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Math Help - Any metric spaces can be viewed as a subset of normed spaces

  1. #1
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    Any metric spaces can be viewed as a subset of normed spaces

    Consider a metric space (X, d). Let (B(X), ||.||) be the vector space of bounded real valued functions on X with the sup norm. Denote fx(y) = d(x, y). Fix point a in X and let gx=fx fa. Show that x-> gx is an isometric embedding of X into B(X). Note that d(x, y) = ||fx fy||

    I uploaded a picture of the problem since I was unable to introduce the subscripts properly (sorry still new in the forum).

    If someone could give me an idea on how to solve it, I would be really greatful.

    Thanks in advance.
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  2. #2
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    Re: Any metric spaces can be viewed as a subset of normed spaces

    [tex]\|f_x-f_y\|[/tex] gives you \|f_x-f_y\|.

    It seems that the claim is obvious once you know that d(x, y) = \|f_x  f_y\|. Do you understand why this fact holds?
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  3. #3
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    Re: Any metric spaces can be viewed as a subset of normed spaces

    Thank you for the tip

    I do not understand why that becomes obvious. What does that property denote, or imply that can be related to the rest of the problem?
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  4. #4
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    Re: Any metric spaces can be viewed as a subset of normed spaces

    You need to show that for all x,y\in X, d(x,y)=\|g_x-g_y\|. But by the definition of g_x and g_y, \|g_x-g_y\|=\|(f_x-f_a)-(f_y-f_a)\|=\|f_x-f_y\|.
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  5. #5
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    Re: Any metric spaces can be viewed as a subset of normed spaces

    I think I understood what you mean. Thank you very much for your help, time and patience ^^
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  6. #6
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    Re: Any metric spaces can be viewed as a subset of normed spaces

    It is just slightly more complicated to show d(x, y) = \|f_x  f_y\|.
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