Separation order

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 X be a connected topological space. Fix a base point 0 and define the separation order on X as follows: x<y iff the points 0 and y lies in different components of X\setminus \{x\}.

Theorem: at most one of x<y and y<x holds.

Proof. Suppose x<y. Let Y be the connected component of y in X\setminus \{x\}. Note that 0,x\in X\setminus Y\subset X\setminus \{y\}. If we can show that X\setminus Y is connected, then 0 and x are in the same component of X\setminus \{y\}, which is to be proved. Suppose to the contrary that X\setminus Y has a nonempty proper closed-open subset A. We may assume x\notin A. Note two things about A:

  • \overline{A}\subset A\cup Y, because A is closed in X\setminus Y
  • A is open in X. Indeed, A open in X\setminus Y. The closure of Y is contained in Y\cup \{x\} and therefore does not meet A. Hence A is open in X\setminus \overline{Y}, and therefore in X.

Since Y\subsetneq A\cup Y\subset X\setminus \{x\}, the set A\cup Y is not connected. So it has a nonempty proper closed-open subset Z. Then either Y\subset Z or Y\cap Z=\varnothing; by replacing Z with its complement in A\cup Y we may assume the latter holds: that is, Z\subset A. We claim that Z is both closed and open in X, contradicting the connectedness of X. Indeed,

  • Z is closed in A\cup Y and Z\subset \overline{A}\subset A\cup Y. Hence Z is closed in \overline{A}, and in X.
  • Z is open in A\cup Y and Z\subset A, hence Z is open in A. Since A is open in X, it follows that Z is open in X.