Consider the $\ell^2$ complex Hilbert space.


Let $m\in \mathbb{N}^\*$ be a fixed number, and set 
$$
S=\left\{ x=(x\_n)\_n\subset \ell^2\ :\ \sum\_{n=1}^m \frac{|x\_n|^2}{n^2}=1\right\}.$$



> 
> I want to show that $S$ is not homeomorphic to 
>  $$
> S(0,1)=\left\{ x=(x\_n)\_n\subset \ell^2\ :\ \sum\_{n=1}^\infty |x\_n|^2=1\right\}.$$
> 
> 
> 



Why $S$ cannot be homeomorphic to the $1$-sphere of $\ell^2$?
$S$ contains the ellipsoid $\{x\in \mathbb C^m: \sum \frac{|x\_n|^2}{n^2}=1\}$ (which is homeomorphic to $S^{2m-1}$) as a strong deformation retract via
$A\_t(x\_1,\dots,x\_m,x\_{m+1},\dots) = (x\_1,x\_2,\dots,x\_m,(1-t)x\_{m+1},(1-t)x\_{m+2},\dots)$ for $t\in[0,1]$, thus the homotopy group $\pi\_{2m-1}(S) = \mathbb Z$.


$S(0,1)$, however, has $\pi\_{2m-1}(S(0,1)) = 0$ since it is contractible: shift first all coordinates one step to the right and then contract in the stereographic chart opposite $(1,0,0,\dots)$.
Or use the contraction described on page 513 of [here](http://www.mat.univie.ac.at/~michor/apbookh-ams.pdf). 


