Difference between revisions of "Commutativity of intersection"
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<math>A\cap B=B\cap A</math> | <math>A\cap B=B\cap A</math> | ||
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==Note== | ==Note== | ||
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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> | ||
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Latest revision as of 19:32, 28 October 2016
A∩B=B∩A
Note
This is somewhere between a theorem and a definition because at some point you have to accept "and" is commutative or something.
Proof
⟹
x∈A∩B⟹x∈A and x∈B⟹x∈B and x∈A⟹x∈B∩A, thus by the implies and subset relation we see A∩B⊂B∩A
⟸
By the exact same procedure we see B∩A⊂A∩B
Thus we conclude A∩B=B∩A
