论文研究-The studies of a peculiar subset of $\omega$.pdf

所需积分/C币:19 2019-08-28 14:50:10 140KB .PDF
收藏 收藏
举报

关于 $\omega$ 的一个奇特子集的研究,赵峰,,根据 $\mbox{\upshape{ZF}}$ 公理,存在极限序数 $\omega$ 的一个奇特子集 $\bar{\omega}$,它既是无穷的,又是D\,–有穷的。只要 $\mbox{\upshape{ZF}}$ 公理�
国利技论文在线 http://www.paper.edu.cn herein, using the Separation Axioms, and the fact that wto, we have y ve(F(,E)∧A(,E)→B(x,x) 5 台(v8(F(,8)∧A(,E)→B(x,礼) (16 台x((F(,8)AA(4,8)→VEB(x,2) (17) (c and are parametric variables being bound, and by(a)of(7))(18 台(8(F(2E)AA(,C)→VvEB(m,) (19 台3三x(F(4,8)AA(4,C)→ raveD(x,A) (20) 台3E8(F(,8)∧A(,E)→B(x,) (21) (c in 3e BE (F(, E)AA(, &) and (22) E in Vr vavE B(, 2) are duNny variables being bound)(23) where Vr B(r, a) (24) V rann>N(x∈(m) (25) 台 Vcvtnin(>N→x∈(m) (26) 台彐彐VNn(m>N)→ y renin(x∈(m) (27) 台VN3(7>N)→WVWm(x∈(m) (28) Now let us assume that 3]Vm((m)Ew. Since for any C, x∈分∈∧vE(F(A,8)∧A(4,E)→B(x,), we na 0m(0(m)∈) (29) 3Ⅶmn(y(m)∈w∧VvE(F(,8)∧A(,E)→B((m),x))(30 台3(m(3(m)∈)∧mvE(F(4,E)∧A(,C)→B(3(m),)(31) →3m((m)∈)A3mvE(F(,E)∧A(x,E)→B(①(m),)(32) where 30VmvE(F(,E)∧A(2,8)→B(3(m),) (33) 台3m(丑(F(,C)∧AA(12E)→VEB(0(m),A) 34) 兮3(现mBE(F(1,E)AA(4,E)→ Vm saveD(①(m),) 5) 台3m3E(F(,8)∧A(42,8)→3)VmⅤvB(①(m),4), (36 国利技论文在线 http://www.paper.edu.cn hich 3yVmvXVEB((m), x) (37) 台30wmvE彐Nvm>N(0(m)∈孔(m) 38 3y)ⅦmE彐Nm(n>N→y(m)∈孔(m) (39) 分 Vm venin(m>N→y(m)∈礼(m) 彐m彐xN )→ Vmvdanin((m)∈x(m)) 兮VN3(>N)→3 )Vm va vn(0(m)∈(m) (42 Then for some )(which is introduced by the Lemma(1), we can let 2 be ], and so we have ∨N彐n(>N)→WmWm(3(m)∈y(m) (43) 台VN3m3n(7>N)→丑NWmm(3(m)∈(m) (44) >NWmm(7>N→y(m)∈0(m) (45) →m>N→0(m)∈0(m) for some foregoing n and any m, with n=m(46) (by∈Nm(m=m)∧Mm彐n∈N(n=m),(47) which is a contradiction because y(m)e(m) for such a y and every m E dom(O), and so the assumption cannot hold. By the Law of Excluded Middle, if ZF is consistent then we have(6) NB: Assuming 3)Vm((m)Ew), by(40)-(42)and(43)-(45) we have for some J mNvm>N((m)∈0(m)兮彐Nmn>N(m)∈0(m) F rom (6) and Dv (a)=(xQ(x)≡x(=Q(x) (48 (b)(Va Q(a))=3.c(Q(a)) re have (yVm(3(m)∈a)台y(-(m(0(m)∈) (49) 03m((m) ∈a (50 3m(o(m)w) 51 Lemma 2. Let y be the variable function such that y,oo where dom(o)=N and ran(v)Cw, and let m be the variable object such that m E dom(o). Then (ym(0(mn)∈b) (52) 国利技论文在线 http://www.paper.edu.cn Proof. Assume that V)Vm(v(m)Eu). Similar to(29)-(42), we have VNm(m>N)→ vVm v Vn((m)∈x(m) Then for any of the foregoing ys, we can let it be 2, and so we obtain N丑n(>N)→VmVm((m)∈J(m) By(43)-(47), there is a contradiction in it. Therefore the assumption cannot hold, and so by the Law of Excluded Middle we have(52) From(52), by(b)of(48), we have (0wm((m)∈a)分3)(-(m((m)∈) (53) 33m(J(m)∈u) (54) (55) In addition, by(1)-(5) and(48), we can obtain (a)33m((m)∈a); (56) (b)=(m((m)ga) Combining the results with(48) we obtain 303m(0(m)∈a)分-(yWm(y(m)g); (57) (30m(0(m)gu)Ⅵ23m(3(m)∈u) (58) Thus, if y is the variable function such that y>o where dom()=N and ran()c and m is the variable object such that m E dom()), then each of the following cannot hold (a)3m((m)∈a (b)yvm(0(m)∈); (c)3y Vm(l(m)w) (a)tyVm(y(m) w) By(1)- 3),y(m)Ew holds for any of the foregoing )'s and any m E dom(2), and so by (G)and(52)we obtain Theorem 1 60) By(5)and(60), we