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Consider $\omega_2 \times \omega$ and $\omega \times \omega_2$ with ordinal arithmetic.

Then $| \omega_2 \times \omega | =\omega_2$

and $| \omega \times \omega_2 | =\omega_2$

Does this imply that neither $\omega_2 \times \omega$ nor $\omega \times \omega_2$ are cardinals??

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How would this imply that? – Andrés E. Caicedo May 16 '14 at 20:32
@Andres because both have cardinality $\omega_2$ but are not equal to $\omega_2$. So they would be ordinals which fall between $\omega_2$ and $\omega_3$ and therefore not cardinals. – Krzysztof May 16 '14 at 20:33
Both have cardinality $\omega_2$, not "greater than". And why are they not equal to $\omega_2$? – Andrés E. Caicedo May 16 '14 at 20:34
up vote 3 down vote accepted

Recall the inductive definitions of ordinal multiplication:

$$\begin{align} &\alpha\cdot0=0\\ &\alpha\cdot(\beta+1)=\alpha\cdot\beta+\alpha\\ &\alpha\cdot\delta=\sup\{\alpha\cdot\gamma\mid\gamma<\delta\},\ \delta\text{ is a limit ordinal} \end{align}$$

Since $\omega$ and $\omega_2$ are certainly limit ordinals, we have that: $\omega_2\cdot\omega=\sup\{\omega_2\cdot n\mid n<\omega\}$, and since $\alpha\cdot2=\alpha+\alpha>\alpha$, whenever $\alpha>0$, we have that indeed $\omega_2\cdot\omega>\omega_2$ as ordinals.

On the other hand, $\omega\cdot\omega_2=\sup\{\omega\cdot\gamma\mid\gamma<\omega_2\}$. This is a supremum of $\omega_2$ ordinals, and we can show that each of those (of the $\omega\cdot\gamma$'s) has size $\leq\aleph_1$. Therefore $\omega\cdot\omega_2=\omega_2$, which is in fact a cardinal.

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