Commutativity of intersection

$$A\cap B=B\cap A$$

Note

This is somewhere between a theorem and a definition because at some point you have to accept "and" is commutative or something.

Proof

$$\implies$$

$$x\in A\cap B\implies x\in A\text{ and }x\in B\implies x\in B\text{ and }x\in A\implies x\in B\cap A$$ , thus by the implies and subset relation we see $$A\cap B\subset B\cap A$$

$$\impliedby$$

By the exact same procedure we see $$B\cap A\subset A\cap B$$

Thus we conclude $$A\cap B=B\cap A$$