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CANTOR VERSUS CANTOR Antonio Leon Sanchez 0 1 Dpt. Natural Sciences, Inst. Francisco Salinas, Salamanca 0 Phil. Sci. & Math. Interciencia, Salamanca, Spain 2 http://www.interciencia.es r p A Abstract. ThispaperexaminesthepossibilitiesofextendingCantor’stwo argumentsontheuncountablenatureofthesetofrealnumberstooneofits 9 properdenumerablesubsets: thesetofrationalnumbers. Thepaperproves that, unless certain restrictive conditions are satisfied, both extensions are ] M possible. It is therefore indispensable to prove that those conditions are in fact satisfied in Cantor’s theory of transfinite sets. Otherwise that theory G would be inconsistent. . h t a m Part I: Cantor’s 1874 argument [ 3 v Cantor’s first proof of the uncountability of the set R of the real numbers was 4 published in the year 1874 [5], (French edition [6], Spanish edition [10]), in a 7 short paper that also included a proof of the countable nature of the set A of 8 2 algebraic numbers, and then of the set Q of rational numbers. The discussion . 1 that follows examines Cantor’s original argument and then the possibilities of 0 its application to the set of rational numbers. 0 1 : v i X 1. Cantor’s proof r a 1-1. Assume the set R is denumerable. In these conditions there would be a one to one correspondence f between the set N of natural numbers and R. Consequently, the elements of R could be ω-ordered as: r , r , r , ... (I) 1 2 3 being r = f(i),∀i ∈ N. Obviously, the sequence hr i defined by f would i i i∈N contain all real numbers. 1-2. Consider now any real interval (a, b). Cantor’s 1874-argument consists of proving the existence of a real number η in (a, b) which is not in hr i . i i∈N The existence of η would, in fact, prove the falseness of the assumption on the countable nature of R. The proof goes as follows. 1 Cantor versus Cantor 1-3. Starting from r , find the first two elements of hr i within (a, b). 1 i i∈N Denote the smaller of them by a and the larger by b . Define the real interval 1 1 (a , b ). 1 1 1-4. Starting from r , find the first two elements of hr i within (a , b ). 1 i i∈N 1 1 Denote the smaller of them by a and the larger by b . Define the real interval 2 2 (a , b ). Evidently it holds: (a , b ) ⊃ (a , b ). 2 2 1 1 2 2 1-5. Starting from r , find the first two elements of hr i within (a , b ). 1 i i∈N 2 2 Denote the smaller of them by a and the larger by b . Define the real interval 3 3 (a , b ). Evidently it holds: (a , b ) ⊃ (a , b ) ⊃ (a , b ). 3 3 1 1 2 2 3 3 1-6. The continuation of the above procedure (RP from now on) defines a sequence of nested real intervals (RP-intervals): (a , b ) ⊃ (a , b ) ⊃ (a , b ) ⊃ ... (II) 1 1 2 2 3 3 whose left endpoints a , a , a ,... form a strictly increasing sequence, and 1 2 3 whose right endpoints b , b , b ,... form a strictly decreasing sequence, being 1 2 3 every element of the first sequence smaller than every element of the second one. 1-7. From the ω-ordering of hr i and the ordered way R defines the suc- i i∈N cessive RP-intervals, it immediately follows that: • If r defines an endpoint a or b , then it must hold i ≤ n. n i i • If r defines an endpoint a or b , then r will not lie within the succes- n i i n sive intervals (a , b ), (a , b ), (a , b ), ... i i i+1 i+1 i+2 i+2 Inconsequence, wecanensurethat, beingr anyelementofhr i , itwillnever v i i∈N belong to any of the successive intervals (a , b ), (a , b ), (a , b ), ... v v v+1 v+1 v+2 v+2 1-8. Thenumber ofRP-intervalscanbefiniteorinfinite, andbothpossibilities have to be considered. 1-9. Assume the number of RP-intervals is finite1. In this case there would be a last real interval2 (a , b ) in the sequence. This last interval would contain, n n at best, one element r of hr i , otherwise it would be possible to define v i i∈N at least one new real interval (a , b ). Let, therefore, η be any element n+1 n+1 within (a , b ) different from r (in the case that r exist). Evidently η is a n n v v real number within (a, b) which does not belong to the sequence hr i . Thus, i i∈N η proves the falseness of Cantor’s initial assumption on the countable nature of R: the sequence hr i does not contain all real numbers. i i∈N 1Including the case that RP defines no interval. 2Or the whole interval (a, b) in the case that RP defines no interval. 