Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!


share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer

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 $\alpha=\gamma_0$. 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 $\beta=\sup\{\gamma:\omega\cdot\gamma<\beta\}$, 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.

share|cite|improve this answer

Your Answer


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.