# How to prove that $\sf CH$ implies $2^{\aleph_0}=\aleph_1$

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 $n=1,\dots,4$).

• 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. – Andrés E. Caicedo May 15 '15 at 22:09
• @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. – Mario Carneiro May 15 '15 at 23:08
• This might interest you. – Asaf Karagila May 15 '15 at 23:09

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 $2^{\aleph_0}=\aleph_1$.
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).
• 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). – Mario Carneiro May 15 '15 at 23:43