I am interested in how to describe an $n$-dimensional cantor set. I think that it may be a good idea to develop the Cantor Set on the two-dimensional plane at first, but I am having issues figuring out the split-structure even on this particular generalization. I am not sure how the splits come into play differently in the two-dimensional plane; I would have thought that we would develop the cantor set independently in two different sets and then take the cross product. Is this an appropriate method?
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$\begingroup$ Yes, taking the product is the easiest and correct way to build such a thing. $\endgroup$– user147263Jan 12, 2016 at 22:12
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$\begingroup$ See Cantor dust, and perhaps also Sierpinski carpet. $\endgroup$– user856Jan 12, 2016 at 22:13
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$\begingroup$ It's what's commonly called a Menger Sponge: en.wikipedia.org/wiki/Menger_sponge $\endgroup$– user127032Jan 12, 2016 at 22:18
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One of the most common constructions is to take a unit square and split it into four subsquares, and then split those subsquares, and so on. For example, if the ratios are all $1/4$, then we'd get the first generation to be
$$[0, 1/4] \times [0, 1/4]$$ $$[3/4, 1] \times [0, 1/4]$$ $$[0, 1/4] \times [3/4, 1]$$ $$[3/4, 1] \times [3/4, 1]$$
Continuing inside each of these squares gives a set with positive and finite $\mathcal{H}^1$ measure. See, e.g. J. Garnett Positive length but zero analytic capacity from '70 for an example application.
Alternatively, this can be thought of as a Cartesian product of a Cantor set with ratios $1/4$ with itself.
A natural generalization is to replace the ratio $1/4$ with some parameters between $0$ and $1/2$.
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1$\begingroup$ Just for the record, this set it's something referred as "$1/4$ planar Cantor set" in literature. $\endgroup$ Jan 12, 2016 at 22:17
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$\begingroup$ @SilviaGhinassi Right. I've also heard "Garnett Cantor set." $\endgroup$– user296602Jan 12, 2016 at 22:21
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