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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.

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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. –  András Bátkai Feb 13 '13 at 8:18
    
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. –  N Unnikrishnan Feb 23 '13 at 11:00

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up vote 18 down vote accepted

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.
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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. –  Gerald Edgar Feb 13 '13 at 14:33
    
On the other hand, if you are interested in this area, then you cannot find a better place than Bessaga & Pelczynski to learn it. –  Gerald Edgar Feb 13 '13 at 14:35
    
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. –  N Unnikrishnan Feb 23 '13 at 19:37

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