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Continuous functions in metric spaces

  1. Mar 4, 2010 #1
    Hi guy's I know this is more of a homework question, I posted a similar thread earlier on but I think I ended up confusing myself.

    I need to show that a function is continous between metric spaces. I'll post the question and what i've done any tips on moving forward would be great.

    I have any metric spaces
    [itex] (X,\rho) [/itex]
    [itex](Y, \theta)[/itex]

    And a metric space
    \bar\rho:X \times X \Rightarrow R_{0}^{+}, (x,y) \Rightarrow \frac{\rho(x,y)}{1+\rho(x,y)}.

    I have got to show the following

    Let [itex](Y, \theta)[/itex]
    be a metric space.
    Prove that.
    [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 have been given that [itex] f : X \rightarrow Y [/itex]
    is continuous with respect to [itex](X,\rho)[/itex]

    So I know that for some [itex]\delta[/itex] and [itex] \epsilon > 0[/itex]

    [itex]{\rho}(z,b) < \delta \rightarrow \theta(f(z),f(b)) < \epsilon[/itex]

    I need to show that for some[itex] \psi > 0[/itex] that

    [itex]{\bar\rho}(x,a)<\psi \rightarrow \theta(f(x),f(a)) <\epsilon[/itex]

    Can some one please show me how to go about finding [itex]\psi[/itex] ?
  2. jcsd
  3. Mar 4, 2010 #2


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    Gold Member

    Fix a in X. To show continuity at a, given [itex]\epsilon>0[/itex], you need to find a number [itex]\psi(\epsilon)>0[/itex] such that

    [itex]{\bar\rho}(x,a)=\frac{\rho(x,a)}{1+\rho(x,a)}<\psi(\epsilon) \Rightarrow \theta(f(x),f(a)) <\epsilon[/itex]

    knowing that there exists a number [itex]\delta(\epsilon)>0[/itex] such that

    [itex]{\rho}(x,a) < \delta(\epsilon) \Rightarrow \theta(f(x),f(a)) < \epsilon[/itex]

    Notice that, just by algebra, we have that

    [tex]\frac{\rho(x,a)}{1+\rho(x,a)}<\psi (\epsilon)\Leftrightarrow \rho(x,a)<\frac{\psi(\epsilon)}{1-\psi (\epsilon)}[/tex]

    So, surely, if you can chose [tex]\psi(\epsilon)[/tex] such that

    [tex]\frac{\psi(\epsilon)}{1-\psi (\epsilon)}<\delta(\epsilon)[/tex]

    then you will have won, yes?
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