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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?

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  • $\begingroup$ As far as the suggestion goes: set-theory should suffice in contrast to elementary-set-theory. $\endgroup$ – AlexR Feb 21 '15 at 22:47
  • $\begingroup$ By the way: Actually ZFC isn't known to be consistent. See Asafs comment $\endgroup$ – AlexR Feb 21 '15 at 22:50
  • $\begingroup$ @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. $\endgroup$ – peterh Feb 21 '15 at 22:51
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    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Feb 21 '15 at 22:52
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    $\begingroup$ @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. $\endgroup$ – Steven Stadnicki Feb 21 '15 at 23:06
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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.

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  • $\begingroup$ Well, I should have used the word "defined", I fixed! Thank you! $\endgroup$ – peterh Feb 21 '15 at 22:52
  • $\begingroup$ Ok, but under "between" I understood a set, for which $\aleph_0 < |S| < \aleph_1$. $\endgroup$ – peterh Feb 21 '15 at 23:07
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Feb 21 '15 at 23:08
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    $\begingroup$ Then I will say, read my answer. $\endgroup$ – Asaf Karagila Feb 21 '15 at 23:11
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    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Feb 21 '15 at 23:15

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