"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?
Uncountable Sets that can not be expressed as a disjoint union of two Uncountable sets
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set-theory
axiom-of-choice
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2What 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... – 2011-10-06
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0I 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.) – 2011-10-06
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0Suppose 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? – 2011-10-06
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0@six yes, but you don't need to use this here, and there's a simpler solution as hinted to above. – 2011-10-06
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0@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. – 2011-10-06
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2@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. – 2011-10-06
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0Ah 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. – 2011-10-06
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0Consider 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. – 2011-10-06
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0@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?) – 2011-10-06
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0There are models of $ZF$ in which there are no free ultrafilters on any set; of course $AC$ fails in those models. – 2011-10-06
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1See [Asaf's answer here](http://math.stackexchange.com/questions/60281/) 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. – 2011-10-06
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0I'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. – 2011-10-06
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0@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... :-)) – 2011-10-06
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0Ha, 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? – 2011-10-06
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0No, 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). – 2011-10-06
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0I'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? – 2011-10-06
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0@Asaf: Are there any models with an amorphous set of reals? – 2011-10-06
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0@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. – 2011-10-06
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0@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 – 2011-10-06
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0I 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. – 2011-10-06
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0Just 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. – 2011-10-06