Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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]

share|improve this question
I usually hear this referred to as the Schroeder-Bernstein theorem. Regardless of the Cantor/Schroeder switch, I think you're misspelling Bernstein. –  Scott Morrison Oct 18 '09 at 20:59
Ooops for the mispelling. As for the denomination of the theorem, I got that from my lecturer, Peter Johnstone. If I recall correctly, he claimed that Cantor was actually the first to correctly prove the theorem, and that Schroeder got his proof wrong. –  Randomblue Oct 18 '09 at 21:20
OTOH, in any category, if there are monomorphisms $A \hookrightarrow B$ and $B\hookrightarrow A$, then for any $X$ the sets $\operatorname{Hom}(X,A)$ and $\operatorname{Hom}(X,B)$ are isomorphic, by unpacking the word "monomorphism" and applying Schroeder-Cantor-Bernstein (one of those rare results with non-alphabetized names). If this isomorphism could be made natural in $X$, then by Yoneda we would have an isomorphism $A\cong B$. But of course it cannot, in general. –  Theo Johnson-Freyd Jul 19 '10 at 4:15
I think you're still misspelling Cantor-Bernsteinness. –  Tom Goodwillie Sep 21 '10 at 19:01
Maybe it is interesting to mention that this thread was mentioned in the book Proofs of the Cantor-Bernstein Theorem: A Mathematical Excursion by Arie Hinkis, see footnote on page 397. –  Martin Sleziak May 19 '14 at 11:42

7 Answers 7

up vote 53 down vote accepted

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 $\mathbb{Z}_p^\infty$ 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:

  1. 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...

  2. 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.

share|improve this answer
There are monomorphisms of countable abelian torsion groups that are not direct summands, e.g. Z/3 into Z/9. I'm not much of a group theorist, and I don't know Ulm's Theorem. Is it true that monomorphisms in both ways implies an isomorphism, or just split monomorphisms (direct summands)? –  Theo Johnson-Freyd Jul 19 '10 at 4:11
@Theo: no, having monomorphisms $f: G \rightarrow H$ and $g: H \rightarrow G$ both ways does not imply that two countable abelian torsion groups are isomorphic. For instance, you can take $G$ to be the direct sum of a countably infinite number of copies of $Z / {9 Z}$ and take $H$ to be the direct sum of $G$ and a copy of $Z / {3 Z}$. –  John Goodrick Jul 21 '10 at 4:29
How does this categorical argument work in the case of computable sets? (Myhill isomorphism theorem) –  Diego de Estrada Oct 3 '10 at 12:42
@Diego: Offhand, I don't know of a "categorical" proof of the Myhill isomorphism theorem, whatever this would mean (I'm not saying that there isn't one, but I don't know of one). To be clear, I didn't mean that "classifiability by a bounded set of cardinal invariants" is necessary to have S-B in your category, just that it is (with a suitable interpretation of the terms) a sufficient condition for S-B. –  John Goodrick Oct 4 '10 at 18:16
@JohnGoodrick This is a very interesting answer. Is there a good place where the slogan of the very first sentence is given precise expression? –  Todd Trimble Feb 16 '14 at 0:46

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.

share|improve this answer

We discussed this at the Secret Blogging Seminar.

share|improve this answer
Thank you for the link. However, the discussion seems to be mostly case by case analysis, not "high level" category theory. –  Randomblue Oct 18 '09 at 21:29
A new version of this question I liked: the category of factors, the category of II_1 factors. –  Noah Snyder Oct 18 '09 at 23:29

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!)

share|improve this answer

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

Isometries between metric spaces

share|improve this answer

Schroeder-Bernstein Theorem holds for $\sigma$-complete Boolean algebras, $\sigma$-algebras, and for $P$-spaces. Recall that a $P$-space is a completely regular space where every $G_{\delta}$-set is open.


  1. Suppose that $A,B$ are $\sigma$-complete Boolean algebras, $a\in A,b\in B$ and $A\simeq B\upharpoonright b,B\simeq A\upharpoonright a$. Then $A\simeq B$.

