Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question
    
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! –  Asaf Karagila Jun 7 '11 at 13:38
    
Done. I thought $\mathfrak{c}$ would be an even worse choice, so $\mathfrak{a}$ it is. –  LostInMath Jun 7 '11 at 14:55
add comment

2 Answers 2

up vote 19 down vote accepted

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 $ZF$.


Bibliography:

  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.

share|improve this answer
2  
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)"? –  LostInMath Jun 7 '11 at 15:00
    
@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. –  Asaf Karagila Jun 7 '11 at 16:13
    
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? –  LostInMath Jun 8 '11 at 13:04
    
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. –  Asaf Karagila Jun 8 '11 at 13:26
    
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? –  LostInMath Jun 8 '11 at 14:35
show 6 more comments

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.

share|improve this answer
    
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. –  Asaf Karagila Jun 8 '11 at 7:31
    
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. –  JDH Jun 8 '11 at 10:11
    
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. –  Asaf Karagila Jun 8 '11 at 10:39
    
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. –  JDH Jun 8 '11 at 11:01
    
I agree, it can be interesting. I'll put it on my crowded to-do list. –  Asaf Karagila Jun 8 '11 at 12:36
add comment

Your Answer

 
discard

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.