0708-1300/the unit sphere in a Hilbert space is contractible

Let $H=\{(x_{1},x_{2},...)|\sum x_{n}^{2}<\infty \}$ and define $S^{\infty }=\{x\in H|||x||=1\}$

Claim

$S^{\infty }$ is contractible

Proof

Suppose $x=(x_{1},x_{2},...)\in S^{\infty }$ then $\sum x_{n}^{2}=1$

Define $F_{1}:S^{\infty }\times I\rightarrow S^{\infty }$ by $F_{1}(x,t)=(sgn(x_{1})((1-t)|x_{1}|+t{\sqrt {x_{1}^{2}+x_{2}^{2}}}),(1-t)x_{2},x_{3},x_{4},...)/||(sgn(x_{1})((1-t)|x_{1}|+t{\sqrt {x_{1}^{2}+x_{2}^{2}}}),(1-t)x_{2},x_{3},x_{4},...)||$

$F_{2}:S^{\infty }\times I\rightarrow S^{\infty }$ by $F_{2}(x,t)=(sgn(x_{1})((1-t)|x_{1}|+t{\sqrt {x_{1}^{2}+x_{3}^{2}}}),0,(1-t)x_{3},x_{4},x_{5},...)/||(sgn(x_{1})((1-t)|x_{1}|+t{\sqrt {x_{1}^{2}+x_{3}^{2}}}),0,(1-t)x_{3},x_{4},x_{5},...)||$

$F_{3}:S^{\infty }\times I\rightarrow S^{\infty }$ by $F_{3}(x,t)=(sgn(x_{1})((1-t)|x_{1}|+t{\sqrt {x_{1}^{2}+x_{4}^{2}}}),0,0,(1-t)x_{4},x_{5},x_{6},...)/||(sgn(x_{1})((1-t)|x_{1}|+t{\sqrt {x_{1}^{2}+x_{4}^{2}}}),0,0,(1-t)x_{4},x_{5},x_{6},...)||$

and so on ...

applying the homotopy $F_{1}$ in the time interval $[0,1/2]$, $F_{2}$ in the interval $[1/2,3/4]$, $F_{3}$ in $[3/4,5/6]$ etc...

we get the desired contraction to the point $(1,0,0,...)$.

A different way to see this is via the cellular structure of $S^{\infty }$. If $S^{\infty }=C_{0}C_{1}...$ you can always contract $C_{k}$ along $C_{k+1}$ like moving contracting the equator along the surface of the earth.

This proof only works in separable Hilbert spaces. Is the unit ball in a non-separable Hilbert space contractible?