On sequentially closed subsets of the real line in ZF Kyriakos Keremedis June 8, 2018 Abstract We show: (i) CAC i¤ every countable product of sequential metric spaces (sequentially closed subsets are closed) is a sequential metric space i¤ every complete metric space is Cantor complete. (ii) Every in…nite subset X of R has a countably in…nite subset i¤ every in…nite sequentially closed subset of R includes an in…nite closed subset. (iii) The statement “R is sequential” is equivalent to each one of the following propositions: (a) Every sequentially closed subset A of R includes a countable co…nal subset C, (b) for every sequentially closed subset A of R, AnA is a meager subset of A, (c) for every sequentially closed subset A of R, AnA 6= A, (d) every sequentially closed subset of R is separable, (e) every sequentially closed subset of R is Cantor complete, (f) every complete subspace of R is Cantor complete. Mathematics Subject Classi…cation (2010): E325, 54E35. Keywords: Axiom of Choice, weak choice principles, Loeb spaces, complete metric spaces, Cantor complete metric spaces, sequential closure, sequential metric spaces, Fréchet-Urysohn metric spaces, Dedekind in…nite sets.
1
1
Notation and terminology
Let X = (X; d) be a metric space and A X. In the sequel, boldface letters will denote metric spaces and lightface letters will denote the underlying sets. X is said to be Loeb (resp. countably Loeb) i¤ the family of all non-empty closed subsets of X has a choice function (resp. every countable family of non-empty closed subsets of X has a choice function). A well-known example of a Loeb space is the real line R with the usual metric. A will denote the set of all points of X for which there exists a sequence (xn )n2N of points of Anfxg such that limn!1 xn = x. A is called sequentially e = A [ A is the sequential closure of A. closed if A A and A
X is called sequential (resp. Fréchet-Urysohn) if every sequentially closed subset of X is closed (resp.Tfor every F 2 P(X); Fe = F ). X is Cantor complete if fGn : n 2 !g is a singleton for every descending family fGn : n 2 !g of non-empty closed subsets of X with limn!1 (Gn ) = 0, where (Gn ) is the diameter of Gn . Let A be a subset of R. A subset C of A is called co…nal in A if for every a 2 A there exist x; y 2 C with x a and y a. Below we list the choice principles we shall be dealing with in the sequel. 1. Cof (R) : Every sequentially closed subset of R includes a countable co…nal subset. Equivalently, for every sequentially closed A 2 P(R), there exists a countable subset C of A such that for every a 2 A there is a c 2 C with a c. 2. CAC (Form 8 in [5]) : Every countable family of non-empty sets has a choice function. Equivalently, see [8B] p. 17 in [5], every family A = fAi : i 2 !g of non-empty sets has a partial choice. i.e., some in…nite subfamily B of A has a choice set. 3. CAC(R) (Form 94 in [5]) : CAC restricted to countable families of subsets of R. Equivalently, see [4], every family A = fAi : i 2 !g of non-empty subsets of R has a partial choice. i.e., some in…nite subfamily B of A has a choice set. 4. CAC! (R) (Form 5 in [5]) : Every family A = fAi : i 2 !g of disjoint non-empty countable subsets of R has a choice set. Equivalently, see 2
[4], if every family A = fAi : i 2 !g of non-empty countable subsets of R has a partial choice. 5. Form 9 : Every in…nite set X is Dedekind in…nite (X has a countably in…nite subset). 6. Form 13 : Every in…nite subset X of R is Dedekind in…nite.
