It is known from a result of Sierpinski that the generalized continuum hypothesis (GCH) implies the axiom of choice (AC). It is also known from the celebrated results of Cohen that AC is independent of ZF and that GCH is independent of ZFC. But suppose we start with the axioms of ZF and assume they are consistent, and then add both the negation of the axiom of choice and the continuum hypothesis (i.e., CH but not GCH).

Is it known whether the resulting system is consistent or inconsistent?

  • $\begingroup$ So you'd like a model of $ZF + \neg AC$ where $\aleph_1 = 2^{\aleph_0}$? Or where $\aleph_1 \not= 2^{\aleph_0}$? That can probably be arranged, but I'll let the set theorists answer it. $\endgroup$ – David Roberts Feb 21 '13 at 1:18
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    $\begingroup$ Please make the title less ambiguous. $\endgroup$ – S. Carnahan Feb 21 '13 at 5:49

I'm assuming you're asking about the theory $ZF+\neg AC+CH$. The answer is yes - but you have to be clear about what you mean by "$CH$."

First, a trivial example: we can start with a model of $ZFC+CH$, and force a failure of $AC$ via a (symmetric submodel of) a $2^{\aleph_0}$-closed forcing extension. Since our forcing extension is sufficiently closed, it adds no new sets of reals, so $CH$ remains true; essentially, what's going on is that we're adding a failure of $AC$ to our model, but we're doing so at such a high level in the cumulative hierarchy that it doesn't affect the reals.

A more refined version of your question: can we have $ZF+\neg AC(\mathbb{R})+CH$? That is, a model of $ZF$ in which $CH$ holds but the reals are not well-ordered.

This is where we need to be precise about what $CH$ means. The useful version of $CH$ is "every set of reals is either at most countable, or can be bijected onto $\mathbb{R}$." Under this phrasing, we do indeed have the consistency of $ZF+\neg AC(\mathbb{R})+CH$!

To get $AC(\mathbb{R})$ to fail, we just need a model in which there is no injection from $\omega_1$ (which is defined to be the least uncountable ordinal) to $\mathbb{R}$: if $AC(\mathbb{R})$ held, we could send each countable ordinal $\alpha$ to the "least" real which codes a well-order of order type $\alpha$. To force the continuum hypothesis to hold requires a bit more subtlety. My favorite model of $ZF+\neg AC(\mathbb{R})+CH$ is $L(\mathbb{R})$ under the assumption of large cardinals: large cardinals imply that $L(\mathbb{R})\models AD$, the axiom of determinacy, which in turn implies that every set of reals has the perfect set property (so $CH$ holds) and also that there is no injection from $\omega_1$ to $\mathbb{R}$ (so $AC(\mathbb{R})$ fails). However, this model does involve a massive jump in consistency strength, past that of $ZF$, which is not necessary: Truss has a construction, which is a variation of a construction of Solovay (which does require large cardinals, albeit just one small one!), which does not require any more consistency strength than $ZF$ itself and satisfies $ZF+\neg AC(\mathbb{R})+CH$.

  • $\begingroup$ I believe it is even possible to get $GCH$ restricted to alephs (that is, well-orderable cardinalities) and still violate $AC(\mathbb{R})$, without any large cardinal assumptions; I don't remember this clearly, though, so somebody please correct me if I'm wrong. $\endgroup$ – Noah Schweber Feb 21 '13 at 1:26
  • $\begingroup$ I am almost certain that this is impossible for two reasons: If we mimic Truss' work, we see that the result is either at least a proper class of inaccessible cardinals, or that there is a bound on regular cardinals which in turn implies very large cardinals in the background; the second reason is that it sounds quite strange that this would hold for singular cardinals (or at least singular limit cardinals). I can buy the claim that from a very weak assumption of a proper class of inaccessible cardinals you can get something like this. Somehow related: mathoverflow.net/questions/88758 $\endgroup$ – Asaf Karagila Feb 21 '13 at 3:26

First we need to straighten up what does the continuum hypothesis mean. It could mean $2^{\aleph_0}=\aleph_1$, and it could mean "Every uncountable set of real numbers has size $2^{\aleph_0}$". Assuming the axiom of choice those are equivalent, but without it they are not.

If we only want the first variant to hold, it's easy. We can start with a universe where it holds, and violate the axiom of choice high enough in the hierarchy of sets to ensure $2^{\aleph_0}=\aleph_1$ still. But it is possible to have this principle fail, but the second variant to hold.

It is consistent that there is no injection from $\omega_1$ into $\mathbb R$ in $\mathsf{ZF}$. This is easily provable from large cardinal assumptions, but it is also true without them.

Truss proved that if you start with a model of $\mathsf{GCH}$ and collapse any limit cardinal to be $\aleph_1$, and then take permutations with bounded support (nevermind the technical babble right now), the result is a model of $\mathsf{ZF}$ in which:

  1. Every uncountable set of real numbers has cardinality $2^{\aleph_0}$.
  2. There is no injection from $\omega_1$ to $\Bbb R$.
  3. $\omega_1$ is regular if and only if $\mathsf{DC}$ holds if and only if we collapsed an inaccessible cardinal.

Since we are not interested in the third case, it is consistent relative to the consistency of $\mathsf{ZFC}$ (sans large cardinals!) that: $\mathsf{ZF+\lnot AC+CH}$ holds.

You may be interested in this math.SE answer of mine on the topic.


There's a little bit of ambiguity in your question. Cohen's 1963 paper gives a model of $$ZF+\hbox{(not AC)} + CH$$ so this theory is consistent.

But when you say that you want to assume $ZF$ is consistent, I'm not sure whether you're asking about the above theory or the theory $$ZF+\hbox{Con(ZF)}+\hbox{not AC}+ CH$$

If the latter, then I don't know the answer to your question.

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    $\begingroup$ Surely Cohen's original model is an $\omega$ model, which will satisfy Con(ZF)? $\endgroup$ – Carl Mummert Feb 21 '13 at 3:11
  • $\begingroup$ Carl, aren't all standard models $\omega$-models? :) $\endgroup$ – Asaf Karagila Feb 21 '13 at 3:39

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