gn.general topology - A question about homeomorphic subsets of Hilbert space

Yes. All we need is to construct a continuous on $H$ function $f$ that is unbounded on $C$. After that, ${x,f(x)}subset Htimes mathbb C$ is closed, unbounded, and homeomorhic to $C$ (in the obvious way) and $Htimes mathbb C$ is isometric to $H$. Being closed and non-compact in an arbitrary separable metric space $X$ implies the existence of such a function. The simplest construction is to take a countable open cover $U_j$ of $X$ that contains no finite subcover of $C$ and to put $f(x)=min f_j(x)$ where $f_j(x)=frac j{min(1,operatorname{dist}(x,Xsetminus U_j))}$. Separability has actually nothing to do with it but in the non-separable case things become a bit more complicated.

• Next lie algebras - Figure out the roots from the Dynkin diagram
• Previous ac.commutative algebra - Class groups of normal domains over finite fields