I was just curious, is it possible to break the middle third Cantor set $C$ into a finite number of pieces, and after rearranging them, obtaining a "larger" set, i.e. perhaps in terms of Lebesgue measure? Has this even been attempted as yet?
Is decomposing the middle third Cantor set doable paradoxically?
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real-analysis
measure-theory
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0$Y$ou might be interested in the types of sets discussed here: http://en.wikipedia.org/wiki/Kakeya_set namely, sets in R² containing unit line segments of every angle, but with Lebesgue measure 0. – 2012-01-29
2 Answers
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It is not possible as asked, because the Cantor set has measure zero; every subset of a measure zero set is Lebesgue measurable with measure zero; the "rearranging" operations would preserve measure; and a finite or countable union of measure zero sets still has measure zero.
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0@Sachin, unless you state some limitation on the measure anything is possible. For example the Cantor set is equinumerous with a ball in $\mathbb R^3$, so you could carry over the measure (and isometries) from that, and use the Banach-Tarski decomposition. – 2012-01-29
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For the ordinary Cantor set, it is not possible, since the Cantor set has Lebesgue measure $0$. But what about "fat" Cantor sets?
Again, the answer is that it cannot be done, but the reason is different. There is a finitely additive translation-invariant "measure" on all sets of reals that extends Lebesgue measure. That "measure" prevents paradoxical decompositions of the Banach-Tarski type on the line.