# Non-aleph infinite cardinals

I'm now confused with a concept of $\aleph$.

1.$\aleph$ is a cardinal number that is well-ordered in ZF.(Defined as an initial ordinal that is equipotent with). Does that mean $\aleph_x$ in ZF may NOT be equal to $\aleph_x$ in ZFC?

2.I don't know how to define $\aleph$ in ZF. Here's what I tried. Do we call $A$, the class of alephs?

That is, let $[\alpha]$ = {$\beta \in OR$|$\beta \simeq \alpha$} Since OR is well-ordered (class of von-Neumann ordinals), $[\alpha]$ has a least element $\alpha_l$. Let $A$={$\alpha_l \in OR$|$\alpha \in OR$} Let $V$ be the union of every $a \in A$. Since $V$ is a subset of OR, $V$ is well-Ordered. I'm trying to show that power set of $V$ is well-ordered, thus if $V$ is a set, $V$ and the power set of $V$ is equiotent (since every well-ordered set is isomorphich with some $b\in OR$), which is a contradiction to show that $V$ is a proper class, hence $A$ is a proper class. - I don't know how to show that the power set of $V$ is well-ordered.

Last question is, How do I define 'class of cardinals'? , since there might be some sets equipotent with none of alephs in ZF.

• I think it is important to add that your first item is incorrect. Perhaps you meant that "In ZF, an $\aleph$ is a cardinal that is well-ordered." The way you wrote this, it seems you are saying that an $\aleph$ must be provably well-orderable, which needs not be the case. – Andrés E. Caicedo Jul 18 '12 at 14:30

The class of $\aleph$ numbers is the same class of cardinals that you know in a model of ZFC. Namely, initial ordinals. The definitions are exactly the same. Furthermore by definition the $\aleph$ cardinals are ordinals, so the correspond to well ordered sets.

On the other hand, if $A$ is not a well-orderable set, then $|A|$ corresponds to the set $$\{B\mid \exists f\colon A\to B\text{ a bijection}\land\operatorname{rank}(B)\text{ is minimal}\}$$ Where the $\operatorname{rank}$ operator is the von Neumann rank of $B$. This set is not an ordinal, clearly, and it may lack any internal structure.

The class of cardinals, therefore, is combined from two parts:

1. The $\aleph$ numbers which are "ordinal which cannot be put in bijection with any of its elements".

We can see that the $\aleph$ numbers do not form a set directly, suppose that they would, then there was an ordinal $\gamma$ such that the set of $\aleph$ has von Neumann rank $\gamma$. In particular all of its elements have rank $<\gamma$. Let $\kappa$ be the first ordinal above $\gamma$ such that $\kappa$ is not in bijection with any of its elements, then $\kappa$ is an $\aleph$, but its von Neumann rank is $\kappa>\gamma$ in contradiction.

2. Cardinals of sets which are not well-orderable. These are described as sets $A$ such that "Every two members of $A$ have a bijection between them, all the elements of $A$ have the same von Neumann rank, and no set of lower rank has a bijection with any element of $A$, and if there is a $B$ of the same von Neumann rank as a member of $A$, and they are in bijection then $B$ is an element of $A$ as well"

Yes, it is a bit clumsy and unclear, but set theory without choice may get like that often.

It is immediate that the class of cardinals is a proper class since it contains all the $\aleph$-cardinals. Much like in ZFC the cardinals make a proper class, the arguments carry over in this case as well.

Lastly, you cannot prove that a power set of a well-ordered set is well-ordered because if the axiom of choice fails this is simply not true. Furthermore, $A$ itself is a class, as it contains elements of unbounded rank, so we need to be more careful with "the union over $A$" as it is not a set as well, that is $V$ itself is a class.

As $V$ is a class its power "set" is not a set and does not exist, and as I remarked power sets of a well-ordered set need not be well-orderable.

1. Defining cardinality in the absence of choice
2. There's non-Aleph transfinite cardinals without the axiom of choice?
3. How do we know an $\aleph_1$ exists at all? (this asserts that $\aleph_1$ exist, even without choice, and the argument carries over to high cardinals)
• I assumed $V$ is a set to make a contradiction leads to show that $A$ is not a set. If $V$ is a set, $V$ is a subset of OR, thus $V$ is well-ordered. In this case, the power set of $V$ still can't be well-ordered? – Katlus Jul 18 '12 at 9:41
• The axiom of choice is equivalent to the assertion "If $x$ is well-ordered then $P(x)$ can be well-ordered as well". So without the axiom of choice you cannot prove such thing. – Asaf Karagila Jul 18 '12 at 9:44
• For general case, yes. But this is a speial case.. Here's how i tried. 1.Assume that $V$ is a set. Then, since $V$ is a sub'set' of OR, $V$ is well-ordered set. Hence every sub'set' of V is well-ordered and there exists $a\in OR$ that is isomorphic with $z$ for every $z\in P(V)$. Order $P(V)$ as follows. $x<y$iff [least element of $x-y$ is less than least element of $y-x$ and $x$ is isomorphic with $y$] or [ordinal of x is less than ordinal of y]. Then isn't $P(V)$ well-ordered by $≦$? – Katlus Jul 18 '12 at 10:14
• You cannot prove the power set of $\omega$ is well-ordered without some appeal to the axiom of choice. How would you expect proving an arbitrary subset of the ordinals has a well-orderable power set? – Asaf Karagila Jul 18 '12 at 10:30

The well-ordered cardinals are defined by the formula $x\in\text{ON}\land |x|=x$. Of course both conjuncts abbreviate fairly complicated formulae, but they’re familiar ones.

As for the first question, the $\omega_\alpha$ are defined by transfinite recursion and are independent of AC.

• Is defining well-ordered cardinals hard that I can't understand on my level right now? (That is, sounds silly, I can't do it just for a day?) Or at least I want to show the class $A$ above is a proper class. – Katlus Jul 18 '12 at 9:33
• @Katlus: No, it’s easy: I just did it. $x$ is a well-ordered cardinal iff $x$ is an ordinal and the least ordinal equipotent with $x$ is $x$. – Brian M. Scott Jul 18 '12 at 9:35
• Isn't it critical to show OR and the class of well-ordered cardinals are isomorphic to mark well-ordered cardinals as $\aleph_x$ for some $x\in OR$? – Katlus Jul 18 '12 at 9:38
• @Katlus: That’s implicit in the definition of the $\omega_\alpha$: $\omega_0=\omega,\omega_{\alpha+1}=(\omega_\alpha)^+$, and $\omega_\gamma=\sup\{\omega_\alpha:\alpha<\gamma\}$ for limit $\gamma$. – Brian M. Scott Jul 18 '12 at 9:40