Does the space $[0,1]^r$ exist with $r\in\mathbb R$? And the Hausdorff dimension is $r$ Thank you very much
1 Answers
Not directly:
In the notation $[0,1]^n$ with $n\in\mathbb{N}$, we mean $[0,1]\times[0,1]\times\cdots\times[0,1]$, i.e. the $n$-times cartesian product.
There is no $r$-times cartesian product for $r\notin\mathbb{N}$, because you cannot repeat an operation (taking cartesian product) for a non-natural number amount of times.
Of course, you could construct a space with hausdorff dimension $r$ for any $r\in\mathbb{R}_{>0}$, but this would not be related directly to the notation $[0,1]^n$ for $n\in\mathbb{N}$, and there is probably more than one way to do it.
One such construction would be to make $[0,1]^r = [0,1]^{\lfloor r\rfloor}\times S_{\mathrm{frac}(r)}$ where $\mathrm{frac}(r)=r-\lfloor r\rfloor$ is the fractional part of $r$ and $S_{\mathrm{frac}(r)}$ is your favorite set of Hausdorff dimension $\mathrm{frac}(r)$, e.g. some generalised Cantor set.
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[+1] Thorough answer. – Jean Marie Nov 23 '22 at 11:20
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Is there a measure on $S_{frac(r)}$ that make this space unique up to measure isomorphism? – Iyari Rojas Nov 24 '22 at 02:06
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No, there's loads of fractals with Hausdorff dimension $\mathrm{frac}(r)$, not one single example to rule them all so to say – student91 Nov 25 '22 at 12:54