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"Can you to any uncountably infinite set $M$ find an uncountably infinite family $F$ consisting of pairwise disjoint uncountably infinite subsets of $M$?"

Intuitively, I feel like it should be possible for the real numbers at least: you simply split the real numbers into two intervals, and since there are uncountably many points where you can do that, there are uncountably many ways to split the reals into two disjoint subsets. But this question isn't asking specifically about the real numbers, so how can I prove this more generally?

Peter Taylor
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Andrea
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Your approach does not work even for $\Bbb R$: you’ve merely shown that there are uncountably many different ways to split $\Bbb R$ into two disjoint uncountable sets. Your task is to split $M$ into uncountably many pairwise disjoint uncountable sets.

HINT: $|M\times M|=|M|$ (assuming the axiom of choice).

Brian M. Scott
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  • Sorry if my answer is similar to the one you intended; I had the idea and then I read your post. It may allow me to practice my rusty set theory. – user99680 Nov 25 '13 at 01:51
  • @user99680: What you’ve done works for $\Bbb R$ and is the application to $\Bbb R$ of the idea that I had in mind, but your attempt to extend it to arbitrary uncountable $S$ doesn’t work. – Brian M. Scott Nov 25 '13 at 01:58
  • Doesn't it work assuming $AC$, so that cardinality is well-defined? AFAIK, a set being uncountable is equivalent ( under AC, I think) to being in bijection with $\mathbb R$ , just as being countably-infinite means being in bijection with $\mathbb N$. – user99680 Nov 25 '13 at 02:00
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    @user99680: No, absolutely not. $\wp(\Bbb R)$ is an uncountable set admitting no bijection with $\Bbb R$. Uncountable simply means not countable, i.e., not finite and not countably infinite. – Brian M. Scott Nov 25 '13 at 02:06
  • I'm not sure I know what to do with this information but I'll give it a try. I guess I start by considering pairs of elements $(x,y)$ where both $x$ and $y$ are in $M$. Then for a fixed $x$ there are uncountably many $y$, forming an uncountable subset of $M\times M$, and since there are uncountably many $x$, there are uncountably many such uncountable sets (wow, this is hard to wrap your head around) and they are all disjoint. So I've solved the problem for $M\times M$. But finally, since |M×M|=|M|, there is a bijection that gives me a corresponding result for $M$ itself. Did I get it right? – Andrea Nov 25 '13 at 02:15
  • @Brian Scott: I was referring to uncountable , as having cardinality $\aleph 1$ . I'm aware that under CH, $P(\mathbb R)$ and $\mathbb R$ do not have the same cardinality. I will edit my post to reflect this. – user99680 Nov 25 '13 at 02:16
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    @Andreas: I think so, though the description is awkward enough that I’m not quite sure. See if this is what you have in mind. For each $x\in M$ let $M_x={x}\times M$; then ${M_x:x\in M}$ is a partition of $M\times M$ into uncountably many uncountable sets — specifically, into $|M|$ sets each of cardinality $|M|$. Then if $h:M\times M\to M$ is a bijection, ${h[M_x]:x\in M}$ is a partition of $M$ into uncountably many uncountable sets. – Brian M. Scott Nov 25 '13 at 02:18
  • @user99680: But that’s simply not what uncountable means, ever, and $\mathsf{CH}$ has nothing to do with whether $\Bbb R$ and $\wp(\Bbb R)$ have the same cardinality: they don’t, full stop. See Cantor’s theorem. – Brian M. Scott Nov 25 '13 at 02:19
  • @Brian M. Scott: I never claimed the two had the same cardinality. – user99680 Nov 25 '13 at 02:26
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    @user99680: You said that under $\mathsf{CH}$ they do not have the same cardinality; mentioning $\mathsf{CH}$ in this connections makes no sense unless you thought that in the absence of $\mathsf{CH}$ they might have the same cardinality. I’m simply pointing out that they don’t, ever, so that the mention of $\mathsf{CH}$ is completely irrelevant. – Brian M. Scott Nov 25 '13 at 02:38
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I think this should work; maybe it is what Brian Scott had in mind, this works for sets of the same cardinality as $\mathbb R$. Use the fact that $|\mathbb R \times \mathbb R|= |\mathbb R|$. Now, $\mathbb R^2$ can be split into an uncountable set of copies of $\mathbb R $ , almost by definition, since $\mathbb R^2=\mathbb R \times \mathbb R $ (setwise). Now, for any set of the same cardinality as $\mathbb R$ , use a bijection between $\mathbb R^2$ and any set $S$ with $|S|=\aleph_1$ , to do a partition for $S$.

