User:IssaRice/Isometry in metric spaces

when playing around with metric spaces, one might notice that certain metric spaces can be "modeled" by other metric spaces. For instance, let $X=\{1,2,3\}$ be a set, and let $(X,d_{\mathrm {disc} })$ be the discrete metric on X. Then we can "model" this metric space by the familiar Euclidean metric on $\{(0,0),(1,0),(1/2,{\sqrt {3}}/2)\}$ (the set looks like an equilateral triangle with edge length 1). Similarly, with $Y=\{1,2,3,4\}$, the metric space $(Y,d_{\mathrm {disc} })$ can be modeled by $\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}$ with the sup norm metric, by $\{(1/2,0,0,0),(0,1/2,0,0),(0,0,1/2,0),(0,0,0,1/2)\}$ with the taxicab metric, or by $\{({\sqrt {1/2}},0,0,0),(0,{\sqrt {1/2}},0,0),(0,0,{\sqrt {1/2}},0),(0,0,0,{\sqrt {1/2}})\}$ with the Euclidean metric.

What, precisely, do we mean by "modeling" metric spaces? Let $(X,d_{X})$ be a metric space, and let $(Y,d_{Y})$ be a metric space. Then it seems like we want to say that given points $x,x'\in X$, we have $d_{X}(x,x')=d_{Y}(y,y')$, where $y,y'$ are the corresponding points in $Y$. We want some bijection f that maps these points for us. So our final notion is this: Y models X iff there exists some bijection $f:X\to Y$ such that $d_{X}(x,x')=d_{Y}(f(x),f(x'))$ for all $x,x'\in X$.

And the above is just the definition of isometric metric spaces.

Questions:

• is the discrete metric always isometric to R^something with the taxicab/sup norm/Euclidean metric? Given $X$, we can consider $Y=\{f\mid f:X\to \mathbf {R} \}$ and define $d(f,g)=\sum _{x\in X}|f(x)-g(x)|$ or something, where $x\mapsto f$ such that $f(x)=1/2$ and zero everywhere else. (and similarly for the other metrics)
• call X more powerful than Y if X can model more metric spaces by taking appropriate subspaces of itself. are the euclidean metric, taxicab metric, and sup norm metric equally powerful?
• is there a metric space that cannot be modeled by the Euclidean metric?