# Is each of the infinite statements of the Generalized Continuum Hypothesis independent?

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 $$k. This is because if $$\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$$.)

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(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?

• 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.
– C7X
Commented Jun 11 at 6:21
• 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 $\kappa$ Commented Jun 11 at 8:16
• @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. Commented Jun 11 at 13:45
• @SarcasticSully Regarding the principle in your edit, see mathoverflow.net/a/6594/1946. Commented Jun 11 at 13:48
• If you are tempted to ask further questions about this topic, I would suggest that math.stackexchange may be a more suitable forum. Commented Jun 11 at 15:02

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