Difference between revisions of "Commutativity of intersection"
From Maths
m |
m |
||
Line 1: | Line 1: | ||
<math>A\cap B=B\cap A</math> | <math>A\cap B=B\cap A</math> | ||
− | |||
==Note== | ==Note== |
Revision as of 19:54, 14 February 2015
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
[Expand]Set Theory