# How big can the continuum be without choice?

I've heard an argument before (although I can't remember where) that the continuum hypothesis is false, since the powerset operation is a something much more 'powerful' than the mere cardinal successor operation, so that $\mathfrak{c}>\aleph_1$, $\mathfrak{c}>\aleph_2$ and so on. (Obviously, that's really sketchy as I've stated it, bit it'll do for motivation.)

(Edit: After some googling, this argument is from Cohen:

A point of view which the author feels may eventually come to be accepted is that the continuum hypothesis is obviously false ... The continuum c is greater than $\aleph_n$,$\aleph_\omega$,$\aleph_\alpha$ where $\alpha=\aleph_\omega$ etc. This point of view regards c as an incredibly rich set given to us by one bold new axiom'

taken from this question )

We know that any particular sentence like $\mathfrak{c}>\aleph_\alpha$ is consistent with ZFC. But the kind of reasoning above suggests something like \begin{equation} \tag{$*$}\forall\alpha (\mathfrak{c}>\aleph_\alpha), \end{equation}

which is provably false in ZFC.

But, if we drop the axiom of choice, the continuum doesn't have to have an aleph as a cardinality, so ($*$) is not obviously provably false. It also has a number of possible formulations.

Definitions:

• $A \preceq B$ iff there is an injection $f:A\to B$
• $A \preceq^* B$ iff there is a surjection $f:B\to A$

Then there are four statements which correspond roughly to ($*$):

$$\tag{1} \forall \alpha, \aleph_\alpha \preceq \mathcal{P}(\omega)$$ $$\tag{2} \forall \alpha, \aleph_\alpha \preceq^* \mathcal{P}(\omega)$$ $$\tag{3} \forall \alpha, \mathcal{P}(\omega) \npreceq \aleph_\alpha$$ $$\tag{4} \forall \alpha, \mathcal{P}(\omega) \npreceq^* \aleph_\alpha$$

Question: Are any of (1)-(4) consistent with ZF?

Related question: How to formulate continuum hypothesis without the axiom of choice?

## 1 Answer

Both statements are provably false without the axiom of choice.

The first one fails because of Hartogs theorem: each set $X$ has an ordinal $\alpha$ such that there is no injection from $\alpha$ into $X$. In particular $\mathcal P(\omega)$ has such $\alpha$, and the least such $\alpha$ is an $\aleph$ number by definition (whose index can be anything larger than $0$).

For a similar reason, the second statement is provably false. You can either prove directly that for every set $X$ there is an ordinal such that $X$ is not mapped onto it; or you can exploit Hartogs theorem, and the fact that if $f\colon X\to Y$ is a surjection, then $f^{-1}\colon Y\to\mathcal P(X)$ is an injection.

Remark. It might be worth pointing out that the least ordinal which cannot be injected into $\mathcal P(\omega)$ can be strictly smaller than the least ordinal that $\mathcal P(\omega)$ cannot be mapped onto.

The third and fourth one can in fact hold, though. They are in general equivalent. Namely, $X$ can be mapped injectively into an ordinal $\alpha$ if and only if $\alpha$ can be mapped onto $X$ (or if $X$ is empty). One direction is obvious, and if $f\colon\alpha\to X$ is a surjection, then $x\mapsto\min\{\beta\mid f(\beta)=x\}$ is an injection.

Moreover if $X$ can be mapped injectively into an ordinal, then it is well-orderable, simply by picking such injection $f\colon X\to\alpha$ and defining $x_1\prec x_2\iff f(x_1)<f(x_2)$.

So in any model where $\mathcal P(\omega)$ cannot be well-ordered, then there is no ordinal which can be mapped onto $\mathcal P(\omega)$, and it is not mapped injectively into any ordinal as well.

As to the question in the title, the continuum can be pretty darn big. One can notice that if $\lambda$ is an aleph number, then $\operatorname{Add}(\omega,\lambda)$ is a well-ordered forcing, therefore a well-behaved one even in models of $\sf ZF+\lnot AC$. So we can always push the continuum to be very large.

I am unaware of a nice classification of non-aleph cardinals which can be the continuum when the axiom of choice fails. But from the top of my head, I can't recall too many cases where adding Cohen reals causes too much problems with the cardinal structure (although it certainly can change things, e.g. if $X$ cannot be linearly ordered, then adding $X$-Cohen reals will definitely change $X$ somehow).

• Thanks! Looking at it, my second formulation's clearly wrong, and I should have had the surjection going the other way around. I hope you don't mind if I edit it? – J.P. Oct 2 '14 at 16:51
• @J.P.: Not at all. I'll change my answer in the meantime. – Asaf Karagila Oct 2 '14 at 16:52
• Done. Realised I'd also mixed up two different factors that can change (injections vs. surjections and existence vs. non-existence of these), so there are now four options (some of which might be equivalent even without choice?) – J.P. Oct 2 '14 at 17:03