Special Aleph Hypothesis AH(0) is the claim $2^{\aleph_0}=\aleph_1$, i.e. there is a bijection from $2^{\aleph_0}$ to $\aleph_1$.

Continuum Hypothesis CH is the claim $\aleph_0 \leq \mathfrak{a}< 2^{\aleph_0} \Rightarrow \mathfrak{a}=\aleph_0$, i.e. if there is an injection from $\aleph_0$ to $\mathfrak{a}$, an injection from $\mathfrak{a}$ to $2^{\aleph_0}$ and no injection from $2^{\aleph_0}$ to $\mathfrak{a}$, then there is a bijection from $\aleph_0$ to $\mathfrak{a}$.

AH(0) and CH are known to be equivalent under ZFC. What if we don't assume the Axiom of Choice? Under ZF, is it known that AH(0) $\nRightarrow$ CH or CH $\nRightarrow$ AH(0)?

  • $\begingroup$ I would use something other than $\mathfrak b$, as it has a meaning in PCF theory (dominating number). It sure got me confused for a moment! $\endgroup$ – Asaf Karagila Jun 7 '11 at 13:38
  • $\begingroup$ Done. I thought $\mathfrak{c}$ would be an even worse choice, so $\mathfrak{a}$ it is. $\endgroup$ – LostInMath Jun 7 '11 at 14:55

January 25, 2012: I have found a mistake in one of the arguments, it is not an important one for the answer (in fact it is flat out irrelevant) but I should rewrite this answer anyway.

The short answer is that $AH(0)$ implies $CH$ under $ZF$, but is unprovable from $CH$ under $ZF$ alone.

$\boxed{ZF\vdash AH(0)\rightarrow CH}$:

Suppose $2^{\aleph_0} = \aleph_1$, suppose $\frak a$ is a set whose cardinality is between $\aleph_0$ and $2^{\aleph_0}=\aleph_1$. $\frak a$ can be well ordered (it can be injected into an ordinal) and therefore has the cardinality of some aleph number. If it is not $\aleph_1$, then it has to be $\aleph_0$.

So in $ZF$ you have that $AH(0)\rightarrow CH$.

From this we have that it is not consistent with $ZF$ that $CH\rightarrow\lnot AH(0)$. We also have that it is consistent relatively to $ZF$ that $AH(0)\land CH$ is true.

$\boxed{ZF\nvdash CH\rightarrow AH(0)}$:

We exhibit a model in which $CH$ is true, but $AH(0)$ is false. Note that this implies that $\aleph_1\nleq2^{\aleph_0}$. Such result can be achieved either when the continuum to be a countable union of countable sets or the presence of an inaccessible cardinal (which is a slight increase in consistency strength).

Surprisingly enough, in the Feferman-Levy model in which the continuum is a countable union of countable sets $CH$ fails. There exists a set of real numbers which is uncountable but there is no injection from $2^\omega$ into the set [1, Remark 3.4].

Consider now the Solovay model of $ZF$, we start with an inaccessible cardinal and end up with a model in which every subset of reals is Lebesgue measurable. It is also a model of the assertion "Every uncountable set of reals has a perfect set", where perfect sets always contain a copy of the Cantor set and therefore have cardinality continuum. In fact, Truss proved in [2] that repeating the Solovay construction from any limit cardinal results in a model where the perfect set property holds, and $AH(0)$ fails, so the inaccessible is redundant for this proof.

Every set of reals, in such model, is either countable or of cardinality continuum. In particular $CH$ holds, but $AH(0)$ not.

Therefore in a model of $ZF$ (without choice) exactly one of the options holds:

  1. $CH\land AH(0)$,
  2. $CH\land\lnot AH(0)$ (Solovay's model, and Truss models),
  3. $\lnot CH\land\lnot AH(0)$ (Feferman-Levy model).

This shows that $CH$ cannot prove or disprove $AH(0)$. Note, by the way, that the first and third options can be found in models of choice, such as Godel's constructible universe and Cohen's construction where the continuum hypothesis fails; the latter can be even shown in models like Cohen's first model where there is a dense Dedekind-finite set of real numbers. However we see more here: the assertion $\aleph_1\leq2^{\aleph_0}$ is unprovable from $ZF$.


