aa r X i v : . [ m a t h . L O ] A p r Cantor-Bernstein implies Excluded Middle
Pierre Pradic ∗ Chad E. Brown † April 22, 2019
Abstract
We prove in constructive logic that the statement of the Cantor-Bernstein theorem impliesexcluded middle. This establishes that the Cantor-Bernstein theorem can only be provenassuming the full power of classical logic. The key ingredient is a theorem of Martín Escardóstating that quantification over a particular subset of the Cantor space N , the so-called one-point compactification of N , preserves decidable predicates. The
Cantor-Bernstein theorem is an elementary statement CB of set theory: for any two sets A and B , if there are injections f : A ֒ → B and g : B ֒ → A , then there exists a bijection h : A ∼ −→ B .An interesting feature of this theorem is that it may be proven in ZF without assuming the axiomof choice. However, this proof is non-constructive in the sense that it goes through classical logic;while the construction of the bijection h is rather explicit, one needs to appeal to excluded middleto show that it is indeed a bijective function.Models of constructive set theory invalidating CB are known, such as for instance, models basedon Kleene realizability . However, it left open the question of whether the full power of excludedmiddle ( EM ) is really necessary to prove the theorem or if a weaker classical principle would beenough. The purpose of this note is to show that, indeed, full excluded middle is required becauseof the following theorem. Theorem 1. ( CB ⇒ EM ) Over intuitionistic set theory, the Cantor-Bernstein theorem implies theprinciple of excluded middle.
The argument is a straightforward application of a key theorem of Martín Escardó [3] concerningthe one-point compactification N ∞ of N and decidable predicates: there exists a function ε : ( N ∞ → −→ N ∞ selecting a counter-witness for its input when possible. Formally speaking, it means that for anydecidable predicate P : N ∞ → , we have the following equivalence . (cid:0) ∀ p ∈ N ∞ . P ( p ) = 1 (cid:1) ⇐⇒ P ( ε ( P )) = 1 This result is rather striking as it means that there exists an infinite set for which decidablepredicates are stable under quantification, provably in constructive logic . This is to be contrastedagainst the case of N , which admits no recursive selection function.Assuming the existence of any infinite set equipped with a selection function, CB ⇒ EM may beproven by very elementary means. The reader familiar with [3] may content themselves with theproofs of Lemma 7 and Theorem 1. Section 3 as a whole gives a self-contained proof of CB ⇒ EM ,integrating the necessary technical content from [3] about N ∞ . For the more casual reader, we firstgive a preliminary example of an elementary set-theoretic statement implying excluded middle inSection 1 to show how the main theorem and many similar statements are proven by consideringsubsingleton sets. We then give an already-known proof [1] of a weaker variant of Theorem 1 inSection 2 for didactic purposes, before moving on to the proof of the main theorem. ∗ ENS de Lyon, Université de Lyon, LIP and University of Warsaw, MIMUW † Czech Technical University in Prague For instance, consider the subset H ⊆ N corresponding to the halting problem. If CB held in the effective topos,we would have a recursive bijection N ∼ −→ { n | n / ∈ H } ∪ { n + 1 | n ∈ N } ; since it is in particular surjective, wewould be able to build a recursive enumeration of the complement of H , which is absurd. This should not be confused with Hilbert’s ε which dually selects witnesses of existential statements. By which we systematically mean Dedekind-infinite here. oundational and notational preliminaries We work in a constructive set theory suchas IZF or CZF where we do not assume excluded middle upfront. For our purpose, the distinctionbetween these two systems will not be relevant. We shall reason informally and trust that the readerknowledgeable of other foundations be able to adapt the argument to other settings. We assumethat the axioms of Zermelo’s set theory hold. Among other things, this means that our universeof sets is closed under pairing, union, powerset (which we write P ( − ) ) and set comprehension.For sets A and B , function spaces B A and disjoint unions A + B are built as usual. We write inl : A → A + B and inr : B → A + B for the usual injections into disjoint unions. In particular,for every x ∈ A + B there is either an a ∈ A such that x = inl ( a ) or b ∈ B such that x = inr ( b ) .Let ∅ , { } and { , } ∼ = 1 + 1 .In set theory, propositions or truth values can be arranged as a set Ω = P (1) , which is closedunder all logical connective. For a formal proposition p ∈ Ω , we sometimes abbreviate “ p = 1 ” by“ p ” when writing formulas. Excluded middle EM may then be formally written as ∀ p ∈ Ω . p ∨ ¬ p .Note that an equivalent formulation of EM in this setting is Ω ∼ = 2 , which is not the case inconstructive settings. In the sequel, although they are definitionally the same in set theory, weoften make a notational distinction between propositions p ∈ Ω and subsets A ⊆ in order to keepthe arguments more readable.Finally, and crucially , we assume the axiom of infinity: there exists a Dedekind-infinite set N minimal for inclusion. We leave to the reader to disambiguate e.g. addition of numbers anddisjoint union from context. Related works
We do not reprove EM ⇒ CB here; while it is not necessary to read this note,they motivate a strengthening of CB metionned in Definition 4. Any introduction to set theoryshould have a satisfactory proof; for reference, one may look at Theorem 3.2 in [5]. For a historicalperspective, the reader may be interested in [4]. The question of the nonconstructivity of CB fromthe categorical point of view was studied by Banaschewski and Brümmer in [1] and mentioned inJohnstone’s Elephant [6] (Lemma D4.1.12, p. 950). EM Let us start with an example of a lemma involving functions which may proved in an introductionto elementary set theory.
