 Solution

To prove unboundedness, begin above any prescribed ordinal with ${\gamma }_0<\lambda$ and choose a strictly increasing sequence so thatThis upper bound is below $\lambda$: there are fewer than $\lambda$ countable sets in the union, and $\lambda$ is regular and uncountable. Put $\delta ={\sup }_m{\gamma }_m<\lambda$. For any $\alpha <\delta$, choose $m$ with $\alpha <{\gamma }_m$; then $B(\alpha )\subseteq {\gamma }_{m+1}\subseteq \delta$. Thus $\delta \in E$.

For closedness, suppose $\delta <\lambda$ is a limit point of $E$. Given $\alpha <\delta$, choose $\eta \in E$ with $\alpha <\eta <\delta$. Then $B(\alpha )\subseteq \eta \subseteq \delta$. Therefore $\delta \in E$ as well. Hence $E$ is a club set in $\lambda$. This is the club of closure points for countable set-valued functions.