aa r X i v : . [ m a t h . A C ] J u l Cantor sets and fields of reals
Gerald Kuba
Abstract.
Our main result is a construction of four families C , C , B , B which are equipollent with thepower set of R and satisfy the following properties. (i) The members of the families are proper subfieldsof R whose algebraic closures equal C . (ii) Each field in C ∪ C contains a Cantor set . (iii) Each fieldin B ∪ B is a Bernstein set . (iv) All fields in C ∪ B are isomorphic. (v) If K, L are fields in C ∪ B then K is isomorphic to a subfield of L only in the trivial case K = L .
1. Introduction
The cardinal number (the size ) of a set S is denoted by | S | . So | R | = | D | = c = 2 ℵ where D = { P ∞ n =1 a n − n | a n ∈ { , } } is the Cantor ternary set .As usual, a nonempty set C ⊂ R is a Cantor set if and only if C is compact and does notcontain nondegenerate intervals or isolated points. Equivalently, C is a Cantor set if andonly if there is a continuous bijection from D onto C .A fundamental question in descriptive analysis of the reals is whether a set X ⊂ R containsa Cantor set. An answer to this question can be important for problems concerning size andmeasure. For example, if X ⊂ R contains a Cantor set then (due to | D | = c ) the sets X and R are equipollent. This leads to the important observation that, while ℵ < | X | < c cannot be ruled out for arbitrary sets X , we can be sure that ℵ < | A | < c is impossiblefor closed subsets A of the real line. Another important example is the following. Asusual, B ⊂ R is a Bernstein set when neither B noch R \ B contains a Cantor set. ABernstein set is never Lebesgue measurable. (See [1] Theorem 6.3.8.)We are interested in the existence of Cantor subsets from a specific algebraic point of view.Let us call a proper subfield of R a Cantor field if and only if it contains some Cantorset. (Notice that a proper subfield of R cannot contain the Cantor ternary set D becauseit is well-known that { x + y | x, y ∈ D } = [0 ,
2] .) Furthermore, a
Bernstein field is asubfield of R which is a Bernstein set. Trivially, each Bernstein field is a proper subfieldof R and it cannot be a Cantor field. A trivial consequence of | D | = c is that | K | = c for every Cantor field K . We also have | K | = c for every Bernstein field K . Actually,the following is true.(1.1) | B | = c for every Bernstein set B ⊂ R . There are several ways to verify (1.1). For example, (1.1) is an immediate consequence of(2.1) below.For abbreviation let us call fields (taken from some collection) incomparable if no field isisomorphic to a subfield of another field. In particular, incomparable fields are mutuallynon-isomorphic. Our first main result is the following theorem. Notice that the field R is an algebraic extension of a subfield K if K = C . (As usual, K denotes the algebraicclosure of the field K .) Theorem 1.
There exist four families C , C , B , B of subfields K of R with K = C such that |C | = |C | = |B | = |B | = 2 c and all fields in C ∪ C are Cantor fields and allfields in B ∪ B are Bernstein fields and all fields in C ∪ B are isomorphic, whereas thefields in C ∪ B are incomparable. . Preparation of the proof For a ∈ R and S ⊂ R we write a + S := { a + x | x ∈ S } . The following observationimplies (1.1) and is also important for the proof of Theorem 1.(2.1) Every Cantor set can be partitioned into c Cantor sets.