2
Introduction and some preliminary results
Proposition 1 Let X be a metric space and F 2 P(X). Then: e (i) Fe = Fe i¤ Fe is sequentially closed. e (ii) (ZFC) Fe = Fe. (iii) If F has a well ordered dense subset then Fe = F . In particular, Fe 6= F implies F has no well ordered dense subsets. Proof. We leave the proof as an easy exercise for the reader. Proposition 2 Let X = (X; d) be a metric space. (i) Every closed subset of X is sequentially closed. T (ii) Let A be a family of sequentially closed subsets of X. Then A is sequentially closed. In particular the intersection of a closed and a sequentially closed subset of X is sequentially closed. (iii) If X is countably Loeb and for every family F = fFi : i 2 !g of nonempty sequentially closed subsets of X there exists a family G = fGi : i 2 !g of non-empty well-orderable sets such that for all i 2 !; Gi Fi then X is sequential. Proof. (i) and (ii) are straightforward. (iii) Fix X a countably Loeb metric space and let A be a sequentially closed subset of X. We show that A is closed. Fix x 2 A and let F = fFn : n 2 Ng; Fn = D(x; 1=n) \ A. By part (i) of this proposition, F is a family of non-empty sequentially closed subsets of X. Thus, by our hypothesis, there exists a family G = fGn : n 2 Ng of non-empty well-orderable sets such that for all n 2 N; Gn Fn . 3
We claim that for all n 2 N; Gn Fn . Indeed, …x a well ordering of Gn and let y 2 Gn . Then, there exists a sequence (ym )m2N Gn Fn with limm!1 ym = y. Thus, y 2 Fn and Gn Fn . Since X is countably Loeb, it follows that the family fGn : n 2 Ng has a choice function (gn )n2N . Clearly, (gn )n2N A and limn!1 gn = x. Since A is sequentially closed, x 2 A. Thus, A = A and A is closed as required …nishing the proof of the proposition. Corollary 3 The statement: “For every family A = fAi : i 2 !g of nonempty sequentially closed subsets of R there exists a family B = fBi : i 2 !g of non-empty countable sets such that for all i 2 !, Bi Ai ” implies “R is sequential”. Proof. The proof follows at once from the Proposition 2 and the fact that R is a Loeb space. By Proposition 1, the sequential closure, in the realm of metric spaces, is idempotent in ZFC. However, as the following result from [2] demonstrates, this is no more true in ZF. e Theorem 4 (i) [2] CAC(R) i¤ for every subset F of R, Fe = Fe. (ii) [2] A metric space X is Fréchet-Uryshon i¤ X is sequential and for every e subset F of X; Fe = Fe. (iii) [[3], p. 74] CAC(R) i¤ R is Fréchet-Uryshon. (iv) [2] The following are equivalent: (a) CAC. (b) Every metric spaces is sequential. e (c) For every metric space X, for every subset F of X, Fe = Fe. (c) Every metric spaces is Fréchet-Urysohn. (v) The following are equivalent: (a) CAC(R). (b) Every subspace of R is sequential. (c) For every A 2 P(R) there exists a countable subset C of A such that for every a 2 A there is a c 2 C with a c. (d) For every A 2 P(R)nf?g there exists a sequence (xn )n2N A with sup(A) if A is bounded limn!1 xn = . 1 if A is unbounded 4
