• 検索結果がありません。

A proof of the independence of the Axiom of Choice from the Boolean Prime Ideal Theorem

N/A
N/A
Protected

Academic year: 2022

シェア "A proof of the independence of the Axiom of Choice from the Boolean Prime Ideal Theorem"

Copied!
4
0
0

読み込み中.... (全文を見る)

全文

(1)

Comment.Math.Univ.Carolin. 56,4 (2015) 543–546 543

A proof of the independence of the Axiom of Choice from the Boolean Prime Ideal Theorem

Miroslav Repick´y

Abstract. We present a proof of the Boolean Prime Ideal Theorem in a transitive model of ZF in which the Axiom of Choice does not hold. We omit the argument based on the full Halpern-L¨auchli partition theorem and instead we reduce the proof to its elementary case.

Keywords: Boolean Prime Ideal Theorem; the Axiom of Choice Classification: Primary 03E35, Secondary 03E25, 03E40, 03E45

Let us recall the following result.

Theorem 1 (Halpern and L´evy [2]). There is a transitive model of ZF in which the Boolean Prime Ideal Theorem holds and the Axiom of Choice fails.

In the paper, we assume V ZFC and we consider the following transitive model M (see [3, pp. 184–187] or [4, pp. 221–223]). Let P be the set of finite functionspsuch that dom(p)⊆ω×ωand rng(p)⊆ {0,1}. LetG⊆P be a generic set of conditions. Fori∈ω let

ai(n) =

(1, if (∃p∈G)p(i, n) = 1, 0, otherwise,

A={ai:i∈ω}, M = HODV[G](A).

ThenM is a transitive model of ZF andA∈M. The Axiom of Choice does not hold inM because the setAis infinite and has no countable subset inM (see [3]).

We prove the Boolean Prime Ideal Theorem inM = HODV[G](A). The present proof uses the same ideas as the proof in [2] but its exposition relies on [3]. We also omit the argument from [2] based on the full Halpern-L¨auchli partition theorem [1]

and instead we reduce the proof to its elementary case substantiated in [2].

Recall that [u] ={x∈ω2 :u⊆x}for any finite functionusuch that dom(u)⊆ ω and rng(u) ∈ {0,1}. For t∈ m(ω2) andk ∈ω, [t↾↾k] =Q

i<m[t(i)↾k] denotes a basic clopen set inm(ω2).

DOI 10.14712/1213-7243.2015.138

The author was supported by grants VEGA 1/0002/12 and APVV-0269-11.

(2)

544 Repick´y M.

Lemma 2(Schema of continuity). Letϕ(x1, . . . , xn, s, A)be a formula of ZF with no free variables other thanx1, . . . ,xn,s,A. Ifx1, . . . , xn∈V,m∈ω,s∈mAis a sequence of distinct members ofA, andϕ(x1, . . . , xn, s, A)holds inV[G], then there is a basic clopen setU ⊆m(ω2)with pairwise disjoint projections inω2such thats∈U andϕ(x1, . . . , xn, t, A)holds inV[G]for everyt∈U ∩mA.

Proof: Let W be the set of all one-to-one functions in mω. For h ∈ W let hmAbe defined byh(i) =ah(i). Forh∈W let

b(h) =kϕ(x1, . . . , xn,h˙,A)k,˙ c(h) =W

k∈ω

V

z∈W − kz˙∈[ ˙h↾↾k]k ∨ kϕ(x1, . . . , xn,z˙,A)k˙

=k(∃k∈ω)(∀z∈W) ˙z∈[ ˙h↾↾k]→ϕ(x1, . . . , xn,z˙,A)k˙

where ˙h, ˙z, and ˙A denote the canonical names for h, z, and A constructed by means of the canonical names ˙ai fori∈ω. The inequality b(h)≤c(h) means that ifϕ(x1, . . . , xn, s, A) holds in V[G] fors=h, then there isk∈ω such that the conclusion of the lemma holds for the clopen setU = [s↾↾k]. Then, sincesis one-to-one, the projections ofU are pairwise disjoint if kis sufficiently large. We proveb(h)≤c(h) for allh∈W.

Letp ∈P satisfyp ≤b(h) and we findp≤p such that p≤c(h). Extendp to a conditionp⊇p so that dom(p) =k×kfor somek∈ω, rng(h)⊆k, and for alli < j < k there isl < k such thatp(i, l)6=p(j, l). For every q∈P letqi be defined byqi(j) = q(i, j). Then pik2 fori < k are pairwise incompatible and p[ ˙h↾↾k] =Q

i<m[ph(i)]. We prove that p≤c(h).

