# If AC is false, does that mean there exist a set $A$ which has different cardinality from any ordinals?

If a set $$A$$ has the same cardinality as an ordinal $$\alpha$$, then there exists a bijection $$f:\alpha\to A$$, so $$A$$ is indexed by $$\alpha$$ and hence well-ordered. Therefore a choice function $$g:\mathcal{P}(A)\to A$$ exists.

Therefore, if Axiom of Choice(AC) is false, then there must exist a set $$A$$, such that no ordinals have the same cardinality of $$A$$.

So the equivalence class of sets with the same cardinal is larger if AC is false than if AC is true.

As a result, the truth or falsity of the Continuum hypothesis(CH) needs to be considered separately with AC and without AC.

Is the reasoning above correct? and is there really two cases to consider for CH?

EDIT: Check my idea about "larger": Let two sets $$X\sim Y$$ iff they have the same cardinality. $$\sim$$ is an equivalence relation. Denote the class of all the equivalence classes under AC $$S_1$$, and the class of all the equivalence classes without AC $$S_2$$. Clearly $$S_1\subseteq S_2$$, since $$S_2$$ has elements not in the set Ord(ordinals). This is really not formal, since it depends on the model we use.

• Have you searched the site for this kind of things? I have written many words on cardinals without choice, and CH with and without AC. – Asaf Karagila Jun 4 at 15:30
• @AsafKaragila Yes, I have found many things about CH without AC. But as you can see above, my main focus is not that. My point is to decide whether the statement "there must exist a set A, such that no ordinals have the same cardinality of A" is true or not. – Jethro Jun 4 at 22:13
• Perhaps I say too much about CH in the question, but that's not what I mean to do. In fact, I just want to check if my line of reasoning is correct. – Jethro Jun 4 at 22:15
• I have found posts for CH-AC: for example math.stackexchange.com/questions/472957/… – Jethro Jun 4 at 22:24
• Read my comment more carefully. I wrote a lot on CH, but generally about cardinals without choice. See math.stackexchange.com/questions/53752/… or math.stackexchange.com/questions/172316/…, for example. – Asaf Karagila Jun 4 at 23:19

## 1 Answer

It is a bit misleading to say that it is "larger" (how do you compare them ?) but you are right about the idea, and the formal statement : AC is indeed equivalent to "for all $$A$$, there is an ordinal $$\alpha$$ and a bijection $$\alpha\to A$$".

You are also right about CH, with AC all the statements that could reasonably called CH are equivalent, while this is no longer true without AC.

To give an example of a statement that amused me, under the Axiom of Determinacy (AD - it's incompatible with AC, but under some large cardinal hypothesis, if ZF is consistent, then so is ZF+AD), any subset of $$\mathbb{R}\sim 2^\omega$$ is either countable, or has the same size as $$\mathbb{R}$$. One could say that this means that CH is satisfied under AD.

On the other hand though, $$\mathbb{R}$$ is not equipotent to $$\omega_1$$, in fact the two have incomparable cardinalities. One could say that this means that CH fails under AD.

Without AC, you have to be more careful about what you mean with CH, and one may even say that CH is not really meaningful without AC.

• Check my idea about "larger": Let two sets $X\sim Y$ iff they have the same cardinality. $\sim$ is an equivalence relation. Denote the class of all the equivalence classes under AC $S_1$, and the class of all the equivalence classes without AC $S_2$. Clearly $S_1\subseteq S_2$, since $S_2$ has elements not in the set Ord(ordinals). This is really not formal, since it depends on the model we use. – Jethro Jun 4 at 11:09
• @Jethro You're implicitly assuming that all cardinalities are comparable - and that's not the case if choice fails. (Indeed, without choice we have to be careful how we define "cardinality" in the first place! It's often better to forget all about it and just talk about bijections - so $X\sim Y$ iff there is a bijection between them. The point then is that without choice there may be a set $S$ which is not $\sim\alpha$ for any ordinal $\alpha$; but in no sense does that mean that $S$ is bigger than every ordinal, rather it's "off to the side.") – Noah Schweber Jun 4 at 11:13
• It depends on the model, and a fixed model either satisfies AC or does not, and so you can't actually compare $S_1$ and $S_2$. And if you take two different models, one with AC and one without you have no way to compare their classes in general – Max Jun 4 at 11:13
• @AlexKruckman : ah ! well if I remember correctly the consistency of AD was proved with the assumption of one measurable cardinal, so one could say a marge cardinal is a measurable large cardinal ! ;) – Max Jun 4 at 18:22
• @Akerbeltz : if I remember correctly, Jech says some things about it in his Set theory, so does Moschovakis in Notes on Set theory, and probably Ralf Schindler as well in Set theory - exploring independence and truth. If you read mathematical french, I would recommend also Yann Pequignot's Sur la détermination des jeux infinis (like a bachelor's thesis, but quite interesting introduction). Still if you read (mathematical) french, I could send you some (very imperfect) notes I had written for a student seminar – Max Jun 4 at 22:47