User:IssaRice/Isometry in metric spaces: Difference between revisions

Latest revision as of 07:37, 6 July 2019

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 $(X, d_{d i s c})$ 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 $(Y, d_{d i s c})$ 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 $x, x^{'} \in X$ , 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 $Y = {f ∣ f : X \to R}$ 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?

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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 <math>X = \{1,2,3\}</math> be a set, and let <math>(X, d_\mathrm{disc})</math> be the discrete metric on X. Then we can "model" this metric space by the familiar Euclidean metric on <math>\{(0,0), (1,0), (1/2, \sqrt{3}/2)\}</math> (the set looks like an equilateral triangle with edge length 1). Similarly, with <math>Y = \{1,2,3,4\}</math>, the metric space <math>(Y, d_\mathrm{disc})</math> can be modeled by <math>\{(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)\}</math> with the sup norm metric, by <math>\{(1/2,0,0,0), (0,1/2,0,0), (0,0,1/2,0), (0,0,0,1/2)\}</math> with the taxicab metric, or by <math>\{(\sqrt{1/2},0,0,0), (0,\sqrt{1/2},0,0), (0,0,\sqrt{1/2},0), (0,0,0,\sqrt{1/2})\}</math> with the Euclidean metric.
-More generally, given <math>X</math>, we can consider <math>Y = \{f \mid f : X \to \mathbf R\}</math> and define <math>d(f,g) = \sum_{x \in X} |f(x)-g(x)|</math> or something, where <math>x \mapsto f</math> such that <math>f(x) = 1/2</math> and zero everywhere else. (and similarly for the other metrics)
 What, precisely, do we mean by "modeling" metric spaces? Let <math>(X, d_X)</math> be a metric space, and let <math>(Y, d_Y)</math> be a metric space. Then it seems like we want to say that given points <math>x,x' \in X</math>, we have <math>d_X(x,x') = d_Y(y,y')</math>, where <math>y,y'</math> are the corresponding points in <math>Y</math>. 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 <math>f : X \to Y</math> such that <math>d_X(x,x') = d_Y(f(x),f(x'))</math> for all <math>x,x' \in X</math>.
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 Questions:
-* is the discrete metric always isometric to R^something with the Euclidean metric?
+* is the discrete metric always isometric to R^something with the taxicab/sup norm/Euclidean metric? Given <math>X</math>, we can consider <math>Y = \{f \mid f : X \to \mathbf R\}</math> and define <math>d(f,g) = \sum_{x \in X} |f(x)-g(x)|</math> or something, where <math>x \mapsto f</math> such that <math>f(x) = 1/2</math> 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?