  1. Miller, A. A Dedekind Finite Borel Set. Arch. Math. Logic 50 (2011), no. 1-2, 1--17.

  2. Truss, J. Models of set theory containing many perfect sets. Ann. Math. Logic 7 (1974), 197–219.

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    $\begingroup$ Thank you for your answer. I'm aware of the fact ZFC does not prove the existence of an inaccessible, but still I can't see how Kanamori's Thm. 11.6 implies that ZF does not prove "CH implies AH(0)". Could you elaborate a bit on "This means that from ZF alone you cannot prove CH $\Longrightarrow$ AH(0)"? $\endgroup$ – LostInMath Jun 7 '11 at 15:00
  • $\begingroup$ @LostInMath: I have added a sketch of a proof, it was slightly harder that I thought, so it was very good of you to point that out. I hope that you understand the idea behind the proof in the addendum. $\endgroup$ – Asaf Karagila Jun 7 '11 at 16:13
  • $\begingroup$ Thank you for the clarification. But again, I have to ask because I could not figure it out by myself. Although this is probably obvious to everyone but me. This is how I understand your proof: The aim is to prove that CH $\Longrightarrow$ AH(0) is not a theorem of ZF. So suppose it is. Then the theory ZF+$\omega_1\nleq 2^{\aleph_0}$+"every subset of reals has the perfect set property" is inconsistent. Then Thm. 11.6 implies that the theory $ZFC+\exists$ inaccessible is inconsistent too. But from this we can't deduce that (ZFC and) ZF are inconsistent and reach a contradiction, can we? $\endgroup$ – LostInMath Jun 8 '11 at 13:04
  • $\begingroup$ No. What I proved is that $CH$ cannot prove from $ZF$ alone either $AH(0)$ or its negation. I.e. $AH(0)$ is independent from $ZF+CH$, and we cannot prove anything further without assuming more (e.g. assuming $AC$ they are in fact equivalent). Nothing in my addendum was pointing out an inconsistency. I will try to clarify things further a bit more. $\endgroup$ – Asaf Karagila Jun 8 '11 at 13:26
  • $\begingroup$ I think I understand it now. What confused me was that when demonstrating the consistency of $ZF+CH+\neg AH(0)$ you assumed the existence of a certain model. (In comparison, when demonstrating the consistency of, for example, $ZFC+\neg CH$, usually the model is constructed by extending an existing model of $ZFC$.) But now I see that the justification behind this assumption is that we have assumed the existence of an inaccessible cardinal. Is it possible to prove the consistency of $ZF+CH+\neg AH(0)$ without any extra assumptions? $\endgroup$ – LostInMath Jun 8 '11 at 14:35

It might also be worth mentioning also that it is known to be relatively consistent with $ZF+\neg AC$ that there are infinite Dedekind finite sets of reals, that is, a set $A\subset\mathbb{R}$ that is infinite, but which has no countable subset. Such a set is uncountable, with cardinality strictly less than the continuum, but is incomparable in cardinality with $\aleph_0$. In other words, it is relatively consistent with $ZF+\neg AC$ that there is a cardinality $\frak{a}$ with $\frak{a}\lt 2^{\aleph_0}$ and $\frak{a}$ is not finite (and even $\frak{a}$ is uncountable), yet $\aleph_0\not\leq\frak{a}$. In other words, just knowing a set $A$ is uncountable, one cannot conclude in ZF (unless inconsistent) that $\aleph_0\leq |A|$. Many would regard the existence of such cardinalities as even worse than the kind of counterexamples for which your question is asking.

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  • $\begingroup$ Joel: In the formulation of CH given in the question there might be other infinite cardinalities below the continuum, but they cannot be comparable with $\aleph_0$. Despite being "even worse than the kind of counterexamples" they still don't serve as counterexamples. $\endgroup$ – Asaf Karagila Jun 8 '11 at 7:31
  • $\begingroup$ Yes, I agree with that, and I intended this answer only as an interesting aside, rather than as a counterexample to the precise formulation of the question that was asked. Nevertheless, the formulation of CH that the OP gives may not be exactly what is desired in the $\neg AC$ context. For example, an alternative formulation would assert that every uncountable set of reals is bijective with $\mathbb{R}$, and the infinite Dedekind finite sets violate this. $\endgroup$ – JDH Jun 8 '11 at 10:11
  • $\begingroup$ Of course, I have some trouble in settling this in my mind prior to writing my answer on whether or not this is the "correct" way to write CH. I actually deleted my original answer (which turned into the comment on the original question) in which I suggested that CH will be formulated as "If $A$ is an infinite set whose cardinality is less than the continuum, then $A$ is countable", similarly to the formulation you gave here. $\endgroup$ – Asaf Karagila Jun 8 '11 at 10:39
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    $\begingroup$ Indeed, I guess there are a large variety of statements, and presumably many of the them are independent over ZF. It would be an interesting project to sort them all out into a big hierarchy. $\endgroup$ – JDH Jun 8 '11 at 11:01
  • $\begingroup$ I agree, it can be interesting. I'll put it on my crowded to-do list. $\endgroup$ – Asaf Karagila Jun 8 '11 at 12:36

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