Let $F(M,k)=\{(x\_1,\cdots,x\_k)\mid x\_1\cdots,x\_k\in M,x\_i\neq x\_j, \text{ for } i\neq j \}$. It is known that $F(\mathbb{R}^\infty,k)$ is contractible for each $k$. 


My question: is $F(S^\infty,k)$ contractible for each $k$? For what kind of space $X$ can we have $F(X,k)$ is contractible for each $k$?



contractible configuration spaces
The space $S^\infty$ is actually homeomorphic to $\mathbb{R}^\infty$. To see this, put 
\begin{align\*}
 B\_n &= \{x\in\mathbb{R}^\infty\::\: \|x\|\leq n, x\_k = 0\text{ for } k \geq n\} \\
 C\_n &= \{x\in S^\infty\::\: x\_n \leq 1/2,\; , x\_k = 0\text{ for } k > n\}.
\end{align\*}
Then $\mathbb{R}^\infty$ is the colimit of the spaces $B\_n$, and $S^\infty$ is the colimit of the spaces $C\_n$. Moreover, $B\_n$ and $C\_n$ are both homeomorphic to the $n$-ball, and they are contained in the interiors of $B\_{n+1}$ and $C\_{n+1}$ respectively. One can cook up compatible homeomorphisms $f\_n\:B\_n\to C\_n$, and pass to the colimit to obtain a homeomorphism $\mathbb{R}^\infty\to S^\infty$.


One can prove along similar lines that various other standard contractible spaces are also homeomorphic to $\mathbb{R}^\infty$, for example $\mathbb{R}^\infty\setminus A$ (for any finite subset $A$), or $F(\mathbb{R}^\infty\setminus A,n)$, or the Stiefel manifold $V\_n(\mathbb{R}^\infty)$. However, the linear isometry space $L(\mathbb{R}^\infty,\mathbb{R}^\infty)$ is homeomorphic to $\prod\_{i=0}^\infty\mathbb{R}^\infty$, which is different. 


