8

Is it explicitly possible to take $ \mathbb R$ and to divide it into two sets (say $\mathbb A$ and $\mathbb B$), which are : disjoint ($\mathbb A \cap \mathbb B = \emptyset$), totally disconnected (i.e. contain no open intervals) (EDIT:) and for all intervals $(a,b)$, $|\mathbb A \cap (a,b)| = |\mathbb B \cap (a,b)| = \mathfrak c$

I asked my measure theory professor, and he said he couldn't think of why such a construction would be impossible, but whenever I try to create such two sets I always end up with undefinable points, nor did an internet search reveal anything about this.

Asaf Karagila
  • 393,674
  • 1
    For an example that satisfies the added "local measure" condition, perhaps the union of all the rational translates of the Cantor set (and the complement of that union) does the trick? – Travis Willse Mar 13 '17 at 00:22
  • 2
    ... or, a simpler representation of @Travis's idea: let $A$ be the set of numbers whose decimal representation eventually consists only of the digits $2$ and $7$. – hmakholm left over Monica Mar 13 '17 at 00:25
  • @Henning-Makholm Thats the Smith-Volterra-Cantor set, its compliment contains intervals –  Mar 13 '17 at 00:28
  • 2
    @EdvinOrlov: No it isn't. My $A$ contains numbers in every interval -- note that its elements can have a finite number of digits that are not $2$ or $7$. In particular, $A$ contains every number of the form $\frac n{10^m}+\frac29$; those are a dense set. – hmakholm left over Monica Mar 13 '17 at 00:31
  • @Henning-Makholm my bad. I think that is the solution to the question, thank you! –  Mar 13 '17 at 00:41
  • Just to make it seem more psychologically equal, I'll go with $A$ is the set of all reals, whose decimal representation contains only $0,2,4,6,8 $ :D –  Mar 13 '17 at 00:48
  • 1
    @EdvinOrlov That's not right - it's the set of reals whose decimal representations eventually consist only of ---s. E.g. in Henning's example, $0.123452727272727272727272727...$ would be in the set, even though it has a few non-$2$-or-$7$ digits. The set you've defined has a complement containing lots of intervals. – Noah Schweber Mar 13 '17 at 01:11
  • @Noah-Schweber if $(a,b)$ is such an interval in $A^{c}$, there exists a rational $r$ with a finite decimal expansion in the interval ( $ a < r < b $). Since $A$ was the set of numbers whose representation contains only numbers from the set ${0,2,4,6,8}$, the complement contains numbers whose repre. is from ${1,3,5,7,9}$. Trivially there exists a natural number $m$ such that $2 \times 10^{-m} + r $ is in the interval, yet such a number is not in $A^{c}$, hence $(a,b)$ is not an interval in $A^{c}$ –  Mar 13 '17 at 01:47
  • 1
    @EdvinOrlov "Since $A$ was the set of numbers whose representation contains only numbers from the set ${0,2,4,6,8}$, the complement contains numbers whose representation is from ${1,3,5,7,9}$." That's incorrect. For example, the number $0.11222222...$ has a "$1$" in it, so is not in $A$ - if you're not in $A$, you're in the complement of $A$! In particular, every number in the interval $(0.11, 0.12)$ has a $1$ in its decimal expansion, so is not in $A$ - so that whole interval is in the complement of $A$. In fact, $A$ is nowhere dense. – Noah Schweber Mar 13 '17 at 01:51
  • 2
    That's why the word "eventually" is crucial in Henning's answer - that way, given any finite decimal sequence (say, $113125$) we can find a number that begins with that sequence but is still in the set (say, $0.113125222222222...$), in particular implying that $A$ is dense. Do you see why this is important, now? – Noah Schweber Mar 13 '17 at 01:54
  • 1
    Yes, I see now. Thanks for taking the time to point that out! –  Mar 13 '17 at 01:56
  • 2
    In your revised question, "totally disconnected" is redundant. – bof Mar 13 '17 at 02:40
  • 2
    @HenningMakholm's set $A$ is clearly still a Lebesgue measure null set (statistically, "all" numbers are normal). What would happen if one strengthen that "local" measure requirement, and say that for each interval $(a,b)$ the intersection $A\cap (a,b)$ must possess a Lebesgue measure, and that measure is exactly half of the interval length, $(b-a)/2$? That seems hard to come up with. But why would it not exist? – Jeppe Stig Nielsen Mar 13 '17 at 11:33
  • 1
    @JeppeStigNielsen: Good question -- I've considered asking exactly that as a separate question (but didn't get around to it yet). – hmakholm left over Monica Mar 13 '17 at 11:41
  • 1
    That is a very similar question to this one that I just recently found - http://math.stackexchange.com/questions/2123343/everywhere-super-dense-subset-of-mathbbr/2123347 –  Mar 13 '17 at 11:50
  • Edvin, nice reference. In that thread Robert Israel says no: a set that locally carries exactly half the "weight" of the real numbers in the sense of the latest comment above by myself and @HenningMakholm, does not exist. – Jeppe Stig Nielsen Mar 13 '17 at 12:19

3 Answers3

11

Let $$\mathbb A=C+\mathbb Q=\{x+y:x\in C,\ y\in\mathbb Q\}$$ where $C$ is the Cantor set, and let $\mathbb B=\mathbb R\setminus\mathbb A.$

Since $\mathbb A$ and $\mathbb B$ are Borel sets ($F_\sigma$ and $G_\delta$ respectively), it will suffice to show that $\mathbb A\cap(a,b)$ and $\mathbb B\cap(a,b)$ are uncountable.

Since $\mathbb A$ has measure zero (as the union of countably many translates of $C$), $\mathbb B\cap(a,b)$ has positive measure, so it's uncountable.

Choose $q\in\mathbb Q\cap(a,b).$ Since every neighborhood of $0$ contains uncountably many points of $C,$ every neighborhood of $q$ contains uncountably many points of the set $C+q;$ in particular, the interval $(a,b)$ contains uncountably many points of the set $C+q\subseteq C+\mathbb Q=\mathbb A.$

bof
  • 78,265
5

Consider the set $$\big( (-\infty, 0) \cap \Bbb Q \big) \cup \big([0, \infty) \cap (\Bbb R - \Bbb Q) \big)$$ ---that is, the union of the negative rationals and the nonnegative irrationals---and its complement.

Travis Willse
  • 99,363
  • I forgot an important nuance in the definition, but your example works for what I wrote at first –  Mar 13 '17 at 00:17
3

Yes:

$$A= \{ x=b.b_1b_2..b_n... | 0.b_2b_4b_6... \in \mathbb Q \} \\ B= \{ x=b.b_1b_2..b_n... | 0.b_2b_4b_6... \notin \mathbb Q \} $$

For any interval $(a,b)$ it is easy to construct onto functions from $A \cap(a,b) , B \cap(a,b) $ to $(0,1)$.

Indeed, pick some $a < \frac{k}{10^{2n}} < \frac{k+1}{10^{2n}} < b$ and show that $$ x=b.b_1b_2..b_n... \to 0.b_{2n+1}b_{2n+3}...$$ is onto function from $A \cap(a,b)$ and $B \cap(a,b) $ to $(0,1)$.

This example has the roots in the example of a discontinuous function with the IVP that appears in Sierpiński (I think).

bof
  • 78,265
N. S.
  • 132,525