 # Cantor-Bernstein implies Excluded Middle

We prove in constructive logic that the statement of the Cantor-Bernstein theorem implies excluded middle. This establishes that the Cantor-Bernstein theorem can only be proven assuming the full power of classical logic. The key ingredient is a theorem of Martín Escardó stating that quantification over a particular subset of the Cantor space 2^N, the so-called one-point compactification of N, preserves decidable predicates.

## Authors

##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 An elementary set-theoretic statement implying EM

Let us start with an example of a lemma involving functions which may proved in an introduction to elementary set theory.

###### Proposition 2.

Let and be sets and an injective function. Suppose that is non-empty. If excluded middle holds, then there exists a surjection .

###### Proof.

Since is non-empty, one can pick a default element . By excluded middle, we know for every either there exists an such that or no such exists. Hence we can define by cases as follows:

 g(y):={x if f(x)=yd if no such x exists

Note is well defined since is injective, and is clearly surjective. ∎

Notice that in this little proof, one needs to make a case analysis using excluded middle. As in the case of Cantor-Bernstein, one can ask if this is necessary. In fact, it is necessary as we now prove.

###### Proposition 3.

Suppose for all sets and there is a surjective function whenever is nonempty and there is an injective function . Then excluded middle holds.

###### Proof.

Let be given and consider . There is clearly an injection given by for and . Note that is non-empty, as . Applying our assumption, we obtain a surjection . Note that holds because of the universal property of disjoint unions. Hence we have two cases.

• Suppose for some . In this case and so holds.

• Suppose and . We prove holds. To this end, assume holds. Hence . Since is surjective there must be some such that , contradicting our assumption.

Here, the strategy was fairly simple: take the associated to the proposition, and try to make it fit in the hypothesis of the lemma using disjoint unions and singletons. The situation in the proof can be visualized as follows:

## 2 Reversing Cantor-Bernstein-Banaschewski-Brümmer

One can try to adopt a similar strategy for proving that Cantor-Bernstein implies excluded middle.

Provided some , one can start building an injection .

However, we need to have an injection going back and we are unsure of the existence of an element in . So let us consider the obvious injection .

Again we need a new value to be the image under of this latest element, which leads us to consider . Since we still do not have two injections, one might be tempted to iterate this process.

This informal discussion suggests using and the following injections.

then provides a bijection . In fact, in elementary proofs of the theorem this bijection can be seen as a perfect matching of the above graph. Note however that the usual statement conceals this relationship between and . Banaschewski and Brümmer  studied the corresponding strengthened version of , which we dub , in a categorical setting and proved that it implied excluded middle.

###### Definition 4.

We say holds if the following statement holds: given sets and and injections and , there is a bijection such that for all and , or whenever .

Let us remark that it is obvious that . Let us also stress that can be easily obtained by adapting elementary proofs of .

###### Theorem 5 (Proposition 4.1 in ).

If holds, then excluded middle holds.

###### Proof.

Assume holds. Let a proposition be given, seen as a subset . Take and to be the injections described above.

By there is a bijection such that or whenever for and . We know either or for some .

• If , then and so holds.

• Suppose for some . We will prove holds. To this end, assume holds, so that . Since is surjective, there is some such that . Either or . The first case is impossible since by the definition of and . Therefore, and, by definition of , we must have . This is also impossible since it implies .

## 3 Reversing Cantor-Bernstein

Let us pause a moment and consider why we failed to prove the analogue of Proposition

3. In that proof of that proposition lemma, we did not use any information about the surjection . Instead, we resorted to exhasutively enumerating the set to check whether we had some such that , which is a decidable property. This feature of of being searchable may be formalized using the notion of omniscience.

###### Definition 6 (Omniscient sets).

We say a set is omniscient if for every if either there exists such that is equal to , or is constantly equal to . That is,

 ∀p∈2O.(∃x∈O.p(x)=0)∨(∀x∈O.p(x)=1)

In classical logic, all sets are clearly omniscient, but this is not necessarily true in constructive logics. However, all finite sets, and in particular , are omniscient. This concept allows us to isolate the actual core of the proof of Proposition 3.

###### Lemma 7.

Suppose that we have an omniscient set and some sets and . If there exists a surjection , then either is inhabited or it is empty.