Proof. (v) (a) $ (b) has been established in [2] and (c) ! (d) is straightforward. (b) ! (c) Fix A 2 P(R)nf?g. Since R is order isomorphic with (0; 1), we may assume that A is bounded. If A has a last element c then, C = fcg satis…es trivially the conclusion. So, assume that A has no last element and let X = A [ fag; a = sup(A). By our hypothesis, the subspace X of R is 0 sequential. Since a 2 A nA; it follows that A is not closed in X and consequently A is not sequentially closed. Thus, there exists a sequence (cn )n2N A with limn!1 cn = c 2 XnA. i.e., limn!1 cn = a. It is straightforward to verify that C = fcn : n 2 Ng satis…es the conclusion of (c). (d) ! (a) Fix A = fAn : n 2 Ng a disjoint family of non-empty subsets of R. We show that A has a partial choice. Assume that each An (n; n + 1). (R is e¤ectively homeomorphic with each of its non-empty open intervals.) S Put A = fAn : n 2 Ng and let, by our hypothesis, (xn )n2N A satisfy limn!1 xn = 1. On the basis of C, one can easily de…ne a partial choice for A. Proposition 5 (i) [3] The statement: “R is sequential” implies Form 13. (ii) For every sequentially closed subset A of R, A A is sequentially closed. (iii) The statement: “Every non-well-orderable subset of R includes a perfect subset” implies CAC! (R). Proof. (i) Assume, aiming for a contradiction, that A is an in…nite Dedekind …nite subset of R (A has no countably in…nite subset). Then every convergent sequence (an )n2N of points of A has a …nite range and consequently (an )n2N is eventually constant. Thus, A is sequentially closed. Hence, by our hypothesis, A is closed. Since R is Loeb and second countable and A is closed, it follows that A is separable. i.e., A is Dedekind in…nite. Contradiction! (ii) This is straightforward. (iii) Assume on the contrary and let A = fAi : i 2 !g be a family of countably in…nite subsets of R having no choice set. Assume that for every i 2 !; Ai (i; i + 1). By our hypothesis, there exists a perfect subset F of X = [A. Since, jF j = jRj it follows that F \ Ai 6= ? for in…nitely many i 2 !. Since R is a Loeb space, it follows that F has a countable dense subset D = fdn : n 2 !g. Use the well-ordering of D and fact that for in…nitely many i 2 !, D \ Ai 6= ? to de…ne a choice set of an in…nite subfamily of A. Thus, A has a choice set as required. 5
In [3] p. 77 and in [1] p. 298 it is asked whether the converse of Proposition 5 (i) is true. i.e., Question 1. Does Form 13 imply “every sequentially closed subset of R is closed”? The research in this paper is motivated by Question 1. We show in the forthcoming Theorem 11 that the statement Cof (R), which resembles Form 13, is equivalent to the proposition “every sequentially closed subset of R is closed”. Clearly, Cof (R) implies Form 13 (any in…nite Dedekind …nite subset of R is sequentially closed). So, in view of Theorem 11, Question 1 may be rephrased as: Question 2. Does Form 13 imply Cof (R)? Proposition 6 (i) The real line R with the usual metric is complete and Cantor complete. (ii) Cantor complete metric spaces are complete. (iii) A Cantor complete subset of a metric space is closed. (iv) Complete subspaces of metric spaces need not be closed in ZF. Proof. (i) The usual proof of the fact that R is complete in ZFC applies in ZF. The fact that R is Cantor complete follows easily from the observation that R is Loeb. (ii) This is is straightforward. (iii) Let X be a metric space and A be Cantor complete subset of X. We show that A is closed. Fix x 2 A. Clearly, G = fGn : n 2 Ng where, Gn = D(x; 1=n) \ A is a descending T family of non-empty closed subsets of A with lim (G ) = 0. Thus, G = fyg for some y 2 A. Since, fyg = n T T n!1 G fD(x; 1=n) : n 2 Ng = fxg we see that x = y and consequently x 2 A. Hence, A = A and A is closed as required. (iv) This has been established in [1].
