TITLE: A question about  homeomorphic subsets of  Hilbert  space
QUESTION [5 upvotes]: Let H be a an infinite dimensional and separable Hilbert space. Let C be a closed and
bounded subset of H that is not compact. Does there always exist a closed and unbounded
subset of H which is homeomorphic to C?

REPLY [5 votes]: Yes. All we need is to construct a continuous on $H$ function $f$ that is unbounded on $C$. After that, $\{x,f(x)\}\subset H\times \mathbb C$ is closed, unbounded, and homeomorhic to $C$ (in the obvious way) and $H\times \mathbb C$ 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 $f_j(x)=\frac j{\min(1,\operatorname{dist}(x,X\setminus U_j))}$. Separability has actually nothing to do with it but in the non-separable case things become a bit more complicated.