I have $2$ questions. I want to ask here because I could not find an answer myself. If the continuum hypothesis is wrong, in other words there exist such a set $X$, which is the cardinality give us $\aleph_0<X<2^{\aleph_0}$.

What will the definition of cardinality be? Countable or uncountable?

And if the continuum hypothesis is wrong, will $X$ be the second cardinal number?

  • 1
    $\begingroup$ Maybe no, If someone can create a new set with $\aleph_0<X<2^{\aleph_0}$, a similar construction will give $\aleph_0<Y<X<W<2^{\aleph_0}$ $\endgroup$
    – L F
    Aug 2, 2019 at 21:39
  • $\begingroup$ @LuisFelipe What is $Y$ and $W$? $\endgroup$
    – user548054
    Aug 2, 2019 at 21:44
  • $\begingroup$ another sets that you can create if this is a wrong hypotesis $\endgroup$
    – L F
    Aug 2, 2019 at 21:46
  • $\begingroup$ @Elementary It depends. Maybe we have $Y=\aleph_1,\,X=\aleph_2,\,W=\aleph_3,\,2^{\aleph_0}=\aleph_4$. $\endgroup$
    – J.G.
    Aug 2, 2019 at 21:47
  • 3
    $\begingroup$ Well $X$ would be uncountable as there is not a bijection between $X$ and $\mathbb N$ (well, more accurately no surjection of $\mathbb N$ into $X$). In this case $2^{\aleph_0}$ is not the smallest uncountable cardinality. But $|X|$ may not be the smallest one either. There maybe a $Y$ so that $|\mathbb N| < |Y| < |X| < |\mathbb R|$As long as we are being hypothetical.. It may be that there may be infinitely, even uncountably infinitely many $|\mathbb N| < ...<|Y|<|X| < |\mathbb R|$. $\endgroup$
    – fleablood
    Aug 2, 2019 at 21:59

2 Answers 2


Throughout this answer I'm assuming ZFC is consistent.

You're essentially asking what the interval between $\aleph_0$ and $2^{\aleph_0}$ can be. E.g. can there be two "intermediate cardinalities"?

The answer is basically: the continuum can be anything whatsoever, with two exceptions: obviously it has to be uncountable, and it turns out it also has to have uncountable cofinality.

For example, it's consistent with ZFC that $2^{\aleph_0}=\aleph_{17}$, or that $2^{\aleph_0}=\aleph_{\omega^2+42}$. It's even consistent with ZFC that $2^{\aleph_0}$ is a "fixed point of the $\aleph$-function," that is, it's consistent with ZFC that $2^{\aleph_0}=\aleph_{2^{\aleph_0}}$. (Fine, more carefully: it's consistent with ZFC that $2^{\aleph_0}=\aleph_\lambda$ where $\lambda$ is the initial ordinal of $2^{\aleph_0}$.)

This isn't very precise, of course, and has some obvious issues - e.g. we clearly can't have $2^{\aleph_0}$ be the smallest cardinal of uncountable cofinality bigger than the continuum, even though the latter is obviously a cardinal of uncountable cofinality! Unfortunately, it takes real work to make the above precise: we need to talk about models of set theory, and in particular forcing:

Suppose $M$ is a model of ZFC + CH and $\kappa$ is a cardinal of uncountable cofinality in the sense of $M$ (or more generally, $M$ is any model of ZFC and $\kappa$ is a cardinal of uncountable cofinality in the sense of $M$ which is bigger than $M$'s continuum). Then there is a forcing extension $N$ of $M$ with the same cardinals and cofinalities satisfying $2^{\aleph_0}=\kappa$.

The requirement that $\kappa$ have uncountable cofinality is necessary as a consequence of Konig's theorem: ZFC proves that the continuum has uncountable cofinality. The above result is due to Solovay, and has a terrific strengthening to the powersets of all regular cardinals due to Easton. Singular cardinals, meanwhile, turn out to be wildly more complicated.

Let me make a couple footnotes about forcing:

  • First, note that I'm playing fast and loose with the assumption that forcing extensions exist, above: we either need to assume that $M$ is countable, or use the Boolean models approach to forcing. But that's a secondary issue.*

  • Second, remember that forcing is definable: the bolded fact above can be turned into a ZFC theorem saying roughly "For any cardinal $\kappa$ of uncountable cofinality, there is a cardinality- and cofinality-preserving forcing making $\kappa$ the continuum." (And in fact we know what forcing does the job.) While talking about models makes everything much easier to understand, there is an actual "internal theorem" here.


$X$ will certainly be uncountable. That just means it is bigger than $\aleph_0$, which comes directly from your definition.

If the continuum hypothesis is wrong...

As far as our current understanding goes, that is the wrong way to look at it. The axioms of Set Theory that we use for general mathematics are incapable of deciding the issue: it is consistent with the axioms for the Continuum Hypothesis to be true, and it is consistent with the axioms for the Continuum Hypothesis to be false.

So it is better to ask:

In a model of Set Theory where the Continuum Hypothesis is false...

And the answer is: it depends. We can construct models where $X$ is the only cardinal between $\aleph_0$ and $2^{\aleph_0}$, and we can construct models where there are an infinite number of such cardinals.

  • $\begingroup$ Do you happen to know the least transfinite ordinal $\alpha$ for which we can have $\aleph_\alpha=2^{\aleph_0}$? It famously can't be $\omega$. $\endgroup$
    – J.G.
    Aug 2, 2019 at 21:49
  • $\begingroup$ @J.G.: I have no idea, I'm afraid. I haven't worked on Set Theory in 35 years, so all I remember is the basics. $\endgroup$
    – TonyK
    Aug 2, 2019 at 21:52
  • 3
    $\begingroup$ @J.G. It's $\omega+1$. $\endgroup$ Aug 2, 2019 at 21:53
  • 2
    $\begingroup$ @J.G. You might also like to see Easton's Theorem which tells you exactly which $\alpha$ can have $2^{\aleph_0} = \aleph_\alpha$ (and, in fact, what is possible for the function $\kappa \mapsto 2^\kappa$ when $\kappa$ is regular) en.wikipedia.org/wiki/Easton%27s_theorem $\endgroup$ Aug 2, 2019 at 21:56
  • $\begingroup$ Thanks everyone. I know the theorem, I've just never calculated the cofinalities involved. $\endgroup$
    – J.G.
    Aug 3, 2019 at 5:26

You must log in to answer this question.