3
Main results
Our …rst result in this section shows, in ZF, that “sequentiality”is not countably productive in the class of metric spaces and, complete metric spaces need not be Cantor complete. 6
Theorem 7 The following are equivalent: (i) CAC. (ii) For every countable family fXn = (Xn ; dn ) : n 2 Ng of sequential metric Q P spaces, X = ( n2N Xn ; d) is sequential where, d(x; y) = n2N n (x2nn;yn ) and for all n 2 N, n (a; b) = minf1; dn (a; b)g. (iii) For every metric space X, every family A = fAn : n 2 Ng of non-empty sequentially closed subsets of X has a partial choice. (iv) Every complete metric space is Cantor complete. Proof. (i) $ (iii), (i) ! (ii) These are straightforward. (ii) ! (i) Fix A = fAi : i 2 Ng a disjoint family of non-empty sets. It suf…ces to show that A has a partial choice. Assume, aiming for a contradiction, that A has no partial choice. Let, for every i 2 N; Xi = Ai [ f1i g; 1i 2 = Ai be endowed with the discrete metric di . i.e., for all x; y 2 Xi ; di (x; y) = 1 if x 6= y . Clearly, for every i 2 N; Xi is a sequential metric space (every 0 if x 6= y Q subset of Xi is closed in Xi ). Thus, by (iii), X = ( n2N Xn ; d) is a sequential metric space. It is easy to see that the element 1 2 X given by 1(i) = 1i ; i 2 N is a limit point of the set Xnf1g. We claim that no sequence of Y = Xnf1g converges to 1. i.e., Y is sequentially closed. If not then there exists an injective sequence (xn )n2N in Y with limn!1 xn = 1. Since, A has no partial choice, it follows that there exists an element i0 2 N such that for all n 2 N, for all i Qi0 ; xn (i) = 1i . Thus, Z = fxn ji0 : n 2 Ng is a countably in…nite subset of i i0 Xi . Hence, there exists j i0 such that the set fn 2 N : xn (j) 6= 1j g is in…nite. Thus, 1 j (f1j g) is a neighborhood of 1 missing in…nitely many terms of (xn )n2N . Contradiction! Hence, Y is sequentially closed as required. Since X is sequential and Y is sequentially closed, it follows by our hypothesis that Y is closed. Contradiction! (i) ! (iv) Fix X a complete metric space. We show that X is Cantor complete. Fix G = fGn : n 2 Ng a nested family of non-empty closed sets such that limn!1 (Gn ) = 0. Fix, by CAC, a sequence (xn )n2N such that x Tn 2 Gn ; n 2 N. Working as in the proof of Proposition 6 we can show that G is a singleton and conclude that X is Cantor complete. (iv) ! (i) Let A = fAn : n 2 Ng be a disjoint family of non-empty sets. It su¢ ces to show that some in…nite subfamily of A has a choice set. Assume 7
S the contrary and consider the following metric d on X = fAn : n 2 Ng given by: 8 0 if x = y < 1=n if x; y 2 An and x 6= y d((x; y)) = . : maxf1=n; 1=mg if x 2 An and y 2 Am
By our assumption, the range of any sequence (an )n2N of XSis included in some …nite union of members of A, say (an )n2N Xk ; Xk = fAn : n 2 kg for some k 2 N. If, in addition, (an )n2N is Cauchy, then each An ; n 2 k includes …nitely many members of fan : n 2 Ng and, for every x; y 2 Xk with x 6= y one of the sets fm 2 N : am = xg, fm 2 N : am = yg is …nite. So, (an )n2N is eventually constant and consequently convergent. Thus, X is complete. Since dSproduces the discrete topology on X it follows that for all n 2 N; Gn = fAm : m ngTis closed. It is straightforward to verify that limn!1 (Gn ) = 0. Since, fGn : n 2 Ng = ?, it follows that X is not Cantor complete, contradicting (iv). Thus, A has a partial choice set as required. Proposition 8 (i) Every non-well-orderable subset A of R includes a nonwell-orderable, dense-in-itself subset. (ii) Every in…nite sequentially closed subset of R is either well-orderable or includes a non-well-orderable, dense-in-itself, sequentially closed subset. Proof. (i) Fix a non-well-orderable subset A of R. Via a straightforward trans…nite induction we construct a descending well ordered family A = fAi : i 2 g of