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

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