MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Thus, let $\mathrm{OPP}$ be the axiom that $|A|\lt|B| \Rightarrow |2^A|\lt|2^B|$ for any sets $A$ and $B$; and, for any ordinal $\alpha$, let $\mathrm{CH}_\alpha$ be the hypothesis that $\aleph_\alpha=\frak c$ (so that $\mathrm{CH}_1=\mathrm{CH}$) . Define $S$ to be the set of those ordinals $\alpha\in\frak c$ such that $\mathrm{CH}_\alpha$ does not provably (within $\mathrm{ZFC}$) violate $\mathrm{ZFC}$ (for example, it is known that $\omega\backslash${$0$}$\subseteq S$ and $\omega \notin S$); and let $S'$ be the set of those $\alpha\in\frak c$ such that $\mathrm{CH}_\alpha$ does not provably (within $\mathrm{ZFC}$) violate $\mathrm{ZFC}$ $\&$ $\mathrm{OPP}$. Clearly $S'\subseteq S$. But is $S'= S$ ? Or are any elements of $S$ known to be not in $S'$ ?

My guess is that $\mathrm{OPP}$ can't restrict the possibilities for violations of $\mathrm{CH}$ because the sets it talks about in the consequent---especially $2^B$--- are too big to be relevant; but I'm not sure of my footing here.

share|cite|improve this question
Related:… – Asaf Karagila Nov 19 '12 at 23:50
I am not sure how to define $S$ formally while still capturing what you are trying to ask. Do you restrict to definable ordinals? Then in what theory does your definition take place? (I am not nitpicking, this is also an issue when trying to formalize the way that one usually describes Easton's result, for example.) – Andrés E. Caicedo Nov 20 '12 at 3:07
I was wondering the same thing as Andres. I think there is a "semantic" interpretation where the reference point is some fixed ground model $V$. That's fun to think about too... – François G. Dorais Nov 20 '12 at 3:23
@Andres: I think that $S$ is well defined, albeit not within $\mathrm{ZFC}$, as a subset of $\frak c$ in the following sense. From $\frak c$ eliminate any $\alpha$ for which there is a proof in $\mathrm{ZFC}$ that $\frak c$ cannot be $\aleph_\alpha$. What remains is $S$. This seems to me to be as definite a mathematical object as, say, the set of nonrecursive real numbers. – John Bentin Nov 20 '12 at 12:51
John, the objection is that it doesn't make sense to prove a statement with a parameter in it, since that parameter is not a syntactic object and not part of the formal language and proof system. – Joel David Hamkins Nov 20 '12 at 15:02
up vote 4 down vote accepted

First, let me remark that the particular way that you've posed the question has several problematic issues of formalization. One issue, noted by François, Andres and Andreas, is that it doesn't make sense to speak about proving an assertion with an ordinal parameter (one would instead want to speak of definitions of particular ordinals). Another issue is that for all we know, we may be living in a universe with ZFC + $\neg\text{Con}(\text{ZFC})$, and in this universe everything is provable, so even if we are able to resolve the first issue nevertheless the sets $S$ and $S'$ will be empty, since everything is provable.

So let me propose a more semantic, alternative version of the question, which to my of thinking gets at the issue in which I believe you are interested.

Question. If $\alpha$ is an ordinal and the continuum $2^{\aleph_0}$ can be $\aleph_\alpha$ in a forcing extension of the universe, then can the continuum be $\aleph_\alpha$ in a forcing extension of the universe in which also the OPP holds?

The answer is yes, and so in this sense the OPP imposes no additional constraints on the value of the continuum. In this question and in the theorem below, I am speaking about possibly proper class forcing, and this is required, since if the OPP fails unboundedly often, it will require proper class forcing to force OPP again.

Theorem. If the universe $V$ satisifes ZFC, then for any ordinal $\alpha$, the following are equivalent:

  1. There is a forcing extension in which $2^\omega=\aleph_\alpha$.
  2. There is a forcing extension in which $2^\omega=\aleph_\alpha$ and the OPP holds.
  3. Either $\alpha$ is a successor ordinal or $\alpha$ has uncountable cofinality.