To verify (2.1) let C be a Cantor set and f be homeomorphism from D onto C . Nowconsider the subset D := { P ∞ n =1 a n − n | a n ∈ { , } } of D which is clearly a Cantorset. Consequently, P = { x + 3 · D | x ∈ D } is a partition of D consisting of Cantor setsand |P| = | D | = c . Thus { f ( P ) | P ∈ P } is a partition of C as desired, q.e.d. As a consequence of the following lemma, a proper subfield K of R is a Bernstein field if K ∩ C = ∅ for every Cantor set C or, equivalently, if R \ K does not contain a Cantorset. (It is not necessary to check that K does not contain a Cantor set.) Lemma 1. If K is a proper subfield of R which contains a Cantor set then R \ K contains a Cantor set as well.Proof. Let K be a subfield of R and ξ ∈ R \ K . Then ξ + k = k is impossible for all k , k ∈ K and hence the translate ξ + K is disjoint from K . Consequently, if C is aCantor set contained in K then ξ + C is a Cantor set contained in R \ K , q.e.d. From Lemma 1 we immediately derive the following useful observation.(2.2) If B is a Bernstein set and K is a proper subfield of R and B ⊂ K then K isa Bernstein field.Remark. In view of (2.2) Bernstein fields can be defined in complete analogy with Cantorfields. While a proper subfield K of R is a Cantor field if and only if K contains aCantor set , K is a Bernstein field if and only if K contains a Bernstein set .As usual, for Y ⊂ R let Q ( Y ) denote the smallest subfield of R containing the set Y .To cut short, a set S ⊂ R is algebraically independent if and only if for arbitrary n ∈ N one cannot find distinct numbers t , ..., t n in S such that p ( t , ..., t n ) = 0 holds for somenonconstant polynomial p ( X , ..., X n ) in n indeterminates. Naturally, this is equivalentto the statement that t Q ( S \ { t } ) for every t ∈ S , see [6]. Since Q ( S ) = R for everyalgebraically independent S ⊂ R , from Lemma 1 and (2.2) we derive(2.3) If B is a Bernstein set and S ⊂ R is algebraically independent and B ⊂ S then S is a Bernstein set and Q ( S ) is a Bernstein field. To cut short, an algebraically independent set S ⊂ R is a trancendence base if and only iffor each real number x S the set S ∪ { x } is not algebraically independent. It is plainto verify the following observation.(2.4) K = C for a subfield K of R if and only if K contains some transcendence base. It goes without saying that | T | = c for each transcendence base T and that the fields Q ( S ) and Q ( S ) are always isomorphic for algebraically independent sets S , S of equalsize. Therefore, from (2.3) and (2.4) we derive the following statement which gives a hinthow to find an appropriate family B in Theorem 1.(2.5) If B is a Bernstein set and T is a transcendence base and B ⊂ T then T is aBernstein set and Q ( T ) is a Bernstein field and Q ( T ) = C . C and B in Theorem 1 by applying the followingtheorem which is interesting of its own. Theorem 2.
There exist two families Y , Y of transcendence bases with |Y | = |Y | =2 c such that each member of Y is a nowhere dense Lebesgue null set containing a Cantorset, while each member of Y is a Bernstein set, and for i ∈ { , } and for distinct R, S ∈ Y i we have Q ( R ) Q ( S ) and in particular Q ( R ) = Q ( S ) . For the proof of Theorem 2 we need the following lemma, which can easily be verified, andtwo propositions we will prove in Sections 5 and 6.
Lemma 2. If B is a transcendence base and θ is an algebraic irrational number then ( B \ X ) ∪ { xθ | x ∈ X } is a transcendence base for every X ⊂ B . Proposition 2.
There exists an algebraically independent Cantor set lying in D . Proposition 3.
There exist disjoint Bernstein sets
A, B such that A ∪ B is analgebraically independent Bernstein set.
3. Proof of Theorem 2
Fix four subsets S , T , S , T of R of size c such that S ∩ T = S ∩ T = ∅ and bothsets S ∪ T and S ∪ T are transcendence bases and S is a Bernstein set while S is aCantor set and S ∪ T is a subset of D . In view of Propositions 2 and 3 a choice of suchsets is possible since each algebraically independent set is contained in some transcendencebase and since each Cantor set can be split into two Cantor sets and since Q ( D ) = R .(We have Q ( D ) = R due to D + D = [0 ,
