3
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ If $\alpha=\omega^2$, $\omega+\alpha=\alpha$... $\endgroup$ – Steven Stadnicki Jul 5 '17 at 22:31
  • 1
    $\begingroup$ math.stackexchange.com/questions/933450/… $\endgroup$ – Asaf Karagila Jul 5 '17 at 22:37
  • 1
    $\begingroup$ @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. $\endgroup$ – Steven Stadnicki Jul 5 '17 at 22:57
  • 1
    $\begingroup$ (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.) $\endgroup$ – Steven Stadnicki Jul 5 '17 at 23:00
  • 1
    $\begingroup$ Epsilon numbers have this property. $\endgroup$ – DanielWainfleet Jul 6 '17 at 0:11
5
$\begingroup$

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. $\Box$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – Maxime Ramzi Jul 6 '17 at 15:59
  • $\begingroup$ Yes, that is a good idea. I have edited. $\endgroup$ – JDH Jul 6 '17 at 18:37
  • $\begingroup$ There cannot be a better answer than this. Thank you. $\endgroup$ – Bumblebee Jul 6 '17 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.