Is it possible to construct a set that is perfect and nowhere dense that is not done in the same way that a "Cantor like" set is - by removing a certain fixed middle percentage of the initial interval and then iterating. Cantor used 1/3 and others have used various percentiles. Is it possible to construct a set in a way that is not similar to this process?
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4 Answers
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Any perfect set will be closed by definition, so its complement will be open. As any open subset of $\Bbb{R}$ is the countable union of disjoint intervals, any perfect set must be constructible by successively deleting open intervals from $\Bbb{R}$ in much the same way as the Cantor set is.
That said, there's no reason the lengths of those intervals have to follow any kind of nice pattern in the way they do for the Cantor set.
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$\begingroup$ I haven't read the following paper, but according to the title and abstract it looks like it considers what happens to the Hausdorff dimension of the Cantor set that results if we allow the lengths of the middle thirds to vary randomly: ieeexplore.ieee.org/document/5196094 $\endgroup$ Aug 8, 2017 at 15:30
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Every uncountable closed set contains a perfect set. Either take the intersection of all the iterated derivatives (set of limit points), or take the points every neighbourhood of which contains uncountably many points of the set (what are these points called ?). So in a way these sets are ubiquitous in mathematics. Just start with some uncountable nowhere dense closed set and take a perfect subset.
answered Jun 1, 2014 at 23:26
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$\begingroup$ I think the term you're looking for is "condensation point." $\endgroup$– MicahJun 2, 2014 at 5:24
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Every nonempty totally disconnected perfect compact metric space is homeomorphic to the Cantor set.
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$\begingroup$ Could you elaborate on this? This sounds really interesting. $\endgroup$ Oct 3, 2017 at 21:55
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The zero-set of Brownian motion is (almost surely) homeomorphic to the Cantor set. Indeed, most compact sets $\subset \mathbb R$ arising in analysis (and not obviously containing an interval) are Cantor sets as well.