Proposition 2.
Let A and B be sets and f : A → B an injective function. Suppose that A isnon-empty. If excluded middle holds, then there exists a surjection g : B → A .Proof. Since A is non-empty, one can pick a default element d ∈ A . By excluded middle, we knowfor every y ∈ B either there exists an x ∈ A such that f ( x ) = y or no such x exists. Hence we candefine g by cases as follows: g ( y ) := (cid:26) x if f ( x ) = yd if no such x existsNote g is well defined since f is injective, and g is clearly surjective.Notice that in this little proof, one needs to make a case analysis using excluded middle. Asin the case of Cantor-Bernstein, one can ask if this is necessary. In fact, it is necessary as we nowprove. Proposition 3.
Suppose for all sets A and B there is a surjective function g : B → A whenever A is nonempty and there is an injective function f : A → B . Then excluded middle holds.Proof. Let p ∈ Ω be given and consider A := { | p } . There is clearly an injection f : A + 1 → given by f ( inl ( x )) = 0 for x ∈ A and f ( inr (0)) = 1 . Note that A + 1 is non-empty, as inr (0) ∈ A + 1 .Applying our assumption, we obtain a surjection g : 2 → A + 1 . Note that ∀ i ∈ . g ( i ) = inl (0) ∨ g ( i ) = inr (0) holds because of the universal property of disjoint unions. Hence we have twocases. Note that the scheme ∀ x. ϕ ( x ) ∨ ¬ ϕ ( x ) for arbitrary ϕ is no more general since it can be recovered by setcomprehension. Interestingly, one may easily construct models of finitary models satisfying CB but not EM : take the internallogic of the topos Finset C op . If C is not a groupoid, EM is not satisfied, while CB always hold in the internal logic.This indicates that assuming the axiom of infinity is essential here. g ( i ) = inl (0) for some i ∈ . In this case ∈ A and so p holds.• Suppose g (0) = inr (0) and g (1) = inr (0) . We prove ¬ p holds. To this end, assume p holds.Hence ∈ A . Since g is surjective there must be some i ∈ such that g ( i ) = inl (0) ,contradicting our assumption.Here, the strategy was fairly simple: take the A ⊆ associated to the proposition, and try tomake it fit in the hypothesis of the lemma using disjoint unions and singletons. The situation inthe proof can be visualized as follows: + A ( • / / • > / / /o/o/o • ) o o > • o o or • • v v ♥♥♥♥♥♥ > • h h PPPPPP or • • o o > • h h PPPPPP
One can try to adopt a similar strategy for proving that Cantor-Bernstein implies excluded middle.Provided some A ⊆ , one can start building an injection f : A → . > / / /o/o/o • However, we need to have an injection going back and we are unsure of the existence of an elementin A . So let us consider the obvious injection → A + 1 . > / / /o/o/o • v v ♥♥♥♥♥♥ • Again we need a new value to be the image under f of this latest element, which leads us toconsider . Since we still do not have two injections, one might be tempted to iterate this process. > / / /o/o/o • v v ♥♥♥♥♥♥ • / / • > / / /o/o/o • v v ♥♥♥♥♥♥ • / / • v v ♠♠♠♠♠♠ • > / / /o/o/o • v v ♥♥♥♥♥♥ • / / • v v ♠♠♠♠♠♠ • / / • u u ❦❦❦❦❦❦ ...This informal discussion suggests using N and the following injections. A + N > / / /o/o/o • v v ♥♥♥♥♥♥ • / / • v v ♠♠♠♠♠♠ • / / • u u ❦❦❦❦❦❦ ...... N f : N −→ A + N n inr ( n ) g : A + N −→ N inl (0) inr ( n ) n + 1 CB then provides a bijection h : N → A + N . In fact, in elementary proofs of the theoremthis bijection can be seen as a perfect matching of the above graph. Note however that the usualstatement CB conceals this relationship between f, g and h . Banaschewski and Brümmer [1] studiedthe corresponding strengthened version of CB , which we dub CBBB , in a categorical setting andproved that it implied excluded middle.
Definition 4.
