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 $(X,d_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}})$ 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 $(Y,d_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}})$ 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 $x,x^{\prime }\in X$, we have $d_X(x,x^{\prime })=d_Y(y,y^{\prime })$, where $y,y^{\prime }$ 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 $d_X(x,x^{\prime })=d_Y(f(x),f(x^{\prime }))$ for all $x,x^{\prime }\in X$.

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