Suppose $\alpha$ is an ordinal endowed with the order topology, i.e. its basic open sets are generalized open intervals. Given $C \subseteq \alpha$, I want to show (1) implies (2) :
(1) For all nonempty $A \subseteq C$ with sup $A \in \alpha$, sup $A \in C$
(2) $C$ is closed
What I have so far is that I want to show that $\alpha \setminus C$ is open, i.e. $\alpha \setminus C$ is a union of open intervals $\cup_{a,b \in \alpha\cup\{-\infty, \infty\}} (a,b)$. This is what I want to arrive at, but I don't know where to begin. It may help that union of ordinals is the same as sup. Any ideas?
Thank you.