Rank of ordinal number

Question

How do I show that $$ \operatorname{rank}(\alpha)=\alpha $$ for all ordinals $\alpha? I've attempted to solve this via transfinite induction but I could not get it right.

Any help will be appreciated.

Answer 1

Fix an ordinal $\alpha$, and assume that $\mathrm{rank}(\beta)=\beta$ for all $\beta < \alpha$.

• If $\alpha$ is a successor ordinal, then $\alpha=\beta^+$ for some ordinal $\beta < \alpha$. But then $\alpha = \beta \cup \{ \beta \}$. Apply the definition of rank, using the assumption that $\mathrm{rank}(\beta)=\beta$.
• If $\alpha$ is a limit ordinal, then $\alpha = \bigcup_{\beta < \alpha} \beta$. Now, again, use the definition of rank, together with the assumption that $\mathrm{rank}(\beta)=\beta$ for all $\beta < \alpha$.

The result then follows by transfinite induction.