TITLE: contractible configuration spaces
QUESTION [7 upvotes]: 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 $k$?

REPLY [15 votes]: 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.