2] , and for every Cantor set C both C ∩ ] −∞ , t ]and C ∩ [ t, ∞ ] are Cantor sets for some real t C .) We define for every set U ⊂ T i g i ( U ) := S i ∪ U ∪ { x √ | x ∈ T i \ U } . By virtue of Lemma 2, g i ( U ) is always a transcendence base. (In view of (4.1) in the nextsection we use the factor √ √ g ( U ) containsa Cantor set. As a union of three nowhere dense null sets g ( U ) is a nowhere dense nullset. By (2.3), g ( U ) is a Bernstein set. Therefore Theorem 2 is settled by defining Y i := { g i ( U ) | U ⊂ T i } ( i ∈ { , } )because in both cases i ∈ { , } the following statement is true.(3.1) For
V, W ⊂ T i the equality V = W follows from Q ( g i ( V )) ⊂ Q ( g i ( W )) . In order to verify (3.1) we show a bit more in view of the next section and Section 7.If Y ⊂ R is algebraically independent, then every algebraic number in the field Q ( Y )is rational and hence √ Q ( Y ) . Therefore (3.1) is an immediate consequence of thefollowing statement.(3.2) Let i, j ∈ { , } and V ⊂ T i and W ⊂ T j . Furthermore let L be a subfield of R with √ L and g j ( W ) ⊂ L . Then g i ( V ) ⊂ L implies V = W . In order to verify (3.2) we firstly point out that the inclusion g i ( V ) ⊂ L enforces theidentity i = j . Indeed, due to g i ( V ) ⊂ L and g j ( W ) ⊂ L , from i = j we derive that L contains both a Cantor set and a Bernstein set which is impossible in view of Lemma 1since L = R . Thus we may assume i = j ∈ { , } and g i ( V ) ⊂ L and g i ( W ) ⊂ L and3 √ L . To conclude the proof of (3.2) by verifying V = W assume indirectly that thereis a real x ∈ ( V \ W ) ∪ ( W \ V ) . Then the pair ( x, x √
2) lies in g i ( V ) × g i ( W ) or in g i ( W ) × g i ( V ) . Consequently the field L contains both reals x, x √ √
4. Proof of Theorem 1
For every subfield K of R let K ∗ denote the intersection of all fields L satisfying K ⊂ L ⊂ R and the property that p | x | ∈ L for every x ∈ L . Of course, K ⊂ K ∗ ⊂ R ∩ K and p | x | ∈ K ∗ for all x ∈ K ∗ or, equivalently, K ∗∗ = K ∗ . Alternatively, K ∗ isobtained from K by successively adjoining square roots. Define inductively W = K and W n = Q ( { p | x | | x ∈ W n − } ) for n ∈ N in order to finally obtain K ∗ = S ∞ n =1 W n .A fortiori Q ( T ) ∗ = C for every transcendence base T , where Q ( T ) ∗ = R due to thefollowing observation.(4.1) If Y ⊂ R is an algebraically independent set then √ Q ( Y ) ∗ . Obviously (4.1) is a consequence of the following two statements.(4.2) If K is a subfield of R then every number in K ∗ \ K is algebraic over K of degree n with n ∈ N . (4.3) For every algebraically independent set Y ⊂ R the degree of √ over the field Q ( Y ) equals . We obtain (4.2) from the well-known fact that K ∗ is the field of all reals which are constructible over K by a finite sequence of ruler and compass constructions.In order to verify (4.3) let Y ⊂ R be algebraically independent and consider distinct y , ..., y n ∈ Y for arbitrary n ∈ N . Naturally the field Q ( y , ..., y n ) is isomorphic withthe quotient field Q ( X , ..., X n ) of the polynomial ring Z [ X , ..., X n ] which is a uniquefactorization domain and in which 2 is irreducible. Consequently, by Eisenstein’s Criterionthe polynomial X − Q ( X , ..., X n ) . Thus X − Q ( y , ..., y n ) and hence X − √ Q ( Y ) and (4.3) is proved.Now we take the transcendence bases g i ( · ) from the previous section and define C := { Q ( g ( U )) ∗ | U ⊂ T } and B := { Q ( g ( U )) ∗ | U ⊂ T } .Then all fields in C resp. B are Cantor fields resp. Bernstein fields. Furthermore,concluding the proof of Theorem 1, the following statement shows that |C | = |B | = 2 c and the fields in C ∪ B are incomparable.(4.4) For i, j ∈ { , } and V ⊂ T i and W ⊂ T j the field Q ( g i ( V )) ∗ can be embeddedinto the field Q ( g j ( W )) ∗ only if V = W . The following lemma shows that if Q ( g i ( V )) ∗ can be embedded into Q ( g j ( W )) ∗ thenalready Q ( g i ( V )) ∗ ⊂ Q ( g j ( W )) ∗ holds. Consequently and in view of (4.1) we obtain (4.4)from (3.2) with L := Q ( g j ( W )) ∗ . Lemma 3. If K is a subfield of R with K ∗ = K and ϕ is a monomorphism from K to R then ϕ ( x ) = x for all x ∈ R . roof. It goes without saying that ϕ ( r ) = r for all r ∈ Q . Due to K ∗ = K we have p | a | ∈ K for every a ∈ K . Now let x ∈ K and r ∈ Q . If r < x then ϕ ( x ) − ϕ ( r ) = ϕ ( x − r ) = ϕ ( √ x − r ) > r > x then ϕ ( r ) − ϕ ( x ) = ϕ ( r − x ) = ϕ ( √ r − x ) > ϕ restricted to the domain Q is the identity, for all r, s ∈ Q and x ∈ K the implication r < x < s = ⇒ r < ϕ ( x ) < s holds and this enforces ϕ ( x ) = x for all x ∈ K , q.e.d.
