Difference between revisions of "Cartesian product"

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(Created page with "{{Todo|Find references}} __TOC__ ==Definition== Given two sets, {{M|X}} and {{M|Y}} their ''Cartesian product'' is the set: * {{M|1=X\times Y:=\{(x,y)\ \vert\ x\in X\wedge y\i...")
 
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Given two sets, {{M|X}} and {{M|Y}} their ''Cartesian product'' is the set:
 
Given two sets, {{M|X}} and {{M|Y}} their ''Cartesian product'' is the set:
 
* {{M|1=X\times Y:=\{(x,y)\ \vert\ x\in X\wedge y\in Y\} }}, note that {{M|(x,y)}} is an ''[[ordered pair]]'' traditionally this means
 
* {{M|1=X\times Y:=\{(x,y)\ \vert\ x\in X\wedge y\in Y\} }}, note that {{M|(x,y)}} is an ''[[ordered pair]]'' traditionally this means
** {{M|1=(x,y):=\{x,\{x,y\}\} }} or indeed
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** {{M|1=(x,y):=\{\{x\},\{x,y\}\} }} or indeed
** {{M|1=X\times Y:=\Big\{\{x,\{x,y\}\}\ \vert\ x\in X\wedge y\in Y\Big\} }}
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** {{M|1=X\times Y:=\Big\{\{\{x\},\{x,y\}\}\ \vert\ x\in X\wedge y\in Y\Big\} }}
 
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===Set construction===
==Set construction==
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{{Todo|Build a set that contains {{M|\{x,y\} }}s, then build another that contains ordered pairs, then the Cartesian product is a subset of this set}}
 
{{Todo|Build a set that contains {{M|\{x,y\} }}s, then build another that contains ordered pairs, then the Cartesian product is a subset of this set}}
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===[[Projection|Projections]]===
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With the ''Cartesian product'' of {{M|X}} and {{M|Y}} come two maps:
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# {{M|\pi_1:X\times Y\rightarrow X}} given by {{M|\pi_1:(x,y)\mapsto x}} and
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# {{M|\pi_2:X\times Y\rightarrow Y}} given by {{M|\pi_2:(x,y)\mapsto y}}
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{{Todo|Give explicitly}}
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==Properties==
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The ''Cartesian product'' has none of the usual<ref group="Note">By usual I mean common properties of binary operators, eg associativity, commutative sometimes, so forth</ref> properties:
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{| class="wikitable" border="1"
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|-
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! Property
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! Definition
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! Meaning
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! Comment
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|-
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! [[Associative|Associativity]]
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| {{M|1=X\times(Y\times Z)=(X\times Y)\times Z}}
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| style="background-color:#F23E3E" | No
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| We can side-step this with obvious mappings
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|-
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! [[Commutative|Commutativity]]
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| {{M|1=X\times Y=Y\times X}}
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| style="background-color:#F23E3E" | No
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|}
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===Associativity===
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Given {{M|X}}, {{M|Y}} and {{M|Z}} notice the two ways of interpreting the ''Cartesian product'' are:
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* {{M|(X\times Y)\times Z}} which gives elements of the form {{M|((x,y),z)}} and
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* {{M|X\times (Y\times Z)}} which gives elements of the form {{M|(x,(y,z))}}
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It is easy to construct a [[bijection]] between these, thus it rarely matters.
 +
==Notes==
 +
<references group="Note"/>
 
==References==
 
==References==
 
<references/>
 
<references/>
 
{{Definition|Set Theory|Abstract Algebra}}
 
{{Definition|Set Theory|Abstract Algebra}}

Revision as of 21:57, 8 December 2015


TODO: Find references


Definition

Given two sets, [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] their Cartesian product is the set:

  • [ilmath]X\times Y:=\{(x,y)\ \vert\ x\in X\wedge y\in Y\}[/ilmath], note that [ilmath](x,y)[/ilmath] is an ordered pair traditionally this means
    • [ilmath](x,y):=\{\{x\},\{x,y\}\}[/ilmath] or indeed
    • [ilmath]X\times Y:=\Big\{\{\{x\},\{x,y\}\}\ \vert\ x\in X\wedge y\in Y\Big\}[/ilmath]

Set construction


TODO: Build a set that contains [ilmath]\{x,y\} [/ilmath]s, then build another that contains ordered pairs, then the Cartesian product is a subset of this set


Projections

With the Cartesian product of [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] come two maps:

  1. [ilmath]\pi_1:X\times Y\rightarrow X[/ilmath] given by [ilmath]\pi_1:(x,y)\mapsto x[/ilmath] and
  2. [ilmath]\pi_2:X\times Y\rightarrow Y[/ilmath] given by [ilmath]\pi_2:(x,y)\mapsto y[/ilmath]

TODO: Give explicitly


Properties

The Cartesian product has none of the usual[Note 1] properties:

Property Definition Meaning Comment
Associativity [ilmath]X\times(Y\times Z)=(X\times Y)\times Z[/ilmath] No We can side-step this with obvious mappings
Commutativity [ilmath]X\times Y=Y\times X[/ilmath] No

Associativity

Given [ilmath]X[/ilmath], [ilmath]Y[/ilmath] and [ilmath]Z[/ilmath] notice the two ways of interpreting the Cartesian product are:

  • [ilmath](X\times Y)\times Z[/ilmath] which gives elements of the form [ilmath]((x,y),z)[/ilmath] and
  • [ilmath]X\times (Y\times Z)[/ilmath] which gives elements of the form [ilmath](x,(y,z))[/ilmath]

It is easy to construct a bijection between these, thus it rarely matters.

Notes

  1. By usual I mean common properties of binary operators, eg associativity, commutative sometimes, so forth

References