# ZFC + not-CH, can a set with cardinality between $\aleph_0$ and $2^{\aleph_0}$ be defined?

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?

• As far as the suggestion goes: set-theory should suffice in contrast to elementary-set-theory. Feb 21, 2015 at 22:47
• By the way: Actually ZFC isn't known to be consistent. See Asafs comment Feb 21, 2015 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. Feb 21, 2015 at 22:51
• 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. Feb 21, 2015 at 22:52
• @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. Feb 21, 2015 at 23:06

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.
• Ok, but under "between" I understood a set, for which $\aleph_0 < |S| < \aleph_1$. Feb 21, 2015 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. Feb 21, 2015 at 23:08
• 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. Feb 21, 2015 at 23:15