Given: $\displaystyle f:\omega_{1} \rightarrow \omega_{1}, S=\{\alpha \in \omega_{1}: \alpha $ is closed under $\displaystyle f\}$.

Show S is closed and unbounded in $\displaystyle \omega_{1}$.

If S were not closed, then $\displaystyle \exists A \in S|sup(A)> S$. Thus, S must be closed since sup(A) must be in S or equal to S (by the definition of closed).

If S were bounded, then $\displaystyle \exists \alpha \in \omega_{1}| \forall \beta \in S, \alpha > \beta$. Choose $\displaystyle \beta \in S|sup(\beta)=\omega_{1}$, then there does not exist $\displaystyle \alpha \in \omega_{1}| \alpha > \beta$, since $\displaystyle \alpha = \beta$ in this case. Thus, a contradiction and S must be unbounded. Therefore, S is closed and unbounded.

NOW, the problem... The second part dealing with bounded does not cover the case where there is no $\displaystyle \beta \in \omega_{1}$ that has $\displaystyle sup(\beta)=\omega_{1}$. So, I need help showing S is unbounded for this possibility. Any takers on helping? Let me know if something is unclear.