> every x in Q is in R but there exists at least one x in R not in Q, and no amount of correspondence shenanigans can get around that so |R|>|Q|. This also app

> every x in Q is in R but there exists at least one x in R not in Q, and no amount of correspondence shenanigans can get around that so |R|>|Q|. This also applies to evens and naturals: evens are a strict subset of naturals and there exists naturals which are not evens. A counter example in the reals is (0,1) and R have the same cardinality, despite there being real numbers not in the interval (0,1). You have the fact incorrect: for infinite sets, all B being a strict subset of A shows is that |A| >= |B|.