sequentially closed subsets such that for all i 2 T , the set Ii = Iso(Ai ) of all isolated points of Ai is non-empty and B = A is the required non-well-orderable, dense-in-itself, sequentially closed subset of A. For i = 0 we let A0 = A. For i = v + 1 a non-limit ordinal we let Iv = Iso(Av ). If Iv = ? then the trans…nite induction terminates and a = i. If Iv 6= ? then we let Av+1 = Av nIv . T For i a limit ordinal, we let Bi = fAj : j 2 ig and Ii = Iso(Bi ). If Ii = ? then a = i. If Ii 6= ? then we put Ai = Bi nIi . Since A is a set it follows that the above trans…nite induction terminates at some at some ordinal . Since for each X R, BX = f(p; q)\X : p; q 2 Qg 8
is a base for the subspace X of R, it follows that each Ii ; i 2 is wellorderable. We consider T the following two cases: S (a) B = A = ?. If this is the case then, A = fIi : i 2 g being a well ordered union of well-orderable sets is well ordered. So, this cannot be the case. T (b) B = A = 6 ?. Clearly, in thisScase Iso(B) = ? and B is densein-itself. S Since A is not well-orderable, fIi : i 2 g is well-orderable and A = B [ fIi : i 2 g it follows that B is not well-orderable as required. (ii) Fix an in…nite sequentially closed subset A of R. If A is well-orderable then we have nothing to show. So, assume that A is not well-orderable. Let 2 On; Ai ; Bi ; i 2 I and B be as in the proof of part (i). By Proposition 2 and part (i), it follows that B is is the required non-well-orderable, dense-initself, sequentially closed subset of A. Corollary 9 Every scattered subspace A of R (A contains no non-empty dense-in-itself subset) is well-orderable. Theorem 10 The following are equivalent: (i) Form 13. (ii) For every dense-in-itself subset A of R, A = A. (iii) For every dense-in-itself subset A of R, A 6= ?. (iv) Every in…nite sequentially closed subset of R includes an in…nite closed subset. (v) Every in…nite sequentially closed subset of R includes a countably in…nite subset. Proof. (i) ! (ii) Fix a dense-in-itself subset A of R. We show that A is dense in A. To see this, …x p; q 2 Q with (p; q) \ A 6= ?. Since A is dense-in-itself, it follows that (p; q) \ A is in…nite. Let, by Form 13, (xn )n2N be an injective sequence in (p; q) \ A. Since [p; q] is compact, (xn )n2N has a subsequence (xkn )n2N converging to some point x 2 [p; q]. Without loss of generality we may assume that x 2 = fxkn : n 2 Ng. Since (xkn )n2N Anfxg it follows that x 2 A . Thus, x 2 [p; q] \ A . Fix p; q 2 Q such that (p; q) \ A 6= ? and …x a; b 2 (p; q) \ A with a < b. Choose p1 ; q1 2 Q such that p < p1 < a and b < q1 < q. It follows, by the above argument that ? 6= [p1 ; q1 ] \ A (p; q) \ A . Thus, every non-empty basic open set of A the form (p; q) \ A; p; q 2 Q meets non-trivially A . Hence, A is dense in A as required. 9
(ii) ! (iii) Fix A a dense-in-itself subset of R. If A = ? then ? = A 6= A contradicting our hypothesis. (iii) ! (iv) Fix A an in…nite sequentially closed subset of R. If A includes an in…nite well-orderable subset B then B A is the required in…nite closed subset of A. So, assume that A is Dedekind …nite. Then, by Proposition 8, A includes a dense-in-itself sequentially closed subset B. By our hypothesis, B 6= ?. Hence, there exists a point x 2 B and an injective sequence (xn )n2N Bnfxg with limn!1 xn = x. Contradiction! (iv) ! (v) Fix A an in…nite sequentially closed subset R and let by our hypothesis, K be an in…nite closed subset of A. Since R is Loeb, it follows easily that K has is separable. Thus, A has a countably in…nite subset as required. (v) ! (i) Fix A an in…nite set of R. If A is Dedekind …nite, then A is sequentially closed. Hence, by our hypothesis, A includes a countably in…nite subset K. Thus, A is Dedekind in…nite and Form 13 holds. Theorem 11 The following are equivalent: (i) R is sequential. (ii) AC(R; scl) : The