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"There does not exist an uncountable subset of the real numbers which can not be expressed as the union of two uncountable sets which are disjoint from one another" is obviously true assuming choice. However, I'm looking for a proof of this statement that does not involve choice. I believe I have come up with one myself, but it's a little complicated. Is there any easy way to go about this?

six
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    What about something like this : you have an uncountable subset of reals $S\subset\mathbb{R}$. Define for all $x\in\mathbb{R}$ the following subsets : $S_{\leq x}=S~\cap~ ]-\infty,x]$ and $S_{>x}=S~\cap~ ]x,+\infty[$. Then for some $x$ both should be uncountable... – Olivier Bégassat Oct 06 '11 at 02:58
  • I up-voted the question, but I feel it will be nice if you share with us what you already know. For instance, (1.) Assuming choice, what is the obvious partition into disjoint uncountable sets? (2.) And what is your construction that doesn't involve choice? (Atleast a gist of the constructions if not the full details.) – Srivatsan Oct 06 '11 at 03:00
  • Suppose U is our uncountable set, then assuming choice, we have a bijection between U and and the union of two disjoint sets (not necessarily subsets of U) with the same cardinality as U, right?

    Also Olivier, are you not assuming choice using that argument? Aren't you associating a set with every element x in R?

     – six Oct 06 '11 at 03:36
  • @six yes, but you don't need to use this here, and there's a simpler solution as hinted to above. – Olivier Bégassat Oct 06 '11 at 03:38
  • @six I use no choice. I define for every $x$ two subsets as intersctions of sets, there is no choice made anywhere. FInally, for finding an $x$ that works, you must realize that the range of such $x$ is a non empty interval, and depending on what kind of interval it is (wether it extends to infinity on two sides, one side or is bounded) you can define an element inside of it without choice. – Olivier Bégassat Oct 06 '11 at 03:41
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    @Olivier: You don’t need AC to define the sets $S_{\le x}$ and $S_{>x}$, but I’m not yet convinced that you can show that there is an $x$ for which both are uncountable without using at least countable choice. – Brian M. Scott Oct 06 '11 at 03:53
  • Ah right, I see now. Is it possible that we can actually generalize this to any arbitrary set? My approach used Ultrafilters. I will go into more detail in a bit. – six Oct 06 '11 at 03:57
  • Consider a set U with cardinality K. We know that the Ultrafilters have the property that for all subsets E of U, either E or U\E is in the Ultrafilter. What I did was, I proved that there must exist at least one Ultrafilter that had the property we were looking for. – six Oct 06 '11 at 04:19
  • @six: And so, presumably, you are using AC (in its Zorn's Lemma guise) to assert that ultrafilters exist (unless you are working with principal ones exclusively?) – Arturo Magidin Oct 06 '11 at 04:28
  • There are models of $ZF$ in which there are no free ultrafilters on any set; of course $AC$ fails in those models. – Brian M. Scott Oct 06 '11 at 04:31
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    See Asaf's answer here for a discussion of amorphous sets and other pathologies with cardinalities that may occur without choice. It doesn't address the question whether those can be realized as subsets of the reals but it should be of interest anyway. – t.b. Oct 06 '11 at 04:38
  • I'm not using the Ultrafilter lemma, I don't believe. Are you not allowed to define a Ultrafilter by saying that for all subsets E, either E is in U, or X/E is in U? I don't believe you need choice in order to prove that a filter is an Ultrafilter if and only if the filter has that given property.

    I think I see what you're saying though. I suppose the only Ultrafilters that I can define without choice are the fixed ones. Which in that case, my proof isn't going to work without choice. Thanks for the help.

     – six Oct 06 '11 at 04:41
  • @six: There are many restricted choice principles which are equivalent to the ultrafilter lemma. Even if you don't use it per se, it is possible that some choice sneaked into your proof in an equivalent form (you'd be surprised how many times it happens to me, and I work in choiceless environments all the time... :-)) – Asaf Karagila Oct 06 '11 at 06:40
  • Ha, I see you have no choice! Well tell me this, suppose for any uncountable set U, I am allowed to construct a fixed ultrafilter M (I believe this is allowed without choice). Now, given any subset E of U, I know that either E is in M, or U\E is in M. Again, this can be proved without choice, I believe. Thus, if U could not be split into two uncountable sets that are disjoint from one another, then we see that either E or U\E is countable. If I were to be able to show that M must be able to be split into two uncountable sets, then I would be done, right? – six Oct 06 '11 at 07:23
  • No, you cannot. There are models in which there are no free ultrafilters. Defining a non-principal ultrafilter requires choice, not full axiom of choice, but it still needs some (for example, compactness theorem in logic; completeness theorem; both equivalent to the Ultrafilter lemma as well). – Asaf Karagila Oct 06 '11 at 07:28
  • I'm working on an answer for the past hour, and slowly eliminating the possible models in which this cannot be consistent in. I will post my answer as the day progresses. Tell me, how familiar are you with forcing? – Asaf Karagila Oct 06 '11 at 07:30
  • @Asaf: Are there any models with an amorphous set of reals? – Brian M. Scott Oct 06 '11 at 07:49
  • @Brian: Amorphous sets cannot be linearly ordered, so no. That means, also, that every Dedekind-finite set of reals can be split into two uncountable sets, otherwise it would have been amorphous. – Asaf Karagila Oct 06 '11 at 07:52
  • @Asaf: Ah, I see. There can be at most finitely many $x$ such that $(\leftarrow,x]$ is finite and only finitely many such that $[x,\to)$ is finite, but that leads to an immediate contradiction – Brian M. Scott Oct 06 '11 at 08:05
  • I am familiar enough, I would love to hear your answer though. Thank you for your help. Also, I was talking about fixed or principal ultrafilters. – six Oct 06 '11 at 08:20
  • Just posting a small update, there might be a way to prove by contradiction that if such $X$ (that cannot be decomposed) exists then we can force it to have an amorphous subset; since $X$ is still linearly ordered by $\mathbb R$'s usual ordering this would be impossible. I have to iron out quite a bunch of details though. It may take some time. – Asaf Karagila Oct 06 '11 at 12:25

