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.

More precisely, is either of the following two statements consistent with ZF:

  1. $2^{\aleph_0}\geq\aleph_{\alpha}$ for every ordinal number $\alpha$,

  2. $2^{\aleph_0}\leq\aleph_{\alpha}\implies 2^{\aleph_0}=\aleph_{\alpha}$ for every ordinal number $\alpha$?

I'm asking mainly out of curiosity.

share|improve this question
    
If you use 1. instead of (1) the software will parse this as an ordered list automatically... –  Asaf Karagila Mar 2 '12 at 16:11
    
@AsafKaragila: Thanks, I didn't know that. –  Dejan Govc Mar 2 '12 at 16:32
    
add comment

3 Answers 3

up vote 13 down vote accepted

Both are inconsistent with ZFC, and the first is inconsistent with ZF as well.

The axiom of infinity tells us that $\mathbb N$, the collection of natural numbers (or finite ordinals) is a set. The axiom of power set tells us therefore that every set has a power set, in particular $\mathbb N$.

We know that the size of $P(\mathbb N)=\mathbb R$ is $2^{\aleph_0}$, however this can be $\aleph_1$ or $\aleph_2$ or even higher. Without the axiom of choice it might not even be an $\aleph$ number.

So we have that $\mathbb R$ is a set, therefore so is $\mathbb R\times\mathbb R$. It therefore has as power set, from which we can take all the subsets of $\mathbb R\times\mathbb R$ which are order relations on some subset of $\mathbb R$, and we can take all those which are well ordered.

Each is order isomorphic to a unique ordinal, so mapping every relation $R$ from this collection to the ordinal is a function defined by a formula (possibly with parameters), whose domain is a set. By the axiom of replacement the image is a set of ordinals, since isomorphism goes "both ways" we have that this is the same set as $\{\beta\in\mathrm{Ord}\mid\ \exists f\colon\beta\to\mathbb R\text{ injective}\}$

Since $\mathbb R$ is a set there can only be set many ordinals of this property, the least ordinal above them is called Hartogs number of $\mathbb R$ and it cannot be injected into $\mathbb R$, denote it by $\aleph(\mathbb R)$, we have if so that $\aleph(R)\nleq\mathbb R$.


As for the second question, if we assume the axiom of choice then the previous argument shows that for some $\aleph_\alpha$ we have that $2^{\aleph_0}<\aleph_\alpha$, and therefore for all $\beta>\alpha$. The second assertion implies, if so, that $\aleph_\alpha=\aleph_\beta=2^{\aleph_0}$ for almost all ordinals.

However without the axiom of choice it is consistent to have that for every ordinal (finite or not) we have either $\alpha=0$ and then $\aleph_0<2^{\aleph_0}$ and otherwise $\aleph_\alpha\neq 2^{\aleph_0}$. This would make the assumption in the implication of the second assertion false and the assertion itself vacuously true.


Further reading:

  1. How do we know an $ \aleph_1 $ exists at all?
  2. Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
  3. Implications of continuum hypothesis and consistency of ZFC
share|improve this answer
    
@Austin: Thanks! –  Asaf Karagila Mar 2 '12 at 15:36
    
In the "Each is order isomorphic" paragraph, I think you're arguing the wrong direction. What you need to say here is that if there is an injection from $\beta$ into $\mathbb R$, then $\beta$ is in the image of such-and-such function from $\mathbb R\times \mathbb R$ to $\mathbf{ON}$. –  Henning Makholm Mar 2 '12 at 15:55
    
Oops -- I meant to write "from $\mathscr P(\mathbb R\times \mathbb R)$ to $\mathbf{ON}$". My point is that you appear to have swapped premise from conclusion in that paragraph -- you want to argue that every $\beta\preceq \mathbb R$ is the image of an $R$, not that every $R$ produces a $\beta\preceq \mathbb R$ (though both directions are true). –  Henning Makholm Mar 2 '12 at 16:02
    
@Henning: I'm talking about $R\subseteq\mathbb R\times\mathbb R$ which is a well order of a subset of $\mathbb R$. I've edited that part to something which I hope is clearer. –  Asaf Karagila Mar 2 '12 at 16:04
    
Thank you, answer accepted. –  Dejan Govc Mar 3 '12 at 8:47
add comment

I am reading your statement "$2^{\aleph_0} \geq \aleph_\alpha$" to say that there is an injection from $\omega_\alpha$ into $\mathbb{R}$. (And similarly for "$2^{\aleph_0} \leq \aleph_\alpha$".)

