Conclusion about cardinalty. Assume that:
$$\left| T \right| > {\aleph _0}$$
Why can't one assume immediately that:
$$\left| T \right| \cdot \left| T \right| > \left| T \right| \cdot {\aleph _0}$$ 
 A: Because assuming the axiom of choice, if $T$ is infinite then $|T\times T|=|T|\cdot|T|=|T|$.
A: 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.
A: For infinite cardinals $\kappa, \lambda$, we have:
$$
\kappa \times \lambda = \max(\kappa, \lambda) \text{ and, therefore } \\
\kappa \times \kappa = \kappa \text{ for $\kappa$ infinite}
$$


In this case, $|T|\cdot |T| = |T|$. Also, $|T|\cdot \aleph_0 = \max(|T|,\aleph_0) = |T|$. So in particular, $|T|\cdot |T| \not > |T|\cdot \aleph_0$, $|T|\cdot |T| = |T|\cdot \aleph_0$, 
A: All that follows is that $\left| T \right| \cdot \left| T \right| \ge \left| T \right| \cdot {\aleph _0}$.
For example, if $|T| = 2^{\aleph_0}$ then $\left| T \right| \cdot \left| T \right| = \left| T \right| \cdot {\aleph _0}$.
