Is the closed unit ball of the Hilbert space (or, for that matter, of the Hilbert cube, in some metric) homeomorphic to the unit sphere (viz., its own boundary) ? This is clearly uncharacteristic of finite-dimensional cubes. This question is motivated by general considerations in dimension theory. If there is such a homeomorphism, the small inductive dimension, generalised verbatim to infinite cardinals, cannot exist for such spaces (whose "dimension" is a "strange" cardinal like w).


The initial question with 'open' ball was unwittigly typed.



Is the closed unit ball of the Hilbert space homeomorphic to the unit sphere ?
The answer to the question in the title is yes. 


In Bessaga and Pelczynski, *Selected topics in infinite-dimensional topology*, Chapter VI, §2 there is a proof of the following:



> 
> **Theorem.** Each of the following sets is homeomorphic to the countable product $\mathbb{R^N}$ of the real line:
> 
> 
> 1. The separable Hilbert space $\ell\_2$.
> 2. The closed unit ball in $\ell\_2$.
> 3. The unit sphere in $\ell\_2$.
> 4. The "upper half space" in $\ell\_2$: those vectors with non-negative first entry.
> 
> 
> 


