User:IssaRice/Isometry in metric spaces - Revision history

IssaRice at 07:37, 6 July 2019

2019-07-06T07:37:53Z

← Older revision		Revision as of 07:37, 6 July 2019
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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.		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>.		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:		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?		* 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?		* is there a metric space that cannot be modeled by the Euclidean metric?

IssaRice at 07:36, 6 July 2019

2019-07-06T07:36:39Z

← Older revision		Revision as of 07:36, 6 July 2019
Line 1:		Line 1:
	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.		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</math> and zero everywhere else.		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>.		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>.

IssaRice at 07:36, 6 July 2019

2019-07-06T07:36:18Z

← Older revision		Revision as of 07:36, 6 July 2019
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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.		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</math> and zero everywhere else.

	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>.		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>.

IssaRice at 07:32, 6 July 2019

2019-07-06T07:32:28Z

← Older revision		Revision as of 07:32, 6 July 2019
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	And the above is just the definition of isometric metric spaces.		And the above is just the definition of isometric metric spaces.

			Questions:

			* is the discrete metric always isometric to R^something with the Euclidean metric?
			* 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?

IssaRice at 07:29, 6 July 2019

2019-07-06T07:29:35Z

← Older revision		Revision as of 07:29, 6 July 2019
Line 1:		Line 1:
	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.		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.

			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>.

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

IssaRice: Created page with "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..."

2019-07-06T07:24:58Z

Created page with "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..."

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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.

User:IssaRice/Isometry in metric spaces - Revision history

IssaRice at 07:37, 6 July 2019

IssaRice at 07:36, 6 July 2019

IssaRice at 07:36, 6 July 2019

IssaRice at 07:32, 6 July 2019

IssaRice at 07:29, 6 July 2019

IssaRice: Created page with "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..."

IssaRice: Created page with "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..."