# Prove that ordinal multiplication is left distributive

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.