[ionfish]

Yes, it does rely implicitly on Cantor–Schröder–Bernstein. That might be a downside, but I think when working informally (that is to say, when not teaching a set theory course) one can simply assert that if there exist injective functions from sets A and B into one another then they are equinumerous.

That being said, although important in the theory of the order of the cardinal numbers, Cantor–Schröder–Bernstein doesn't show that the cardinals are totally ordered. That statement is actually equivalent to the Axiom of Choice, whereas as far as I'm aware Cantor–Schröder–Bernstein holds in ZF.

[pndmnm]

That's absolutely correct -- trichotomy for arbitrary cardinals (any two cardinals are cardinal-size-comparable) requires AC, but SB doesn't require AC. Trichotomy for the cardinal numbers of well-ordered sets (e.g. ordinals) doesn't require AC.

It's a little irrelevant to this thread... but as long as I'm quoting non-proofs that require lots of extra machinery, I'll give my favorite appeal-to-intuition equivalent of choice: the product of non-empty sets is non-empty (any point in the product of a collection of non-empty sets is a choice function on those sets).

[ionfish]

AC is equivalent to a lot of things. There's a collection of them on the Wikipedia page.

http://en.wikipedia.org/wiki/Axiom_of_choice#Equivalents

Something I find pretty interesting is that some of these equivalences break down in weak systems.

http://www.math.uchicago.edu/~antonio/RM11/RM%20talks/mummer...

[JadeNB]

> That being said, although important in the theory of the order of the cardinal numbers, Cantor–Schröder–Bernstein doesn't show that the cardinals are totally ordered.

Good point, thanks; I've corrected my post accordingly.