Levels of Ultrafilters with Extension Divisibilities

Transcription

1 arxiv: v1 [mathgn] 12 Jul 2018 Levels of Ultrafilters with Extension Divisibilities Salahddeen Khalifa Department of Mathematics and Computer Sience University of Missouri- St Louis StLouis, MO USA July 16, 2018 Abstract To worore accurately with elements of the semigroup of the Stone Cech compactification (βn,) of the discrete semigroup of natural numbers N under multiplication We divided these elements into ultrafilters which are on finite levels and ultrafilters which are not on finite levels For the ultrafilters that are on finite levels we prove that any element is irreducible or product of irreducible elements and all elements on higher levels are -divided by some elements on lower levels We characterize ultrafilters that are not on finite levels and the effect of -divisibility on the ultrafilters which are not on finite levels 1 Introduction If we given the discrete semi-group (N,) of the discrete space of natural numbers with multiplication, then (βn,) is a semigroup of the Stone-Cech compactification βn of the discrete space N with operation() that is defined as: For any x,y βn,a xy iff {n N : A/n y} x where A/n = {m N : mn A} The topology on βn has base that contains the basic (clopen) sets that are defined as : for any A N,A = {x βn : A x} For each n N the 1

2 principal ultrafilter is defined by element {n} N = βn N anda = A A for A N βn = 2 c where c = R The collection of upward closed subsets of N is µ = {A N : A = A } where A = {n N : a A,a n} andthecollection of downward closed subsets of N is ν = {A N : A = A } where A = {n N : a A,n a} For every f : N N there is a unique continuous extension function f : βn βn such that for every x βn, f(x) = {A N : f 1 (A) x} If A x, then f(a) f(x) and if B f(x), then f 1 (B) x For x βn,if A x, then x A = {B A : B x} is an ultrafilter on A Let x,y βn Then: y is left-divisible by x, x l y if there is z βn such that y = zx y is right-divisible by x,x r y, if there is z βn such that y = xz y is mid-divisible by x,x m y, if there are z,w βn such that y = zxw y is -divisible by x, x y if for all A N,A x implies A = {n N : a A,a n} y When x = n N, we write n y,(y = nz,z βn) Lemma 11 ([1]Theorem 415)Assume x,y βn and A N Then A xy if and only if there exist B x and an indexed family < C n > n B in y such that n B nc n A Lemma 12 ([3]Lemma 14)The following conditions are equavalent (a) x y (b) x µ y µ (c) y ν x ν Lemma 13 ([4]Lemma 21)Let x βn, A x, and f : N N (a) If f(a) a for all a A, then f(x) x (b) If a f(a) for all a A, then x f(x) In Lemma 1-3, if A x, in order to determine f(x) it is enough to know values of f(a) for a A, ie we will sometimes define the function only on a set in x An element p βn {1} is prime with respect to l, r, m or divisibility if it is divisible only by 1 and itself We call p βn irreducible in X βn if it can not be represented in the form p = xy for x,y X {1} Lemma 14 (a) p βn is prime with respect to l, r and m divisibilities if and only if p is irreducible 2

3 (b) ([4]Lemma 22) p βn is prime with respect to -divisibility if and only if p P (P the set of prime numbers in N) (c) ([2]Lemma73) If p βn and p P, then p is irreducible in βn (d) If p βn is prime with respect to - divisibility Then p is prime with respect to l, r and m divisibilities Proof (a) This follows immediately from definition of prime and irreducible (d)by (a),(b),(c) 2 Ultrafilters on finite levels in βn In this section we consider the set of ultrafilters which are elements in all basic open sets,i = 0,1,2, where L 0 = {1} and L n = {a 1 a 2 a n : a 1,a 2,,a n P}, P the set of prime numbers in N Definition 21 ([4] definition 24)Let P be the set of prime numbers in N, and let L 0 = {1} L 1 = {a : a P} L 2 = {a 1 a 2 : a 1,a 2 P} L n = {a 1 a 2 a n : a 1,a 2,,a n P} Then, the ultrafilter x is called on finite level if it is in exactly one of the follwing sets:,i = 0,1,2, where = {y βn : y} Definition 22 Let A P We denote A n = {a n : a A} and A (n) = {a 1 a 2 a n : all a i Aaremutuallyprimenumbers} Lemma 23 (1) L j = for any i j (2) = P i P i 1 P P (i) for any i 2, and P i,p i 1 P,,P (i) are disjoint (3) has 2 c elements for all i 0 (4) All P i,p i 1 P,, P (i) has 2 c elements (5) Any principal ultrafilter in βn is ultrafilter on finite level ie N 3

