Let $X$ be a topological space. Given an integer $n\ge 2$, let $F_n(X)$ be the set of all ordered $n$-tuples $(x_1,x_2,\dots,x_n)\in X^n$ such that $x_i\ne x_j$ whenever $i\ne j$. Being a subset of the product $X^n$, the set $F_n(X)$ is a topological space. The symmetric group $S_n$ acts on it by permutations of coordinates. Let $C_n(X)=F_n(X)/S_n$ with the quotient topology. The space $C_n(X)$ is called the configuration space of $X$.

Question: if $X$ is connected, is $C_n(X)$ connected?

Here is a simple example to illustrate the problem: if $X=[0,1]$, then $F_2(X)$ is the square minus its diagonal, hence not connected. However, $C_2(X)$ is connected, being a triangle.

When $X$ is a manifold, the statement is true: in one dimension it is verified directly, and in more than one dimension even $F_n(X)$ is connected.

This question is motivated by Homeomorphism preserving distance. For that problem, it would suffice to prove that $C_2(X)$ is connected when $X$ is a compact connected metric space.

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    $\begingroup$ Dear Leonid, I just wanted to let you know that I have edited my answer, in response to Niels Diepeveen's comments, and it seems to give a positive answer for connectedness of $C_2(X)$ when $X$ is a compact connected metric space. Regards, $\endgroup$ – Matt E Jun 2 '12 at 4:09

Here is a proof that $C_2(X)$ is connected when $X$ is a connected and locally connected Hausdorff space. I would guess that it generalizes to the case of $C_n$, but I didn't think through the combinatorics in the general case.

To show that $C_2(X)$ is connected, it evidently suffices to show that $(x_1,x_2)$ and $(x_1',x_2)$ are in the same connected component, for any $x_1,x_1',x_2 \in X$ such that $x_1,x_1' \neq x_2$.

Suppose first that $x_1,x_1'$ lie in the same connected component of $X \setminus \{x_2\}$, say $A$. Then the map $A \to C_2(X)$ defined by $a \mapsto (a,x_2)$ contains both $(x_1,x_2)$ and $(x_1',x_2)$ in its image, and we are done.

Suppose instead that $x_1,x_1'$ do not lie in the same connected component of $X \setminus \{x_2\}$. Because $X,$ and so $X\setminus \{x_2\}$, is locally connected, the connected components of $X\setminus \{x_2\}$ are open as well as closed, and so if we let $A$ denote the connected component of $X\setminus \{x_2\}$ containing $x_1$, then $\overline{A}$ (the closure of $A$ in $X$) contains $x_2$ (because $X$ is connected), but does not contains $x_1'$. Similarly, if we let $A'$ denote the connected component of $X\setminus \{x_1\}$ containing $x_1'$, then $\overline{A}'$ contains $x_2$, but not $x_1$.

Then $a \mapsto (a,x_1')$ is a continuous map $\overline{A} \to C_2(X)$, whose image contains $(x_2,x_1') = (x_1',x_2)$ and $(x_1,x_1')$. Similary $a' \mapsto (x_1,a')$ is a continuous map $\overline{A}' \to C_2(X)$ whose image contains $(x_1,x_2)$ and $(x_1,x_1')$. Thus again we see that $(x_1,x_2)$ and $(x_1',x_2)$ lie in the same connected component of $C_2(X)$.

Probably locally connected is an unnecessarily strong assumption; what is being used is that the closure in $X$ of any connected component of $X\setminus \{x_2\}$ contains $x_2$. My general topology is too rusty to be sure how generally this property holds.

Also, I guess I'm not using the full strength of the Hausdorff assumption; I'm just using the fact that points of $X$ are closed (so that $X\setminus \{x_2\}$ is open in $X$, and thus inherits the property of being locally connected).

Added in response to Niels Diepeveen's comments below:

Suppose that $X$ is compact, Hausdorff and connected. If $C$ is a component of $X \setminus \{x_2\}$, and one takes $Y = X \setminus \{x_2\}$ and $Z = C \cup \{x_2\}$ in the main result of this answer of Niels Diepeveen, then one finds that $C$ is not a component of $Z$, and hence that the closure of $C$ in $X$ contains $x_2$. The above argument goes through, and so we conclude that $C_2(X)$ is connected.

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  • $\begingroup$ Points being closed is equivalent to $T_1$, so you can use that instead of Hausdorff. $\endgroup$ – Niels J. Diepeveen Jun 2 '12 at 1:25
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    $\begingroup$ Any component $C$ of $X \setminus \{x_2\}$ is closed in $X \setminus \{x_2\}$, so either $C$ or $C \cup \{x_2\}$ is closed in $X$. Clearly $x_2$ is a limit point of $C$ iff $C$ is not closed in $X$. My first answer to math.stackexchange.com/q/22032/3457 shows that this is not always true, but the second answer seems to imply that it is true in compact connected Hausdorff spaces. $\endgroup$ – Niels J. Diepeveen Jun 2 '12 at 1:39
  • $\begingroup$ @Niels: Dear Niels, Thanks very much for your comments. I have edited my answer to take them into account; hopefully I have understood the logic correctly. Best wishes, $\endgroup$ – Matt E Jun 2 '12 at 4:08
  • $\begingroup$ This is very interesting! So the connectedness of $C_2(X)$ follows from three pieces put together (quasicomponent=component, @NielsDiepeveen answer, and Matt's argument). I'll try to put these together in a streamlined way, since the part on components is needed only in a special case. And will also try to see if Neils' example can be modified to yield a non-compact connected metric space $X$ with disconnected $C_2(X)$... Thanks to both of you! $\endgroup$ – user31373 Jun 2 '12 at 18:02
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    $\begingroup$ This might be useful: if we define $x|yz$ to mean that $y$ and $z$ are not in the same quasi-component of $X\setminus x$, then for distinct $x, y, z$ in a connected space at most one of $x|yz$, $y|xz$ and $z|xy$ can be true. Intuitively this means that if the diagonal gets in the way, there is a way around it. $\endgroup$ – Niels J. Diepeveen Jun 3 '12 at 16:03

I'm making this a CW since (a) the ideas are due to Matt and Niels, I'm only messing them up; (b) the answer is still partial: the proof works only for $n=2$.

