Difference between revisions of "Ordered pair"

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{{Definition|Set Theory}}
 
{{Definition|Set Theory}}
{{Theorem|Set Theory}}
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{{Theorem Of|Set Theory}}

Latest revision as of 07:22, 27 April 2015

Kuratowski definition

An ordered pair [math](a,b)=\{\{a\},\{a,b\}\}[/math], this way [math](a,b)\ne(b,a)[/math].

Ordered pairs are vital in the study of relations which leads to functions

Proof of existence

It is easy to prove ordered pairs exist
Suppose we are given [math]a,b[/math] (so we can be sure they exist).

By the axiom of a pair we may create [math]\{a,b\}[/math] and [math]\{a,a\}=\{a\}[/math], then we simply have a pair of these, thus [math]\{\{a\},\{a,b\}\}[/math] exists.

The axioms may be found here

Proof of uniqueness

Before we may write [math](a,b)[/math] we must make sure this is not ambiguous.

Proof that [math](a,b)=(a',b')\iff[a=a'\wedge b=b'][/math]


[math]\impliedby[/math]

Clearly if [math]a=a'[/math] and [math]b=b'[/math] then [math](a,b)=\{\{a\},\{a,b\}\}=\{\{a'\},\{a',b'\}\}=(a',b')[/math] and we're done.

[math]\implies[/math]

Assume [math](a,b)=\{\{a\},\{a,b\}\}=\{\{a'\},\{a',b'\}\}=(a',b')[/math].

If [math]a\ne b[/math] then we must have [math]\{a\}=\{a'\}[/math] and [math]\{a,b\}=\{a',b'\}[/math] (as clearly [math]\{a\}=\{a',b'\}[/math] is false, there are either 2 or 1 elements not contained in [math]\{a\}[/math] that are in [math]\{a',b'\}[/math] - namely [math]a'[/math] and [math]b'[/math])

Clearly [math]a=a'[/math], then [math]\{a,b\}=\{a',b'\}\implies b=b'[/math].

If [math]a=b[/math] then [math](a,a)=\{\{a\},\{a,a\}\}=\{\{a\}\}[/math], we know [math]\{\{a\}\}=\{\{a'\},\{a',b'\}\}[/math] so again using the Set theory axioms (namely Extensionality) we see [math]a=a'=b'[/math] so [math]a=a'[/math] and [math]b=b'[/math] holds here too. This completes the proof.