Does there exist a dense set of reals which is uncountable in every interval, whose complement is also dense and uncountable in every interval? So, for example, the union of positive rationals and negative irrationals would not work for my question.
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1https://math.stackexchange.com/questions/506423/a-question-about-uncountable-dense-sets-in-r – Batrachotoxin Dec 12 '20 at 03:46
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3Does this answer your question? A question about uncountable, dense sets in R – PNDas Dec 12 '20 at 03:48
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That's not exactly a duplicate - it doesn't get the "on every interval" requirement. The answer is still the same though (and if I recall correctly this has been asked before). – Noah Schweber Dec 12 '20 at 03:53
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It follows for example from here. – Noah Schweber Dec 12 '20 at 03:56
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I vividly remember this exact question being asked before, but I can't seem to find it at the moment. So here goes:
There are in fact natural examples of such pairs of sets - in particular, the axiom of choice isn't necessary. For example, the set $N$ of normal numbers has full measure (= complement has measure zero) but is meager (= union of countably many nowhere dense sets); in particular, for every nontrivial interval $I$ both $N\cap I$ and $I\setminus N$ have size continuum (which is a bit stronger than what you ask for). Besides providing an answer to the problem, $N$ and its variants are interesting objects in their own right.
What if we don't want to use measure/category ideas? Well, we can still get an explicit and choice-free construction, albeit a bit less natural: let $A$ be the set of all reals whose decimal expansion uses only finitely many $7$s (say). Then both $A$ and $\mathbb{R}\setminus A$ have size-continuum intersection with every nontrivial interval. This is definitely easier to construct and verify, but it's less interesting as an object in its own right; offhand I don't know of any natural examples which do not require at least a little bit of analysis to verify.
Noah Schweber
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Using the axiom of choice, well order the reals in type $2^{\aleph_0}$ and go through and chose a real to belong in set $A$ and which has not already been chosen for $A^c$ and belongs in a given interval. Do this then for set $A^c$. Organize so that each interval with rational endpoints occurs uncountably many times.
Rene Schipperus
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Note that the axiom of choice is not actually needed for this problem, and in fact there are natural examples. – Noah Schweber Dec 12 '20 at 04:02
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Just consider the unit interval, and take the decimal expansion of the reals. Let A be the set of reals which, from some point on, all even decimal positions are zero. Then given an interval there will be a finite expansion so that all completions of the finite decimal to an infinite decimal lie in the given interval. There are unaccountably many in $A$, as the odd places can be chosen at will. $A^c$ clearly has the same property.
Rene Schipperus
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This is basically equivalent to the second example in my answer. – Noah Schweber Dec 12 '20 at 04:36