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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. :)

Re: Any metric spaces can be viewed as a subset of normed spaces

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

It seems that the claim is obvious once you know that $\displaystyle d(x, y) = \|f_x – f_y\|$. Do you understand why this fact holds?

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?

Re: Any metric spaces can be viewed as a subset of normed spaces

You need to show that for all $\displaystyle x,y\in X$, $\displaystyle d(x,y)=\|g_x-g_y\|$. But by the definition of $\displaystyle g_x$ and $\displaystyle g_y$, $\displaystyle \|g_x-g_y\|=\|(f_x-f_a)-(f_y-f_a)\|=\|f_x-f_y\|$.

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 ^^

Re: Any metric spaces can be viewed as a subset of normed spaces

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