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Of course, most of you will, upon reading the title, exclaim "But isn't that the definition of the continuum hypothesis?" So I need to be a little more careful about the exact definitions.

Let ${\sf CH}(\frak m)$ be the statement that either $\frak m$ is a finite cardinal or there is no cardinality $\frak n$ such that ${\frak m}<{\frak n}<2^{\frak m}$. The standard continuum hypothesis is then ${\sf CH}(\aleph_0)$, and the GCH is $\forall{\frak m}\,{\sf CH}({\frak m})$.

The statement I am interested in is the question of whether ${\sf CH}(\aleph_\alpha)$ implies $2^{\aleph_\alpha}=\aleph_{\alpha+1}$. I suspect the following lemma, used in the proof of $\sf GCH\to AC$, may be useful:

If ${\sf CH}(\frak m)$ and $\frak m+m=m$ and ${\frak n}\le 2^{\frak m}$, then $\frak m,n$ are comparable, which is to say $\frak m\le n$ or $\frak n\le m$.

One side of the inequality is easy - if $2^{\aleph_\alpha}<\aleph_{\alpha+1}$ then we would have $\aleph_\alpha<2^{\aleph_\alpha}<\aleph_{\alpha+1}$ which violates the properties of the $\aleph$ function. But usually proving $\aleph_{\alpha+1}\le 2^{\aleph_\alpha}$ is done using some kind of choice, and I don't have enough $\sf GCH$ here to prove that $2^{\aleph_\alpha}$ is well-orderable (Sierpinski's proof gives the result assuming ${\sf CH}({\cal P}^n(\aleph_\alpha))$ for $n=1,\dots,4$).

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    $\begingroup$ You need to use choice: It is consistent without choice that $\mathsf{CH}(\aleph_0)$ but $\mathbb R$ and $\aleph_1$ are incomparable. On the other hand, if you have enough choice to argue that $\mathbb R$ is well-orderable, you are done. $\endgroup$ – Andrés E. Caicedo May 15 '15 at 22:09
  • $\begingroup$ @AndresCaicedo What if you assume ${\sf CH}(\aleph_0)+{\sf CH}(\aleph_1)$? I think this is a result of Specker, but I can't find the proof. $\endgroup$ – Mario Carneiro May 15 '15 at 23:08
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    $\begingroup$ This might interest you. $\endgroup$ – Asaf Karagila May 15 '15 at 23:09
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Assuming the axiom of choice, first prove that $\Bbb R$ can be well-ordered. Since Cantor's theorem says that $\aleph_0<2^{\aleph_0}$ this means that $\aleph_1\leq 2^{\aleph_0}$, and so $\sf CH$ implies $2^{\aleph_0}=\aleph_1$.

Without assuming the axiom of choice, Solovay's model as well the various Truss models (which don't use an inaccessible cardinal for the construction) are models in which the perfect set property holds. Namely every uncountable set of reals has a subset which is homeomorphic to the Cantor set.

This implies two immediate things, the first is that $\sf CH(\aleph_0)$ holds, since every uncountable set of reals has a subset of size $2^{\aleph_0}$; and the second is that $\aleph_1\neq2^{\aleph_0}$ since if $\aleph_1\leq 2^{\aleph_0}$ then there is an uncountable set of reals without a perfect subset (either $\aleph_1<2^{\aleph_0}$ and then this is a cardinality argument; or they are equal and the Bernstein set construction follows without needing choice).

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  • $\begingroup$ One last side question: How do people usually get around the apparent ambiguity in the meaning of CH in a teaching environment? As I mention in the first sentence (corroborated by many of the related links) many people believe CH to mean $\aleph_1=2^{\aleph_0}$, while the facts you point out here show that this is in fact a strictly stronger statement than what I am calling ${\sf CH}(\aleph_0)$ (which also has some literature precedent, for example in the Kanamori/Pincus paper you linked). $\endgroup$ – Mario Carneiro May 15 '15 at 23:43
  • $\begingroup$ Well, either you adopt the axiom of choice explicitly in which case it's fine; or you don't and then you have to decide which formulation to use, and then stick with that. $\endgroup$ – Asaf Karagila May 15 '15 at 23:47
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    $\begingroup$ Also, this for some background. $\endgroup$ – Asaf Karagila May 15 '15 at 23:51

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