2 Cantor versus Cantor 1-10. Consider now the number of RP-intervals is infinite. Since the sequence ha i is strictly increasing and upper bounded by every element of hb i , i i∈N i i∈N the limit L of ha i does exist. On its part, the sequence hb i is strictly a i i∈N i i∈N decreasing and lower bounded by every element of ha i , in consequence the i i∈N limit L of this sequence also exists. Taking into account that every a is less b i than every b it must hold: L ≤ L . i a b 1-11. Assume that L < L . In this case, any of the infinitely many elements a b within the real interval (L , L ) is a real number within (a, b) which does not a b belong to the sequence hr i , and then a proof of the falseness of Cantor’s i i∈N initial hypothesis on the countable nature of R. 1-12. Finally, assume that L = L = L. It is immediate that L is a real a b number within (a, b) which is not in hr i . In fact, assume that L is an i i∈N element r of hr i . According to 1-7, r does not belong to any of the v i i∈N v successive intervals (a , b ), (a , b ), (a , b ), ..., while L belongs to v v v+1 v+1 v+2 v+2 all of them. Therefore, L is not r . The limit L is a real number in (a, b) v which is not in hr i , and then a proof of the falseness of Cantor’s initial i i∈N assumption on the countable nature of R. 2. Cantor’s 1874-argument applied to rational numbers Next paragraphs 2-1/2-14 extend Cantor’s 1874-argument to the set Q of rational numbers. Both arguments are identical except in the last step. 2-1. Assume thesetQofrationalnumbersisdenumerable. Intheseconditions there would be a one to one correspondence f between the set N of natural numbers and Q so that the elements of Q could be ω-ordered as: q , q , q , ... (III) 1 2 3 being q = f(i),∀i ∈ N. Obviously, the sequence hq i defined by f would i i i∈N contain all rational numbers. 2-2. Consider now any rational interval (a, b). We will try to prove the existence of a rational number η in (a, b) which is not in hq i . Evidently, η i i∈N would prove the falseness of our initial assumption on the countability of Q. The proof goes as follows. 2-3. Starting from q , find the first two elements of hq i within (a, b). 1 i i∈N Denote the smaller of them by a and the larger by b . Define the rational 1 1 interval (a , b ). 1 1 3 Cantor versus Cantor 2-4. Starting from q , find the first two elements of hq i within (a , b ). 1 i i∈N 1 1 Denote the smaller of them by a and the larger by b . Define the rational 2 2 interval (a , b ). Evidently it holds: (a , b ) ⊃ (a , b ). 2 2 1 1 2 2 2-5. Starting from q , find the first two elements of hq i within (a , b ). 1 i i∈N 2 2 Denote the smaller of them by a and the larger by b . Define the rational 3 3 interval (a , b ). Evidently it holds: (a , b ) ⊃ (a , b ) ⊃ (a , b ). 3 3 1 1 2 2 3 3 2-6. The continuation of the above procedure (QP from now on) defines a sequence of nested rational intervals (QP-intervals): (a , b ) ⊃ (a , b ) ⊃ (a , b ) ⊃ ... (IV) 1 1 2 2 3 3 whose left endpoints a , a , a ,... form a strictly increasing sequence, and 1 2 3 whose right endpoints b , b , b ,... form a strictly decreasing sequence, being 1 2 3 every element of the first sequence smaller than every element of the second one. 2-7. From the ω-ordering of hq i and the ordered way QP defines the i i∈N successive QP-intervals, it immediately follows that: (1) If q defines an endpoint a or b , then it must hold i ≤ n. n i i (2) If q defines an endpoint a or b , then q will not lie within the succes- n i i n sive intervals (a , b ), (a , b ), (a , b ), ... i i i+1 i+1 i+2 i+2 Inconsequence, wecanensurethat,beingq anyelement ofhq i ,itwillnever v i i∈N belong to any of the successive intervals (a , b ), (a , b ), (a , b ), ... v v v+1 v+1 v+2 v+2 2-8. Thenumber ofQP-intervalscanbefiniteorinfinite, andbothpossibilities have to be considered. 2-9. Assume the number of QP-intervals is finite3. In this case there would be a last rational interval4 (a , b ) in the sequence. This last interval would n n contain, at best, one element q of hq i , otherwise it would be possible to v i i∈N define at least one new rational interval (a , b ). Let, therefore, η be any n+1 n+1 element within (a , b ) different from q (in the case that q exist). Evidently n n v v η is a rational number within (a, b) which does not belong to the sequence hq i . Thus, η proves the falseness of the initial assumption on the countable i i∈N nature of Q: the sequence hq i does not contain all rational numbers. i i∈N 3Including the case that QP defines no interval. 4Or the whole interval (a, b) in the case that QP defines no interval. 