The intersection of sets is a subset of each set

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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)[1]

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 (we have shown that if [math]x\in A\cap B[/math] then [math]x\in A[/math], this is what [math]\implies[/math] means)

See also

References

  1. Alec's (my) own work

TODO: Tidy up this page!