I follow up on the previous post to report that the question stated there was answered by Dejan Govc. I’ll try to put together a self-contained version of his proof.
Let be a connected topological space. Fix a base point and define the separation order on as follows: iff the points and lies in different components of .
Theorem: at most one of and holds.
Proof. Suppose . Let be the connected component of in . Note that . If we can show that is connected, then and are in the same component of , which is to be proved. Suppose to the contrary that has a nonempty proper closed-open subset . We may assume . Note two things about :
- , because is closed in
- is open in . Indeed, open in . The closure of is contained in and therefore does not meet . Hence is open in , and therefore in .
Since , the set is not connected. So it has a nonempty proper closed-open subset . Then either or ; by replacing with its complement in we may assume the latter holds: that is, . We claim that is both closed and open in , contradicting the connectedness of . Indeed,
- is closed in and . Hence is closed in , and in .
- is open in and , hence is open in . Since is open in , it follows that is open in .