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Let greek letters be ordinals. I want to prove $\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$ by induction on $\gamma$ and I already know it holds true for $\gamma = \emptyset$ and $\gamma$ a successor ordinal. Let $\gamma$ be a limit ordinal. I found $$ \alpha(\beta + \gamma) = \alpha \cdot \sup_{\epsilon < \gamma} (\beta + \epsilon) = \sup_{\epsilon < \gamma} (\alpha(\beta + \epsilon)) = \sup_{\epsilon < \gamma} (\alpha\beta + \alpha\epsilon) = \alpha\beta + \alpha\gamma, $$ but I am suddenly doubting if the second equality is justified.

Question: Is the second equality correct?

Stefan Mesken
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1 Answers1

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Yes, it is correct. It may be more obvious, if we arrange things a bit differently: If $\gamma$ is a limit ordinal, then $\beta+\gamma$ is a limit ordinal as well. So by definition of ordinal multiplication we have $$ \begin{align*} \alpha \cdot (\beta + \gamma) &= \sup_{\delta < \beta + \gamma} \alpha \cdot \delta \\ &= \sup_{\epsilon < \gamma} \alpha \cdot (\beta + \epsilon). \end{align*} $$

Stefan Mesken
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