Theorem 1 [ZFC, classical logic]: If $A,B$ are sets such that $\textbf{2}\times A\cong \textbf{2}\times B$, then $A\cong B$.

That's because the axiom of choice allows for the definition of cardinality $|A|$ of any set $A$, and for $|A|\geq\aleph_0$ we have $|\textbf{2}\times A|=|A|$.

Theorem 2: Theorem 1 still holds in ZF with classical logic.

This is less trivial and explained in Section 5 of Division by Three - however, though the construction does not involve any choices, it does involve the law of excluded middle.

Question: Are there intuitionistic set theories in which one can prove $$\textbf{2}\times A\cong \textbf{2}\times B\quad\Rightarrow\quad A\cong B\quad\text{?}$$

For example, is this statement true in elementary topoi or can it be proved in some intuitionistic type theory?

In his comment below Kyle indicated that the statement is unprovable in some type theory - does somebody know the argument or a reference for that?

Edit See also the related question Does $A\times A\cong B\times B$ imply $A\cong B$? about 'square roots'

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  • Perhaps I'm thinking about this naively, but couldn't it be approached from the contrapositive? $A\ncong B \Rightarrow 2\times A \ncong 2\times B$. That seems to me like a trivially true statement. – JMoravitz Nov 16 '14 at 17:18
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    JMoravitz: In intuitionistic logic, the contrapositive of a statement does not imply the statement itself; i.e. $\neg B \Rightarrow \neg A$ does not imply $A \Rightarrow B$. (It only implies $A \Rightarrow \neg\neg B$, which is intuitionistically a strictly weaker statement.) Also, I don't understand in which way the contrapositive is "trivially true"; note that $A$ and $B$ may be arbitrary (maybe infinite) sets. – Ingo Blechschmidt Nov 17 '14 at 09:37
  • If I recall correctly, you can probably prove something like this in type theory. – Kyle Gannon Nov 17 '14 at 17:31
  • @KyleGannon: that would be great! Do you remember where you saw that? – Hanno Nov 18 '14 at 07:35
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    @Hanno: Nevermind, I lied. I was told today that it was impossible. – Kyle Gannon Nov 18 '14 at 21:04
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    @KyleGannon: Ah ok - and *why*? :-) – Hanno Nov 18 '14 at 21:16
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    $\aleph_0$ does not exist in constructive mathematics. – Han de Bruijn Jan 03 '15 at 20:41
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    @HandeBruijn That depends a lot on the exact variety of constructive mathematics. In particular, intuitionistic set theories often make use of infinity in some way. – Mario Carneiro Jan 04 '15 at 11:24
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    @MarioCarneiro: L.E.J. Brouwer is the founder of intuitionism. Quote from his thesis (Over de grondslagen der wiskunde, 1907): _De tweede getalklasse van Cantor bestaat niet_. Translated in English: _Cantor's second number class does not exist_. – Han de Bruijn Jan 04 '15 at 13:34
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    @HandeBruijn: I don't quite understand your objection - the question does not refer to $\aleph_0$. – Hanno Jan 04 '15 at 13:38
  • @Hanno: The symbol is in there at $|A|\geq\aleph_0$ : I can read (though understand nothing of the transfinite). – Han de Bruijn Jan 04 '15 at 13:42
  • @HandeBruijn: ? This was about how to settle the question affirmatively in the classical framework of ZFC - the problem itself does not involve $\aleph_0$. – Hanno Jan 04 '15 at 15:17
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    $\aleph_0$ certainly does exist in constructive mathematics, which although inspired in part by Brouwer is not very closely related to his exact form of intuitionism. The natural reading of the question is how to prove the result in a theory like IZF of CZF, both of which can handle transfinite ordinals just fine. – Carl Mummert Jan 05 '15 at 00:22
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    @Hanno: if no answer is obtained here, I believe this question would fit on MathOverflow, where it would reach a different set of eyes. – Carl Mummert Jan 07 '15 at 16:21
  • Couldn't this be proved with a proof that starts out with $A\cong B$, operate $2\times$ on $A$ and $B$, then end with $2\times A\cong 2\times B$? – Conor O'Brien Jan 08 '15 at 20:51
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    @ConorO'Brien: Yes that's fine and proves $\Leftarrow$, but the question asks whether one can also show $\Rightarrow$ from intuitionistic principles. – Hanno Jan 08 '15 at 21:17
  • @Hanno Oh, that makes sense. Thanks! – Conor O'Brien Jan 08 '15 at 21:33
  • @Carl Mummert: Yes, that's a good idea, maybe someone on MO can help. Thank you for your interest and for offering the bounty! – Hanno Jan 10 '15 at 07:46
  • I can move it if you want. – Mariano Suárez-Álvarez Jan 22 '15 at 08:58
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    @Mariano: As a first step, I've posted this on the meta.MO thread for "Interesting questions on MSE". The link is [here](http://meta.mathoverflow.net/a/2097/7206). This seems like a logical step before migrating the question like a European swallow, carrying a coconut on a strand of creeper (under the dorsal guiding feathers). – Asaf Karagila Jan 23 '15 at 00:28
  • You should be careful with the use of $A \cong B$ in a topos: this can mean (1) externally, that $A$ and $B$ are isomorphic (as in any category), or (2) internally, that the statement "there exists a bijection between $A$ and $B$" is true. Now (1) implies (2), but the converse is not valid in general, for the same reason that an object can be internally [inhabited](https://ncatlab.org/nlab/show/inhabited+object) without being externally so (we can make an object of bijections between $A$ and $B$ which is externally resp. internally inhabited iff $A$ and $B$ are isomorphic ditto). (continued.) – Gro-Tsen Jan 12 '16 at 18:45
  • (cont.) For a simple example, consider a group $G$, and in the topos of $G$-sets, $A=G$ with trivial action while $B=G$ with left action: then $A$ and $B$ are not (externally) isomorphic, but they are internally so. So your question of whether $2\times A \cong 2\times B$ implies $A \cong B$ has two (or even three) different meanings according to what is meant by $\cong$. Meaning 1(external) is perhaps more sensible in categorical terms, meaning 2(internal) if you're thinking in terms of the internal logic / set theory of the topos. – Gro-Tsen Jan 12 '16 at 18:51

