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Locally non-enumerable dense subsets of R

This may be a standard result... I don't know... anyway...

Does there exists an uncountable dense subset of $\mathbb{R}$ whose complement is also uncountable and dense?

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marked as duplicate by Carl Mummert, Asaf Karagila, t.b., Ross Millikan, Willie Wong Jul 19 '11 at 19:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

This is much like…. As the asker there notes, you can take the rationals $<0$ and the irrationals $>0$ as one set, and its complement as the other. – Nick Strehlke Jul 19 '11 at 15:36
A similar problem that I find really fascinating is to find a dense set (in $\mathbb{R}$) of rational numbers whose compliment in the rationals is still dense, i.e. a partition of the rationals into two sets, both of which are dense. – Matt Jul 19 '11 at 15:51
@Matt: I think the set of dyadic rationals works. – LostInMath Jul 19 '11 at 15:57
The dyadic rationals (numbers that can be expressed as an integer divided by a nonnegative power of 2) and the remaining rationals are each dense in the reals and pairwise disjoint. By considering an analogous decomposition into all possible "prime-adic" rationals and the remaining rationals, you can get infinitely many pairwise disjoint sets of rationals, each of which is dense in the reals. Interestingly, if we weaken "pairwise disjoint" to "pairwise finite intersection", it's possible to get uncountably many dense sets of rationals. – Dave L. Renfro Jul 19 '11 at 15:57
My favorite example is fairly obscure. If $\sigma(n)$ is the sum of divisors of $n$, then the set of rational numbers of the form $\sigma(n)/n$ is dense, and the set of rationals not of that form is also dense! (This is sometimes called the "abundancy index" of an integer). Oops. I guess strictly speaking this is not true, they are only dense in $(-\infty, -1]\cup [1, \infty)$, still kind of weird though. – Matt Jul 19 '11 at 16:06
up vote 9 down vote accepted

Funny, I asked myself the same question at one point! This should work:

$$A:=((-\infty,0]\cap\mathbb{Q})\cup ((0,+\infty)\cap(\mathbb{R}\setminus\mathbb{Q})),$$ $$B:=((-\infty,0]\cap(\mathbb{R}\setminus\mathbb{Q}))\cup ((0,+\infty)\cap\mathbb{Q}) $$ They are both uncountable and dense, and form a partition of $\mathbb{R}$.

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It doesn't get much simpler than that! (Note that intersecting $(-\infty,0]$ with $\mathbb{R}$ is a bit redundant.) – François G. Dorais Jul 19 '11 at 16:13
thanks, I'll correct that, I forgot to put parantheses... – Olivier Bégassat Jul 19 '11 at 16:16
It's even simpler to write $(-\infty,0] \setminus \mathbb{Q}$... – François G. Dorais Jul 19 '11 at 16:18
Absolutely, but I prefer the above notation. – Olivier Bégassat Jul 19 '11 at 16:21
As I commented at another (now deleted?) answer, these can be nicely written as $A = (0, \infty) \vartriangle \mathbb Q$ and $B = \mathbb R \setminus A = (-\infty, 0] \vartriangle \mathbb Q$, where $\vartriangle$ is symmetric set difference. – Ilmari Karonen Jul 19 '11 at 19:33

Take any uncountable set $A_0$ with measure $0$, e.g., the Cantor set. Let $$A = \bigcup_{q \in \mathbb{Q}} (A_0 + q) = \{a + q : a \in A_0, q \in \mathbb{Q}\}.$$ Then $A$ is uncountable, dense and it still has measure $0$ since it is the union of countably many translates of $A_0$. Since the complement of a measure $0$ set is always uncountable and dense, this set $A$ fits the bill.

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Probably not the best approach, but you can also try this: Let $B$ be a basis of $\mathbb R$ over $\mathbb Q$. Write $B=B_1 \cup B_2$ with $B_1 \cap B_2 =\emptyset$ and $B_1, B_2$ of the same cardinality.

Then $V= span_{\mathbb Q} B_1$ will do it. It is uncountable and dense ($B_1$ in uncountable, thus it contains two elements with irrational ration) and $\mathbb R \backslash V$ contains $W= span_{\mathbb Q} B_2$.

Added a Second example

Here is another example:

$$A:= \{ x=m.x_1x_2...x_n...| 0.x_2x_4x_6... {\rm \, is \, periodic \,} \}\,.$$

More exactly, $A$ consists of all those real numbers for which, the digits after $.$ of even order form a periodic sequence....

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The only drawback is that you need some Axiom of Choice to get a basis for $\mathbb{R}$ over $\mathbb{Q}$... – François G. Dorais Jul 19 '11 at 16:15
True... And I am pretty sure one also needs the Axiom of choice to split $B$ that way... – N. S. Jul 19 '11 at 16:25

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