4 Cantor versus Cantor 2-10. Consider now the number of QP-intervals is infinite. Since the sequence ha i isstrictly increasing andupper bounded by every element of hb i , the i i∈N i i∈N real limit L of ha i does exist. On its part, the sequence hb i is strictly a i i∈N i i∈N decreasing and lower bounded by every element of ha i , in consequence the i i∈N real limit L of this sequence also exists. Taking into account that every a is b i less than every b it must hold: L ≤ L , being L and L two real (rational or i a b a b irrational) numbers. 2-11. Assume that L < L . In this case, any of the infinitely many rationals a b within the real interval5 (L , L ) is a rational number within (a, b) which does a b not belong to the sequence hq i , and then a proof of the falseness of the i i∈N initial hypothesis on the countable nature of Q. 2-12. Finally, assume that L = L = L. It is immediate that L is a real a b number within the real interval (a, b) which is not in hq i . In fact, If L is i i∈N irrational then it is clear that it will not in hq i ; assume then L is rational, i i∈N and assume also it is an element q of hq i . According to 2-7, q does not v i i∈N v belong to any of the successive intervals (a , b ), (a , b ), (a , b ), ..., v v v+1 v+1 v+2 v+2 while L belongs to all of them. Therefore, L is not q . The limit L is a real v number (rational or irrational) in the real interval (a, b) which is not in hq i . i i∈N Thus, if L were rational then our initial assumption on the countable nature of Q would be false. 2-13. Under the assumption 2-12, let hc i be the sequence of elements i i∈I of hq i within (a, b) which are not endpoints of the QP-intervals. If there i i∈N would exist an v such that: (a , a )∩hc i 6= ∅ (V) v v+1 i i∈I then there would exist at least a rational η within (a , a ) which is not v v+1 an element of hc i and such that a < η < a . Now then, being a i i∈I v v+1 v the immediate predecessor of a in ha i , and being this sequence strictly v+1 i i∈N increasing it would be impossible for η to be a left endpoint of any QP-interval. And being all right points hb i greater than a , and then than η, this last i i∈N v+1 number could not be an element of hb i either. Thus, η would be a rational i i∈N number within (a, b) which is not in hq i , and then a proof of the falseness i i∈N of our initial assumption on the countability of Q. 2-14. The same argument 2-13 applies to each of the successive intervals (b , b ) defined by the successive hb i , the right endpoints of the QP- i+1 i i i∈N intervals. 5That any real interval contains infinitely many rationals is an elementary result that can easily be proved. 5 Cantor versus Cantor 3. Conclusion on Cantor’s 1874 argument We have just proved in 3-1/3-14 the alternatives of Cantor 1874-argument on the cardinality of the real numbers can be applied to the set Q of rational numbers, except the last one, that applies only if the common limit of the sequences of left and right endpoints of the QP-intervals is rational. Evidently, If Cantor’s 1874-argument could be extended to the rational numbers we would have a contradiction: the set Q would and would not be denumerable. Ac- cordingly, in order to ensure the impossibility of that contradiction, each of the following points have to be proved: 3-1. Whatsoever be the rational interval (a, b) and whatsoever be the reordering of hq i , it must holds: i i∈N (1) The number of QP-intervals can never be finite. (2) The sequences of endpoints ha i and hb i can never have different i i∈N i i∈N limits. (3) The common limit of ha i and hb i can never be rational. i i∈N i i∈N (4) Being hc i the sequence defined in 2-13, it must hold: i i∈I (a, a )∩hc i = ∅; (b , b)∩hc i = ∅ (VI) 1 i i∈I 1 i i∈I (a , a )∩hc i = ∅ ∀i ∈ N : i i+1 i i∈I (VII) (b , b )∩hc i = ∅ ( i+1 i i i∈I Until those proofs be given, Cantor’s 1874-argument should be suspended, and the possibility of a contradiction involving the foundation of (infinitist) set theory should be considered. 4. A short epilog to Cantor’s 1874-argument The following short argument proves the existence of rational numbers within any rational interval (a, b) which are not in the sequence hq i of the above i i∈N extension of Cantor’s 1874-argument. 