A question about open balls in Hilbert space. Let $S$ be a finite dimensional Euclidean space and let $B$ be an open ball of $S$. If $f$ is any homeomorphism of $S$ onto itself, then (it is easy to see that) $f(B)$ is a bounded and connected open subset of $S$. Is this still true if S is an infinite dimensional and separable Hilbert space? In particular, is $f(B)$ always still bounded in this case?
 A: Of course, topological properties (being open and connected) are preserved under homeomorphisms. 
Metric properties, such as boundedness, are not: the infinite-dimensional spaces are very stretchy under homeomorphisms. Exaggerated rule of thumb: everything we know about homeomorphisms of $\mathbb R^n$ is false for infinite-dimensional spaces. 
To construct an example, take an  unbounded continuous positive function $\varphi$ on the unit sphere: for example, $$\varphi(\xi)=1+\sum_{n=1}^\infty n(1-2\|\xi-e_n\|)^+, \quad \|\xi\|=1 $$ 
where $e_n$ are the standard basis vectors. Define a map $F:\ell^2\to\ell^2$ by 
$$F(x)= x + (\|x\|-1)^+ \varphi(x/\|x\|)\frac{x}{\|x\|} \tag1$$ 
Clearly, $F$ is continuous. So is the inverse map $G$, which is given by 
$$G(y)= y + (\|y\|-1)^+ \frac{1}{1+\varphi(y/\|y\|)}\frac{y}{\|y\|} \tag2$$ 
(To check that $F$ and $G$ are inverses of each other, use the fact that both of them leave the half-lines emanating from the origin invariant: this reduces the computation to simple algebra.)  
The image of the ball $\|x\|<2$ under $F$ is clearly unbounded.
