# Is the class of cardinals totally ordered?

In a Wikipedia article

http://en.wikipedia.org/wiki/Aleph_number#Aleph-one

I encountered the following sentence:

"If the axiom of choice (AC) is used, it can be proved that the class of cardinal numbers is totally ordered."

But isnt't the class of ordinals totally ordered (in fact, well-ordered) without axiom of choice? Being a subclass of the class of ordinals, isn't the class of cardinals obviously totally ordered?

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I think is issue is with the actual definition of "cardinal" when the axiom of choice is not assumed. It seems like the axiom is invoked somewhere in the process of the typical definition of cardinals. – Bill Cook Oct 4 '11 at 13:06

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The axiom of choice is equivalent to the statement that all cardinals have a corresponding ordinal. So if AC is not true, there are cardinals which cannot be well-ordered. – Thomas Andrews Oct 4 '11 at 13:41

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The following questions (in particular the fist one) seems to be related to this one: math.stackexchange.com/questions/53770/… math.stackexchange.com/questions/53752/… – Martin Sleziak Oct 4 '11 at 13:42

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@Bill: Only if you "insist". We can define $\aleph$ cardinals just the same without the axiom of choice. For example, $|X|=\aleph_\alpha$ if and only if there is a bijection between $X$ and the $\alpha$-th initial ordinal. – Asaf Karagila Oct 4 '11 at 16:06

## 3 Answers

If I understand the problem correctly, it depends on your definition of cardinal. If you define the cardinals as initial ordinals, then your argument works fine, but without choice you cannot show that every set is equinumerous to some cardinal. (Since AC is equivalent to every set being well-orderable.)

On the other hand, if you have some definition which implies that each set is equinumerous to some cardinal number, then without choice you cannot show that any two sets (any two cardinals) are comparable. (AC is equivalent to: For two sets $A$, $B$ there exists either an injective map $A\to B$ or an injective map $B\to A$. It is listed as one of equivalent forms if AC at wiki.)

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The statement that the class of cardinals is a subclass of the class of ordinals is equivalent to the axiom of choice.

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The ordinals are well ordered regardless to any assumption of choice. It is defined that the class of ordinal numbers is the smallest transitive class that can be well ordered, and they form a backbone of transitive models, that is if $M,N$ are two transitive models have the same ordinals then $L^M=L^N$.

Without the axiom of choice there are non-well orderable sets. Their cardinality, if so, is not an $\aleph$ number. We define the cardinality of $X$ as either finite, or $\aleph_\alpha$ for some ordinal $\alpha$, in case that $X$ can be well ordered; or as a definable subset of the class of $A$'s such that there is a bijection between $X$ and $A$.

For example, it is consistent with ZF that there are infinite sets that cannot be split into two infinite sets (every partition into two disjoint sets will yield one of them finite). Such set does not even have a countable subset and therefore incomparable with $\aleph_0$.

To see how total order of cardinals is equivalent to the axiom of choice:

If the axiom of choice holds, then every set can be well ordered and is finite or equivalent to some $\aleph$-cardinal. Therefore all cardinals are $\aleph$'s and so cardinals are totally ordered (and well ordered too).

On the other hand, if cardinals are totally ordered, given a set $X$ denote $H(X)$ the least ordinal $\alpha$ such that there is no injective function from $\alpha$ into $X$.

It can be shown that $\alpha$ is an $\aleph$ cardinal (by the fact it has no bijection with smaller ordinals which are injectible into $X$), and $\alpha\nleq|X|$.

By the assumption that cardinalities of all sets are comparable $|X|$ is comparable with $\alpha$, therefore $|X|<\alpha$, and we have that $X$ can be injected into $\alpha$ and so it inherits a well order from such injection.

Therefore all sets can be well ordered, which is equivalent to the axiom of choice.

(The term for which I used for $H(X)$ is also known as Hartogs number)

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