(1) is not consistent, since given any set $X$, there are only set-many well-orderings of subsets of $X$. (Without Choice, define $WO(X) = \{ R : R\text{ is a well-ordering of a subset of }X \}$. This is clearly a set by Power Set, Pairing and Separation). Using Replacement the family of order-types of elements of $WO(X)$ is also a set. From this it follows that for every set $X$ there must be an $\alpha$ such that "$\aleph_\alpha \leq | X |$" does not hold.

(2) would only be consistent vacuously: where there is no well-ordering of the reals and so the statement $2^{\aleph_0} \leq \aleph_\alpha$ never holds. (If there is an injection from $\mathbb{R}$ into $\omega_\alpha$, then there is also an injection from $\mathbb{R}$ into $\omega_{\alpha+1}$, and clearly $\aleph_\alpha \neq \aleph_{\alpha+1}$.)

share|improve this answer
    
Thanks, this is a very concise answer. Much appreciated. –  Dejan Govc Mar 3 '12 at 8:45
add comment

Being careful about quantifiers, the answer to your first question is yes, and to your second one is no. More precisely, for any ordinal $\alpha$ there is a forcing extension of the universe where (not just ZF but even ZFC holds and) $2^{\aleph_0} \ge\aleph_\alpha$. A simple modification of Cohen's original argument works. What Cohen proved is that one can "add lots of (Cohen) reals" to the universe without changing any cardinals, meaning that if and ordinal $\alpha$ was a cardinal, then it is a cardinal after adding all those reals. So, if you add at least $\aleph_\alpha$ reals, in the forcing extension the continuum has size at least $\aleph_\alpha$.

However, equality is not always possible, at least under some choice. There is a basic restriction: Koenig proved that $\kappa^{cf(\kappa)}>\kappa$ for any infinite cardinal $\kappa$. Since $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$, we conclude that $2^{\aleph_0}$ cannot be a cardinal of cofinality $\omega$.

Solovay proved (right after Cohen's result) that this is basically the only objection: Cohen's construction mentioned above not only "preserves cardinals" but in fact preserves all cofinalities. Solovay showed that if we start in Goedel's L, or more generally in a model of the GCH, then adding $\aleph_\alpha$ Cohen reals gives a model where $2^{\aleph_0}$ is precisely $\aleph_\alpha$, provided that $\aleph_\alpha$ did not have countable cofinality to begin with.

For example, if one adds $\aleph_\omega$ Cohen reals, then somehow one has actually added $\aleph_{\omega+1}$ reals.

If the original model is not a model of GCH, then there is an obvious additional restriction, given by monotonicity of cardinal exponentiation: If $2^\kappa >\lambda > \kappa$, then of course we cannot preserve all cardinals and make $2^{\aleph_0}=\lambda$, regardless of the cofinality of $\lambda$. But this situation does not occur if GCH holds.


I see that you meant literally to have the quantifiers the way you wrote them. The answer to the first question is now no: Hartog proved (in ZF) that for any set $X$ there is an upper bound on the ordinals that can inject into $X$. This is easy to see: If $\alpha$ injects into $X$, then there is a subset $Y$ of $X$ and a binary relation $R$ on that subset such that $(Y,R)$ is isomorphic to $(\alpha,\in)$. But then the collection of $\alpha$ that inject into $X$ is a set (the image of the collection of such pairs $(Y,R)$).

As for the second question, the answer is no if the reals are well-orderable, because they are always larger than $\aleph_0$, since Cantor theorem does not use choice. The answer is yes (vacuously) if the reals cannot be well-ordered, and this is consistent. (Cohen's proof that choice is independent of ZF actually gives models where the reals are not well-orderable.)

share|improve this answer
1  
He asked about ZF, not ZFC. The second answer is "no if and only if there is a well ordering of the continuum." –  Asaf Karagila Mar 2 '12 at 15:06
1  
The question said "either of the two following statements", so $\forall \alpha\in\mathbf{ON}: 2^{\aleph_0}\ge\aleph_\alpha$ must count as one statement, and this statement is false under ZF, per Hartogs' construction. –  Henning Makholm Mar 2 '12 at 15:49
    
@Bruce: Thanks, even though this is not what I was asking, it still does contain useful information. –  Dejan Govc Mar 2 '12 at 16:37
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.