User:IssaRice/Isometry in metric spaces

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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,ddisc) be the discrete metric on X. Then we can "model" this metric space by the familiar Euclidean metric on {(0,0),(1,0),(1/2,3/2)} (the set looks like an equilateral triangle with edge length 1). Similarly, with Y={1,2,3,4}, the metric space (Y,ddisc) 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 {(1/2,0,0,0),(0,1/2,0,0),(0,0,1/2,0),(0,0,0,1/2)} with the Euclidean metric.

What, precisely, do we mean by "modeling" metric spaces? Let (X,dX) be a metric space, and let (Y,dY) be a metric space. Then it seems like we want to say that given points x,xX, we have dX(x,x)=dY(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:XY such that dX(x,x)=dY(f(x),f(x)) for all x,xX.

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={ff:XR} and define d(f,g)=xX|f(x)g(x)| or something, where xf 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?