To get a contradiction assume that for some z ∈ W there is r ≤ p such that r z˙ ∈ [ ˙h↾↾k] and r ¬ϕ(x1, . . . , xn,z˙,A); the former assumption is˙ equivalent to saying that rz(i)↾k = ph(i) for all i < m. If z(i) 6= h(i), then z(i)> h(i) because pj forj < k are pairwise incompatible. Letπbe the permu- tation ofω that interchangesh(i) andz(i) for alli < mandπ(j) =j otherwise.

The permutationπinduces an automorphism ofP and an automorphism ofVP, i.e., for p, q ∈ P, q = π(p) if q(π(i), j) = p(i, j). By the symmetry lemma π(r) ¬ϕ(x1, . . . , xn, π( ˙z), π( ˙A)) which is impossible because π(r) and p are compatible,π( ˙z) = ˙h,π( ˙A) = ˙A, andpϕ(x1, . . . , xn,h˙,A). This contradic-˙ tion proves that there is no suchrand hencep≤c(h).

LetF ∈[A]m. We say that a sequencehUi :i < miof pairwise disjoint basic open sets inω2 distinguishesF, if|F∩Ui|= 1 for alli < m.

Corollary 3. Letϕ(x1, . . . , xn, F)be a formula of ZF with no free variables other than x1, . . . , xn, F. Ifs ∈ A, x1, . . . , xn ∈ ODV[G][A, s], F ⊆A\rng(s) is a finite set,m=|F|, andϕ(x1, . . . , xn, F)holds inV[G], then there is a sequence of basic open sets hUi : i < mi in ω2 disjoint from rng(s) and distinguishing members ofF such thatϕ(x1, . . . , xn, F)holds inV[G]for everyF∈[A]msuch that |F∩Ui|= 1for alli < m.

(3)

A proof of the independence of the Axiom of Choice 545 Proof: Assume |s| = k and let t : m → F be any one-to-one enumeration.

There is a formulaψ such that for some ordinalsα1, . . . ,αr,

V[G](∀t)ψ(α1, . . . , αr, st, A)→ϕ(x1, . . . , xn,rng(t)), and V[G]ψ(α1, . . . , αr, st, A).

By Lemma 2 there is a disjoint sequence of basic open sets hVi : i < k+mi inω2 such thatst ∈Q

i<k+mViandψ(α1, . . . , αr, t, A) holds inV[G] for every t∈Q

i<k+mVi. TakeUi=Vk+i fori < m.

Now we prove the Boolean Prime Ideal Theorem inM = HODV[G](A).

Let (B,∨,∧,−,0,1) be a Boolean algebra in M. Then there is f ∈ A such thatB∈ODV[G][A, f]. The class ODV[G][A, f] has a well-ordering ordinal- definable fromAandf. Using this well-ordering by transfinite recursion we can define a proper idealI⊆B maximal ordinal-definable fromAandf. Hence, for everyx ∈ B which is ordinal-definable from A and f, either x∈ I or −x∈ I.

ClearlyI∈M becauseI⊆B ⊆M. We prove thatI is a prime ideal ofB inM. Suppose thatI is not prime and letk ∈ω be the least natural number such that for somehk+1Athere is anx∈ODV[G][A, fh] such thatx∈B\Iand

−x∈ B\I. Let a = h(k) and h= h↾k. Then B ∈ODV[G][A, fh] and by minimality ofk it is obvious thata ∈/ rng(f)∪rng(h) andI is a maximal ideal of B in ODV[G][A, fh] because I is a prime ideal there. There is a formula ϕ such that

x={u∈V[G] :V[G]ϕ(u, α1, . . . , αn, fh, a, A)}

for some ordinalsα1, . . . , αn. Sincefh, α1, . . . , αn are fixed throughout the proof we shall denote

d(a) ={u∈V[G] :V[G]ϕ(u, α1, . . . , αn, fh, a, A)}.

Henced(a)∈B\I and−d(a)∈B\I. By Corollary 3 there is a basic open set U ⊆ω2 such thata∈U,U∩rng(fh) =∅, and

(∀a∈U ∩A)−d(a)∈B\I and d(a)∈B\I.

