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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?

ashpool
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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 04 '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 04 '11 at 13:41
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    The following questions (in particular the fist one) seems to be related to this one: http://math.stackexchange.com/questions/53770/defining-cardinality-in-the-absence-of-choice http://math.stackexchange.com/questions/53752/theres-non-aleph-transfinite-cardinals-without-the-axiom-of-choice – Martin Sleziak Oct 04 '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 04 '11 at 16:06
  • @ThomasAndrews By "there are cardinals which cannot be well-ordered" do you mean that exist cardinalities s.t. sets of this cardinality can't be well-ordered, or that there exist cardinalities that are not ordered relative to each other? – Stewart Dec 20 '14 at 18:43
  • The former. Or, more specifically, there exists a set for which no well-ordering exists. That is essentially the same as to former - a cardinality for which no representative can be well-ordered. @Stewart – Thomas Andrews Dec 20 '14 at 18:47
  • The existence of that set would also mean that no set of comparably larger cardinality could be well-ordered, because a well-ordering of $X$ means a well-ordering of any $Y\subseteq X$. – Thomas Andrews Dec 20 '14 at 18:53

4 Answers4

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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.)

Martin Sleziak
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A cardinal number is a non-negative integer or an $\aleph$ number, and yes, the cardinal numbers are totally ordered. By this definition, which is fairly standard as far as I know, a cardinal number is any ordinal number $\sigma$ from which no injective function exists to any ordinal number $\tau\subsetneqq\sigma$. Thus $\aleph_0=\omega_0$, $\aleph_1=\omega_1$, etc.

The cardinality of a set is a slightly more subtle concept: we say that the cardinality of a set A is less than or equal to that of a set B if there exists a one-to-one (or injective) function f from A to B. We express this mathematically as follows: $$|A|\le|B|\leftrightarrow \exists f:A\xrightarrow{1-1}B$$

Clearly the existence of such a one-to-one function is a reflexive, transitive relation. By the Cantor-Bernstein theorem,

$$\left(\exists f:A\xrightarrow{1-1}B\right)\land\left(\exists g:B\xrightarrow{\rm 1-1}A\right)\rightarrow \left(\exists h:A\xrightarrow{\rm 1-1,\ onto}B\right).$$

Therefore if A and B are sets, such $|A|\le|B|$ and $|B|\le|A|$, then there is a one-to-one and onto function (called a bijection) from A to B. We say that $|A|=|B|$ in this case, and the class of all sets is partitioned into equivalence classes by this relation. The cardinality of a set is defined to be its equivalence class under this relation $|\cdot|=|\cdot|$. These equivalence classes are always partially ordered by the relation $|\cdot|\le|\cdot|$ on the underlying sets.

So far I have not assumed the axiom of choice: if we accept the axiom of choice, then the equivalence classes of cardinality are totally ordered, one and only one cardinal number being an element of each class. Without the axiom of choice, the equivalence classes of cardinality are not totally ordered.

See proof by Asaf and my question The axiom of choice in terms of cardinality of sets.

Annorlunda
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    I would add, however, that it is not universal to take your definition as the definition of "cardinal number" if you are not assuming the axiom of choice. – Eric Wofsey Nov 03 '16 at 21:07
  • @EricWofsey Is the "cardinal number" then the entire equivalence class? (No doubt some say that it is.) – Annorlunda Nov 03 '16 at 21:15
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    More or less, except that the equivalence class isn't a set. But you can make them sets using Scott's trick (instead of taking the whole equivalence class, take only its elements of minimal rank). – Eric Wofsey Nov 03 '16 at 23:31
  • See also my comment here: http://math.stackexchange.com/questions/1985905/is-cardinality-a-well-defined-function/1985909#comment4078525_1985909 – Asaf Karagila Nov 04 '16 at 10:24
  • @AsafKaragila Is it right to say that the statement "Given two sets $A,B$ there exists an injection from $A$ to $B$ or an injection from $B$ to $A$" is equivalent to the axiom of choice ? – Gabriel Romon Sep 14 '19 at 14:03
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    @GabrielRomon: Yes. – Asaf Karagila Sep 14 '19 at 22:48
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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.

aaa
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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)

Asaf Karagila
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