Prove using ordinal induction that for any two ordinals $\alpha$ and $\beta$, if $\alpha \in \beta$ then $S(\alpha)$ $\in$ $S(\beta)$

Question

So I am trying to use ordinal induction to prove a rather trivial fact about ordinals.

That is for any two ordinals $\alpha$ and $\beta$, if $\alpha \in \beta$ then $S(\alpha)$ $\in$ $S(\beta)$. So I fixed any $\alpha$ and induct on $\beta$.

I cleared the base case and the successor case. But I am stuck on the limit case.

Any insights on how to proceed is appreciated.

Answer 1

Let us assume that $\beta$ is a limit ordinal, in other words $\beta=\bigcup_{\beta'\in \beta}\beta'$. The inductive hypothesis is that, for every $\beta' \in \beta$, if $\alpha \in \beta'$ then $S(\alpha) \in S(\beta')$.

Let us assume that $\alpha \in \beta$. We are left to prove that $S(\alpha) \in S(\beta)$. As $\alpha \in \beta$, there is $\beta'\in \beta$ such that $\alpha \in \beta'$. By the inductive hypothesis $S(\alpha) \in S(\beta')=\beta' \cup \{\beta'\}$. Since $\beta' \in \beta$, $\beta' \subseteq \beta$ because $\beta$ is transitive. Hence: $$S(\alpha) \in S(\beta')=\beta' \cup \{\beta'\}\subseteq \beta \subseteq \beta \cup \{\beta\}=S(\beta)$$ and we are done.

Answer 2

As $S(\alpha)$ is the least ordinal $>\alpha$, and $\beta>\alpha$, we either have $S(\alpha)<\beta$ or $S(\alpha)=\beta$.