5. Proof of Proposition 2
As usual, ZF means ZFC set theory minus the Axiom of Choice. In this section we aregoing to prove Proposition 2 without applying the Axiom of Choice. This has the benefitthat the following trivial consequence of Theorem 1 turns out to be a theorem in ZF.
Corollary 1.
There exist two families C , C equipollent with the power set of R such that C consists of isomorphic Cantor fields while C consists of incomparable Cantor fields.Remark. There are two reasons why Theorem 1 is not a theorem in ZF. Firstly, theexistence of a proper subfield K of R satisfying K = C is unprovable in ZF because of(2.4) and the well-known fact that the existence of transcendence bases is unprovable inZF. Secondly, also the existence of a Bernstein set (and in particular the existence of aBernstein field) cannot be derived from the axioms of ZF only. Additionally, equationslike |F | = 2 c need to be interpreted in a specific way in ZF since | X | is not defined inZF for arbitrary sets X . (But notice that sets like R , C and D are well-defined objectsin ZF. Notice also that in ZF the field K ∩ R is well-defined for each subfield K of R constructed in ZF.)If C is an algebraically independent Cantor set in ZF then a ZF-proof of Corallary 1 caneasily be obtained by adopting parts of the ZFC-proof of Theorem 1. Simply split C constructively into two disjoint Cantor sets C , C and define the sets S and T in theproof of Theorem 1 via S = C and T = C . Then, never considering transcendencebases, C ∪ U ∪ { x √ | x ∈ C \ U } is algebraically independent for all U ⊂ C andthis is enough. Alternatively, there is a direct and much shorter ZF-proof of Corollary 1as follows. With C , C as above put C := { Q ( C ∪ X ) | X ⊂ C } and C := { R ∩ Q ( C ∪ X ) | X ∈ F } where F is a family equipollent with the power set of R suchthat X ⊂ C for every X ∈ F and X Y whenever X, Y ∈ F are distinct. (Such afamily F can easily be constructed in ZF. For example take a bijection f from R onto C and put F := { f ( T ∪ ([2 , \ (2+ T ))) | T ⊂ [0 , } .) The family C obviously fitsand the family C does the job by virtue of Lemma 3.There is a more algebraic way and a more topological way to prove Proposition 2 in ZF.The algebraic way is far from being elementary since an appropriate modification of thefamous von Neumann numbers (see [5]) is used.As usual, [ ξ ] denotes the largest integer k ≤ ξ for ξ ∈ R . We put ψ ( x, n ) := 2 n − [ nx ] and observe that 0 < ψ ( x, n ) < ψ ( x, m ) whenever n, m ∈ N and 0 < x < n < m .Consequently, σ ( x ) := 2 · P n>x − ψ ( x,n ) defines a function from ]0 , ∞ [ into D for alle x ∈ R . (Notice that the von Neumannnumbers are not contained in D .) In the same way as carried out in [5] one can verify5hat σ ( x ) Q ( σ (]0 , x [)) for every x > σ is injective and the set σ (]0 , ∞ [) isalgebraically independent. It goes without saying that σ is continuous at each positiveirrational. Now consider the uncountable partition P of D defined in the proof of (2.1)and obviously well-defined in ZF. Then P ∗ := { X ∈ P | X ∩ Q = ∅ } is a countablesubfamily of P and hence the family P \ P ∗ is not empty. Hence we can select an element C from this family. Automatically, C is a Cantor set containing only irrational numbers.Consequently, σ ( C ) is a Cantor set lying in an algebraically independent set and thisconcludes the ZF-proof of Proposition 2.Alternatively we present now an elementary ZF-proof of Proposition 2. In doing so weshow a bit more. Let D be an arbitrary Cantor set defined in ZF. (For example, D = D .)Then the following statement (which cannot be verified using some modification of the vonNeumann numbers) is a theorem in ZF.(5.1) The set D contains an algebraically independent Cantor set. For n ∈ N consider the ring R n = Z [ X , ..., X n ] of all integral polynomials in n indeterminates. Of course, R n is a countable set. For every n ∈ N let g n be a bijectionfrom N onto R n \ Z . For ( n, m ) ∈ N let A ( n, m ) denote the set of alle n -tuples( t , ..., t n ) of reals that are annihilated by the polynomial g n ( m ) . Naturally, A ( n, m )is a closed subset of the space R n . Furthermore, A ( n, m ) is nowhere dense because foran integral polynomial p ( X , ..., X n ) the equation p ( t , ..., t n ) = 0 holds for all points( t , ..., t n ) in a nonempty open subset of R n only if p ( X , ..., X n ) is the zero polynomial.For a ≤ b put λ ([ a, b ]) := b − a . For every k ∈ N choose 2 k pairwise disjoint compactintervals I k,j (1 ≤ j ≤ k ) such that I m,j ⊃ I m +1 , j ∪ I m +1 , j − whenever m, j ∈ N and j ≤ m and that lim k →∞ max { λ ( I k,j ) | ≤ j ≤ k } = 0 . Certainly, in eachparallelepiped of positive volume there lies a parallelepiped of positive volume disjointwith the closed, and nowhere dense point set A ( n, m ) . Therefore, step by step it can beaccomplished that the following condition B [ m ] holds for all m ∈ N as well.