###### Proof.

Let be given and define by

 p(x)={0 if ∃a∈A.f(x)=inl(a)1 if ∃b∈B.f(x)=inr(b)

Since is omniscient either or . If , then is clearly inhabited. Suppose for every . We will prove is empty. Suppose . Since is surjective there must be some such that , contradicting . ∎

From (instead of the stronger ) we could use the injections from the proof of Theorem 5 to obtain a surjection . If were omniscient, then we could use this surjection with Lemma 7 to . However, omniscience of correspond to the axiom of limited principle of omniscience () a well-known constructive taboo  , which can be thought of as a (strictly weaker) version of 666 Remark that, at this point, we have over constructive set theory. This observation is however not necessary to carry out the subsequent argument.. This means that deriving from the existence of bijections by way of Lemma 7 is unreasonable. Luckily, Escardó proved that there exists an infinite subset of the Cantor space, , which is omniscient  and can be used to prove .

In order to keep the argument self-contained, we reproduce his argument below before deriving the main result.

###### Definition 8.

We define to be the set of non-increasing sequences in , i.e.777For more categorically-inclined people, is the final coalgebra for the functor . This justifies calling the set of conatural numbers. The induced topology from in Definition 8 also justifies calling the one-point compactification of .,

 N∞={p∈2N∣∣∀n∈N. (p(n)=1⇒∀m∈N. (m

Let denote the constant function. We define an injection taking to by taking

 n––(m)={1 if m

Finally we define by cases taking and .

The set is infinite as witnessed by and .

###### Lemma 9.

The function is injective and for all .

###### Proof.

Suppose . Since for every we know , as desired. The fact that for all follows from . ∎

Classically every element of is either or of the form . The corresponding disjunction is equivalent to , and so is unprovable constructively. However, for decidable predicates, it is sufficient to show that they hold over all elements and to show they hold everywhere888This constitutes a particular case of Lemma 3.4 in ..

###### Lemma 10.

Let be given. If and , then .

###### Proof.

Let such that and . Let be given. To prove , it is enough to prove . Assume . Under this assumption we can prove by strong induction. Assume and . This is enough information to infer , contradicting and . To end the proof we note that must be (since ), contradicting and . ∎

While the desired function is rather straightforward to define, Lemma 10 is critical in allowing to prove constructively that it is indeed a selection function.

###### Theorem 11 (Theorem 3.15 in ).

There is a function such that for every , if , then .

###### Proof.

For , take to be

 ε(Q)(n)={1 if Q(k––) for each k≤n0 otherwise.

This may be well-defined recursion over . It is easy to check that as well.

Assume and let be given. If and , then and so , contradicting . Consequently, for every by induction. Thus and so holds as well. Hence for every by Lemma 10. ∎

###### Corollary 12 (Corollary 3.6 in ).

The set is omniscient.

###### Proof.

Let be given. If , then . If , then by Theorem 11. ∎

This completes the part of the construction we obtained following Escardó . We can now easily put it together with Lemma 7 to conclude.

###### Proof of Theorem 1.

Assume holds. We know is omniscient by Lemma 12. We know is injective and for every by Lemma 9. Let be a proposition and take . Analogously to Section 2, we consider the following functions:

 f:N∞⟶A+N∞aaaaaaag:A+N∞⟶N∞x↦inr(x)inl(0)↦0–inr(x)↦S––(x)

Both and are clearly injective, so we can apply to obtain a bijection . Lemma 7 now implies that either is inhabited (so holds) or is empty (so holds). ∎

## References

•  G. C.L. Banaschewski, B.; Brümmer. Thoughts on the Cantor-Bernstein Theorem. Quaestiones Mathematicae, 9, 01 1986.
•  E. Bishop. Foundations of constructive analysis. McGraw-Hill, 1967.
•  M. Escardó. Infinite sets that satisfy the principle of omniscience in any variety of constructive mathematics. J. Symb. Log., 78(3):764–784, 2013.
•  A. Hinkis. Proofs of the Cantor-Bernstein Theorem: A Mathematical Excursion. Science Networks. Historical Studies, Vol. 45. Birkhäuser (Springer Basel), 2013.
•  T. Jech. Set theory. Springer Science & Business Media, 2013.
•  P. T. Johnstone. Sketches of an Elephant: A Topos Theory Compendium: 2 Volume Set. Oxford University Press UK, 2002.