(1)

The ideal ofB generated byI∪ {d(a) :a∈U∩A} is in ODV[G][A, fh] and it coincides withB by maximality ofI. Therefore for some finite setF1 ⊆U∩Awe haveV

a∈F1−d(a)∈I. Similarly, if we consider the ideal generated byI∪{−d(a) : a∈U∩A}we obtain a finite set F2 ⊆U∩Asuch that V

a∈F2d(a)∈I. Denote F=F1∪F2 andm=|F|. Then

V

a∈F−d(a)∈I and V

a∈Fd(a)∈I.

By Corollary 3, there is a sequence of basic open sets hUi : i < midistinguish- ingF, such that each setUi is a subset ofU (this is possible becauseF ⊆U),

(4)

546 Repick´y M.

hence disjoint from rng(fh), and for every F ∈ [A]m such that (∀i < m) F∩Ui 6=∅,

V

a∈F−d(a)∈I and V

a∈Fd(a)∈I.

(2)

For everyi < m, (1) holds withU replaced with Ui becauseUi⊆U. Replacing U with Ui in the argument that leads to (2) we obtain a sequence of pairwise disjoint basic open setshUi,j:j < miiwhich are subsets ofUisuch that for every i < m, and for everyF⊆A∩U with (∀j < mi)F∩Ui,j6=∅, we have

V

a∈F−d(a)∈I and V

a∈Fd(a)∈I.

(3)

The systemS={Ui,j:i < mand j < mi} is a pairwise disjoint system of basic clopen sets inω2 andA is a dense subset ofω2. Lety⊆A∩U be a finite set of the size|S|such that (∀V ∈S)|y∩V|= 1. Then for everyz⊆y,

V

a∈zd(a)∧V

a∈y\z−d(a)∈I.

(4)

To prove this let us consider these two possibilities.

(i) For everyi < m,z∩Ui6=∅. Then by (2),V

a∈zd(a)∈Iand hence (4) holds.

(ii) There is i < m such thatz∩Ui =∅. Then (∀j < mi) (y\z)∩Ui,j 6=∅, and by (3),V

a∈y\z−d(a)∈I, and hence (4) holds.

Using (4) we obtain a contradiction as follows: 1 = V

a∈y(d(a)∨ −d(a)) = W

z⊆y[V

a∈zd(a)∧V

a∈y\z−d(a)]∈I. This contradiction proves thatI is prime inM.

References

[1] Halpern J.D., L¨auchli H.,A partition theorem, Trans. Amer. Math. Soc.124(1966), 360–

367.

[2] Halpern J.D., L´evy A., The Boolean Prime Ideal Theorem does not imply the Axiom of Choice, In: Axiomatic Set Theory, Proceedings of Symposia in Pure Mathematics, vol. XIII, Part I, pp. 83–134, AMS, Providence, 1971.

[3] Jech T.,Set Theory, Academic Press, New York-London, 1978.

[4] Jech T.,Set Theory, the third millennium edition, revised and expanded, Springer Mono- graphs in Mathematics, Springer, Berlin, 2003.

Mathematical Institute, Slovak Academy of Sciences, Greˇs´akova 6, 040 01 Koˇsice, Slovak Republic

E-mail: [email protected]

(Received December 3, 2014, revised April 16, 2015)

参照

関連したドキュメント

For a class of reversible PCA dynamics on {−1, +1} Z d , with a naturally associated Gibbsian potential ϕ , we prove that a (spatial-) weak mixing condition (WM) for ϕ implies

Abstract The binomial ideal associated with the intersection axiom of conditional probability is shown to be radical and is expressed as an intersection of toric prime ideals..

For example, a maximal embedded collection of tori in an irreducible manifold is complete as each of the component manifolds is indecomposable (any additional surface would have to

In this paper, we study the uniform stability of mutidimensional planar travelling waves for the nonlocal Allen-Cahn equationc.

The limiting distribution µ of the normalized number of key comparisons required by the Quicksort sorting algorithm is known to be the unique fixed point of a certain

The measure σ p,n of Theorem 1 assigns to measurable subsets of S p,n (1) their Minkowski surface area, an intrinsic area in that it depends on geodesic distances on the surface..

In this paper we obtain existence results for the positive solution of a singular elliptic boundary value problem.. Our study is motivated by the works of Shu [17], Arcoya,

If all the corners of a 2–cell are labelled by elements of a group, then a word can be read around the 2–cell boundary by composing these elements either unchanged or inverted