We say
CBBB holds if the following statement holds: given sets A and B andinjections f : A → B and g : B → A , there is a bijection h : A → B such that for all x ∈ A and y ∈ B , f ( x ) = y or x = g ( y ) whenever h ( x ) = y . Let us remark that it is obvious that
CBBB ⇒ CB . Let us also stress that EM ⇒ CBBB can beeasily obtained by adapting elementary proofs of EM ⇒ CB . Theorem 5 (Proposition 4.1 in [1]) . If CBBB holds, then excluded middle holds. roof. Assume
CBBB holds. Let a proposition p ∈ Ω be given, seen as a subset A = { | p } ⊆ .Take f : N → A + N and g : A + N → N to be the injections described above.By CBBB there is a bijection h : N → A + N such that f ( x ) = y or g ( y ) = x whenever h ( x ) = y for x ∈ N and y ∈ A + N . We know either h (0) = inl (0) or h (0) = inr ( n ) for some n ∈ N .• If h (0) = inl (0) , then ∈ A and so p holds.• Suppose h (0) = inr ( n ) for some n ∈ N . We will prove ¬ p holds. To this end, assume p holds,so that ∈ A . Since h is surjective, there is some m ∈ N such that h ( m ) = inl (0) . Either f ( m ) = inl (0) or g ( inl (0)) = m . The first case is impossible since f ( m ) = inr ( m ) by thedefinition of f and inl (0) = inr ( m ) . Therefore, g ( inl (0)) = m and, by definition of g , we musthave m = 0 . This is also impossible since it implies inl (0) = h ( m ) = h (0) = inr ( n ) . Let us pause a moment and consider why we failed to prove the analogue of Proposition 3. Inthat proof of that proposition lemma, we did not use any information about the surjection g .Instead, we resorted to exhasutively enumerating the set to check whether we had some x ∈ such that g ( x ) = inr (0) , which is a decidable property. This feature of of being searchable maybe formalized using the notion of omniscience . Definition 6 (Omniscient sets) . We say a set O is omniscient if for every p ∈ O if either thereexists x ∈ O such that p ( x ) is equal to , or p is constantly equal to . That is, ∀ p ∈ O . ( ∃ 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 constructivelogics. However, all finite sets, and in particular , are omniscient. This concept allows us to isolatethe actual core of the proof of Proposition 3. Lemma 7.
Suppose that we have an omniscient set O and some sets A and B . If there exists asurjection f : O → A + B , then either A is inhabited or it is empty.Proof. Let f : O → A + B be given and define p ∈ O by p ( x ) = (cid:26) if ∃ a ∈ A. f ( x ) = inl ( a )1 if ∃ b ∈ B. f ( x ) = inr ( b ) Since O is omniscient either ∃ x. P ( x ) = 0 or ∀ x ∈ O. P ( x ) = 1 . If ∃ x ∈ O. P ( x ) = 0 , then A isclearly inhabited. Suppose P ( x ) = 1 for every x ∈ O . We will prove A is empty. Suppose a ∈ A .Since f is surjective there must be some x ∈ O such that f ( x ) = inl ( a ) , contradicting P ( x ) = 1 .From CB (instead of the stronger CBBB ) we could use the injections from the proof of Theorem 5to obtain a surjection N → A + N . If N were omniscient, then we could use this surjection withLemma 7 to EM . However, omniscience of N correspond to the axiom of limited principle ofomniscience ( LPO ) a well-known constructive taboo [2], which can be thought of as a (strictlyweaker) version of EM . This means that deriving EM from the existence of bijections N ∼ = 1 + N by way of Lemma 7 is unreasonable. Luckily, Escardó proved that there exists an infinite subsetof the Cantor space, N ∞ , which is omniscient [3] and can be used to prove CB ⇒ EM .In order to keep the argument self-contained, we reproduce his argument below before derivingthe main result. Definition 8.
We define N ∞ to be the set of non-increasing sequences in N , i.e. , N ∞ = (cid:8) p ∈ N (cid:12)(cid:12) ∀ n ∈ N . (cid:0) p ( n ) = 1 ⇒ ∀ m ∈ N . ( m < n ⇒ p ( m ) = 1) (cid:1)(cid:9) . Remark that, at this point, we have
LPO ∧ CB ⇒ EM over constructive set theory. This observation is howevernot necessary to carry out the subsequent argument. For more categorically-inclined people, N ∞ is the final coalgebra for the functor X X . This justifiescalling N ∞ the set of conatural numbers. The induced topology from N in Definition 8 also justifies calling N ∞ the one-point compactification of N . et ω ∈ N ∞ denote the constant function. We define an injection taking n ∈ N to n ∈ N ∞ bytaking n ( m ) = (cid:26) if m < n otherwise.Finally we define S : N ∞ → N ∞ by cases taking S ( p )(0) = 1 and S ( p )( n + 1) = p ( n ) . The set N ∞ is infinite as witnessed by and S . Lemma 9.