4 (6) If A N is finite, then all ultrafilters x A are on finite level Proof (1) Assume L j for some i j, so there exist x βn such that x L j, so,,l j x and L j = x a contradictionthus L j = for any i j (2) To avoid cumbersome notation we prove this in case of i = 3 Ultrafilters inthe3rdlevelcontainsl 3 = {a 1 a 2 a 3 : a 1,a 2,a 3 P} = {8,12,18,20,27,} and we can partition L 3 as: L 3 = {8,27,125,} {12,18,20,28} {30,42,66} = P 3 P 2 P P (3) So L 3 = P 3 P 2 P P (3) = P 3 P 2 P P(3) Since P 3,P 2 P,P (3) are disjoint, then P 3,P 2 P and P (3) are disjointed (3) Since is infinite set and = L i so by([5]theorem 33) has 2 c ultrafilters (4) If we prove this in case of i = 3 All P 3,P 2 P,P (3) are infinite sets Thus P 3,P 2 P and P (3) have 2 c elelments (5) Since for any n N is prime number or product of prime numbers So there exist L n such that n L n and since L n = L n L n, so n L n as principal ultrafilterthus N (6) Let A = {n 1,n 2,,n m } be a finite subset of N For any x A, we have {n 1 n 2,,n m } x, so x = n i for exactly of one of 1 i mthus x is principal ultrafilter, and by (5) x is on finite level Definition 24 (1)([4]Definition 51) We call ultrafilters of the form p k for some p P and k N basic Let B be the set of all basic ultrafilters and let A be the set of all functions α : B N {0} with finite support ( {b B : α(b) 0} is finite) ie α = {(b 1,n 1 )(b 2,n 2 )(b m,n m )}, α(b) = 0 for α / {b 1,b 2,,b m } Let α = {(p 1,n 1 ),(p 2,n 2 ),,(p m,n m )} A, (p i P) Set F α = {(A 1 ) (n 1) (A 2 ) (n 2) (A m ) (nm) : A i p i P,A i = A j if p i = p j A i A j = otherwise} where (A 1 ) (n 1) (A 2 K 2 ) (n 2) (A m ) (nm) = { m 4 n i i=1j=1 a k i i,j : a i,j A i

5 for all i,j and a i,j are distinct} (2)([4]Definition 31) If pow n : N N is defined by pow n (a) = a n then for x N, pow n (x) is generated by sets A n for A x we will denote pow n (x), with x n Example 25 If α = {(p 2,1)}, then F α = {A 2 : A p P} If α = {(p,2)} then F α = {A (2) : A p P} If α = {(p 3,2),(q 2,2)}, then F α = {(A 3 ) (2) (B 2 ) (2) : A p P,B q P,A B = } k Definition26 ([4]Definition54)Ifα = {(p 1,n 1 ),(p 2,n 2 ),,(p m m,n m )} m A,p i P, we denote σ(α) = k i n i i=1 Theorem 27 F α x for some α A if and only if x is n-th level L n (for n N) such that σ(α) = n k Proof ( ) Let α = {(p 1,n 1 ),(p 2,n 2 ),,(p m m,n m )} A k with σ(α) = n and F α x, so (A 1 1 ) (n1) k (A 2 2 ) (n2) (k (A m) m ) (nm) x, and k since (A 1 1 ) (n1) k (A 2 2 ) (n2) k (A m m ) (nm) L n So L n x and x L n ( ) By ([4] Theorem 55) Theorem 28 (a) If x L m and y L n Then xy L m+n (b) If F α x and F β y Then F α+β xy where α+β = {(b,n+n ) : (b,n) α,(b,n ) β} Proof (a) Let x L m and y L n,, then L m x,l n y In (Lemma 1-1) if we put B = L m and for all n L m,c n = L n,then we have L m L n = nl n xy, xy L m L n Thus xy L m+n n L m (b) Let F α x,f β y and α,β A, so x is the m-th level L m and y is the n-th level L n, where σ(α) = m and σ(β) = n, so by (Theorem 27) and (a) xy L m+n where m+n = σ(α+β), and also by ( Theorem 2-7 ) we have F α+β xy Lemma 29 : n = n 5