( B [ m ]) A ( n, m ) ∩ Q ni =1 ϕ ( i ) = ∅ whenever n ∈ N and ϕ is an injection from { , ..., n } into {I m, , ..., I m, m } . In comparison with the standard construction of the ternary Cantor set it is clear that C := ∞ T k =1 2 k S j =1 I k,j is a Cantor set. (Alternatively it is plain that C is compact, dense in itself and does notcontain isolated points.) We claim that C is algebraically independent.Assume indirectly that C is not algebraically independent. Then we can find distinctnumbers t , t , ..., t n in C such that the n -tuple ( t , t , ..., t n ) is annihilated by somenon-zero polynomial p in the ring Z [ X , ..., X n ] . Trivially, this n -tuple is also annihilatedby the polynomial k · p ( X , ..., X n ) for every k ∈ N . Hence there is an infinite set M ⊂ N such that for all m ∈ M the point ( t , ..., t n ) lies in the set A ( n, m ) . Now choose m ∈ M sufficiently large such that each one of the 2 m intervals I m,j contains at mostone of the numbers t , ..., t n . This provides us with an injection ϕ from { , ..., n } into {I m, , ... I m, m } such that the point ( t , ..., t n ) lies in the parallelepiped Q ni =1 ϕ ( i ) . Thisis a contradiction to B [ m ] since ( t , ..., t n ) ∈ A ( n, m ) .6o we have proved that C is algebraically independent. In doing so also Proposition 2 issettled because if D = D then in the first place we can choose the intervals I k,j so that C ⊂ D . Moreover, an appropriate choice of these intervals is also possible for an arbitraryCantor set D and hence (5.1) is settled in view of the following proposition about Cantorsets and because f (cid:0) T k S j I k,j (cid:1) = T k S j f ( I k,j ) and f ([ u, v ]) = [ f ( u ) , f ( v )] for everyincreasing bijection f : R → R . Proposition 4.
A nonempty set C ⊂ R is compact, totally disconnected and dense initself if and only if C = f ( D ) for some increasing bijection f from R onto R .Proof. Naturally, every stictly increasing function f from R onto R is a homeomorphism.Hence as the topological space D the set f ( D ) is compact, totally disconnected and densein itself. Conversely let D = C similarly as D = D be a compact, totally disconnectedand dense-in-itself subspace of R . In the following, i ∈ { , } .The set D i has a maximum and a minimum due to compactness. Let a i = min D i and b i = max D i . (In particular, a = 0 and b = 1 .) Since the set [ a i , b i ] \ D i is open, wecan define a family U i of pairwise disjoint open intervals such that S U i ⊂ [ a i , b i ] and D i = [ a i , b i ] \ S U i .In a natural way the family U i is linearly ordered by declaring U ∈ U i smaller than V ∈ U i if and only if sup U ≤ inf V . Since D i is totally disconnected, S U i is dense in[ a i , b i ] . And since D i has no isolated point, between two intervals from U i there alwayslie infinitely many intervals from U i . Finally, the linearly orderd set U i has neither asmallest nor a largest element.Taking all these observations into account, by virtue of a well-known classical theorem byCantor, the linearly ordered set U i is order-isomorphic to the naturally ordered set Q . Inparticular the two families U and U are order-isomorphic. Since for u < v and r < s the interval ] u, v [ can easily be mapped onto the interval ] r, s [ by a strictly increasingfunction, with the help of an order-isomorphism from U onto U we obtain an increasingbijection g from S U onto S U . This function g can be extented to a continuousfunction f from R to R by defining f ( t ) : = a + t for t ≤ f ( t ) := b + t − t ≥ f ( t ) := sup { g ( x ) | x ∈ S U ∧ x ≤ t } for 0 < t < f is strictly increasing and surjective and f ( D ) = C , q.e.d.Remark. Proposition 4 can be regarded as a purely algebraic characterization of Cantorsets. Because firstly in the definition of D the series P a n − n need not refer to thetopological concept convergency but can be regarded as the depiction of a sequence of digits.And secondly, monotonicity of real functions can be characterized purely algebraicallybecause a mapping f from R into R is increasing if and only if for arbitrary u, v ∈ R the equation ( u − v )( f ( u ) − f ( v )) = x has a real solution x .
