Let $\alpha$ an ordinal and $A\subseteq\alpha$. We say that $A$ is closed in $\alpha$ iff for all $\gamma<\alpha$ we have that $$ \bigcup(A\cap\gamma)\in A $$ I need prove that $A$ is closed iff for any limit ordinal $\gamma<\alpha$ and for each strictly increasing sequence $\{\alpha_\tau\}_{\tau<\gamma}\subseteq A$ we have that $\sup_{\tau<\gamma}\alpha_\tau\in A$.
The ($\Rightarrow$) direction is done. Can someone give me a hint for solve the another direction?
Edit: This is the excercise. The book is "Constructability" of Devlin.
Suppose for every limit ordinal $\gamma$ and for each strictly increasing sequence $\{\alpha_\tau\}_{\tau < \gamma} \subseteq A$ we have $\sup_{\tau<\gamma}{\alpha_\tau} \in A$.
Suppose $\beta < \alpha$ is arbitrary. We wish to show that $$ \sup(A \cap \beta) = \bigcup{(A \cap \beta)} \in A$$ Now, $A\cap \beta$ has order type $\leq \beta$, say $\gamma$. Then $\gamma < \alpha$, and the elements of $A\cap \beta$ can be enumerated as a strictly increasing sequence $\{\alpha_\tau\}_{\tau < \gamma} \subseteq A$. Then $$ \sup_{\tau< \gamma}{\alpha_\tau} = \sup(A\cap \beta) = \bigcup(A\cap \beta) \in A$$