Proof. Clearly 2 implies 1, and 1 implies 3. Suppose 3 holds, and I argue for 2. Fix any ordinal $\alpha$ as in $3$. First, we may simply force the GCH by the canonical forcing of the GCH. This forcing (which may be a proper class), is countably closed and hence preserves the property of having uncountable cofinality. So $3$ still holds about $\alpha$ in the extension with GCH. We may now simply apply Easton's theorem, using an Easton function $E$ that takes $\aleph_0$ to the current $\aleph_\alpha$, and more generally which takes $\aleph_\beta$ to $\aleph_{\alpha+\beta+1}$. (But any strictly increasing Easton function will do, and there are many variations.) Note that $\alpha+\beta=\beta$ once $\alpha\cdot\omega\leq\beta$, and so this pattern is eventually the GCH pattern. By Easton's theorem, there is a further forcing extension in which $2^{\aleph_0}=\aleph_\alpha$ and the continuum function is given by $E$, which is strictly increasing, so the OPP holds. QED

In particular, for any ordinal $\alpha$ that you care to define, then you can provably force the continuum to become $\aleph_\alpha$ if and only if you can do so while also ensuring the OPP.

Notice that in the proof of the theorem, the value of $\aleph_\alpha$ may have changed, during the forcing of the GCH, since this will collapse cardinals if the GCH did not already hold. So there is another version of the question, which is about cardinals, rather than about ordinals. If we start with the GCH, then a similar conclusion can be made.

Theorem. If $V$ is a model of ZFC+GCH, then for any cardinal $\delta$ the following are equivalent:

  1. There is a forcing extension $V[G]$ in which the continuum is $\delta$.
  2. There is a forcing extension $V[G]$ in which the continuum is $\delta$ and the OPP holds.
  3. The cardinal $\delta$ has uncountable cofinality.

The proof is essentially the same as above. The nontrivial part is 3 implies 2, which can be achieved via Easton's theorem by using a strictly increasing Easton function $E$ with the property that $E(\aleph_0)=\aleph_\alpha$. There are many choices of such $E$, such as $E(\aleph_\beta)=\aleph_{\alpha+\beta+1}$, as above. Any such $E$ will ensure the right value for $2^{\aleph_0}$ and, because it is strictly increasing, will also achieve the OPP. In this case, since we started with the GCH, one requires only set-sized forcing.

By the way, there seems to be alternative terminology to refer to what you call the OPP. For example, in my paper, "Is the dream solution of the continuum hypothesis attainable?", I refer to the power set size axiom, denoted, PSA, and this is the same as what you call OPP. This axiom also appears in the MO question on reasonable-sounding statements that are independent of ZFC.

share|cite|improve this answer
Joel, if $\beta>\alpha$ then $\alpha+\beta=\beta$. Your given $E$ does not satisfy the requirements for Easton forcing. You should change that to include the case where $\alpha+\beta=\beta$, and then just take $\beta+1$. – Asaf Karagila Nov 20 '12 at 5:42
I got stuck on the last paragraph... maybe because I'm just too tired today. What if GCH fails badly in $V$? Can we force GCH? Can we force OPP? I think so but, as always, the singular case is bugging me... – François G. Dorais Nov 20 '12 at 6:58
@Francois, I thought that you can take some sort of an Easton product over collapses and force GCH to hold (for regular cardinals, singulars get that for free). – Asaf Karagila Nov 20 '12 at 8:47
Asaf, it isn't true that $\beta\gt\alpha\implies\alpha+\beta=\beta$, since $\omega+omega+1\neq\omega+1$. But you are right, I should be saying $\alpha+\beta+1$ here. – Joel David Hamkins Nov 20 '12 at 11:05
François, there is the canonical forcing of the GCH: at stage $\gamma$, if $\gamma$ is a cardinal, force GCH at $\gamma$, and continue with Easton support. This is progressively closed, and so preserves ZFC, and forces GCH. – Joel David Hamkins Nov 20 '12 at 11:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.