Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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??

share|improve this question
1  
How would this imply that? –  Andres Caicedo May 16 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. –  fafddf May 16 at 20:33
    
Both have cardinality $\omega_2$, not "greater than". And why are they not equal to $\omega_2$? –  Andres Caicedo May 16 at 20:34

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.