Unbounded subsets of ordinals with countable confinality

Question

Let $\alpha$ and $\beta$ be limit ordinals such that $\alpha > \beta$ and $cf (\beta) > cf(\alpha) = \omega$. It is not true that we can always find an unbounded subset of $\alpha$ with order type $\beta$: For an uncountable successor cardinal $\kappa$ we can find a counterexample by setting $\alpha = \kappa + \omega$ and $\beta = \kappa$. But is it possible to find other values for $\alpha$ and $\beta$, such that there is an unbounded subset of $\alpha$ with order type $\beta$?

Answer 1

No: if $S\subset\alpha$ is an unbounded subset with order type $\gamma$, then always $cf(\alpha)=cf(\gamma)$. The inequality $cf(\alpha)\leq cf(\gamma)$ is trivial, since any unbounded sequence in $S$ is an unbounded sequence in $\alpha$. For the converse, let $\kappa=cf(\alpha)$ and $f:\kappa\to\alpha$ be strictly increasing and unbounded. For each $\xi<\kappa$, let $g(\xi)$ be the least element of $S$ which is greater than or equal to $f(\xi)$. Then $g:\kappa\to S$ is nonstrictly increasing and unbounded. Letting $A\subset\kappa$ be the set of $\xi$ such that $g(\xi)>g(\theta)$ for all $\theta<\xi$, $g:A\to S$ is strictly increasing and unbounded. Since $A$ has order-type at most $\kappa$, this proves that $cf(\gamma)=cf(S)\leq \kappa=cf(\alpha)$.