clearly have w-≠andu-≠ neoren va∈a彐O∈w(∈O (61) 国利技论文在线 http://www.paper.edu.cn .By(1)-(4) or any a a∈a分a∈Nv8(F(A,E)∧A(A,E)→B(a,x) 62 a∈wA彐Nm>N((m)∈ua∈(m) →N=m>N(a∈uA孔(m)∈u∧a∈(m) (64) (because a is a parametric variable being bound and (65 le property that q A3 S()= 3 i (qA S(r))) (66) →彐(∈∧a∈) (67) 分∈(a∈日) 68 Or, put more briefly, for any ce 分6(a∈日∧白∈) (71) 彐6∈u(a∈O) (72) Hence we have(61) Theorem 3 (VO∈u(Oca) Proof. Assume that VeEw(ec w). Since for any a O∈u兮a∈Ju 台彐(a∈O∧O∈ (75 →((a∈6A白c) 76) ∈ we have wC w. And by (5), we would have w= w, which is contradictory with(60). Thus we have(73 From(73), by(b)of(48), we have (V∈u(ca)兮彐6∈(-(c) (78) 分=6∈(gc) pare a wi va(a∈→彐((∈6∧日∈u); (80 va(a∈分O(∈6∧日∈) (81) 国利技论文在线 http://www.paper.edu.cn Theorem 4 p Proof. Firstly, for any (x 台3(a∈∧0∈a) (84) →(a∈∧∈AME(F(,8)∧A(4,8)→B(,x) 85 台3(c∈0∧0∧a∈”∧v8(F(,CAA(4,E)→B(,)(86) →a∈N彐(a∈AVE(F(,E)∧A(,E)→B(0,2) (87) where for any &, 8, if F(, )and A(, 8)hold then we have for any y Ew B(0,) (88) →彐NWn>N(∈(m) Nm(m>N→?∈x(m) (90 Nn(>N)→彐NW(∈(m) →VNn(7>N)→Wm(0∈:(n) (92) →VN彐n(m>N)→m(c(n) (93 冷VN丑n(m>N)→W的(6∈0→6∈x(m) (94) →彐NM(m>N→(B∈→B∈(n), 95) and therefore (a∈∧!VE(F(4,E)∧A(4,C)→B(,x) (96) →MvE(P(,C)∧A(,E)→B(a,2 (97 hence we have n∈∪a→a∈ wave(F(x,E)AA(,E)→B(a,x) 98 Then it follows that 100 condly, by(2)-(4), for any a E w we have Wr vE(F(a, &)AA(r,&)= B(a, x) (10 台V"vE(F(x,C)AA(,E)→三NW>N(a∈x(m) (102) 台V"VE(F(H",C")AA(4,C)→彐NWmn(n>N→a∈(m),(103) 国利技论文在线 http://www.paper.edu.cn and so for any such function with some N there is some Na N such that ∈x(Na) (104) Since No E N and 0 I'(NaE w, by (2) and (3), for any 2n with some E/we have EM=E(Y(Na)in>M(e(n)>'(Na))if F(I ", 8)and A(Y",8 hold, and so for any such l there is some m such that n>M(x"(Na)∈B"(m); (105) thus by(4), we have l'(Na)E w. Therefore a∈3∈a(a∈0), (106) that is. w has no the maximal element and hence supa (107) CoInbining(100) with(107), it follows that(82) Since w+0 is a set of finite ordinals, w is an ordinal. Rather oddly enough, by( 82) and(60),w=Ua is exactly a limit ordinal, as distinguished from the limit ordinal w, and thus we have Corollary 1 (a)va(a∈a铃彐(a∈?∧?∈) 108 (b)c∈a(aCa); (c)LetP(u) be the power set of. Then WCp(ω),i.e,Va∈a(a∈P(a) This also shows that in ZF the linit of a sequenCe based on(2)and (3 ) will be fundamentally different than the limit of a sequence based on the axiom of Infinity. However, the natural question to ask is, whether w E w or not. Obviously, from(1) it follows that Theorem 5 Theorem 6 is infinite but is D-fi2 (110 Proof. On the one hand, by(106), there is no a one-to-one mapping of w onto a natural number, that is. w is infinite On the other hand, if we let J be an arbitrary proper subset of w, then J C w and J +w and thus∪TcUa. Obviously,∪丁and∪ w are ordinals, and we have: 国利技论文在线 http://www.paper.edu.cn (1)I∪≠u, then since∪u= w is a limit ordinal and∪Jg,∪ g is not a limit ordinal, and so J has the maximal element; hence there is no a one-to-one mapping of w onto such a proper subset J of w (2)If∪T=∪a,then∪J= w is a limit ordinal. Now we let2> oo be such that dom(z)= and ran(2)=u, and let m be a variable such that m E dom(2). By(51)and (79), there is some m such that Z(m)Ew