The function S : N ∞ → N ∞ is injective and S ( p ) = 0 for all p ∈ N ∞ .Proof. Suppose S ( p ) = S ( q ) . Since p ( n ) = S ( p )( n +1) = S ( q )( n +1) = q ( n ) for every n ∈ N we know p = q , as desired. The fact that S ( p ) = 0 for all x ∈ N ∞ follows from S ( p )(0) = 1 = 0 = 0(0) .Classically every element of N ∞ is either ω or of the form n . The corresponding disjunctionis equivalent to LPO , and so is unprovable constructively. However, for decidable predicates, it issufficient to show that they hold over all elements n and ω to show they hold everywhere . Lemma 10.
Let Q ∈ N ∞ be given. If Q ( ω ) = 1 and ∀ n ∈ N . Q ( n ) = 1 , then ∀ p ∈ N ∞ . Q ( p ) = 1 .Proof. Let Q ∈ N ∞ such that Q ( ω ) = 1 and ∀ n ∈ N . Q ( n ) = 1 . Let p ∈ N ∞ be given. To prove Q ( p ) = 1 , it is enough to prove Q ( p ) = 0 . Assume Q ( p ) = 0 . Under this assumption we can prove ∀ n ∈ N . p ( n ) = 1 by strong induction. Assume ∀ k ∈ N . ( k < n ⇒ p ( k ) = 1) and p ( n ) = 0 . Thisis enough information to infer p = n , contradicting Q ( p ) = 0 and Q ( n ) = 1 . To end the proof wenote that p must be ω (since ∀ n ∈ N . p ( n ) = 1 ), contradicting Q ( p ) = 0 and Q ( ω ) = 1 .While the desired function ε : 2 N ∞ → N ∞ is rather straightforward to define, Lemma 10 iscritical in allowing to prove constructively that it is indeed a selection function. Theorem 11 (Theorem 3.15 in [3]) . There is a function ε : 2 N ∞ → N ∞ such that for every Q ∈ N ∞ , if Q ( ε ( Q )) = 1 , then ∀ p ∈ N ∞ . Q ( p ) = 1 .Proof. For Q ∈ N ∞ , take ε ( Q ) ∈ N to be ε ( Q )( n ) = (cid:26) if Q ( k ) for each k ≤ n otherwise.This may be well-defined recursion over n . It is easy to check that ε ( Q ) ∈ N ∞ as well.Assume Q ( ε ( Q )) = 1 and let p ∈ N ∞ be given. If ∀ k < n. Q ( k ) = 1 and Q ( n ) = 0 , then ε ( Q ) = n and so Q ( n ) = Q ( ε ( Q )) = 1 , contradicting Q ( n ) = 0 . Consequently, Q ( n ) = 1 for every n by induction. Thus ε ( Q ) = ω and so Q ( ω ) = Q ( ε ( Q )) = 1 holds as well. Hence Q ( p ) = 1 forevery p ∈ N by Lemma 10. Corollary 12 (Corollary 3.6 in [3]) . The set N ∞ is omniscient.Proof. Let Q ∈ N ∞ be given. If Q ( ε ( Q )) = 0 , then ∃ x ∈ N ∞ .Q ( x ) = 0 . If Q ( ε ( Q )) = 1 , then ∀ x ∈ N ∞ . Q ( x ) = 1 by Theorem 11.This completes the part of the construction we obtained following Escardó [3]. We can noweasily put it together with Lemma 7 to conclude. Proof of Theorem 1.
Assume CB holds. We know N ∞ is omniscient by Lemma 12. We know S is injective and S ( x ) = 0 for every x ∈ N ∞ by Lemma 9. Let p ∈ Ω be a proposition and take A = { | p } ⊆ . Analogously to Section 2, we consider the following functions: f : N ∞ −→ A + N ∞ g : A + N ∞ −→ N ∞ x inr ( x ) inl (0) inr ( x ) S ( x ) Both f and g are clearly injective, so we can apply CB to obtain a bijection h : N ∞ → A + N ∞ .Lemma 7 now implies that either A is inhabited (so p holds) or A is empty (so ¬ p holds). This constitutes a particular case of Lemma 3.4 in [3]. eferences [1] G. C.L. Banaschewski, B.; Brümmer. Thoughts on the Cantor-Bernstein Theorem. QuaestionesMathematicae , 9, 01 1986.[2] E. Bishop.
Foundations of constructive analysis . McGraw-Hill, 1967.[3] M. Escardó. Infinite sets that satisfy the principle of omniscience in any variety of constructivemathematics.
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Proofs of the Cantor-Bernstein Theorem: A Mathematical Excursion . ScienceNetworks. Historical Studies, Vol. 45. Birkhäuser (Springer Basel), 2013.[5] T. Jech.
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