6 Proof Let x βn,x n then x for some i, so x, n x, and x n Thus n n If x n, then n x, so x for some i and x Thus x n, so n n Therefore n = n Since the irreducible elements are the prime elements with respect to l, r and m -divisibilities, then the notion that any natural number is either a prime number or a product of prime numbers transfer from N to the set of ultrafilters which are on finite levels Theorem 210 Any ultrafilter x βn on finite level (x ) where i 1 is irreducible or product of irreducible elements Proof (1) If x L 1, then by (Lemma 1-4 (c)) x is irreducible (2) If x L 2 : For x L 2, x is product of two prime numbers, so by (Lemma 1-4 (c)) x is product of irreducible elements Let x L 2,x = yz L 2, (L 2 x = yz), then {m N : L 2 /m z} y, and since L 2 /m = {r N : rm L 2 } So we have four cases for values of L 2 /m such as : L 2 /m = {r N : rm L 2 } = 1 when m is in L 2 (1) = L 2 when m = 1 (2) = L 1 when m is in L 1 (3) = when m is otherwise (4) Now : By (1) and (2) we have either y = 1 or z = 1, so x is irreducible And by (3) x is product of two prime elements so by (Lemma 14 (c) ) x is product of irreducible elements (3) If x L 3 : For x L 3, x is product of three prime numbers, so by (Lemma 1-4 (c)) x is product of irreducible elements Let x L 3, x = yz L 3 (L 3 x = yz), then {m N : L 3 /m z} y and sine L 3 /m = {r N : rm L 3 } So we have five cases for values of L 3 /msuch 6

7 as: L 3 /m = {r N : rm L 3 } = 1 when m is in L 3 (1) = L 3 when m = 1 (2) = L 2 when m is in L 1 (3) = L 1 when m is in L 2 (4) = when m is otherwise (5) Now : By (1) and (2) we have either y = 1 or z = 1 Thus, x is irreducible By (3) L 1 y,l 2 z, we have two cases : (a) y is irreducible and z is irreducible Therefore, x is product of irreducible elements (b) y is irreducible, and z is product of irreducible elements Therefore, x is product of irreducible elements Similar to (3), we have by (4) x is product of irreducible elements (4) If we continue in this way and we suppose that any element x, 1 i n 1 is irreducible or product of irreducible elements Then we can prove that for any x L n is irreducible or product of irreducible elements such as : If x L n : For x L n, x is product of n times prime numbers, so by ( Lemma 1-4 (c)) x is product of irreduciblelet x L n, x = yz L n (L n x = yz), then {m N : L n /m z} y, and sine L n /m = {r N : rm L n } So we have n+2 cases for values of L 2 /n such as: L n /m = {r N : mr L n } = 1 when m is in L n (1) = L n when m = 1 (2) = L n 1 when m in L 1 (3) = L n 2 when m is in L 2 (4) = L 2 when m is in L n 2 (n) = L 1 when m is in L n 1 (n+1) = when m is otherwise (n+2) 7

8 Now By (1) and (2) we have either y = 1 or z = 1, so x is irreducible Moreover,in all the other cases we have x is product of irreducible elements Corollary 211 (a) For any n-th level, L n has 2 c irreducible elements (b) For any n-th level, L n there are 2 c ultrafilters in L that are not irreducible ( product of irreducible elements ) Proof (a) Since for any L n N is infinite,so by ( [2] proposition 74 ) there exist infinite set A L n, such that all elements of A are irreducible Also since by ([ 5] Theorem 3-3 ) A has 2 c elements and A L n Thus L n has 2 c irreducible elements (b) Let x and y are distinct elements of L 1 and let p L n 1Since for every disjoint subset A and B of L 1 we have Ap Bp =, then by ( [1] Theorem 811 (5),(3)), xp yp [xp,yp L n by (Theorem 2-8 (a) )] Thus L n has 2 c ultrafilters which are product of irreducible elements Corollary 212 If x L n,n 2 is not irreducible ultrafilter, then there exist at least two ultrafilters x i,x j L j,i,j < n,x = x i x j Proof Let x L n, and x is not irreducible, then by ( Theorem 210 ) x is product of at least two irreducible elements x i,x j,i,j < n,x i, x j L j, L j = L n and x = x i x j L j L j = L n Corollary 213 For any x,i 2 which is not irreducible, there exist at least two ultrafilters x i and x j L j,i,j < n,such that: x i l x, x j r x,and x i m x, x j m x The following theorem shows that the facts that for any m L m there is n L n where n m such that n m, and for any n L n there is m L m such that n m can be transfered to the -divisibility on the ultrafilters that are on finite levels Theorem 214 (a) For every ultrafilter x βn L 0 L 1 L n 1 on finite level, there is an ultrafilter y L n such that y x (b) For any ultrafilter x L m, there exist an ultrafilter y L n,m n such that x y Proof (a)let x βn L 0 L 1 L n 1 is ultrfilter onfinite level and letf : N L 0 L 1 L n 1 N isdefinedbyf(n)bethesmallestfactor 8

