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by ionfish 5081 days ago
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.
2 comments
pndmnm 5081 days ago
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).

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ionfish 5081 days ago
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...

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pndmnm 5081 days ago
Yup, I did a few projects on equivalents of AC back in the day. That's just my favorite "appeal to intuition" one (my favorite "appeal to intuition" against AC is: the identity function is the sum of two periodic functions (though this is a consequence and not equivalent)).

Equivalence breakdown in alternate systems is a wonderful topic. I've been trying for a couple years now to figure out how to get back into set theory now that I'm out of academia. Maybe later this summer...

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JadeNB 5081 days ago
> 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.

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