user99680
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    There are many uncountable sets not in bijection with $\Bbb R$. Under CH you can always biject $\Bbb R$ with part of an uncountable set, then make the rest (if any) another part, but if CH fails there may be uncountable sets strictly smaller than $\Bbb R$ – Ross Millikan Nov 25 '13 at 01:56
  • If $S$ is an arbitrary uncountable set, there need not be a bijection between $\Bbb R^2$ and $S$. – Brian M. Scott Nov 25 '13 at 01:56
  • Well, I'm assumming $C.H$ Maybe I should state that. I may also have to assume C.H too to guarantee the existence of the bijection. – user99680 Nov 25 '13 at 01:58
  • As an example, we know $\mathcal{P}(\Bbb R)$ cannot be put into bijection with $\Bbb R$, hence not with $\Bbb R^n$ for any $n\geq1$. – Clayton Nov 25 '13 at 01:59
  • @Clayton: But under generalized CH, $|P(\mathbb R)| \neq |\mathbb R |$ – user99680 Nov 25 '13 at 02:03
  • @user99680: Precisely. $\mathcal{P}(\Bbb R)$ is uncountable, but it cannot be put into bijection with $\Bbb R^2$. This is the flaw of your last sentence. – Clayton Nov 25 '13 at 02:04
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    @user99680: $|\wp(\Bbb R)|$ is never equal to $|\Bbb R|$; this has nothing to do with CH or GCH. – Brian M. Scott Nov 25 '13 at 02:07
  • @Clayton: But by uncountable, I mean $\aleph_1 $ , and, I believe, in CH, $P(\mathbb R)= \aleph_2$, while $|\mathbb R =\aleph_1|$ . Is this wrong? I don't know if this is circular, but, by uncountable, I mean of the same cardinality as $\mathbb R $ – user99680 Nov 25 '13 at 02:07
  • @Brian M. Scott: Maybe my usage is non-standard, and maybe this is circular, but by uncountable, I – user99680 Nov 25 '13 at 02:14
  • Please see my edit; I specified that it works for sets having cardinality $\aleph_1$. I guess my use of uncountable is non-standard. – user99680 Nov 25 '13 at 02:19
  • $\mathsf{CH}$ does not imply that $\mathcal{P}(\mathbb{R}) = \aleph_2$, although $\mathsf{GCH}$ (the generalized continuum hypothesis) does. And yes, your use to "uncountable" to mean something other than "not countable" is definitely not standard. – Trevor Wilson Nov 25 '13 at 02:53
  • @TrvorWilson: You're right; it is GCH. Still, when you hang out in websites where (horror! ;)) set theory is used only in the background, or results are assumed, "üncountable set" is used almost exclusively to denote that the set has the cardinality of $\mathbb R$ , since cardinalities other than those two just do not come up most of the time. – user99680 Nov 25 '13 at 04:30
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It will be too much work to keep typing "uncountably infinite", so I will just type "uncountable" which means the same thing.

Plainly, if $M$ is an uncountable set, the Cartesian product $M\times M$ contains an uncountable family of pairwise disjoint uncountable sets.

The set $\omega_1$ of all countable ordinals is an uncountable set; its cardinality is called $\aleph_1$. Thus, the set $\omega_1\times\omega_1$ contains an uncountable family of pairwise disjoint uncountable sets. The cardinality of $\omega_1\times\omega_1$ is $\aleph_1\cdot\aleph_1=\aleph_1^2=\aleph_1$. (None of these equalities depends on the Axiom of Choice.) Therefore, every set of cardinality $\aleph_1$ contains an uncountable family of pairwise disjoint uncountable sets.

To conclude from this that every uncountable set contains an uncountable family of pairwise disjoint uncountable sets, we need a weak form of the Axiom of Choice, namely: "Every uncountable set has a subset of cardinality $\aleph_1$." In other words, this says that $\aleph_1$ is the smallest uncountable cardinal.

It can also be shown without the Axiom of Choice that $(2^{\aleph_0})^2=2^{\aleph_0}$, i.e., there is a bijection between $\mathbb R\times\mathbb R$ and $\mathbb R$. Hence the general result follows from the still weaker axiom: "Every uncountable set has a subset whose cardinality is either $\aleph_1$ or $2^{\aleph_0}$."

bof
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