1 Answers1

6

Let $x\in\mathbb{R}$. Define $$S_{\leq x}=S~\cap~(-\infty,x]~~~\mathrm{and}~~~S_{>x}=S~\cap~ (x,+\infty)$$ and define $S_{\leq}=\{$ the set of all $x$ such that $S_{\leq x}$ is uncountable $\}$, and $S_>=\{$ the set of all $x$ such that $S_{>x}$ is uncountable $\}$.

$S_{\leq}$ is a non empty$^{(*)}$ interval that extends to infinity on the right, and is therefore of the form $[a,+\infty)$ or $(a,+\infty)$ for some $a\in\mathbb{R}$ or is equal to the whole of $\mathbb{R}$.

Similarly, $S_>$ is a non empty$^{(*)}$ interval that extends to infinity on the left, and is thus of the form $(-\infty,b]$ or $(-\infty,b)$ for some $b\in\mathbb{R}$ or is equal to the whole of $\mathbb{R}$.

$^{(*)}$ : I have used some form of choice I believe when I say that $S_>$ and $S_{\leq}$ are non empty, because I use the fact that with some form of choice, the countable union of countable sets is still countable. Is it countable choice? Since $$S=\cup_{n\in\mathbb{N}}S_{\leq n}$$ is uncountable, at least one of the $S_{<n}$ must be uncountable, and so for some natural number $n$ you have $n\in S_{<}\neq\emptyset$.

Let's show that $S_>\cap S_{\leq}$ is non empty. If either one of $S_>$ or $S_{\leq}$ is the whole real line, there is no problem. So let's suppose none of them is the whole of $\mathbb{R}$.

Suppose $x\in S_{\leq}$ that is $S_{\leq x}=S~\cap~(-\infty,x]$ is uncountable. Since $$\bigcup_{n\in\mathbb{N}^*}S_{\leq x-\frac{1}{n}}\subset S_{\leq x}=S~\cap~(-\infty,x]\subset \{x\}\cup \bigcup_{n\in\mathbb{N}^*}S_{\leq x-\frac{1}{n}}$$ the already used fact that countable unions of countable sets are countable with some form of choice implies that at least one of the $S_{\leq x -\frac{1}{n}}$ is uncountable, which means that for some $n\in\mathbb{N}^*,~x -\frac{1}{n}\in L$ and thus $S_{\leq}$ is open, that is, $$\mathbf{S_{\leq}=(b,+\infty)}$$ for some real number $b$. Similarly, $$\mathbf{S_{>}=(-\infty,a)}$$ for some real number $a$.

Then $S_>\cap S_{\leq}$ is empty iff $a\leq b$. But this would entail that $$S=S_{\leq a}\cup S_{> a}$$ is countable as the union of two countable sets, since $a \notin (-\infty,a)=S_{\leq}$ means $S_{\leq a}$ is countable, and $a \notin (b,+\infty)=S_{>}$ means $S_{> a}$ countable.

Olivier Bégassat
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    You’ve definitely used some form of choice, because there are models in which $\mathbb{R}$ is a countable union of countable sets. – Brian M. Scott Oct 06 '11 at 05:00
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    After further investigation: you’ve used CUT($\mathbb{R}$) (countable unions of countable subsets of $\mathbb{R}$ are countable), which is strictly weaker than CC($\mathbb{R}$) (axiom of countable choice for subsets of $\mathbb{R}$), which is strictly weaker than CC (every countable family of non-empty sets has a choice function). – Brian M. Scott Oct 06 '11 at 05:09
  • @Brian M. Scott yes. What does 'CUT' stand for? 'C'ountable 'U'nion ... ? – Olivier Bégassat Oct 06 '11 at 05:33
  • Countable Union Theorem. I don’t know how widely used the name and abbreviation are; I took them from Herrlich’s Axiom of Choice (Springer Lecture Notes in Mathematics). – Brian M. Scott Oct 06 '11 at 05:52