When does Cantor-Bernstein hold? - MathOverflow

When does Cantor-Bernstein hold?

2009-10-18T18:35:32Z

The Cantor-Bernstein theorem in the category of sets (A injects in B, B injects in A => A, B equivalent) holds in other categories such as vector spaces, compact metric spaces, Noetherian topological spaces of finite dimension, and well-ordered sets.

However, it fails in other categories: topological spaces, groups, rings, fields, graphs, posets, etc.

Can we caracterize Cantor-Bernsteiness in terms of other categorical properties?

[Edit: Corrected misspelling of Bernstein]

Answer by David Speyer for When does Cantor-Bernstein hold?

2009-10-18T18:43:45Z

We discussed this at the Secret Blogging Seminar.

Answer by John Goodrick for When does Cantor-Bernstein hold?

2009-10-18T23:28:58Z

Whenever the objects in your category can be classified by a bounded collection of cardinal invariants, then you should expect to have the Schroeder-Bernstein property.

For example, vector spaces (over some fixed field K) or algebraically closed fields (of some fixed characteristic) can each be classified by a single cardinal invariant: the dimension of the vector space, or the transcendence degree of the field.

More interesting example: countable abelian torsion groups. Suppose A and B are two such groups, A is a direct summand of B, and vice-versa; are they isomorphic? By Ulm's Theorem, A and B are determined up to isomorphism by countable sequences of cardinal numbers -- namely, the number of summands of Z_p^infty and the "Ulm invariants," which are dimensions of some vector spaces associated with A and B. All of these invariants behave nicely with respect to direct sum decompositions, so it follows that A and B are isomorphic. (See Kaplansky's Infinite Abelian Groups for a very nice, and elementary, proof of all this.)

If you like model theory, I could tell you a lot about when the categories of models of a complete theory have the Schroeder-Bernstein property (under elementary embeddings). If not, at least I can tell you this:

Categories of structures with "definable" partial orderings with infinite chains (e.g. real-closed fields, atomless Boolean algebras) will NOT have the S-B property. Again, I need some model theory to make this statement precise...
Let C be a first-order axiomatizable class of structures (in a countable language) which is "categorical in 2^{aleph_0}" -- i.e. any two structures in C of size continuum are isomorphic. Then C has the S-B property with respect to elementary embeddings. (This generalizes the cases of vector spaces and algebraically closed fields.)

Addendum: A completely different way that a category C might be Schroeder-Bernstein is if every object is "surjunctive" (i.e. any injective self-morphism of an object is necessarily surjective). This covers Justin's example of the category of well-orderings.

Answer by Andreas Blass for When does Cantor-Bernstein hold?

2010-07-18T22:08:48Z

Let me point out a curious (non-categorical) twist to an elementary observation. The elementary observation is that the Cantor-Schroeder-Bernstein property fails for linear orderings, even if we require the injections to map onto an initial segment. That is, you can have two non-isomorphic linear orders, each isomorphic to an initial segment of the other. One example is the same as a familiar example for the topological case, the closed interval [0,1] and the half-open interval [0,1) of real numbers. And of course, the situation is the same for final segments. The curious twist is that, if a linear order A is isomorphic to an initial segment of another linear order B, while B is isomorphic to a final segment of A, then A and B are isomorphic.

Answer by Manny Reyes for When does Cantor-Bernstein hold?

2010-09-21T17:12:22Z

Here's another example where a type of Cantor-Schroeder-Bernstein theorem holds. In this paper of R. Bumby, it was proved that if $A$ and $B$ are injective modules over a ring that can be embedded in each other, then $A \cong B$.

An immediate corollary is that if any two modules over a ring embed in one another, then their injective hulls are isomorphic.

I wonder whether this fits into John Goodrick's answer above. I'm not aware of any bounded collection of cardinal invariants that classify injectives over an arbitrary ring. (If the ring is right noetherian then each injective decomposes into a direct sum of indecomposable injectives, and this allows us to classify the injectives. But Bumby's result holds for arbitrary rings!)

Answer by Roberto Frigerio for When does Cantor-Bernstein hold?

2011-08-21T14:53:30Z

Here is another example where a type of Cantor-Schroeder-Bernstein theorem holds: the category of compact metric spaces, which is in fact surjunctive according to the terminology described by John Goodrick. This remark is motivated by the question

http://mathoverflow.net/questions/73309/isomorphisms-between-metric-spaces