Configuration space, part II

The questions was: if $X$ is a connected topological space, is its configuration space $C_n(X)$ also connected? The discussion at Math.StackExchange led to the following argument, which works for $n=2$ and hopefully can be useful in dealing with $n>2$ .

Let $X$ be a connected topological space (no other assumptions). Suppose that $C_n(X)=R\cup B$ where $R$ and $B$ are disjoint closed subsets of $C_n(X)$ . (I think in terms of coloring each $n$ -subset of $X$ either red or blue.) We want to prove that all $n$ -subsets have the same color. To this end, it suffices to show that for any $(n+1)$ -subset $E=\{x_1,\dots,x_{n+1}\}$ all sets $E\setminus \{x_i\}$ are of the same color.

If not, then by relabeling points we may assume that $E\setminus \{x_i\}$ is red when $1\le i\le k$ and blue when $k<i\le n+1$ . For each pair $(i,j)$ such that $1\le i\le k<j\le n+1$ we have the partition

$X=R_{ij}\sqcup B_{ij}\sqcup E_{ij}$

where $E_{ij}=\{x_\ell \colon \ell\ne i,j\}$ , the set $R_{ij}$ consists of all points $x$ such that $E_{ij}\cup \{x\}$ is red, $B_{ij}$ consists of all points $x$ such that $E_{ij}\cup \{x\}$ is blue. For example, $x_i\in R_{ij}$ and $x_j\in B_{ij}$ . Since both $R_{ij}$ and $B_{ij}$ are closed in $X\setminus E_{ij}$ , their closures in $X$ satisfy

$\overline{R_{ij}}\subset R_{ij}\cup E_{ij}\quad \text{and} \quad \overline{B_{ij}}\subset B_{ij}\cup E_{ij}$

From now on, assume $n=2$ . By exchanging red and blue, we may assume $k=1$ . So,

$X= R_{12}\sqcup B_{12} \sqcup \{x_3\} = R_{13}\sqcup B_{13} \sqcup \{x_2\}$

I claim that $X= (R_{12}\cap R_{13}) \sqcup (B_{12}\cup B_{13})$ where both sets are nonempty and closed, in contradiction to the connectedness of $X$ . Indeed,

$x_1\in R_{12}\cap R_{13}$ and $x_2,x_3\in B_{12}\cup B_{13}$ .
Taking the closure of $R_{12}\cap R_{13}$ can add only the points in $E_{12}\cap E_{13}$ , but this set is empty.
Taking the closure of $B_{12}\cup B_{13}$ can add only the points in $E_{12}\cup E_{13}=\{x_2,x_3\}$ , but these points are already in $B_{12}\cup B_{13}$ .

If $n>2$ , the above still works if $k=1$ (or $k=n$ , which is the same thing) by writing
$\displaystyle X= \left(\bigcap_j R_{1j}\right) \sqcup \left(\bigcup_j B_{1j}\right)$

The simplest of remaining cases is $n=3$ and $k=2$ …

Configuration space, part II

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