Ordinal addition being closed in infinite initial ordinals

I am having real trouble proving that $\alpha + \beta < \omega_\gamma$ for any infinite initial ordinal $\omega_\gamma$ and any ordinals $\alpha < \omega_\gamma$ and $\beta < \omega_\gamma$. That is, that ordinal addtional is closed in any infinite initial ordinal. This is pretty easy to show for $\omega_0 = \omega$ since the addition of any two natural numbers is clearly still a natural number. However, I am not sure how to approach this for larger initial ordinals. I feel like cardinality has to play a role in this since pretty much all I know about infinite initial ordinals is in terms of cardinality. But there is no clear link between ordinal addition and cardinality. Any hints would be very helpful here!

• Show that if $\kappa$ is an infinite cardinal, then there is a bijection $\kappa\times2\to\kappa$. Think of even and odd ordinals. – Andrés E. Caicedo Aug 31 '17 at 1:11
• Anyway, a better result is that an ordinal is an ordinal power of $\omega$ iff it is closed under ordinal addition, and that every infinite cardinal is an ordinal power of $\omega$. – Andrés E. Caicedo Aug 31 '17 at 1:13
• Do you know that $|\alpha+\beta|=|\alpha|+|\beta|$ for ordinal numbers (or order types) $\alpha$ and $\beta?$ – bof Aug 31 '17 at 1:55
• What is your definition of ordinal addition? – Eric Wofsey Aug 31 '17 at 3:07
• @Andrés: It's definitely a better result, but it requires a bit more hands-on ordinal arithmetic and induction. With the case of initial ordinals, you just need to know the fact stated by bof two comments up, which in itself is not very difficult to prove. – Asaf Karagila Aug 31 '17 at 6:38

We prove this by induction on $$\gamma$$. For $$\gamma=0$$, this is just the fact that a union of two finite sets is finite. If $$\gamma$$ is a limit ordinal, then $$\alpha,\beta<\omega_\gamma$$ means that there is some $$\gamma'<\gamma$$ such that $$\alpha,\beta<\omega_{\gamma'}$$ and therefore $$\alpha+\beta<\omega_{\gamma'}$$ by the induction hypothesis.

Finally, suppose that $$\gamma=\gamma'+1$$. Then $$\alpha,\beta<\omega_\gamma$$ means that there are injections $$f\colon\alpha\to\omega_{\gamma'}$$ and $$g\colon\beta\to\omega_{\gamma'}$$. Define the following injection from $$\alpha+\beta$$:

$$h(\xi)=\begin{cases}\delta_{f(\xi)}+2n_{f(\xi)} & \xi<\alpha\\ \delta_{g(\xi')}+2n_{g(\xi')}+1 & \xi=\alpha+\xi' \end{cases}$$ Where $$\delta_\eta$$ is the largest limit ordinal smaller than $$\eta$$, or $$0$$ otherwise; and $$n_\eta$$ is the finite difference between $$\eta$$ and $$\delta_\eta$$ (so if $$\eta<\omega$$, $$n_\eta=\eta$$, for example).

Easily, $$h$$ is well-defined since the two cases are mutually exclusive, and $$h(\xi)\leq f(\xi),g(\xi')$$ so these are always ordinals below $$\omega_{\gamma'}$$. Moreover, it is injective, since $$g$$ and $$f$$ are each injective and $$h(\xi)$$ is always an even ordinal where $$\xi<\alpha$$, whereas $$h(\xi)$$ is always an odd ordinal for $$\alpha\leq\xi$$. So $$h$$ is the union of two injective functions with disjoint domains and disjoint ranges, and thus an injective function on its own.

Therefore $$|\alpha+\beta|\leq\omega_{\gamma'}<\omega_\gamma$$.    $$\square$$

Remark 1. Note that really $$\alpha+\beta$$ is just the disjoint union of a copy of $$\alpha$$ with a copy of $$\beta$$. Therefore by definition, $$|\alpha+\beta|=|\alpha|+|\beta|$$, so by basic cardinal arithmetic results on $$\aleph$$ numbers, $$|\alpha+\beta|=\max\{|\alpha|,|\beta|\}$$ when at least one of them is infinite.

Remark 2. We could have worked harder and defined $$h$$ to be a bijection between $$\alpha+\beta$$ and $$|\alpha+\beta|$$. But this requires separation into cases and whatnot, and I don't see the point of that, as an injection is enough to establish that the cardinality is strictly less than $$\omega_\gamma$$.

• Hi, is it possible that there is a problem in the definition of $h(\xi)$, because $f(\xi)+\omega$ need not be injective? Different $f(\xi)$ may give the same ordinal when adding $\omega$ on the right. – PatrickR Jul 9 at 6:55
• Yes, thank you, you're absolutely right. I will correct this later today. Just need to finish "booting up"... :-) – Asaf Karagila Jul 9 at 6:59
• Okay, this should fix it. Thanks for the attention for details! – Asaf Karagila Jul 9 at 19:43