The questions was: if is a connected topological space, is its configuration space also connected? The discussion at Math.StackExchange led to the following argument, which works for and hopefully can be useful in dealing with .
Let be a connected topological space (no other assumptions). Suppose that where and are disjoint closed subsets of . (I think in terms of coloring each -subset of either red or blue.) We want to prove that all -subsets have the same color. To this end, it suffices to show that for any -subset all sets are of the same color.
If not, then by relabeling points we may assume that is red when and blue when . For each pair such that we have the partition
where , the set consists of all points such that is red, consists of all points such that is blue. For example, and . Since both and are closed in , their closures in satisfy
From now on, assume . By exchanging red and blue, we may assume . So,
I claim that where both sets are nonempty and closed, in contradiction to the connectedness of . Indeed,
- and .
- Taking the closure of can add only the points in , but this set is empty.
- Taking the closure of can add only the points in , but these points are already in .
If , the above still works if (or , which is the same thing) by writing
The simplest of remaining cases is and …