What I know so far is that:

  • The Continuum Hypothesis (CH) states $\nexists\mathbb{S}|\beth_0<|S|<\beth_1$

    • $\beth_0$ being equal to $\aleph_0$, and $\beth_n$ being equal to $2^{\beth_{n-1}}$ for $n \ge 1$
  • The Generalized Continuum Hypothesis (GCH) states that $\forall n\in \mathbb{N} \nexists\mathbb{S}|\beth_n<|S|<\beth_{n+1}$

    • (EDIT: for any ordinal, not just the natural numbers)
  • Both CH and GCH are independent of ZFC. CH obviously follows GCH, but GCH is independent of ZFC+CH

GCH can be divided into an infinite number of statements, those being:

  • $\nexists\mathbb{S}|\beth_0<|S|<\beth_1$

    • This is the Continuum Hypothesis. But I'll be calling it the First Continuum Hypothesis, $1$CH
  • $\nexists\mathbb{S}|\beth_1<|S|<\beth_2$. The Second Continuum Hypothesis, $2$CH.

  • $\nexists\mathbb{S}|\beth_2<|S|<\beth_3$ $3$CH.

In general, for all $n$, $n$CH states $\nexists\mathbb{S}|\beth_{n-1}<|S|<\beth_n$

EDIT 2: The part in italics turns out not to be true, as the statement $|\mathbb{A}|<|\mathbb{B}| \implies |\mathcal{P}(\mathbb{A})|<|\mathcal{P}(\mathbb{B})|$, though intuitive, is actually independent of ZFC.

(EDIT: I thought the below step was valid, but I was assuming that $|\mathbb{A}|<|\mathbb{B}| \implies |\mathcal{P}(\mathbb{A})|<|\mathcal{P}(\mathbb{B})|$ and although that seems intuitive, I've not been able to prove it)

(I can also show that given a statement $n$CH, $k$CH holds for all $k<n$. This is because if $\exists\mathbb{S}|\beth_{k-1}<|S|<\beth_k$, (not-kCH) then $\beth_{k}<|\mathcal{P}(S)|<\beth_{k+1}$ (not-(k+1)CH), and this would cascade upwards to any $n$.)


(My question is, is each $n$CH independent of ZFC+$(n-1)$CH? Or do any of the Continuum Hypotheses imply higher ones?)

(EDIT 2 The original framing of the question made a wrong assumption. The above in italics is the original, the below is the new version. I kept $n$ for a consistent convention even though it represents an ordinal, not necessarily a natural number.)

My question is, is each $n$CH independent of every other under ZFC? Or do some of the Continuum Hypotheses imply some others?

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    $\begingroup$ The generalized continuum hypothesis is the even stronger statement that for all ordinals $\alpha$, there is no set $\mathbb S$ where $\beth_\alpha<\vert\mathbb S\vert<\beth_{\alpha+1}$. About the proof that $k$CH implies $k+1$CH, I claim the proof is not correct: why should $\beth_k<\vert\mathcal P(\mathbb S)\vert$ necessarily hold? I also claim that Easton's theorem shows that you can consistently have any desired pattern of $k$CH successes and failures. $\endgroup$
    – C7X
    Commented Jun 11 at 6:21
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    $\begingroup$ As already mentioned, by Easton's theorem the aleph function can do pretty much anything at regular cardinals. There are some restrictions at singular cardinals, for example if GCH holds below a singular $\kappa$ with uncountable cofinality, then GCH must also hold at $\kappa$ $\endgroup$ Commented Jun 11 at 8:16
  • $\begingroup$ @C7X oh, yeah I'd missed infinite ordinals. To clarify, my claim wasn't that $k$CH implies $k+1$CH, it was the converse: that $k+1$CH should imply $k$CH. I got that from assuming not-$k$CH (that is, there does exist such a set), and proving that its power set must have a cardinality that contradicts $k+1$CH. That said, I did make an assumption listed in an edit that I'm not as sure about as I was. $\endgroup$ Commented Jun 11 at 13:45
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    $\begingroup$ @SarcasticSully Regarding the principle in your edit, see mathoverflow.net/a/6594/1946. $\endgroup$ Commented Jun 11 at 13:48
  • $\begingroup$ If you are tempted to ask further questions about this topic, I would suggest that math.stackexchange may be a more suitable forum. $\endgroup$ Commented Jun 11 at 15:02

1 Answer 1


In the interest of bringing the question to a conclusion, let me say that it is an immediate consequence of Easton's theorem, as mentioned in the comments, that the various GCH assertions at the $\aleph_n$ or indeed at regular cardinals generally are independent of each other—any GCH pattern whatsoever on regular cardinals can be achieved by forcing.

For example, you can have the GCH hold at $\aleph_n$ exactly when $n$ is prime, or a perfect square, or prime power, or any pattern at all. They are independent.

  • $\begingroup$ To clarify, this applies to infinite ordinal $n$ values? Or just natural number $n$ values? $\endgroup$ Commented Jun 11 at 14:11
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    $\begingroup$ It applies to regular cardinals. This includes all infinite successor cardinals, those of the form $\aleph_{\alpha+1}$ for any ordinal $\alpha$. $\endgroup$ Commented Jun 11 at 14:13

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