# how slow can the dimension of a product set grow?

Let us define the following "dimension" of a Borel subet $B \subset \mathbb{R}^k$:

$\dim(B) = \min\{n \in \mathbb{N}: \exists K \subset \mathbb{R}^n, ~{\rm s.t.} ~ B \sim K\}$,

where $\sim$ denotes "homeomorphic to". Obviously, $0 \leq \dim(B) \leq k$.

I have three questions: Given a $B \subset \mathbb{R}$,
1) As $k \to \infty$, how slow can $\dim(B^k)$ grow? Can we choose some $B$ such that $\dim(B^k) = o(k)$ or even $O(1)$?
2) Will it make a difference if we drop the Borel measurability of $B$ or add the condition that $B$ has positive Lebesgue measure?
3) Does this dimension-like notion have a name? The dimension concepts I usually see are Lebesgue's covering dimension, inductive dimension, Hausdorff dimension, Minkowski dimension, etc. I do not think the quantity defined above coincides with any of these, but of course bounds exist.

Thanks!

• So the circle S^1 has dimension 2 in your sense? If so, I don't know if this has a name but I certainly wouldn't recommend "dimension". – Alon Amit Mar 19 '10 at 0:33
• It probably makes more sense to define this dimension locally to avoid the $S^1$ issue? – François G. Dorais Mar 19 '10 at 1:18

As for 1.) $"dim"(\mathbb{Z}^k)=1$ for all $k\in\mathbb{N}$, because all $\mathbb{Z}^k$ are discrete countable and therefore homeomorphic to each other.
The Cantor set satisfies $\dim(C^k) = 1$ for all $k$. You can easily find homeomorphic copies of the Cantor set with positive measure (e.g. at the $n$-th step remove every middle $3^n$-th instead of every middle third).
• I see. So extending this argument, we can find a $B \subset \mathbb{R^k}$ with arbitrarily large Lebesgue measure and homeomorphic to the standard Cantor set. Is this also true for any other Borel measure? – gondolier Mar 19 '10 at 2:40
Of course the point has the desired property, but I guess, this is not the space you are looking for. As François said, $C=\{0;1\}^\omega$ and so we get $C^2\cong C$.
• The fact that $C^2 \cong C$ is easy to see if you think of $C$ as $\{0,1\}^\omega$. – François G. Dorais Mar 19 '10 at 1:13