Set Theory – Understanding Non-Aleph Infinite Cardinals

axiom-of-choicecardinalselementary-set-theory

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.

Best Answer

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.

See also:

  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)
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