and Z(m)E W, and so if we let m be the smallest such m, then Z()∈a∧z(⑨)go, and by(2)-(4)and(82), we have={2(m)∈u: m E dom(z)Am< 33, where m E N will be large enough, at least not less than any a E w thus w has st-l elements even though o is uncertain. Assume that there exists a one-to-one mapping of w onto the proper subset J of w where J w and U=w. Then since there is at least one i E w such that 2(m-i)J, J has at most -2 elements, and so 3-1=m-2 a contradiction; hence there is no a one-to-one mapping of w onto the proper subset T of w II a word, there is 110 a Olle-to-Onle Mapping of w onto a proper subset of w, that is, w is Dedekind-finite(D-finite) Since w is infinite, if we consider l=s cw: s is finite then it follows that w is Tarski- infinite(T-infinite). By(110), w is not just T-infinite, but also D-finite. This result would not be inconsistent in ZF, because olle cannot prove that every D-finite set is finite without the Axiom of choice By ZF, it is common knowledge that w=w=wo=No and a= a for every a< w. Now we consider the cardinality w of w Theorem 7 c∈u(a<|l|<lul). 111 Proof. On the one hand, by(82), w is transitive and has no maximal element, and so by(2)of (108)VaEa(a<al) holds. And by(4), we have 彐a∈u(a<|) (112) On the other hand, by(5), we have u< w. Assume that w= w. Then there exists a one- to-one mapping of w onto w. Let h be a one-to-one mapping of w onto w such that h(a)-9 Then we have a one-to-one mapping 2a+ 2 of the proper subset ( 2a: a Ew of w onto the proper subset (29:9 E w] of w. Since the mapping a i 2a of w onto [2a: a E w is also one-to-one, we have |20:0EWl=ol. However, this is impossible since lal is D-finite. Therefore 113) By(112)and(113), we have(111) neoren (114 国利技论文在线 http://www.paper.edu.cn Proof. By the Power Set and Separation Axioms, we let P(w)=is: scw] PD(u)={s∈P(u):s<|ul∧sisD- finite} and Pwolw)=sc pw):s< wll. Since P(a)c pp(w)(by(110)(113)) and Ppl)c pwo(w), we have P()≤|P≤n() Subsequently, let us consider the cardinality of Pwo(w). Given aSE Pwo(w), we construct a function Ps: w, 0, 1] such that, s(i)=ai, where ai=1 if iE S, and ai=0 otherwise hence S-{∈:(S)∈u∧≤工(S)∧s(2)-1} Ifrs=ax(s)x(s)-1…1·a2a1 o such that T(S)∈uA0≤i≤工(S)∧a;∈{0,1}, then rs∈u∧rs=φs((S)ys(工(S)-1).s().9s(2)s(1)s and so ior every 9∈Pn(u), ≠ Ss if and only if rs。≠ It is clear that there is a one-to-one mapping of pwo(w)into w. Thus 116 Then, since P()=2o, by(115)and(116 )we have(114) Theorem g (117) Proof.(a) Here we let w= w. By(111), we can consider the cofinality of a. Let B<w be a finite ordinal such that cof (a, B) holds. Then there exists an ordinal number B1 such that B= 61, and there is a function l such that cof (w, 6)holds. Let a1 E wl be such that (B1)=a1. Then there is an ordinal number a2 E a such that a1 a2, and thus th here is an ordinal number B2 E B such that y (B2)=a2. However, it is impossible since B2< BI and y is strictly monotonically increasing. Hence we only have B= w. i.e., cof (w, a )and cf()=| (b)By(111). there is some a w such that wll s wl and wi and since d W1=2 o, we have 118

...展开详情
立即下载 低至0.43元/次 身份认证VIP会员低至7折
抢沙发
一个资源只可评论一次,评论内容不能少于5个字
  • 至尊王者

    成功上传501个资源即可获取
关注 私信 TA的资源
上传资源赚积分,得勋章
最新推荐