6. Proof of Proposition 3
Let D be the family of all Cantor sets. We have |D| = c since every set in D is compactand R contains precisely c closed sets and since the translate x + D is obviously a Cantorset for each one of the c real numbers x . 7et ϕ be a bijection from R onto D and let ρ be a choice function defined on thenonempty subsets of R . Thus ρ ( X ) ∈ X for every nonempty set X ⊂ R . Furthermorelet (cid:22) be a minimal well-ordering of the set L := R × { , } , i.e. (cid:22) is a linear orderingof L such that |{ ( u, v ) | ( u, v ) ≺ ( x, y ) }| < | L | = c for every point ( x, y ) ∈ L and thatevery nonempty subset of L contains a smallest element with respect to the ordering (cid:22) .The existence of such a relation (cid:22) is guaranteed by fundamental considerations from settheory, see [1]. (However, in view of [2] there is absolutely no chance to give an explicitdefinition of such a well-ordering (cid:22) .)Now, referring to the well-ordering (cid:22) we define by induction an injective mapping g from L into R . Assume that for ( x, y ) ∈ L real numbers g ( u, v ) are already defined for( u, v ) ≺ ( x, y ) . Then, under consideration that | L | = c = | ϕ ( x ) | for all x ∈ R , we define g ( x, y ) := ρ (cid:0) ϕ ( x ) \ Q ( { g ( u, v ) | ( u, v ) ≺ ( x, y ) } ) (cid:1) . This definition is correct since the choice function ρ is applied to a nonempty set. Indeed,for arbitrary Y, T ⊂ R the set Y \ Q ( T ) certainly is nonempty provided that | Y | = c and | T | < c because the field Q ( T ) is either countable or equipollent with T . From thedefinition of g we immediately obtain the following observation.(6.1) The set { g ( x, y ) | ( x, y ) ∈ L } is algebraically independent. Now we define two subsets H , H of { g ( x, y ) | ( x, y ) ∈ L } by H j := { g ( x, j ) | x ∈ R } for j ∈ { , } .Since g is injective, H , H are disjoint sets of size c . For each pair ( x, j ) ∈ L we have g ( x, j ) ∈ ϕ ( x ) and hence for both j ∈ { , } the set H j meets every Cantor set. Thus inview of (6.1) and (2.3) Proposition 3 is settled with the definitions A = H and B = H .
7. Extremely large fields
Due to the condition K = C the fields depicted in Theorem 1 are rather large. But in acertain sense they are not extremely large. Let K be a subfield of R . Naturally, R is avector space over the field K . As usual, [ R : K ] denotes the dimension of the K -vectorspace R . Naturally, in the nontrivial case K = R the dimension [ R : K ] is always atransfinite cardinal number not greater than c = | R | .For every field K in the family C ∪ C ∪ B ∪ B defined in the proof of Theorem 1 we have[ R : K ] = c because if T is a transcendence base then [ R : Q ( T )] = c (see [3] Theorem 4)and we have [ R : Q ( T ) ∗ ] = c as well. (Of course, [ R : Q ( T ) ∗ ] = c implies [ R : Q ( T )] = c and [ R : Q ( T ) ∗ ] = c can be proved by verifying that the c reals √ t ( t ∈ T ) are linearlyindependent vectors in the vector space R over the field Q ( T ) ∗ .)One may regard a subfield K of R larger than a subfield L of R when [ R : K ] < [ R : L ] .In this sense the following theorem provides extremely many mutually non-isomorphic extremely large Cantor fields and Bernstein fields, respectively. Notice that (as explainedin [3]) we must have K = C for a subfield K of R satisfying the condition [ R : K ] = ℵ Theorem 3.
