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Math Help - Metric Space Isometry

  1. #1
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    Metric Space Isometry

    "Let ( \mathbb{R}, d) be a metric space where d(x,y) = |y-x|.
    Let \phi:\mathbb{R}\rightarrow\mathbb{R} be an isometry from the metric space to itself.
    Consider the image x_0 = \phi(0). Show that for every x in \mathbb{R}, the image \phi(x) equals x_0 + x or x_0 - x.
    Next, show that assuming \phi is continuous, either \phi(x) equals x_0 + x for every x, or it equals x_0 - x for every x."

    First, I don't think I understand what "Consider the image x_0 = \phi(0)" means. I guess it makes sense that each \phi(x) equals x_0 + x or x_0 -  x, but I don't know how to prove it.

    For the second statement, I know continuity means that for every \epsilon > 0, there is a \delta > 0, so d(\phi(x), \phi(x_0)) < \epsilon when  d(x, x_0) < \delta and that it's continuous for each x, but I don't know how to apply this definition to show that it is always either x_0 + x or x_0 -  x for every x.

    I really appreciate any help, even (or especially!) if it's just explaining very basic things that I probably don't understand correctly. Thanks.
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  2. #2
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    Hello Kimberu!

    For the first part, recall that for \phi:\mathbb{R} \to \mathbb{R} to be an isometry, we need to have that d(\phi(x_1),\phi(x_2)) = d(x_1,x_2) ~ \forall x_1,x_2 \in \mathbb{R}.

    Now, if we choose x_1=0 and let x\in \mathbb{R}, then we require that  d(\phi(0), \phi(x)) = d(0,x) = |x|, that is, |\phi(x)-\phi(0)| = |x|. Do you see how to continue?
    Last edited by Defunkt; September 10th 2010 at 04:21 PM.
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  3. #3
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    Oh, now I understand that the question was just defining x_0 = \phi(0)!
    So from what you said, from |\phi(x)-\phi(0)| = |x|, just solving for \phi(x) would lead to either the case of x_0 + x or x_0 - x by the nature of absolute values if I'm not mistaken.

    Still, I don't know how to apply the definition of continuity though.

    Thank you for your help!
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  4. #4
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    Actually, the first question is redundant: Evey isometry is continous (actually, it's Lipschitz continous).

    To prove the second one, assume two values x,y go to x_0+x and x_0-y, can the function be an isometry in this case?
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  5. #5
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    Quote Originally Posted by Jose27 View Post
    Actually, the first question is redundant: Evey isometry is continous (actually, it's Lipschitz continous).

    To prove the second one, assume two values x,y go to x_0+x and x_0-y, can the function be an isometry in this case?
    So \phi(x) = x_0+x and \phi(y)=x_0-y? I assume the answer to your isometry question is 'no', but I don't know why. I tried manipulating the equations for those distances of x and y but I didn't get anywhere with that.

    Sorry to be so dense!
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