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
…