  2. Suppose that $(X,\mathcal{M}),(Y,\mathcal{N})$ are $\sigma$-algebras and there exists $R\in\mathcal{M},S\in\mathcal{N}$ so that $(R,\mathcal{M}|_{R})\simeq(Y,\mathcal{N})$ and $(S,\mathcal{N}|_{S})\simeq(X,\mathcal{N})$. Then $(X,\mathcal{M})\simeq(Y,\mathcal{N})$.

  3. Suppose that $X,Y$ are $P$-spaces, $X$ is homeomorphic to a clopen subset of $Y$, and $Y$ is homeomorphic to a clopen subset of $X$. Then $X$ and $Y$ are homeomorphic.

The proof of the above theorem is a straightforward generalization of the standard proof of the Schroeder-Bernstein Theorem. The Schroeder-Bernstein Theorem also holds for other similar structures that are not as well known such as MV-algebras. In fact, this paper gives an abstract version of the Schroeder-Bernstein Theorem that one can apply to structures such as $\sigma$-complete Boolean algebras and $MV$-algebras.

share|improve this answer

The Schroeder-Bernstein Theorem holds in categories where every endomorphism is an isomorphism. More precisely, if every endomorphism is an isomorphism and there are morphisms $f:A\rightarrow B,g:B\rightarrow A$, then $A$ and $B$ are isomorphic.

It turns out that the category of ultrafilters has no non-trivial endomorphisms and hence the Schroeder-Bernstein Theorem holds for the category of ultrafilters.

To be precise, the objects in the category of ultrafilters are pairs of the form $(X,\mathcal{U})$ where $X$ is a set and $\mathcal{U}$ is an ultrafilter on the set $X$. If $(X,\mathcal{U}),(Y,\mathcal{V})$ are objects in the category of ultrafilters, then a premorphism from $(X,\mathcal{U})$ to $(Y,\mathcal{V})$ is a function $f:X\rightarrow Y$ such that if $R\subseteq Y$, then $R\in\mathcal{V}$ iff $f^{-1}[R]\in\mathcal{U}$. Now let $\mathcal{A}_{(X,\mathcal{U}),(Y,\mathcal{V})}$ be the set of all premorphisms from $(X,\mathcal{U})$ to $(Y,\mathcal{V})$. Then define an equivalence relation $\simeq$ on $\mathcal{A}_{(X,\mathcal{U}),(Y,\mathcal{V})}$ so that $f\simeq g$ iff $\{x\in X|f(x)=g(x)\}\in\mathcal{U}$. Then the morphism between $(X,\mathcal{U})$ and $(Y,\mathcal{V})$ are the equivalence classes in $\simeq$. The composition of morphisms is ordinary composition of functions. Every endomorphism in the category of ultrafilters is the identity mapping.

In fact, the fact that the category of ultrafilters satisfies the Schroeder-Bernstein theorem is one reason why the Rudin-Keisler ordering on ultrafilters is well known while the actual category of ultrafilters is less well known:

Take note that $\mathcal{U}\leq_{RK}\mathcal{V}$ if and only if there is a morphism from $\mathcal{V}$ to $\mathcal{U}$. Since the category of ultrafilters satisfies the Schroeder-Bernstein Theorem, if $\mathcal{U}\leq_{RK}\mathcal{V}\leq_{RK}\mathcal{U}$, then $\mathcal{U}$ and $\mathcal{V}$ are isomorphic ultrafilters. Therefore the Rudin-Keisler ordering on ultrafilters is a satisfactory ordering on the class of ultrafilters because the category of ultrafilters satisfies the Schroeder-Bernstein Theorem (and because the ultrafilters have no non-trivial endomorphisms also).

The Rudin-Keisler ordering has been generalized to other kinds of objects such as filters and Boolean ultrapowers. However, the generalized Rudin-Keisler orderings on the category of filters and the category of Boolean ultrapowers respectively does not satisfy the Schroeder-Bernstein Theorem (presumably), so the generalization of the Rudin-Keisler ordering are less satisfactory in these categories (and yes filters and Boolean ultrapowers do form categories).

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.