# How to reach $\aleph_1$ without power sets?

At the end of Set Theory and the Continuum Hypothesis, Paul Cohen wrote:

A point of view which the author feels may eventually come to be accepted is that CH is obviously false. The main reason one accepts the Axiom of Infinity is probably that we feel it absurd to think that the process of adding only one set a a time can exhaust the entire universe. Similarly with the higher axioms of infinity. Now $\aleph_1$ is the set of countable ordinals and this is merely a special and the simplest way of generating a higher cardinal. The set $\mathfrak c$ is, in contrast, generated by a totally new and more powerful principle, namely the Power Set Axiom. It is unreasonable to expect that any description of a larger cardinal which attempts to build up to that cardinal from ideas deriving from the Replacement Axiom can ever reach $\mathfrak c$. [...]

This seems to imply that Cohen thought that building up to $\aleph_1$ "from ideas deriving from the Replacement Axiom" can reasonably be expected to succeed. But how?

It is definitely not the case that one can prove that $\omega_1$ exists in the theory we get from naively omitting the Power Set Axiom from a standard presentation of ZFC -- for the set of hereditarily countable sets is a model of that theory and contains no set of all countable ordinals.

Cohen was certainly well aware of this, so he must have had something other than that in mind. Short of resurrecting him so we can ask him, is there any evidence that shows which kind of "building-up" he was thinking of here? Is the argument simply that if Power Set had not been invented, we should still have accepted "$\omega_1$ exists" as a new axiom for a reason similar to the one he proposes for accepting Infinity?

• Interesting... Given a transitive model $\mathcal{M}$ of $\operatorname{ZFC}^-$, the existence of $\mathbb{R}^{\mathcal{M}}$ does imply the existence of $\omega_1^{\mathcal{M}}$ - simply because the ordinal witnessing the well-order on $\mathbb{R}^{\mathcal{M}}$ exists. Is the converse true? – Stefan Mesken Dec 24 '16 at 13:56
• @Stefan: More elementarily, if CH fails at the metalevel, we could take $\mathcal M$ to be the set of all sets hereditarily of cardinality $\le\aleph_1$. (Of course, this would only show "not necessarily" rather than "no"). – Henning Makholm Dec 24 '16 at 14:07
• Looking through Kanamori's "In Praise of Replacement", I am not entirely sure that Cohen knew that Power Set is necessary for proving the existence of $\omega_1$ when he wrote that paragraph. – Asaf Karagila Dec 24 '16 at 14:41
• @Stefan Actually, I'm considering $\mathbb{R}=\mathcal{P}(\omega)$. From this you get $\mathcal{P}(\omega\times\omega)$ and that's all you need, since you can separate out the wellorderings of subsets of $\omega$. The map that sends a wellordering to its order type is definable in $ZF^-$, and using Replacement once more you have your uncountable ordinal. – Pedro Sánchez Terraf Dec 25 '16 at 16:14
• @Pedro Yeah, that works. – Stefan Mesken Dec 25 '16 at 16:23

## 1 Answer

I always took that paragraph to mean the following: In the theory (ZFC - Power set) + (Every cardinal has a successor), we cannot prove the existence of the set of reals (for example, this holds in $H_{\kappa}$ for a weakly inaccessible $\kappa \leq \mathfrak{c}$). In this sense, power set axiom is a more powerful principle than "every cardinal has a successor". Foreman and Woodin gave a model where this holds globally: For every $\kappa$, $2^{\kappa}$ is weakly inaccessible.

• That is an interesting interpretation. – Asaf Karagila Dec 27 '16 at 4:02
• How is "every cardinal has a successor" more believable than "every set has a power-set"? After all, if there isn't the set of sets of naturals then there is no reason (Cantor's theorem) to believe that there would be anything that cannot be countable. In any case I can't see how the question is answered, since "Now $\aleph_1$ is the set of countable ordinals and this is merely a special and the simplest way of generating a higher cardinal." is fallacious as others have stated. (Without the power-set axiom, or equivalently higher types, how can we define countable well-orderings?) – user21820 Feb 25 '17 at 7:10
• I don't know what you mean by belief here. All I said was that in ZFC - Power set, "every set has a power-set" implies "every cardinal has a successor" and not vice versa. So power set axiom is provably more powerful than "every cardinal has a successor". Also, I don't claim to have "answered" the question but I am merely suggesting a plausible mathematical result that could explain Cohen's remarks. – hot_queen May 2 '17 at 18:13