# 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!

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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

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).

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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? – mr.gondolier Mar 19 '10 at 2:40
If you have an inner regular measure, then every set of positive measure contains a compact set of positive measure. If the compact set is not already totally disconnected, then you can try to punch tiny open holes out until it is totally disconnected while removing only epsilon measure... – François G. Dorais Mar 19 '10 at 3:06

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.

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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 $C^2\cong C$.

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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