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Assume that:
$$\left| T \right| > {\aleph _0}$$

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

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up vote 3 down vote accepted

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, $|T|\cdot |T| \not > |T|\cdot \aleph_0$,
$|T|\cdot |T| = |T|\cdot \aleph_0$,

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Just mentioning that you assume AC in your explanation; that is not needed to refute the given claim. – Bach Mar 21 '14 at 7:34

Because assuming the axiom of choice, if $T$ is infinite then $|T\times T|=|T|\cdot|T|=|T|$.

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But we do not need choice here: We cannot conclude the inequality, because $\mathfrak c$ contradicts it. – Andrés E. Caicedo Mar 21 '14 at 1:23
@Andres: Also $\aleph_1$. – Asaf Karagila Mar 21 '14 at 1:23
So, definitely not an issue of choice. – Andrés E. Caicedo Mar 21 '14 at 1:24
Of course. But it is consistent with the failure of the axiom of choice that this is true for some $T$. – Asaf Karagila Mar 21 '14 at 1:25

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.

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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 $\left| T \right| \cdot \left| T \right| = \left| T \right| \cdot {\aleph _0}$.

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