I think Mike is right. The sets are Borel sets, so paradoxes involving non-measurable sets and/or failures of the pigeonhole principle don't apply (though constructivist scruples do). Jim Propp On Sunday, June 11, 2017, Mike Speciner <ms@alum.mit.edu> wrote:
If two subsets of the unit interval (say A and B) both have measure > .5, then their complements in the unit interval each have measure < .5, so the union of their complements has measure < 1. So A intersect B = ~(~A union ~B) must have measure > 0, no?
--ms
On 11-Jun-17 13:18, Keith F. Lynch wrote:
According to http://www.math.brown.edu/~rkenyon/openprobs/ it is unknown whether the middle-third and middle-half Cantor sets have any irrational number in common.
The middle-third Cantor set is equivalent to numbers in the unit interval expressible in ternary with only 0s and 2s. The middle-half Cantor set is equivalent to numbers in the unit interval expressible in quaternary with only 0s and 3s.
I considered writing a program to convert every 20-digit (after the radix point) ternary number in the set into quaternary, and see if any of the 2^20 possible sums consist -- within the precision -- entirely of 0s and 3s. My guess is that none would, except for some repeating numbers such as 0.00000000..., 0.02020202..., 0.20202020..., and 0.22222222..., all of which are of course rational. That's because such a small proportion (2^-N) of N-digit quaternary numbers consist only of 0s and 3s.
If none did, that would prove nothing, since there could be a solution that stops repeating after 20 places. And if some did, that would prove nothing either, since they could stop working after 20 places (unless I could find some non-repeating pattern of 0s and 2s that would keep working forever -- I assume someone has already looked). So I didn't bother.
Similarly if I did it the other way around, converting all 2^20 20-digit quaternary numbers in the set into ternary.
My guess is that there's no overlap, since both sets are of measure zero. So an "accidental" overlap would be an astonishing coincidence.
But that raises the question of overlaps in Cantor-like sets that are *not* of measure zero. For instance consider the set in which, instead of removing the middle half each time, you remove the middle 8th, then the middle 16th, then the middle 32nd, etc. This set has measure 0.577.... Or the set in which, instead of removing the middle third each time, you remove the middle third, then the middle ninth, then the middle 27th, etc. This set has measure 0.560....
My intuition says that since those two sets are each in the unit interval, which of course has measure 1, and since they each have a measure of more than 1/2, that by the pigeonhole principle overlaps are inevitable. (Again, I'm talking only about irrational numbers. The rationals have measure 0.)
But reasoning about infinite sets is tricky. Especially uncountable infinite sets. Wikipedia says the pigeonhole principle is valid for infinite sets, but the only examples it gives are of infinities of different cardinalities, which is not the case here.
I think "measure" means that if I choose a large number of random numbers in the unit interval, 58% of them will be in the first set and 56% of them will be in the second set. That certainly seems to imply to me that at least 14% of them must be in both sets.
Of course that doesn't mean I'll be able to actually *find* any numbers in both sets. I don't think generating random real numbers in the unit interval is computable, even given a finite-speed perfect random bit generator. However many digits you generate, a later digit may knock the number out of the target set.
What do my two sets look like in numbers? In the first set, in binary the first three digits must not be 011 or 100, then the next four digits must not be 0111 or 1000, then the next five digits must not be 01111 or 10000, etc. In the second set, in ternary the first digit must not be 1, then the next two digits must not be 11, then the next three digits must not be 111, etc.
You can see how the probability of choosing a random next digit which takes you outside the set drops so quickly that you have a better than even chance of remaining in the set forever. But as a practical matter, you can't actually generate random numbers in the set. Nor can you ever be sure that 1/pi, 1/e, 1/sqrt(2), 1/phi, etc., is in either set, at least not if you're stuck with an algorithm that generates those numbers one digit at a time. (Is Plouffe reading this?) (You can be sure -- eventually -- if it's *not* in the set. This is the halting probem.)
To find a number in the set, you need to have some pattern. Some non- repeating pattern, since we're only interested in irrationals. For instance 0.2202002000200002... with the number of 0s increasing by one each time. Then you have to prove there's a similar pattern when you convert to the other base. Good luck with that.
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