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 $k<i\le n+1$. (Here $k\in \{1,\dots,n\}$). 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 $x_j\in B_{ij}$. Both $R_{ij}$ and $B_{ij}$ are closed in $X\setminus E_{ij}$, hence, their closures in $X$ satisfy $$(*)\qquad \qquad \overline{R_{ij}}\subset R_{ij}\cup E_{ij}\quad \text{and} \quad \overline{B_{ij}}\subset B_{ij}\cup E_{ij}$$

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 $B_{12}\cup B_{13}$. QED.

If $n>2$, the above still works if $k=1$ (or $k=n$, which is the same thing) by writing $$X= \left(\bigcap_j R_{1j}\right) \sqcup \left(\bigcup_j B_{1j}\right)$$ The simplest of remaining cases is $n=3$ and $k=2$.


After due consideration, I have concluded that it is true for any connected space. The proof I have found for this is a little similar to solving a sliding puzzle.

Some notational issues: For convenience we shall pretend that the points of $C_n(X)$ are unordered $n$-tuples (i.e. $n$-element subsets of $X$), rather than equivalence classes of ordered $n$-tuples, as the construction in the question suggests. We will use $a \sim b$ (in $Y$) to mean that $a$ and $b$ are in the same quasicomponent (of $Y$). (in other words, that there is no clopen subset of $Y$ that contains $a$, but not $b$)

Lemma: Let $X$ be a topological space and $P$ a finite subset of $X$. Define a graph $G$ with vertex set $P$ and an edge between $a, b \in P$ if and only if $a \ne b$ and $a \sim b$ in $(X \setminus P) \cup \{a,b\}$. If $X$ is a connected space, then $G$ is a connected graph.

Proof (by contraposition): Suppose $\{ Q, R \}$ is a nontrivial partition of $P$ such that $G$ has no edge between any point of $Q$ and any point of $R$. Then for every $a \in Q, b \in R$ there is a set $A_{ab}$ which is clopen in $(X \setminus P) \cup \{ a, b \}$ such that $a \in A_{ab}, b \notin A_{ab}$. This means there must be an open $U_{ab} \subset X$ such that $A_{ab} \subset U_{ab} \subset A_{ab} \cup P \setminus \{b\}$ and a closed $F_{ab} \subset X$ such that $A_{ab} \subset F_{ab} \subset A_{ab} \cup P \setminus \{b\}$.

If we put $A = \bigcup_{a \in Q}\bigcap_{b \in R} A_{ab}$ and similarly $U$ and $F$, it is easy to see that $Q \subset A$ and $R \cap A = \emptyset$. Moreover, since $A_{ab} \subset U_{ab} \subset A_{ab} \cup P \setminus \{b\}$, we have $A \subset U \subset A \cup Q = A$, therefore $U = A$. By the same reasoning we have $F=A$. Clearly $U$ is open and $F$ is closed, so $A, X \setminus A$ is a separation of $X$.

Theorem: If $X$ is a connected space, then the configuration space $C_n(X)$ is connected.

Proof: It suffices to prove for arbitrary $u, v \in C_n(X)$ that $u \sim v$. We shall do this by induction on $\left| u \setminus v \right|$. Clearly if $\left| u\setminus v \right| = 0$ then $u=v$ and trivially $u \sim v$.

For the induction step we may assume that $u \sim w$ for any $u, w \in C_n(X)$ such that $\left|u\setminus w \right| < m <= n$. In particular, if $\left| u\setminus v \right| = m > 0$, we can choose $p \in u \setminus v$ and $q \in v \setminus u$ and take $w = (u \setminus \{p\}) \cup \{q\}$. Then $\left|v \setminus w \right| = m-1$ and by the induction hypothesis $v \sim w$.

To establish that $u \sim w$, we consider $P = u \cup \{q\} \subset X$ and the associated graph $G$ as defined in the lemma. Since $X$ is connected, $G$ is connected and contains a path $p = p_0, p_1, \ldots, p_k = q$. By the definition of $G$, we have for every $0 \lt i \le k$ that $p_i \sim p_{i-1}$ in $(X\setminus P) \cup \{ p_{i-1}, p_i \}$.

Note that $f_i: (X \setminus P) \cup \{p_{i-1}, p_i\} \to C_n(X)$ defined by $f_i(x) = (P \setminus \{ p_{i-1}, p_i \}) \cup \{x\}$ is continuous, as it is the composition of a homeomorphic embedding $(X \setminus P) \cup \{p_{i-1}, p_i\} \to F_n(X)$ and the quotient map $F_n(X) \to C_n(x)$. It follows that $$ P \setminus \{p_i\} = f_i(p_{i-1}) \sim f_i(p_i) = P \setminus \{p_{i-1}\} $$ in $C_n(X)$, hence $\{ P\setminus \{p_i\} \mid 0 \le i \le k \}$ is a subset of a quasicomponent. Since $P\setminus \{p_k\} = u$ and $P\setminus \{p_0\} = w$, this proves that $u \sim w \sim v$.

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