15 Inconsistency of the Nested Sets

Chapter of the book Infinity Put to the Test by Antonio Le´on available HERE

Abstract.-This chapter examines a denumerable version of the nested-set theo- rem and derives from it a contradiction involving the formal consistency of the actual infinity hypothesis subsumed into the axiom of infinity.

Keywords: Nested Sets Theorem, Nested Sets inconsistency, actual infinity.

A denumerable version of the Nested-Sets Theorem P310 Let A = {a1, a2, a3 . . . } be any ω-ordered set and consider the following recursive definition:

A1 = A − {a1} (1) (Ai = Ai−1 − {ai}; i = 2, 3, 4,... that yields the ω-ordered sequence S = hAni of nested sets A1 ⊃ A2 ⊃ A3 ⊃ ..., being each set An = {an+1, an+2, an+3,... } a denumerable pro- per subset of all its predecessors, as well as a superset of all of its successors. Note that, in order to define the numerable sequence of numerable sets hAii, the possibility of removing one by one all elements of A is assumed, even if there is not a last element to be removed.

A1 A2

A3 ... ?

Figura 15.1 – Venn diagram of the Empty Intersection Theorem: All sets are nested and, being denumerable, each of them occupies a concentric area greater than zero. However the common concentric area is null.

1 2 Inconsistency of the Nested Sets

P311 The following theorem is a denumerable version of the so called Nested Sets Theorem (the original version, also called Cantor’s Intersec- tion Theorem, deals with compact sets, and the conclusion is exactly the contrary, i.e. that the intersection is nonempty [2, p. 98-99]).

Theorema) P311, of the empty intersection.-The sequence S of sets hAni defined in P310 satisfies:

Ai = ∅ (2) i \ Proof.-If an element ak would belong to the intersection then only a finite number (equal or less than k) of sets would have been defined by (1), since ak does not belong to Ak, Ak+1, Ak+2, ....

P312 The Empty Intersection Theorem is a trivial result in modern infini- tist mathematics. It simply states the sets hAni have no common element. As far as I know, the consequences of the fact that each set Ai is a denu- merable proper subset of all its predecessors have never been examined. This chapter discusses some of those consequences.

ta t1 t2 t3 b1 b2 b3

À0 balls À0 balls À0 balls À0 balls

BX BX BX BX ... Figura 15.2 – Removing, one by one, the balls of a box that contains ℵo balls.

P313 Before starting the main discussion that will take place in the next section, let us examine an elementary physical version of the Empty Inter- section Theorem. Let BX be a box containing a denumerable collection hbii of balls indexed as b1, b2, b3, . . . , and let htni be a strictly increasing ω-ordered sequence of instants within the real interval (ta,tb) whose limit is tb. Now consider the following supertask: at each instant ti remove from the box the ball bi, and only the ball bi. The one to one correspondence f between htii and hbii defined by f(ti) = bi, ∀ti ∈ htii proves that at tb all balls will have been removed from BX.

P314 In accordance with the way of removing the balls, one by one and in such a way that between the removal of a ball bn and the removal of the A denumerable version of the Nested-Sets Theorem 3 next one bn+1 an interval of time tn+1 −tn greater than zero always elapses, it could be expected that just before completing the removal of all balls from the box, the box will contain . . . 5, 4, 3, 2, 1 balls. Nothing further from the (infinitist) truth: before it is empty, the box will never contain a finite number n of balls, whatever n, simply because those n balls would be the impossible last n balls of an ω-ordered collection of indexed balls; and the successive instants at which the successive balls were successively removed from the box would be the impossible last n instants of an ω- ordered sequence of instants.

At any t in [ta, tb) Just at t b

Àoballs 0 balls

BX BX

Figura 15.3 – The Aleph-zero or zero dichotomy

P315 Let f(t) be the number of balls within the box at any instant t in [ta,tb], i.e. the number of balls to be removed at the precise instant t. As a consequence of ω-order, we will have the following inevitable dichotomy:

ℵo if t ∈ [ta,tb) ∀t ∈ [ta,tb] : f(t)= (3) 0ift= tb 

Otherwise, if for a t in [ta,tb) it holds f(t) = n, being n any natural number, then there would exist the impossible last n terms of an ω-ordered sequence.