4-1. Let x be a rational variable initially defined as b, the right endpoint of any rational interval (a, b), and hq i the sequence of rational numbers i i∈N defined in 2-1. Then, consider the following procedure P: q ∈ (a,b)∧q < x ⇒ x = q ; i = 1,2,3,... (VIII) i i i that compares x with the successive elements of hq i that belong to (a, b), i i∈N and defines x as the compared element each time the compared element is less than x. 6 Cantor versus Cantor 4-2. Whatsoever be the finite or infinite number of times that P defines x, and whatsoever be the current value of x once performed P, x will be a rational number within (a,b] because, by definition, it is always defined as a rational number within (a,b), and only as a rational number within (a,b). Consider thentherationalinterval (a, x)andanyelement η within(a, x), asforinstance 1/2(a + x). It evidently holds η ∈(a, b) and η < x. But η does not belong to hq i for if that were the case there would exist a q such that η = q , i i∈N v v and then we would have q < x, which is impossible because according to P v it must be x ≤ q , ∀v ∈ N. The rational η proves, therefore, the existence of v rational numbers within (a, b) which are not in hq i , and then the falseness i i∈N of our initial assumption on the countable nature of Q. 4-3. Infinitist mathematics assumes the performance of procedures of infinitely many steps, as the above Cantor’s argument, extension and epilog; and many, many others. Mathematics is not usually concerned with the way those procedures could be, in fact, carried out; it is only concerned with the con- sistency of the involved arguments. When the result of an infinite procedure is an infinite set (or sequence), then the set (or sequence) is assumed to be a complete infinite totality, which implies the completion of the infinitely many steps of the defining procedure. That said, it could be appropriate to recall we dispose of a formal theory whose main objective is just the analysis of the performance of those infinite procedures in a finite or infinite interval of time (supertask and bifurcated supertask respectively6). 4-4. Notice that conclusion 4-2 is not derived from the successive performed operations (Benacerraf criticism of supertasks [2]) but from the own definition of P: x can only be defined as a rational number in (a, b). And notice also that if we assume the actual infinity then we can also assume the existence of a conceptual universe in which supertasks and hypercomputations are possible. Part II: Cantor’s diagonal method Cantor’s diagonal argument makes use of a hypothetical table T containing all real numbers within the real interval (0,1). That table can be easily redefined in order to ensure it contains at least all rational numbers within (0,1). In theseconditions, couldtherowsofT bereorderedsothattheresultingdiagonal and antidiagonal were rational numbers? In that case not only the set of real 6Aprocedureofcountablymanystepscouldbeaccomplishedbyperformingeachstepsi at the precise instant ti of an ω-ordered sequence of instants htiii∈N within a finite interval of time (ta, tb) whose limit is tb. See for instance [3], [11], [17], [12], [18], [1], [16]. 7 Cantor versus Cantor numbers but also, and for the same reason, the set of rational numbers would be non denumerable. And then we would have a contradiction since Cantor also proved the set of rational numbers is denumerable. Should, therefore, Cantor’s diagonal argument be suspended until it be proved the impossibility of such a reordering? Is that reordering possible? The discussion that follows addresses both questions. 5. An elementary previous result 5-5. Let M betheset ofallreal numbersintherealinterval (0, 1)expressed in decimal notation and completed, in the cases of finitely many decimal digits, with infinitely many 0’s in the right side of their decimal expansions. The subset of all rational numbers in M will be denoted by M . Q 5-6. Foreverynaturalnumber nthereareinfinitelymanyelements inM with Q thesamedecimal digitd inthesamen-thpositionofitsdecimalexpansion. In n fact, consider the following element r of M : r = 0.d d ...d 000... where 0 Q 0 1 2 n d , d , ...d are any decimal digits. From r we define the sequence: 1 2 n 0 r = 0.d d ...d 1000... (IX) 1 1 2 n r = 0.d d ...d 11000... (X) 2 1 2 n r = 0.d d ...d 111000... (XI) 3 1 2 n ... r = 0.d d ...d 1 .(.i). 1000... (XII) i 1 2 n ... The function f from N (the set of natural numbers) to M defined according Q to: f(i) = r , ∀i ∈ N (XIII) i proves the existence of a denumerable subset f(N) of M , all of whose ele- Q ments have the same decimal digit d in the same n-th position of its decimal n expansion. 6. Discussion 6-1. Next paragraph 6-2 defines a table T of the real numbers within (0,1) which, according to Cantor [4], contains at least all rational numbers in (0,1). Subsequent paragraphs 6-3/6-5 consider the consequences of a rational diagonal and antidiagonal in T. 8 Cantor versus Cantor 6-2. Cantor’s set M is the union of two disjoint sets: the denumerable set M of all rational numbers in (0,1) and the set M of all irrationals in the Q I same interval (0,1). Assume, as Cantor did in 1891 [7], M is denumerable. In these conditions, it is evident that M would also be denumerable. Let then g I be a bijection between N and M , and h a bijection between N and M . From Q I g and h we define a one to one correspondence f between N and M according to: f(2n−1) = g(n) ∀n ∈ N (XIV) f(2n) = h(n) (cid:27) Wecanthereforeconsidertheω-orderedtableT whosesuccessive rowsr ,r ,r 1 2 3 ...are just f(1), f(2), f(3) .... By construction, and being M denumerable, Q T contains a denumerable subtable with all rational numbers in (0, 1). 