Difference between revisions of "The intersection of sets is a subset of each set"

Revision as of 18:13, 13 February 2015

Theorem

That [math]A\cap B\subset A[/math]

Of course by commutivity of [math]\cap[/math] we have [math]A\cap B\subset B[/math] (as [math]A\cap B=B\cap A[/math] and [math]B\cap A\subset B[/math] by the statement above)

Proof

We will show [math]x\in A\cap B\implies x\in A[/math] then use the implies and subset relation to conclude [math]A\cap B\subset A[/math]

Suppose [math]x\in A\cap B[/math]

Then [math]x\in A[/math] and [math]x\in B[/math]

QED