P316 Taking into account the one to one correspondence f(ti) = bi, all balls hbni are removed one by one from the box BX, one after the other and in in such a way that an interval of time ∆it = ti+1 − ti greater than zero always elapses between the removal of two successive balls bi, bi+1, ∀i ∈ N. But according to the above ℵo or 0 dichotomy (3), this is impossible because the number of balls to be removed from the box has to change directly from ℵo to 0 (without intermediate finite states at which only a finite number of balls remain to be removed), and this is only possible by removing simultaneously ℵo balls. 4 Inconsistency of the Nested Sets

P317 The box BX plays the role of the set A and the successive remo- vals of the balls from BX represent the successive steps of the recursive definition (1). Since the successive elements a1, a2, a3, ...of A are succes- sively removed in order to define the successive terms A1, A2, A2, ...of the sequence S, we could write:

Ai = {a/1, a/2, . . . a/i, ai+1, ai+2,... } (4) where a/1, a/2, . . . a/i simply indicate the successive elements a1, a2,... ai, of A that have been successively removed in order to define the successive sets A1, A2,. . . Ai, of the sequence S.

P318 As in the case of the box BX, and for the same reasons, if we focus our attention on the number of elements that remain unmarked in (4) as the recursive definition (1) progresses, then we will immediately come to the conclusion that that number can only take two values: ℵo and 0.

P319 The ℵo or 0 dichotomy implies the number of unmarked elements in (4) changes directly from ℵo to 0, and this is only possible by marking ℵo elements at once, i.e. by defining simultaneously ℵo sets of the sequence S, which evidently is not compatible with the recursiveness of that definition, in the same way that to remove simultaneously ℵo balls from the box is not compatible with the successiveness of the removals. P320 There is, however, a significant difference between taking away the balls from BX and the recursive definition (1): while the box BX is always the same box BX as the balls are successively removed from it (which makes it evident the fallacy of the removal), the set A originates a sequence of sets: starting from A1, each set Ai originates a new set Ai+1 when the element ai+1 is removed from it in order to define the next term of the sequence. Thus, A dissolves in a complete infinite sequence of sets without a last set completing the sequence, which conceals the fallacy of removing one by one all elements of a collection without ever resting . . . 3, 2, 1 elements to be removed. P321 Faced with the evidence of the fact that by removing one by one the infinitely many balls within the box BX you will inevitably get a box BX that will successively contain . . . , 5, 4, 3, 2, 1, 0 balls, some infinitists claim that you cannot remove one by one the balls from that box because there is not a last ball to be removed. You can remove one by one the elements of a set to define a numerable sequence of sets, such as the above sequence hAii, even if there is no last element to be removed, but you cannot remove one by one the infinitely many balls of a box because there is not a last ball Inconsistency of the nested sets 5 to be removed from the box. What to think of a formal theory that allows to remove elements from a set, but not balls from a box because this would call the theory into question? If that theory assumes the hypothesis of the actual infinity, it is assuming that all elements of an infinite collection exist as a complete totality, with or without a last element. And if all elements of the collections are removed from the collection, the result can only be the empty set, otherwise not all elements of the collection would have been removed from the collection. Be the collection a denumerable set or a box that contains infinitely many balls. In consequence, if a bijection as the above one proves that all elements of a collection have been removed from the collection at a certain instant, at that instant the resulting collection can only be the empty set. Not accepting this conclusion means accepting that after removing all elements from a collection, not all elements of the collection have been removed from the collection. And if the elements of the collection are removed one by one, and all are removed, it is difficult to explain that the container, be it a box or a set, never contains a finite number of elements not yet removed. Inconsistency of the nested sets P322 The above discussion of the Empty Intersection Theorem suggests that this theorem is not as trivial as it seems. It, in fact, motivates the short discussion that follows, whose main objective is to put into question the formal consistency of the actual infinity hypothesis. It seems convenient at this point to recall that Cantor took it for granted the existence of the set of all finite cardinals as a complete infinite totality (a hypothesis now subsumed into the modern Axiom of Infinity), and that from that initial assumption he successfully derived the infinite sequence of the transfinite ordinals of the second class, the smallest of which is ω [1, p. 167, Theorem §15 K]. Thus, any result affecting the formal consistency of ω will affect the whole sequence of transfinite ordinals of the second class as well as the formal consistency of the actual infinity hypothesis. Let us just begin by assuming the Axiom of Infinity and then the existence of ω-ordered sets and ω-ordered sequences as complete infinite totalities.

