Prove $\forall \alpha >2\, (\alpha+\alpha)<\alpha \cdot\alpha <\alpha^\alpha$

Question

Prove $\forall \alpha >2: (\alpha+\alpha)<\alpha \cdot\alpha <\alpha^\alpha$ where $*$ is ordinal multiplication.

Attempt: Proof by induction on $\alpha$, so for the base case let $\alpha=3$ then $\alpha+\alpha=3+3<\alpha\cdot\alpha=9<\alpha^{\alpha}=27$ Hence the case when $\alpha=3$ holds.

Now let $\alpha$ be a successor, then $\alpha=S(\beta)$ then $\alpha+\alpha=S(\beta)+S(\beta)=S(\beta+\beta)<\alpha\cdot\alpha=S(\beta)*S(\beta)=s(\beta\cdot\beta) <\alpha^{\alpha}=S(\beta)^{S(\beta)}=S(\beta^{\beta})$ so the case when $\alpha$ is a successor holds

Now let $\alpha$ be a limit ordinal, then $\alpha+\alpha=\sup_{\lambda<\alpha}(\lambda+\lambda)<\alpha\cdot\alpha=\sup_{\lambda<\alpha}(\lambda\cdot\lambda)<\alpha^{\alpha}=\sup_{\lambda<\alpha}(\lambda^{\lambda})$

So the case when $\alpha$ is a limit ordinal holds,

Thus $\forall \alpha>2:(\alpha+\alpha)<\alpha\cdot\alpha<\alpha^{\alpha}$

I doubt this is correct as i got stuck with the successor and limit steps, any further help would be gratefully appreciated. I haven't done transfinite induction for a while so that might be why

Answer 1

Note that $\alpha+\alpha=\alpha\cdot2$ and $\alpha\cdot\alpha=\alpha^2$.

To prove the result, show that if we fix $\alpha>2$, then for all $\beta<\gamma$:

1. $\alpha\cdot\beta<\alpha\cdot\gamma$, and
2. $\alpha^\beta<\alpha^\gamma$.

Then apply this to the specific case that $\beta=2$ and $\gamma=\alpha$.