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

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

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.

• How does $\gamma_0$ exist Nov 30, 2017 at 10:11
• I think the sentence "Let $\gamma_0$ be the greatest limit ordinal with $\gamma$ with $\gamma<\alpha$" has a typo. What should it say? Also, why is $\gamma_0+\omega$ the least limit ordinal greater than $\gamma_0$?
– cut
May 13 at 17:57
• I don't get the second dot point. If $\alpha$ is $\omega \cdot\ 3$ and $\gamma$ is $\omega$ then there are limit ordinals between $\gamma$ and $\alpha$ and $\{ \gamma \lt \alpha : \gamma$ is an ordinal $\}$ is $\{\omega, \omega \cdot 2 \}$. But the Sup of this set is not $\omega \cdot 3$ Jun 18 at 3:48

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.