Difference between revisions of "Commutativity of intersection"
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(Created page with "<math>A\cap B=B\cap A</math> {{Todo}} {{Theorem|Set Theory}}") |
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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
A∩B=B∩A
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
⟹
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
[Expand]Set Theory
