Let then our set is the preimage of under the map the continuous map given by and so it suffices to prove that is closed in . To see this we merely note that and let . We have two choices, if there exists then set . If there is no point such that then set . Either way this is an open set in with the order topology which contains and is disjoint from . It thus follows that is open and so is closed, so that the conclusion follows from previous comment.