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Continuum hypothesis states, there is no set with cardinality between the integers and the reals.

There is a milestone result, that CH is independent from ZFC. That means, both of ZFC + CH, and ZFC + not-CH are consistent.

What if ZFC and not-CH. Thus, we have an axiom which states, there is a cardinality between $\aleph_0$ and $2^{\aleph_0}$.

Can a such set be defined?

peterh
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  • As far as the suggestion goes: [tag:set-theory] should suffice in contrast to [tag:elementary-set-theory]. – AlexR Feb 21 '15 at 22:47
  • By the way: Actually ZFC isn't known to be consistent. See Asafs comment – AlexR Feb 21 '15 at 22:50
  • @Meelo If somebody shows a set definition whose cardinality would be between them. Maybe I should have used the word "defined". If it passes better, feel free yourself to fix. – peterh Feb 21 '15 at 22:51
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    And by the way, the independence of CH is now over 50 years old. Sure, new compared some things, but not really new in terms of mathematics. – Asaf Karagila Feb 21 '15 at 22:52
  • Seems a little odd to call a fifty-year-old result 'relatively new'! – Brian M. Scott Feb 21 '15 at 22:53
  • How do you feel about 'the set of all countable ordinals'? – Steven Stadnicki Feb 21 '15 at 22:54
  • @BrianM.Scott I am a curious layman. To me, everything is new after around 1900. :-) – peterh Feb 21 '15 at 22:54
  • @StevenStadnicki I think, they are elements of the $\aleph_0$, $\aleph_1$, ... series, whose cardinality is $\aleph_0$. – peterh Feb 21 '15 at 22:55
  • @AsafKaragila Independence means: Both of ZFC + CH and ZFC + not-CH are consistent if and only if ZFC is consistent. Consistence of ZFC isn't needed to be proven. – peterh Feb 21 '15 at 23:00
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    @peterh those are cardinals (alephs), not ordinals, and only the first of those is countable. $\omega_1$ (from Asaf's answer) is the set of countable ordinals, and while that's not a construction per se, it is a somewhat 'graspable' definition. – Steven Stadnicki Feb 21 '15 at 23:06
  • A more interesting question is whether, under the failure of the continuum hypothesis, a set of reals of intermediate cardinality can be "explicitly defined". The answer is that (consistently) this is not the case. – Andrés E. Caicedo Feb 21 '15 at 23:46
  • @AndresCaicedo Well, the brain of a programmer says for that, then CH is de facto true. – peterh Feb 21 '15 at 23:57
  • @AndresCaicedo Can it be proven, that no such set of reals exists? – peterh Feb 22 '15 at 00:00
  • @peterh Under appropriate large cardinal assumptions it can be proved that any set of reals in $L(\mathbb R)$ is either countable or of the same size as $\mathbb R$. "Belonging to $L(\mathbb R)$" is a (more than generous) formalization of "being explicitly definable". The result is stronger, in the sense that $L(\mathbb R)$ can be replaced by larger classes, and we actually get a rich structure theory of set of reals. Of course, the relevant large cardinal assumptions are not provable in $\mathsf{ZFC}$, although many set theorists would agree that they are part of the "correct" axiomatization. – Andrés E. Caicedo Feb 22 '15 at 00:06
  • I do not know what you mean by $L(\mathbb Z)$. Your use of $\mathbb R\setminus L(\mathbb R)$ is also incorrect. Anyway, no, the continuum hypothesis is not a theorem. – Andrés E. Caicedo Feb 22 '15 at 00:24
  • Sorry, I probably misunderstood! But thank you the answer! – peterh Feb 22 '15 at 00:26

1 Answers1

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In some sense, yes, you can always construct a set of size $\aleph_1$. Specifically $\omega_1$ is a set of size $\aleph_1$. And if the continuum hypothesis fails, it serves as a counterexample.

You might want to ask whether or not you can construct a set of real numbers of this particular size, and the answer to that will depend on your notion of "construct", but if you mean define "in a reasonable way" the answer is consistently negative.

Asaf Karagila
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  • Well, I should have used the word "defined", I fixed! Thank you! – peterh Feb 21 '15 at 22:52
  • Ok, but under "between" I understood a set, for which $\aleph_0 < |S| < \aleph_1$. – peterh Feb 21 '15 at 23:07
  • This means that you didn't understand the definition of $\aleph_1$. There is no such set, by definition. Regardless to the continuum hypothesis or otherwise. – Asaf Karagila Feb 21 '15 at 23:08
  • Yes, mea culpa. What if I say: $\aleph_0 < |S| < 2^{\aleph_0}$? Can a such set $S$ be defined from ZFC + not-CH? – peterh Feb 21 '15 at 23:11
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    Then I will say, read my answer. – Asaf Karagila Feb 21 '15 at 23:11
  • Thank you your wonderful answer. My only question what remains: what is $\omega_1$? – peterh Feb 21 '15 at 23:13
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    It is the smallest uncountable ordinal, equivalently the set of all countable ordinals, equivalently the canonical set of cardinality $\aleph_1$. If you look around you could find several answers explaining it in great detail. I'd give you some links but I'm already in bed. And doing so from my phone is a pain in the wrist. – Asaf Karagila Feb 21 '15 at 23:15
  • On the other hand, it is simple to define a partition of $\mathbb R$ into $\aleph_1$ many disjoint nonempty sets. – hmakholm left over Monica Feb 22 '15 at 00:35