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Show that an ordinal is a limit ordinal if and only if it is $\omega\cdot\beta$ for some $\beta$.

To show the first implication, I was trying to use transfinite induction on beta to show that $\omega\cdot\beta$ is always limit, but I got a little stuck with the induction step.

Any ideas would be really appreciated!

Thanks

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Well, $\omega(\beta+1)=\omega\cdot\beta+\omega$. In limit cases it's even simpler because the multiplication of two limit ordinals cannot produce a successor ordinal.

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Hint: Use Cantor Normal Form and the fact that $\alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma$ for any ordinals $\alpha,\beta,\gamma$. With this you can prove the equivalence

Edit: $(\Rightarrow)$ Let us prove by induction on $\alpha$ that if $\alpha$ is a limit ordinal, it has the prescribed form. There are two cases:

  • There is some $\gamma<\alpha$ such that there are no limit ordinals between $\gamma$ and $\alpha$. Let $\gamma_0$ be the greatest limit ordinal with $\gamma$ with $\gamma<\alpha$; which clearly exists in this case. We have that $\gamma_0+\omega$ is the least limit ordinal greater than $\gamma_0$, but so is $\alpha$, hence $\alpha=\gamma_0+\omega$. By the inductive hypothesis, $\gamma_0=\omega\cdot\beta'$ for some $\beta'$, thus $\alpha=\gamma_0+\omega=\omega\cdot(\beta'+1).$

  • For any $\gamma<\alpha$ there always exist limit ordinals between $\gamma$ and $\alpha$. In this case we get that $\alpha=\sup\{\gamma<\alpha:\gamma$ is a limit ordinal$\}$. Let $\beta=\sup\{\gamma:\omega\cdot\gamma<\alpha\}$, then by the inductive hypothesis we obtain $\alpha=\lim_{\gamma\to\beta}\omega\cdot\gamma=\omega\cdot\beta.$

$(\Leftarrow)$ Just as in Asaf's Answer.

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  • $\begingroup$ How does $\gamma_0$ exist $\endgroup$ – Jhon Doe Nov 30 '17 at 10:11

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