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In Bonn we've the discussion about the following topic:

Pretty much what the title says: Suppoose that A and B is are classes and that there are injections from A to B and fom B to A. Does it follow that there is a bijection between A and B?

Example: Let A the class of sets of cardinality one and let B be the class of sets of cardinality two. There is an injection

A -> B sending a to {a, emptyset},

B-> A sending b to {{b}}.

Does it follow that there is a bijection between A and B?

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I'm not sure whether the question makes sense. – Philipp Lampe Oct 19 at 1:12
I think it does make sense. A "bijection" between classes A and B should just be a subclass of A x B (i.e. a collection defined by some formula in the language of set theory) which is a one-to-one correspondence; similarly for "class injections." – John Goodrick Oct 19 at 2:02
Yes, it makes sense (see my answer for elaboration). – Andrew Critch Oct 19 at 2:20

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Ignoring set-theoretic technicalities of formulating the question properly, I see no reason that the usual proof of Schroder-Bernstein wouldn't work.

(Set-theoretic technicalities: In the standard language of set theory, you can't quantify over classes, so you can't quite state this. However, you can prove a metatheorem saying that whenever you exhibit two such injections, you can prove there is also a bijection. Alternatively, you could work in set theory with classes, in which the statement can be made properly and you ought to be able to prove it just like ordinary Schroder-Bernstein. Alternatively, it is a trivial corollary of the "global" axiom of choice (which implies, in particular, that all proper classes have the same size), though this is kind of applying a sledgehammer.)

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Me either, but is that a proof? – Philipp Lampe Oct 19 at 1:21
Sure it is. You can use the Schroder-Bernstein argument to explicitly construct a bijection from A to B given your two injections, for example. – Eric Wofsey Oct 19 at 1:41
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To ask this question, you have to first be clear what you mean by a "class". Do you mean a finite formula in ZFC language with one free variable, P(x)? (Writing P(x) roughly means "x has property P".)

Second, what do you mean by a map from a class P to a class Q? Do you mean a class of ordered pairs?

If "yes" to both of the above, then I think the answer to your question is also "yes", because the Schroeder-Bernstein argument will allow you to explicitly write down a formula F(x,y) from the formulas P( ) and Q( ) that is a "bijection" in the sense that for all x such that P(x) there is a unique y such that Q(y) and F(x,y).

If, however, you mean class in some other sense, like the undefined notion of "class" used in NBG set theory, or something more vague, a different answer will be required.

I recommend reading the wikipedia article on Zermelo-Frenkel Set Theory and browsing related articles until you feel comfortable the precise meanings of the basic terminology :)

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In NBG the same answer still works, it just works even better because you can state it as a theorem rather than a metatheorem. The only thing to check is that you never would need to use comprehension with class variables, but you don't--you only use the specific classes in question as parameters. – Eric Wofsey Oct 19 at 2:24

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