# Ordinals of the form $\alpha=\omega+\alpha$?

Epsilon numbers are the fixed points of ordinal exponentiation. i.e., $\epsilon=\omega^{\epsilon}.$ But I never came up with any ordinal number of the form $\alpha=\beta+\alpha$ for all ordinals $\omega\le\beta\lt\alpha.$ I tried to prove that there is no such ordinal and couldn't success. Any hint?

• If $\alpha=\omega^2$, $\omega+\alpha=\alpha$... – Steven Stadnicki Jul 5 '17 at 22:31
• math.stackexchange.com/questions/933450/… – Asaf Karagila Jul 5 '17 at 22:37
• @Nil Do you understand why $1+\omega = \omega$? The fact that $\omega+\omega^2=\omega^2$ is just literally that fact 'multiplied by $\omega$'; putting another copy of $\omega$ on the left effectively shifts the indices of the other $\omega$ copies of $\omega$ by one. – Steven Stadnicki Jul 5 '17 at 22:57
• (More generally, $(m\cdot\omega+n)+\omega^2 = m\cdot\omega+(n+\omega^2)=m\cdot\omega+\omega^2 = \omega^2$ for all natural numbers $m$ and $n$, so in fact the broader property you're talking about holds there.) – Steven Stadnicki Jul 5 '17 at 23:00
• Epsilon numbers have this property. – DanielWainfleet Jul 6 '17 at 0:11

Regarding the title question, which is different from the question in the body, the answer is:

Theorem. An ordinal $\alpha$ has the form $\alpha=\omega+\alpha$ if and only if $\omega^2\leq\alpha$.

In particular, every ordinal above $\omega^2$ has the property in the title.

Proof. If $\alpha<\omega^2$, then $\alpha=\omega\cdot n+k$ for some finite $n,k<\omega$, and it is easy to see that $\alpha<\omega\cdot(n+1)+k=\omega+\alpha$.

Conversely, if $\omega^2\leq\alpha$, then $\alpha=\omega^2+\eta$ for some ordinal $\eta$. Since $1+\omega=\omega$, it follows by multiplying by $\omega$ on both sides (on the left) that $\omega+\omega^2=\omega^2$, and from this it follows that $\omega+\alpha=\omega+\omega^2+\eta=\omega^2+\eta=\alpha$, as desired. $\Box$

More generally, we can provide a similar criterion for any fixed $\beta$.

Theorem. For any ordinal $\beta$, the ordinals $\alpha$ of the form $\alpha=\beta+\alpha$ are precisely the ordinals with $\beta\cdot\omega\leq\alpha$.

Proof. If $\alpha<\beta\cdot\omega$, then $\alpha=\beta\cdot n+\gamma$ for some $n<\omega$ and some $\gamma<\beta$. It follows that $\alpha<\beta\cdot (n+1)+\gamma=\beta+\alpha$.

Conversely, if $\beta\cdot\omega\leq\alpha$, then we may write $\alpha=\beta\cdot\omega+\eta$ for some ordinal $\eta$. Since $1+\omega=\omega$, it follows that $\beta\cdot(1+\omega)=\beta\cdot\omega$ and therefore $\beta+\beta\cdot\omega=\beta\cdot\omega$. Thus, $\beta+\alpha=\beta+\beta\cdot\omega+\eta=\beta\cdot\omega+\eta=\alpha$, as desired. $\Box$

• For the converse of your first claim, isn't it easier to just say $\alpha\geq \omega^2 \implies \alpha = \omega^2 + \eta$ where $\eta$ is an ordinal and then proceed similarly, so the solution can be more elementary ? – Maxime Ramzi Jul 6 '17 at 15:59
• Yes, that is a good idea. I have edited. – JDH Jul 6 '17 at 18:37
• There cannot be a better answer than this. Thank you. – Bumblebee Jul 6 '17 at 21:30