Some question regarding the proof of Cantor normal form.

Question

So basically I wanted to prove Cantor normal form using transfinite induction. There is a good post here:

How to prove that Cantor's normal form can produce all ordinal numbers.

But I have some question regarding one part of the proof.

In that post, one part reads:

"Note that there is a greatest ordinal $\delta$ such that $\omega^\delta \leq \alpha$ (since the least ordinal $\zeta$ such that $\omega^\zeta > \alpha$ must be a successor ordinal)."

I have no intuition why this is true (Sorry if the statement is too obvious for some) meaning;

1.) Why such a lower bound exist in the first place

2.)why "least ordinal $\zeta$ such that $\omega^\zeta > \alpha$ must be a successor ordinal" (already answered)

hence I was hoping to get some explanation on that point.

Thank you, Cheers

Answer 1

Suppose the least ordinal $\zeta$ such that $\omega^\zeta>\alpha$ is a limit ordinal. Since $\zeta$ is a limit ordinal, $\omega^\zeta$ is the supremum of all the ordinals $\omega^\delta$ for $\delta<\zeta$. By minimality of $\zeta$, $\omega^\delta\leq\alpha$ for all $\delta<\zeta$. But then $\omega^\zeta$ is also at most $\alpha$, since it is the least upper bound of the $\omega^\delta$ and $\alpha$ is an upper bound for them. This is a contradiction.

Therefore, $\zeta$ is a successor ordinal, and $\zeta=\delta+1$ for some $\delta$. This $\delta$ is then the greatest ordinal such that $\omega^\delta\leq\alpha$.

Answer 2

1) Why such a lower bound exist in the first place?

There must exist a bound $<\alpha+1$ because $\omega^{\alpha+1}>\alpha$. And there must be a greatest bound because:

• If $\alpha$ can't be expressed in the form of $\omega^\beta$ you can "count backwards" until the previous power of $\omega$

• And if $α=\omega^\beta$ than $\delta=\beta$