Prove that an ordinal is limit iff it is of this form

Question

Prove that an ordinal is a limit ordinal $\iff$ it is of the form $\omega*v$ for some $v>0$.

My attempt at the proof is the following

Proof: $(\leftarrow)$ By induction on $v$ to show that it is always a limit ordinal of the form $\omega*v$, so the base case is when $v=1$ since $v\ne 0$ [ if $v=0$ then $\omega*0=0$ which is a limit ordinal?] which implies that $\omega*v=\omega*1=\omega$ which is a limit ordinal, hence the base case is satisfied. Now when $v=\delta+1$ we have that $\omega*(\delta+1)=\omega*\delta+\omega=\omega+\omega=\omega$ I dont think this step is correct since $\omega+\omega=\omega*2$ however this is of the form $\omega*v$ [$v=2$]

Now when $v$ is a limit ordinal we have that $\omega*v=\sup_{\epsilon<v}\{\omega*\epsilon\}=\omega*v$ which implies that when $v$ is a limit ordinal it satisfies the claim, that it is of the form $\omega*v$ as required.

Now for ($\rightarrow$) I'm unsure where to start for this case.

Any help would be appreciated, Thanks.

Answer 1

Another way to handle this is to consider that $\omega v$ is, by definition, the unique ordinal that is order-isomorphic to the lexicographic order on $v\times \omega.$

For $x=\cup x\in On$ let $S=\{y<x:y=\cup y\}.$ Since $S$ is a set of ordinals there is a unique $v\in On$ that is order-isomorphic to $S$ and a unique order-isomorphism $f:v\to S. $

For $u\in v$ and $n\in \omega$ let $g(u,n)=f(u)+n.$ Then $g$ is an order-isomorphism from the lexicograpically-ordered $v\times \omega$ to $x.$ So $x=\omega v.$

Answer 2

Let $\beta$ be given. Let $\alpha$ be least such that $$\beta<\omega\cdot\alpha. $$

Then $\alpha$ is a successor ordinal, for otherwise if $\alpha=\lim \gamma_i$, for some $i$, $$\beta<\omega\cdot\gamma_i. $$ Let $\alpha =\delta+1$, then

$$\omega\cdot\delta\leq\beta<\omega\cdot\delta+\omega.$$

So if $\beta$ is a limit, $\beta=\omega\cdot\delta$.