9 of nin L n So we have x / L 0 L 1 L n 1 and L 0 L 1 L n 1 / x andwehaven L 0 L 1 L n 1 x,f(n L 0 L 1 L n 1 ) f(x), and since f(n L 0 L 1 L n 1 ) L n, then we have L n f(x), f(x) L n Since for any n N L 0 L 1 L n 1 by definition of the function f we have f(n) n Then by (Lemma 13 (a)) we have f(x) x, f(x) L n (b) Let x βn, x L m, and let f : L m N be defined by f(n) is the smallest multiple of n in L n, so we have L m x,f(l m ) f(x), and since f(l m ) L n then we have L n f(x), f(x) L n Since for any n L m by definition of the function f we have n f(n) Then by (Lemma13 (b)) we have x f(x), f(x) Ln Corollary 215 (a) For any ultrafilter x L n there are ultrfilters x i, i n 1 such that 1 x 1 x n 1 x (b) For any ultrafilter x m L m there exist a sequence < x n : x n L n,n m > such that x m xn Proof (a)let x L n, then by (Theorem 214 (a)) there are ultrafilters x i i n 1 such that 1 x 1 x n 1 x (b)let x m L m then by (Theorem 214 (b)) there exist a sequence < x n : x n L n,n m > such that x m xm+1 xm+2 3 Ultrafiters that are not on finite levels In order to look for the ultrafilters x that are not on finite levels(x /,i = 0,1,),we use the facts that the set of all basic open sets B = {A : A N} is a base for the space βn and {,i = 0,1,} B From these we can find a basic open set A B such that all nonprincipal ultrafilters x A will not be elements in any basic open set,i = 0,1, Lemma 31 (a) There are 2 c ultrafilters x that are not on finite levels: ie x / (b) There are 2 c irreducible ultrafilters x that are not on finite levels (c) 9

10 Proof Let A = {n 0,n 1,n 2 } where n i Any ultrafilter x that is in finite level in A is principal, because, if x for some i and x A, Then x and A x, so A x, but A = {n i },{n i } x so x = {n i } is principal ultrafilter Now, since A is closed and by ([5]Theorem 33 ) any closed subset of βn has finitely many or 2 c elements, so A has 2 c elements Thus A has 2 c nonprincipal ultrafilters that are not on finite levels (b)if we take A as in (a), then by ([2]proposition 74) there is infinite set B such that B A and all elements x B are irreducible (c) by (a) there are 2 c ultrafilters x such that x / for all i = 0,1,2, so βn = In (Lemma 3-1 (a)) for any ultrafilter x βn, x A, we have / x for all i N,so N x, x N and x N Definition 32 I = N The ultrafilters that are belong to I in (Definition (3-2)) are called the ultrafilters that are not on finite levels Lemma 33 (a) An ultrafilter x I if and only if x / for all i = 0,1,2 (b) βn I = Proof (a) ( ) Let x βn,x I, so x N for all i = 0,1,2,, so N x, / x Thus x / for all i = 0,1,2, ( ) Let x / for all i = 0,1,, so / x,n x, so x N for all i = 0,1, Thus x N (b)by ([1]Lemma 317 (c)) we have βn I = βn = N βn (N ) 10