P323 Consider again the above sequence of sets S = A1, A2, A3,... From ∗ ∗ S, define the sequence S of sets by successively adding to S (that is initially empty) the successive sets A1, A2, A3 ..., of S if, and only if, n ∩i=1Ai 6= ∅: i=n ∗ n = 1, 2, 3, · · · : add An to S iffn=1or Ai 6= ∅ (5) i=1 \ 6 Inconsistency of the Nested Sets

P324 As in previous arguments in this book, it could easily be proved by induction or by Modus Tollens that for any natural number v the first v successive additions (5) can be carried out. The inductive proof is as ∗ follows. According to (5) the set A1 can be added to S . Suppose that for ∗ any natural number n it is possible to add to S the first n sets A1, A2,... An of the sequence S. We will have:

A1 ∩ A2 ∩···∩ An = An 6= ∅ (6)

Since An+1 = {an+2, an+2, an+2,... } is a denumerable subset of An we can write: A1 ∩ A2 ∩···∩ An ∩ An+1 = An+1 6= ∅ (7) ∗ Hence, An+1 can also be added to S , which proves that for every natural ∗ number v it is possible to add the first v elements of S to S . And then, for any natural number v, the first v successive additions (5) can be carried out.

P325 Assume that while the successive additions (5) can be carried out they are carried out. Once all possible successive additions (5) have been ∗ carried out (Principle of Execution), the sequence S will be formed by a certain (finite or infinite) number of sets that by addition have a no- nempty intersection. Let, therefore, av be any element of that intersection. Evidently it holds: av ∈/ Av. In consequence, Av is not a member of the ∗ sequence S .

∗ P326 It is immediate to prove, however, Av is a member of S :

a) The subindex v in Av is a natural number. b) According to P324, for each natural number v the fists v successive additions (5) can be carried out. c) All possible successive additions (5) have been carried out. d) The first v successive additions (5) have been carried out (Principle of Execution). ∗ e) The vth addition (5) adds Av to S because:

A1 ∩ A2 ∩···∩ Av = Av 6= ∅ (8)

∗ f) In consequence Av is a member of S .

P327 We have, therefore, derived a contradiction from our initial assum- ∗ ption: the set Av is and is not in the sequence S . Inconsistency of the nested sets 7

P328 The alternative to the above contradiction is another contradiction even more elemental: after having performed all possible successive addi- tions (5) in accordance with the Principle of Execution, not all possible successive additions (5) have been performed.

∗ P329 It could also be argued that S is defined infinitely many times and ∗ that although each and every addition (5) defines S as a sequence of sets whose intersection is nonempty, the completion of the sequence of succes- ∗ sive additions (5) converts S into a sequence of sets whose intersection is empty. As if the completion of an ω-ordered sequence of additions, as such a completion, had additional arbitrary consequences on the defined object. The same arbitrary consequences could be expected in any other procedure or proof consisting of an ω-ordered sequence of steps. In those conditions any thing could be expected in infinitist mathematics because the Principle of Invariance could be violated.

P330 Moreover, by timetabling the sequence of additions (5) so that each nth step takes place at the precise instant tn of a strictly increasing sequen- ce of instants htii within (ta,tb) whose limit is tb, it could easily be proved ∗ that only at tb, once completed the sequence of additions (5), could S be- come a sequence of sets whose intersection is empty. This would confirm, on the one hand that the completion of an ω-ordered sequence of additions, as such a completion, has additional arbitrary effects on the resulting object; and on the other that those arbitrary effects takes place at the instant tb, ∗ the first instant after completing the sequence S of additions; an instant in which no step of the addition is carried out; an instant when nothing ∗ happens that can justify the empty intersection of the sequence of sets S defined by (5). 8 Inconsistency of the Nested Sets Chapter References

[1] Georg Cantor, Contributions to the founding of the theory of transfinite numbers, Dover, New York, 1955. [2] Enrique Lin´es, Principios de An´alisis Matem´atico, Revert´e, Barcelona, 1991.

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