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I know that $\omega^\omega\omega$ is $\omega$ copies of $\omega^\omega$, and $\omega\omega^\omega$ is $\omega^\omega$ copies of $\omega$. But how can I tell which one is a member of the other?

$\omega^\omega\omega\in\omega\omega^\omega$ or $\omega\omega^\omega\in\omega^\omega\omega$? And why?

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  • $\begingroup$ what is $\omega$ here? $\endgroup$ – MAN-MADE Aug 22 '17 at 9:57
  • $\begingroup$ $\omega$ is the set of all finite ordinals, it's just like $\mathbb{N}$ $\endgroup$ – Sid Caroline Aug 22 '17 at 10:00
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The ordinal $\omega^\omega\cdot\omega$ is equal to $\omega^{\omega+1}$.

The ordinal $\omega\cdot\omega^\omega$ is just equal to $\omega^\omega$.

So $\omega^\omega\cdot\omega$ is the bigger ordinal while $\omega\cdot\omega^\omega$ is the smaller ordinal.

One way of understanding $\omega.\omega^\omega$ might be that: After $\omega$ copies of $\omega$ are placed the ordinal reached is $\omega^2$. After $\omega ^2$ copies of $\omega$ are placed the ordinal reached is $\omega^3$. After $\omega ^3$ copies of $\omega$ are placed the ordinal reached is $\omega^4$. After $\omega ^4$ copies of $\omega$ are placed the ordinal reached is $\omega^5$. And so on...

So when we have placed $\omega^ \omega$ copies of $\omega$ we are just at $\omega^ \omega$. And not surprisingly this is supremum (least-upper-bound) of :

$$\{\,\omega ^2,\omega ^3,\omega ^4,\omega^5,...\}$$

Edit: Note that partly why this kind of question may arise is that for any finite ordinal $n$ we actually have: $$\omega\cdot\omega^n=\omega^n\cdot\omega=\omega^{n+1}$$

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  • $\begingroup$ So how do you interpret $\omega^\omega\omega$? $\endgroup$ – Sid Caroline Aug 22 '17 at 10:20
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    $\begingroup$ Like you mentioned, $\omega^\omega.\omega$ can be interpreted as $\omega$ copies of $\omega^\omega$. $\endgroup$ – SSequence Aug 22 '17 at 10:23

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