Difference between revisions of "Commutativity of intersection"

Revision as of 19:54, 14 February 2015 (view source) Alec (Talk | contribs) m ← Older edit		Revision as of 07:21, 27 April 2015 (view source) Alec (Talk | contribs) m Newer edit →
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	Thus we conclude <math>A\cap B=B\cap A</math>		Thus we conclude <math>A\cap B=B\cap A</math>
−	{{Theorem|Set Theory}}	+	{{Theorem Of|Set Theory}}

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Revision as of 07:21, 27 April 2015

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

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

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