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!
 A: 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 $\omega_\gamma$.
