Ordinal Arithmetic Identity

Question

Let $\alpha$ be an Ordinal and $S$ a set of Ordinals. Is it the case that

$$\alpha\bigcup\limits_{x\in S} x=\bigcup\limits_{x\in S} \alpha x$$

and if so how could one prove this?

Answer 1

Yes, it’s true.

Let $\beta=\bigcup S$. If $\beta\in S$, the result is clear. If not, $\beta$ is a limit ordinal, so $\alpha\cdot\beta=\bigcup_{\xi<\beta}(\alpha\cdot\xi)$, and we need only verify that

$$\bigcup_{\xi<\beta}(\alpha\cdot\xi)=\bigcup_{\xi\in S}(\alpha\cdot\xi)\;.$$

It’s clear that

$$\bigcup_{\xi<\beta}(\alpha\cdot\xi)\supseteq\bigcup_{\xi\in S}(\alpha\cdot\xi)\;,$$

but on the other hand for each $\xi<\beta$ there is an $\eta\in S$ such that $\xi\le\eta$, so

$$\bigcup_{\xi<\beta}(\alpha\cdot\xi)\subseteq\bigcup_{\xi\in S}(\alpha\cdot\xi)$$

as well, and the result is proved.