2 Answers2


There was a paper recently posted to arXiv about this question: Swan, On Dividing by Two in Constructive Mathematics.

It turns out that there are examples of toposes where you can't divide by two.

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(The following is not really an answer, or just a very partial one, but it's definitely relevant and too long for a comment.)

There is a theorem of Richard Friedberg ("The uniqueness of finite division for recursive equivalence types", Math. Z. 75 (1961), 3–7) which goes as follows (all of this is in classical logic):

For $A$ and $B$ subsets of $\mathbb{N}$, define $A \sim B$ when there exists a partial computable function $f:\mathbb{N}\rightharpoonup\mathbb{N}$ that is one-to-one on its domain and defined at least on all of $A$ such that $f(A) = B$. (One also says that $A$ and $B$ are computably equivalent or recursively equivalent, and it is indeed an equivalence relation, not to be confused with "computably/recursively isomorphic", see here.) Then [Friedberg's theorem states]: if $n$ is a positive integer then $(n\times A) \sim (n \times B)$ implies $A\sim B$ (here, $n\times A$ is the set of natural numbers coding pairs $(i,k)$ where $0\leq i<n$ and $k\in A$ for some standard coding of pairs of natural numbers by natural numbers).

To make this assertion closer to the question asked here, subsets of $\mathbb{N}$ can be considered as objects, indeed subobjects of $\mathcal{N}$, in the effective topos (an elementary topos with n.n.o. $\mathcal{N}$ such that all functions $\mathcal{N}\to\mathcal{N}$ are computable), in fact, these subobjects are exactly those classified by maps $\mathcal{N} \to \Omega_{\neg\neg}$ where $\Omega_{\neg\neg} = \nabla 2$ is the subobject of the truth values $p\in\Omega$ such that $\neg\neg p = p$; moreover, to say that two such objects are isomorphic, or internally isomorphic, in the effective topos, is equivalent to saying that $A$ and $B$ are computably isomorphic as above. So Friedberg's result can be reinterpreted by saying that if $A$ and $B$ are such objects of the effective topos and if $n\times A$ and $n\times B$ are isomorphic then $A$ and $B$ are.

I'm not sure how much this can be internalized (e.g., does the effective topos validate "if $A$ and $B$ are $\neg\neg$-stable sets of natural numbers and $n\times A$ is isomorphic to $n\times B$ then $A$ is isomorphic to $B$" for explicit $n$? and how about for $n$ quantified inside the topos?) or generalized (do we really need $\neg\neg$-stability?). But this may be worth looking into, and provides at least a positive kind-of-answer to the original question.

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