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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 $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$.

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