# $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$ for all ordinals $\alpha,\beta,\gamma$

I have come up with a proof of this theorem from my textbook Introduction to Set Theory by Karel Hrbacek and Thomas Jech. Does it look fine or contain gaps? Thank you for your help!

$$(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$$ for all ordinals $$\alpha,\beta,\gamma$$.

My attempt:

We proceed by transfinite induction on $$\gamma$$

• If $$\gamma=0$$, then $$(\alpha+\beta)+\gamma=\alpha+\beta=\alpha+(\beta)=\alpha+(\beta+\gamma)$$.

• If $$\gamma=\delta+1$$, then by IH $$(\alpha+\beta)+\delta=\alpha+(\beta+\delta)$$. We have $$(\alpha+\beta)+(\delta+1)=((\alpha+\beta)+\delta)+1=(\alpha+(\beta+\delta))+1=\alpha+((\beta+\delta)+1)=\alpha+(\beta+(\delta+1)).$$

• If $$\gamma$$ is a limit ordinal, then $$(\alpha+\beta)+\gamma=\sup\{(\alpha+\beta)+\delta\mid\delta<\gamma\}=$$ $$\sup\{\alpha+(\beta+\delta)\mid\delta<\gamma\}$$.

We have 3 observations.

First, $$\beta+\gamma=\sup\{\beta+\delta\mid\delta<\gamma\}$$.

Second, $$\beta+\gamma$$ is a limit ordinal. This is because $$\xi<\beta+\gamma\implies\xi\le\beta+\delta$$ for some $$\delta<\gamma\implies\xi+1\le(\beta+\delta)+1=\beta+(\delta+1)<\beta+\gamma$$.

Third, $$\sup\{\alpha+(\beta+\delta)\mid\delta<\gamma\}=$$ $$\sup\{\alpha+\xi\mid\xi<\beta+\gamma\}$$. This is because $$\alpha+\xi$$ for some $$\xi<\beta+\gamma$$ $$\iff\alpha+\xi$$ for some $$\xi=\beta+\delta$$ and $$\delta<\gamma$$ $$\iff\alpha+(\beta+\delta)$$ for some $$\delta<\gamma$$.

It follows that $$(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$$.