contractible configuration spaces 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 $k$?
 A: I believe each of these arguments will work.
Argument 1: Consider $S^n \subset \Bbb R^{n+1} \subset S^{n+1}$, where the last inclusion is
given by the upper hemisphere (which is homeomorphic to $\Bbb R^{n+1}$. This gives rise to a string of inclusions
$$
\cdots \subset F(S^n,k) \subset F( \Bbb R^{n+1},k) \subset F(S^{n+1};k) \subset F(\Bbb R^{n+2},k)\subset \cdots
$$
The colimit this system is identified with both $F(S^\infty,k)$ and $F(\Bbb R^\infty,k)$.
Argument 2: The projection onto the last coordinate defines a Hurewicz fibration
$$
F(S^\infty,k) \to S^\infty
$$
whose fiber is identified with $F(\Bbb R^\infty,k)$. But $S^\infty$ is contractible, which shows that the fibration is trivializable. In particular, one has a homotopy equivalence
$$
F(S^\infty;k) \simeq F(\Bbb R^\infty,k)\times S^\infty\, .
$$
Now use the fact that $F(\Bbb R^\infty,k)$ is contractible.
A: 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. 