11 = βn (βn ) = Lemma 34 If x I Then (a) x / n (b) x Proof (a) Let x I, then x / n and by (Lemma 2-9) we have x / n (b) Since x, so x,l 0 ( ) x, so x Again L 1 ( ) x, and we have x If we continuous in this we will get i=2 x Thus x i=2 In particular, any union of infinite elements,i = 0,1,2, is element in any ultrafilter x I This fact leads us to prove that the elements in I are -divisible by elements of any finite level,i = 0,1,, as the following theorem shows Theorem 35 (a) For any ultrafilter x I, there exist an ultrafilter y L n such that y x (b) For any ultrafilter x L n there exist an ultrafilter y I such that x y (c) There exist an ultrafilter x I divided by an ultrafilter y I, (y x) Proof (a) Let x βn,x I, so by (Lemma 3-4 (b)) we have x and x Let f : N is defined by f(n) is the smallest factor of n in 11

12 L n, so f( ) L n, and since f( ) f(x), so L n f(x), f(x) L n i=1 Therefore by (Lemma 13 (a)) f(x) x, f(x) L n (b) Let x βn, x L n and let A = {m n,m n+1,} N where m i,i n such that for any n L n has multiple in A Let f : L n N be surjective function and it is defined by f(n) is the smallest multiple of n in A If x L n, so L n x,f(l n ) f(x), and since f(l n ) A, so A f(x), f(x) A [ f(x) I, because x is nonprincipal ultrafilter,so any element of x is infinite subset of N, since f is injective, then also any subset of f(x) is infinite subset of NThus f(x) is nonprincipal ultrafilter, so by definition of f(x) and (Lemma 3-1 (a)) f(x) I]Therefore by (Lemma 13 (b)) x f(x), f(x) I (c)letx βn andleta 1 = {32 n : n = 1,2,},A 2 = {2 n : n = 1,2,}, and x A 1,so A 1 x, and by (Lemma 3-1 (a)) we have x I Let f : A 1 N is defined by f(32 n ) = 2 n, n = 1,2, f(a 1 ) f(x) and since f(a 1 ) A 2, so A 2 f(x), f(x) A2 Then similar to analogous in (c) we have f(x) I Therefore, by (Lemma 13 (a) ) we have f(x) x, f(x) I Corollary 36 For any ultrafilter x I, there exist a sequence < x n : n N > of ultrafilters such that x n L n and x 1 x2 x Proof By (Theorem 3-5 (b)) for any x I there exist an ultrafilter x n L n for any finite level such that x n x,and by (Thmeorem 2-14 (a)) for anyx n L n thereexistanultrafilterx n 1 L n 1 suchthatx n 1 xn Therefore, there exist a sequence < x n : n N > such that x 1 x2 x Theorem 37 (a) If x,y βn and x,y I Then xy I and yx I (b) If x,y βn and x I,y / I Then xy I and yx I Proof (a) If we assume that xy / I,then xy L n for some n NSo by (Theorem 27) we have F α xy for some α A such that σ(α) = n where m k α = {(p 1,n 1 ),(p 2,n 2 ),(p m m,n m )},n = k i n i, F α = {(A 1 ) (n 1) (A 2 ) (n 2) (A m ) (nm) : A i p i P,A i A j = if p i p j } xy i=1 12

13 so (A 1 ) (n 1) (A 2 ) (n 2) (A m ) (nm) xy, and we have {r N : (A 1 ) (n 1) }(A 2 ) (n 2) (A m ) (nm) /r y} x, but (A 1 ) (n 1) (A 2 ) (n 2) (A m ) (nm) /r = {s N : rs (A 1 ) (n 2) (A m ) (nm) } so (A 1 ) (n 1) (A 2 ) (n 2) (A m ) (nm) /r, and {r N : (A 1 ) (n 2) (A m ) (nm) /r y} L j where i,j n such that L j = +j = L n, So y,y for some i n and L j x,x L j for some j n, so we have a contradiction Thus xy I Same analogues for prove yx I (b) Similar to (a) Corollary 38 For any n N there exist x I such that n x Proof Let n N and y I, then by (Theorem 37)we have ny I, and since n ny then n ny,ny I References [1] N Hindman,D Strauss: Algebra in the Stone-Cech compactification,theory and applications 2nd revised and extended, De Gruyter,2012 [2] BSobot, Divisibility in the Stone-Cech compactification, Rep Math Logic 50(2015),53-66 [3] B Sobot, Divisibility orders in the Stone-Cech compactification,mathlogic(2016) [4] BSobot, -divisibility of ultrafilters,mathlogic,submitted [5] R Walker, The Cech-Stone compactification Springer,