In Theorem 1 it can be accomplished that [ R : K ] = ℵ for every field K in the family C ∪ B . In order to settle Theorem 3 we consider the families C , B defined in Section 4, expandeach field K in C ∪ B to an appropriate proper subfield ˆ K , and replace C and B { ˆ K | K ∈ C } and { ˆ K | K ∈ B } . In view of (3.2) this is enough provided that foreach subfield K of R with √ K we can find a subfield ˆ K of R with √ ˆ K andˆ K ⊃ K and [ R : ˆ K ] = ℵ .So let K be a subfield of R with √ K . Then the family F of all subfields L of R with √ L and L ⊃ K is not empty. Clearly, S G ∈ F for every chain G of fieldsin F . Consequently, by applying Zorn’s Lemma, the partially ordered family ( F , ⊂ ) hasa maximal element ˆ K . Such a field ˆ K is a subfield K of R satisfying for the irrationalnumber θ = √ Proposition 5.
Let θ ∈ R be irrational and let K be a subfield of R such that θ K but θ ∈ L for every field L with K ⊂ L ⊂ R and L = K . Then [ R : K ] = ℵ . The best way to verify Proposition 5 is to apply methods from infinite-dimensional Galoisas carried out in [4]. To keep this article self-contained, we write down the following proofof Proposition 5 which is identical with the proof of [4] Proposition 3.Let θ, K be as depicted in Proposition 5. First it is plain that R lies algebraically over K . Hence C is an algebraic Galois extension of K . Let G be its Galois group. Let θ ′ = θ be a K -conjugate of θ and choose τ ∈ G such that τ ( θ ) = θ ′ . Further, let σ be the complex conjugation z z . Then σ ∈ G since K ⊂ R . Now let H bethe smallest closed subgroup of G with σ, τ ∈ H . Then K must be the fixed field F = { z ∈ C | ∀ ϕ ∈ H : ϕ ( z ) = z } because σ ∈ H implies K ⊂ F ⊂ R and F = K would contradict the maximality of K since θ F . Consequently, G = H . Therefore G is the closure of a finitely generated subgroup and hence it is easy to verify that thefamily N of all closed normal subgroups N of G with finite index [ G : N ] is countable.Since the family L of all normal field extensions L ⊂ C of K with finite degree [ L : K ] isequipollent to N because of the Galois correspondence, L is countable too. Since everyfinite extension lies under a suitable normal finite extension, C = S L . Now let B be a basis of the vector space C over K . Then B ∩ L is finite for all L ∈ L andhence B = B ∩ C is a countable union of the finite sets B ∩ L ( L ∈ L ) . Consequently, ℵ = | B | = [ C : K ] ≥ [ R : K ] ≥ ℵ .
8. Applications and questions
For a cardinal number κ let N ( κ ) denote the total number of all fields of size κ upto isomorphism . More precisely, N ( κ ) is the unique cardinal number such that thereexists a family F of fields of size κ such that |F | = N ( κ ) and every field of size κ isisomorphic to precisely one field in the family F . It goes without saying that N ( κ ) ≤ κ for every transfinite κ . (Besides, N ( κ ) ≤ κ also holds for every finite cardinal κ due to N ( k ) ≤ k ∈ N .) Therefore, as an immediate consequence of Theorem 1 weobtain the equation N ( c ) = 2 c . Furthermore, in view of the considerations in Section 5 itis straightforward to track down 2 κ incomparable subfields K of R with | K | = κ for eachtransfinite cardinal κ ≤ c . Consequently, we obtain the following enumeration theorem.(Notice that it is consistent with ZFC that there exist c cardinals κ with ℵ < κ < c .) Corollary 2. N ( κ ) = 2 κ for every transfinite cardinal κ ≤ c . N ( κ ) = 2 κ is true also for κ > c . Our proofof Corollary 2 is based on an application of Lemma 3. More generally, it is essentialto consider fields K such that K carries a natural ordering compatible with the fieldstructure where the elements of the prime field of K lie dense with respect to this ordering.Trivially this leads to the restriction that K is of characteristic 0 and of size not greaterthan c . Furthermore, the well-known fact that there exist 2 κ mutually non-isomorphicabelian groups of size κ for each transfinite κ is of no help in order to settle N ( κ ) = 2 κ for arbitrary κ because for two uncountable fields K , K of equal size and identicalcharacteristic the additive groups ( K , +) and ( K , +) are isomorphic. It is not onlyin question whether N ( κ ) = 2 κ also holds for κ > c , at first sight it is not even clearwhether the function κ N ( κ ) is unbounded. This, however, is true because the followingtheorem implies the existence of a proper class K of cardinal numbers such that N ( κ ) ≥ κ for each κ ∈ K . Theorem 4. N ( ℵ ξ ) ≥ | ξ | for every ordinal number ξ . The estimate in Theorem 4 can be improved for certain ordinals ξ . Actually we canaccomplish the following lower bound of N ( · ) .(8.1) N ( κ ) ≥ min { κ , c } for every transfinite cardinal κ . Obviously, (8.1) is an immediate consequence of the following result we can derive fromTheorem 1.
