Suppose $\alpha$, $\beta$ and $\gamma$ are ordinals. Prove the distributive law $\alpha \cdot ( \beta + \gamma) = \alpha \cdot \beta + \alpha \cdot \gamma$. The following is my proof:
Proof: We use transfinite induction on $\gamma$.
Zero Case: If $\gamma=0$, then $\alpha \cdot ( \beta + 0 ) = \alpha \cdot \beta = \alpha \cdot \beta + 0$.
Successor Case: Suppose $\alpha \cdot ( \beta + \gamma) = \alpha \cdot \beta + \alpha \cdot \gamma$ holds. We want to prove $\alpha \cdot [ \beta + (\gamma +1)]= \alpha \cdot \beta + \alpha \cdot ( \gamma +1)$. Note that $$\alpha \cdot [ \beta + ( \gamma +1)] = \alpha \cdot [ (\beta + \gamma) +1]= \alpha \cdot ( \beta + \gamma) + \alpha = \alpha \cdot \beta + \alpha \cdot \gamma + \alpha =\alpha \cdot \beta + \alpha \cdot ( \gamma +1)$$
Limit Case: Suppose $\gamma \neq 0$ is a limit ordinal. For all $\delta < \gamma$, we have $\alpha \cdot (\beta +\delta)= \alpha \cdot \beta + \alpha \cdot \delta$. We want to prove $\alpha \cdot ( \beta + \gamma) = \alpha \cdot \beta + \alpha \cdot \gamma$. By definitions, we have $$\alpha \cdot ( \beta + \gamma) = \alpha \cdot \sup\{ \beta + \delta \mid \delta < \gamma \}= \sup\{ \alpha \cdot ( \beta + \delta) \mid \delta < \gamma \},\\ \alpha \cdot \beta + \alpha \cdot \gamma = \alpha \cdot \beta + \sup\{ \alpha \cdot \delta \mid \delta < \gamma \}= \sup\{ \alpha \cdot \beta + \alpha \cdot \delta \mid \delta < \gamma \}.$$ Hence, from the induction hypothesis, for all $\delta < \gamma$, we have $$\alpha \cdot (\beta +\delta)= \alpha \cdot \beta + \alpha \cdot \delta \implies\\ \implies \sup\{ \alpha \cdot ( \beta + \delta) \mid \delta < \gamma \} = \sup\{ \alpha \cdot \beta + \alpha \cdot \delta \mid \delta < \gamma \} \implies \\ \implies \alpha \cdot ( \beta + \gamma)= \alpha \cdot \beta + \alpha \cdot \gamma.$$
I want to check whether or not my proof is correct, especially the limit case. It would be nice if someone can help me to verify the accuracy of my proof.
Yes, your proof is correct. The limit case has also been discussed here.