# There's non-Aleph transfinite cardinals without the axiom of choice?

I can't find anything on this anywhere. The book I'm largely using at the moment is based around ZFC, so it makes no mention of anything other than the Aleph numbers, but according to Wikipedia on the Cardinal Numbers page, if you're not using the axiom of choice then there are non-Aleph ones.

I don't like putting too much stock in Wikipedia, but it usually tends to be decent enough for Maths so it seems worth asking. I do have a book on Naive Set Theory, but it doesn't seem to mention anything outside of the Aleph numbers either. Originally I thought it was just talking about the Continuum hypothesis, but that doesn't seem to be the case, and google just brings up the Aleph numbers whatever I try search for anything because everything is based around the ZFC axioms, so any chance anyone could shed any light on this with an example, or has a link or something?

• If you would like me to explain more, or add anything to my answer do let me know. – Asaf Karagila Jul 27 '11 at 7:02

The $\aleph$-numbers are cardinalities which can be well ordered. Since each well-ordered set can be identified with an ordinal, each $\aleph$-number is the cardinality of some ordinal. If we consider the least ordinal of this cardinality we will have an initial ordinal (i.e. an ordinal which cannot be bijected with any ordinal strictly smaller than itself).

The axiom of choice is equivalent to the assertion that every set can be well ordered, therefore every infinite cardinality is an $\aleph$-number.

Suppose that the axiom of choice does not hold, then we have some sets which cannot be well-ordered. Cardinalities which cannot be well-ordered cannot be $\aleph$-numbers.

This gets worse. It is consistent that every set can be linearly ordered, and the axiom of choice still fails. It is even consistent that there is a set that cannot be linearly ordered at all (an amorphous set cannot be linearly ordered, for example). In Thomas Jech's book The Axiom of Choice he has a chapter about cardinal arithmetics without choice. It shows how wild the universe of ZF can be with respect to cardinals, when the axiom of choice is absent.

A few examples:

1. It is consistent that there is no "canonical" representative to the cardinalities. When the axiom of choice is absent, we are reduced to define cardinalities as equivalent classes in the relation of "there exists a bijection", by using Scott's trick we have that this is a well defined relation. Where as the axiom of choice allows us to pick a representative from each equivalence class (namely, initial ordinals) it is consistent that without the axiom of choice there is no way to choose such representative from all cardinalities at once. (i.e. there is no definable function which returns exactly one element of each cardinality)

2. We can have that for all infinite sets $|\{0,1\}\times A|=|A|$ but the axiom of choice still does not hold (this in contrast to $|A|\times|A| = |A|$ for every infinite $A$ implies choice)

3. The continuum hypothesis is independent of choice, that is there is a model in which the axiom of choice does not hold, and there are no sets of cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$.

These are just examples to show how cardinals can behave like a bunch of unsupervised teenagers in a house full of alcohol.

• To check my understanding: Without AC, cardinalities are equivalent classes of "sets that have bijections between any two of them". The cardinality of two sets might not be comparable, so if this is the case, then the class of cardinals are not well-ordered or even linearly ordered. Is this the correct understanding? Many thanks. – Jethro Jun 5 at 0:52