# Commutativity of intersection

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[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]