A question about  homeomorphic subsets of  Hilbert  space 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?
 A: Yes. Let $A$ be the set in question. We may assume that $0\notin A$ and moreover that $A$ is outside the unit ball centered at the origin.
Since $A$ is closed (in a complete space) and not compact, it contains an infinite set $\{p_i\}_{i\in\mathbb N}$ of points whose radial projections to the unit sphere are $\varepsilon$-separated for some $\varepsilon>0$. For each $i$, consider a ray $R_i=\{tp_i:t\ge 1\}$. These rays are also $\varepsilon$-separated away from one another. Let $U_i$ denote the $(\varepsilon/3)$-neighborhood of $R_i$. It is easy to construct a homeomorphism $f_i:U_i\to U_i$ which maps $p_i$ to $i\cdot p_i$ and is the identity away from the $(\varepsilon/5)$-neighborhood of $R_i$. The union of these maps $f_i$ and the identity map of $C\setminus \bigcup U_i$ is a homeomorphism from $C$ to itself that sends $A$ to an unbounded set.
A: 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.