Corollary 3.
For every transfinite cardinal κ there exist min { κ , c } incomparable fieldsof size κ and characteristic .Proof of Theorem 4. Let K be an algebraically closed field of size ℵ ξ . For every ordinalnumber α < ξ choose a set X α of precisely ℵ α indeterminates and define the field K α by K α := K ( X α ) . Trivially, the transcendental degree of K α over K equals ℵ α . Furthermorethe field K can be recovered from K α because t ∈ K α lies in K if and only if for every n ∈ N the polynomial X n − t has a zero in the field K α . (Let t ∈ K α \ K , whence t = p ( X , ..., X n ) /q ( X , ..., X n ) for some polynomials p, q and indeterminates X i ∈ X α where the degree N of p · q is positive. Now, every zero of the polynomial X N +1 − t in K α is of the form ˜ p ( X , ..., X n ) / ˜ q ( X , ..., X n ) for some polynomials ˜ p, ˜ q and with the sameindeterminates X i as above since X α is algebraically independent over K . Therefore andby comparing degrees of polynomials we conclude that X N +1 − t cannot have a root in K α .) Thus the | ξ | fields K α ( α < ξ ) are mutually non-isomorphic fields of size ℵ ξ , q.e.d.Remark. Obviously, two fields in the family { K α | α < ξ } are never incomparable. Onthe other hand there is no restriction concerning the characteristic of the fields K α . Proof of Corollary 3.
Put λ := min { κ, c } , whence 2 λ = min { κ , c } . Let L be a familyof incomparable subfields L of R with |L| = 2 λ and | L | = λ and L = L ∗ for each L ∈ L . (Recall that L = L ∗ means that p | x | ∈ L for each x ∈ L .) Such a family L exists in view of the proof of Theorem 1. Now let X be a set of precisely κ indeterminatesand consider the field extension L ( X ) of size κ for each L ∈ L . The field L can alwaysbe recovered from the field L ( X ) because, similarly as in the previous proof, t ∈ L ( X )lies in L if and only if for every n ∈ N the polynomial X n − t has a zero in the field L ( X ) . Consequently, the 2 λ fields L ( X ) ( L ∈ L ) of size κ are incomparable, q.e.d.
10f course, Corollary 3 and (8.1) and Theorem 4 lead to the following
Challenge.
Compute N ( κ ) for each cardinal number κ > c . References [1] Ciesielski, K.:
Set Theory for the Working Mathematician.
Cambridge 1997.[2] Feferman, S.:
Some applications of the notion of forcing and generic sets.
Fund Math , 325-345 (1965).[3] Kuba, G.: ber den Grad unendlicher Krpererweiterungen. Mitt. Math. SeminarUniv. Hamburg , 117-124 (2007)[4] Kuba, G.: Counting fields of complex numbers.
Am. Math. Monthly , No. 6,541-547 (2009)[5] von Neumann, J.:
Ein System algebraisch unabh¨angiger Zahlen.
Math. Ann. (1928) 134-141.[6] van der Waerden, B.L.: Algebra I.
Springer-Verlag, 1971.
Author’s address:
Institute of Mathematics.University of Natural Resources and Life Sciences, Vienna, Austria.
E-mail: gerald.kuba(at)boku.ac.atgerald.kuba(at)boku.ac.at