set theory - 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?

Answer

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.