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.

  • $\begingroup$ Welcome here, interesting question. A remark: You can always edit your own questions and it is better to do so. It is a bit confusing to have both questions here. $\endgroup$ Feb 13, 2013 at 8:18
  • $\begingroup$ Apologies for the duplication in question, more the so after the answer by Martin which showed that the original question was as relevent and, in some sense, equivalent. $\endgroup$ Feb 23, 2013 at 11:00

1 Answer 1


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.
  • $\begingroup$ The fact that 2 and 3 are homeomorphic is easier than this $\mathbb R^{\mathbb N}$ result. And can be found in more elementary textbooks. $\endgroup$ Feb 13, 2013 at 14:33
  • $\begingroup$ On the other hand, if you are interested in this area, then you cannot find a better place than Bessaga & Pelczynski to learn it. $\endgroup$ Feb 13, 2013 at 14:35
  • $\begingroup$ Can we expect this phenomenon to happen for every limit ordinal? To be precise, does every R^α contain open sets which are homeomorphic to their boundaries whenever α is a limit ordinal? Or does regularity of α have a role there? Unfortunately I am not able to procure Bessaga & Pelczynski. $\endgroup$ Feb 23, 2013 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.