Thread: functions of topologies

Let X be a topological space and let Y be a simply ordered set in the order topology. Let be two continuous functions.

Show that the set is closed in X.

Since f is continuous, for every open set , is an open subset of X. Similar, for every open set , .

How do I go from this to showing what needs to be shown?

Originally Posted by dwsmith

Let X be a topological space and let Y be a simply ordered set in the order topology. Let be two continuous functions.

Show that the set is closed in X.

Since f is continuous, for every open set , is an open subset of X. Similar, for every open set , .

How do I go from this to showing what needs to be shown?

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.