Difference between revisions of "Commutativity of intersection"

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Thus we conclude <math>A\cap B=B\cap A</math>
 
Thus we conclude <math>A\cap B=B\cap A</math>
  
{{Theorem|Set Theory}}
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{{Theorem Of|Set Theory}}

Revision as of 07:21, 27 April 2015

AB=BA

Note

This is somewhere between a theorem and a definition because at some point you have to accept "and" is commutative or something.

Proof

xABxA and xBxB and xAxBA, thus by the implies and subset relation we see ABBA

By the exact same procedure we see BAAB

Thus we conclude AB=BA