## 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 iff the points $0$ and $y$ lies in different components of $X\setminus \{x\}$.

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

Proof. Suppose $x. 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