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> | ||
{{Todo}} | {{Todo}} | ||
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| + | ==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|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> | ||
{{Theorem|Set Theory}} | {{Theorem|Set Theory}} | ||
Revision as of 19:53, 14 February 2015
[math]A\cap B=B\cap A[/math]
TODO:
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]
Set Theory
