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}} | + | {{Theorem Of|Set Theory}} |
Latest revision as of 19:32, 28 October 2016
[math]A\cap B=B\cap A[/math]
Note
This is somewhere between a theorem and a definition because at some point you have to accept "and" is commutative or something.
Proof
[math]\implies[/math]
[math]x\in A\cap B\implies x\in A\text{ and }x\in B\implies x\in B\text{ and }x\in A\implies x\in B\cap A[/math], thus by the implies and subset relation we see [math]A\cap B\subset B\cap A[/math]
[math]\impliedby[/math]
By the exact same procedure we see [math]B\cap A\subset A\cap B[/math]
Thus we conclude [math]A\cap B=B\cap A[/math]
