I need prove that:
There is a single limit point of type (o Cantor-Bendixson rank) $\alpha$ in $\omega^\alpha + 1$.
I try this:
Since $\omega^\alpha + 1 = \omega^\alpha \cup \{ \omega^\alpha \}$, then its sufficient prove that if $\beta < \omega^\alpha$ then $CB(\beta) < \alpha$.
If we proceed by induction on $\alpha$, then we get:
- If $\alpha = 0$ its true, clearly.
- Suppose that $\alpha = \lambda + 1$. Let $\beta < \omega^\alpha$, then $\beta < sup\{ \omega^\lambda \cdot n \colon n \in \mathbb{N} \}$. Thus, $\beta < \omega^\lambda \cdot k$ for a some $k \in \mathbb{N}$.
Therefore, if $CB(\omega^{\lambda} \cdot k) = \lambda$ then it would be.
So I would like some suggestion to prove that $CB(\omega^{\lambda} \cdot k) = \lambda$ or another way to prove what I want.
Let the ordinal $\beta=\gamma+\omega^\alpha$ where every nonzero term of $\gamma$ is $\omega^\alpha$ or larger. We will show $CB(\beta)=\alpha$.
If $\alpha=0$, then $\beta$ ends with a positive finite ordinal. Thus $\alpha$ is isolated and $CB(\alpha)=0$.
Otherwise assume the result holds for $\lambda<\alpha$. For any open set around $\beta$, there are ordinals ending in $\omega^\lambda$ for each $\lambda<\alpha$. Thus $\beta$ is not isolated at level $\lambda<\alpha$.
Furthermore, $(\gamma,\beta+1)$ is a neighborhood of $\beta$ for which every other ordinal ends with some $\omega^\lambda$ with $\lambda<\alpha$. Thus $\beta$ is isolated at level $\alpha$ and $CB(\beta)=\alpha$.