Conclusion about cardinalty. Assume that:
$$\left| T \right| > {\aleph _0}$$
Why can't one assume immediately that:
$$\left| T \right| \cdot \left| T \right| > \left| T \right| \cdot {\aleph _0}$$ 

The simpler counterexample: we have $2\gt 1$, but $|2\cdot\aleph_0|\not\gt |1\cdot\aleph_0|$.  The main 'reason' behind this is that the usual proofs that $a\gt b\implies ac\gt bc$ for $a, b, c\in\mathbb{N}$ use the finiteness of $a, b, c$ in an essential fashion: they assume that a whole number cannot be put into 1-1 correspondence with any of its proper subsets (this is what '$\gt$' denotes, after all).  This assumption breaks down when we get out of the finite realm; in fact, it's one of the definitions of infinitude.