6-3. The diagonal of T is a real number D = 0.d d d ... whose n-th 11 22 33 decimal digit d is the n-th decimal digit of the n-th row r of T. Cantor nn n successfully proved [7] the existence of another real number in M derived from D,theantidiagonalD−, whichbyconstructioncannotbeinT. Inconsequence, M cannot be denumerable as was assumed to be (Cantor’s diagonal argument, an impeccable Modus Tollens7 [7]). 6-4. Since D− is a real number in (0, 1), it will be either rational or irrational. But if it were rational then, and for the same reason as in the case of M, the subset M of all rational numbers in M would also be non denumerable. The Q problem here is that Cantor proved the set Q of all rational numbers, and therefore M , is denumerable [5]. Q 6-5. According to 6-4, if it were possible to reorder the rows of T in such a way that a rational antidiagonal could be defined, then we would have two contradictory results: the set Q of rational numbers would and would not be denumerable. Both results can be considered as proved by Cantor, although the second one only as an unexpected (and so far unknown) consequence of his famous diagonal method. Accordingly, we can state our first conclusion: ◮ Cantor’s diagonal argument and all its formal consequences should be suspended until it be proved the impossibility of defining a rational antidiagonal in all possible reorderings of T’s rows. 6-6. Next paragraphs 6-7/6-17 examines the possibilities and consequences of reordering the rows of T in the sense indicated in 6-5. 7The critiques of Cantor’s diagonal argument are invariably related to constructionist aspects which are not pertinent with the formal structure of Cantor’s demonstration. 9 Cantor versus Cantor 6-7. A formal consequence of the existence of complete infinite totalities (actual infinity) is the existence of ω-ordered sequences [8], [9]. In an ω- ordered sequence, as our table T, every element -whatever it be- will always beprecededbyafinitenumberofelementsandsucceededbyaninfinitenumber of such elements. We will see now a conflictive consequence of this immense and suspicious asymmetry. 6-8. Arowr ofT willbesaidn-modularifitsn-thdecimaldigitis(n mod 10). i Thismeansthatarowis, forinstance, 2348-modularifits2348-thdecimaldigit is 8; or that it is 453-modular if its 453-th decimal digit is 3. If a row r is n- n modular (being n in n-modular the same as n in r ) it will be said D-modular. n For instance, the rows: r = 0.1007647464749943400034577774413... (XV) 1 r = 0.2200045667778943000000000000000... (XVI) 2 r = 0.0030000000000000000000000000000... (XVII) 3 r = 0.1004000001111111100000000000000... (XVIII) 4 r = 0.1234567899999999666666666666333... (XIX) 9 are all of them D-modular. 6-9. Consider now the following permutation8 P of the rows hr i of T: i i∈N • For each successive row r in T: i – If r is D-modular then let it unchanged. i – If r is not D-modular then exchange it with any succeeding i- i modular row r , provided that at least one of its succeeding j,j>i rows be i-modular. Notice that, thanks to the condition j > i (in r ), once a row r has been j,j>i i exchanged with a succeeding i-modular row r , it remains D-modular and j,j>i unaffected by the subsequent exchanges. And notice also the exchanges do not alter the nature ω-ordered of the table: P does not modify the ω-ordered set N of indexes but, in any case, the real numbers indexed by its elements. 6-10. It is immediate to prove that each and every row of T becomes D- modular as a consequence of permutation P. In fact, let us assume that a 8Infinitist mathematics assumes that procedures of infinitely many successive steps, as, forinstance,ω-recursivedefinitions,canbeperformedwithoutworryingaboutthewaythey could be carriedout. Supertask theory provides,on the other hand, a way by which any ω- orderedsequence of theoreticalactions couldbe accomplishedin a finite intervalof time, or even in an infinite interval of time in the case of bifurcated supertasks (hypercomputation) [14], [13], [16]. But this is not a problem infinitist mathematics is usually concerned with. 10