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.

Topological puzzle: configuration space

I came across a problem which looks like an exercise from Topology by Munkres, yet I can’t figure it out.

Let X be a connected topological space. The product X\times X is also a topological space, and is connected as well. (Why? Because each “horizontal” fiber X\times \{x\} and each “vertical” fiber \{x\}\times X must lie within some connected component, but since they intersect, there is just one component.) However, if we remove the diagonal D=\{(x,x)\colon x\in X\} from the product, the connectedness may be lost. For example, (X\times X)\setminus D is disconnected when X is a line.

So far so easy. Now let’s take a quotient of (X\times X)\setminus D, identifying each pair (x_1,x_2) with the reordered pair (x_2,x_1). In other words, the symmetric group S_2 acts on (X\times X)\setminus D by permutation of coordinates, and we take the space of orbits. This new space is denoted C_2(X) and is called the configuration space of X.

Is C_2(X) connected?

It is in all examples that I can think of. Of course, this is not much evidence (for one thing, my thinking does not go beyond Hausdorff spaces). I’m happy to assume that X is metrizable, in which case C_2(X) carries the Hausdorff metric d(\{a,b\},\{a',b'\}) = \min(\max(|a-a'|,|b-b'|),\max(|a-b'|,|b-a'|)). Here I’m using the |\cdot -\cdot | notation for the metric to reduce the clutter.

Discussion on Math.SE