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.