I call a set PECULIAR, if its elements are uncountable, pairwise disjoint subsets of R (the real number system). As for example, the set {(0,1),(3,5),[8,9]\Q},where Q denotes the set of rationals, is peculiar. My first question may have a very trivial answer that I cannot see immediately,that does there exist an uncountable, peculiar set? Next, I define the STRENGTH of a peculiar set to be the union of all its elements (which are sets). My question is that what are the subsets of R (if any), that are the strength of some uncountable peculiar set? In case complete specification seems impossible or meaningless, some examples will do.
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2$\begingroup$ i do not think that this question is appropriate for math overflow. Every uncountable set is the disjoint union of uncountably many uncountable sets. $\endgroup$– GoldsternCommented Dec 17, 2011 at 16:34
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1$\begingroup$ Take a bijection $f\colon\mathbb R\to\mathbb R\times\mathbb R$ and define the peculiar collection $\{\{f^{-1}(a,x)\mid a\in\mathbb R\}\mid x\in\mathbb R\}$. $\endgroup$– Asaf Karagila ♦Commented Dec 17, 2011 at 16:41
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2 Answers
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Since the OP asked about sets of reals, and since these cannot be amorphous (because no amorphous set admits a linear ordering), let me point out that it is consistent with ZF that there is a Dedekind finite (hence not countable, as in Joel's answer) set of reals that cannot be partitioned into uncountably many many infinite pieces. One model in which this happens is the basic Cohen model, as defined in Section 5.3 of Jech's book "The Axiom of Choice". The relevant Dedekind finite set is the set of Cohen reals explicitly added by the forcing (the set Jech calls $A$). The fact that it cannot be split into uncountably many uncountable sets follows from work of Halpern and Levy ("The Boolean prime ideal theorem does not imply the axiom of choice," Axiomatic Set Theory, Part I, Proc. Symp. Pure Math. XIII, Part I (1971) pp. 83-134).
answered Dec 17, 2011 at 17:52
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$\begingroup$ This set $A$ cannot even be split into infinitely many infinite sets, regardless to uncountability of either the partition or the parts. $\endgroup$– Asaf Karagila ♦Commented Dec 17, 2011 at 19:41
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$\begingroup$ @Asaf: If you view the elements of $A$ as subsets of $\omega$, won't there be infinitely many whose first element is $n$, for each $n\in\omega$? $\endgroup$ Commented Dec 17, 2011 at 22:25
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$\begingroup$ @Andreas: While it does sound as a very convincing argument, I have run into several proofs that this set is a counterexample to the assertion that every set has a group operation, by the fact that every partition has only finitely many non-singletons. At this moment I am too tired to verify all the details of either arguments, and it will have to wait for tomorrow. $\endgroup$– Asaf Karagila ♦Commented Dec 17, 2011 at 22:38
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$\begingroup$ See Justin's answer in this link: mathoverflow.net/questions/12973/… In Cohen's basic model the D-finite set of reals is "strong" in the amorphous sense of the word (no partition into infinitely many non-singletons exists). $\endgroup$– Asaf Karagila ♦Commented Dec 18, 2011 at 23:05
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$\begingroup$ @Asaf: Justin wrote that $A$ can't be partitioned into finite sets, and that proof looks OK to me. But I don't see how that leads to your claim that $A$ can't be partitioned into infinitely many infinite sets. Specifically, if I try to carry out a proof like Justin's in the case where the pieces are allowed to be infinite, there's no condition like Justin's $q$ that decides which $a_i$'s (with a ground-model set of indices $i$) constitute a particular piece of the partition. $\endgroup$ Commented Dec 19, 2011 at 0:39
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The comments have answered the question in the context of ZFC, where the axiom of choice is available.
But there is something interesting to say when the axiom of choice fails (and this answer was too long for a comment). Namely, it is consistent with ZF that not every uncountable set can be partitioned into uncountably many disjoint uncountable sets. Indeed, it is consistent with ZF that an uncountable set cannot even be partitioned into two disjoint infinite sets. This is because it is consistent with ZF that there is an amorphous set, an infinite set all of whose subsets are either finite or the complement of a finite set. It follows that amorphous sets cannot be countable, since if there were an enumeration we could produce bad subsets by taking every other element. Thus, amorphous sets are uncountable sets with no partition into two or more disjoint infinite sets.
answered Dec 17, 2011 at 17:00