_{\bar\rho}(x,\psi) < \rho(f(x),f(b)) \leq B_{\rho}(a,\delta) \leq \theta(f(x),f(a)) < \epsilon[/itex]
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