family of all non-empty sequentially closed subsets of R has a choice set. (iii) CAC(R; scl) : Every countable family of non-empty sequentially closed subsets of R has a choice function. (iv) PCAC(R; scl) : Every countable family of non-empty sequentially closed subsets of R has an in…nite subfamily with a choice function. (v) Cof (R). (vi) Every sequentially closed subset of R is separable. (vii) For every sequentially closed subset A of R, AnA is a meager subset of A. (viii) For every sequentially closed subset A of R, AnA 6= A. Proof. The implications (i) ! (ii) ! (iii) ! (iv) are straightforward and (iv) ! (i) can be derived as in the proof of Proposition 2 (ii). In fact, the implications (i) ! (ii) ! (iii) ! (i) are also mentioned in [1]. (ii) ! (vi) Clearly, in view of Proposition 2, A = fApq : p; q 2 Q; p < qgnf?g where, Apq = [p; q] \ A;
10
is a family of sequentially closed subsets of R. Let, by our hypothesis, D be a choice set of A. Clearly, D is a countable dense subset of A and A is separable as required. (vi) ! (v) Fix a sequentially closed subset A of R. Let D be a countable dense subset of A. It is easy to see that D [ fa; bg where, a; b are the …rst and last element of A (if they exist) is the required co…nal subset of A. (v) ! (i) Fix A a sequentially closed subset of R. We show that A is closed. Assume on the contrary and …x x 2 AnA. Assume that [x 1; x] \ A is in…nite (if [x 1; x] \ A is …nite then [x; x + 1] \ A is in…nite and we work with [x; x + 1] \ A). By Proposition 2, [x 1; x] \ A is sequentially closed. Let, by our hypothesis, C be a countable co…nal subset of [x 1; x] \ A. Via an easy induction we can construct a sequence (xn )n2N of points of C such that limn!1 xn = x. Thus, x 2 A and A is closed as required. (vi) ! (vii) Fix A a sequentially closed subset of R. By our hypothesis, e = A. Thus, AnA = ? is trivially A is separable and by Proposition 1, A = A a meager subset of R. (vii) ! (viii) Fix A 6= ? a sequentially closed subset of R. Assume, aiming for a contradiction, that AnA = A. Fix x 2 AnA. Assume that x 2 [x 1; x] \ A. Otherwise x 2 [x; x + 1] \ A and we work accordingly. Let by our hypothesis, M = fM Si : i 2 Ng be a family of closed nowhere M. dense subsets of A with AnA By Proposition 2, for all n 2 N, An = [x 1=n; x 1=(n + 1)] \ A is a sequentially closed subset of R. Since x 2 AnA we may assume that for all n 2 N, (x 1=n) 2 = A. Claim. For all i; n 2 N, Min = Mi \ An is a closed nowhere dense subset of An . Proof of the claim. Fix i; n 2 N. We assume that x 1=n; x 1=n+1 2 A. The rest of the cases (x 1=n 2 = A, x 1=n + 1 2 A etc) can be treated similarly. Clearly, O = (x 1=n; x 1=(n+1))\An = (x 1=n; x 1=(n+1))\A is a non-empty open subset of A and Mic \ O is a dense open subset of An . Thus, An n(Mic \ O) = Min [ fx 1=n; x 1=n + 1g and cosequently Min is a closed nowhere dense subset of An . Since An is a compact metric space and compact metric spaces are Baire in ZF, S it follows that An is a Baire space. Hence, by the previous claim, An n fMin : i 2 Ng is an uncountable subset of An . Since, An = (An nAn ) [ 11
S S An and An nAn fMin : i 2 Ng it follows that An n fMin : i 2 Ng is an uncountable subset of An . Since every partial choice set of A = fAn : n 2 Ng is a sequence of points of A converging to x 2 AnA, we may assume that A has no partial choice. In particular, we may assume that for all n 2 N, An is dense-in-itself and An \ Q = ?. For every n 2 N, we construct, via a straightforward induction, a nested sequence of closed intervals with rational endpoints ([pnm ; qnm ])m2N such that for all m 2 N, qnm pnm < 1=m; Mmn \ [pnm ; qnm ] = ? and [pnm ; qnm ] \ An 6= ?. Since, R is Cantor T complete and limm!1 (qnm pnm ) = 0, it follows by Proposition 6 that F = fyn g where, F = fFm = [pnm ; qnm ] \ An : m 2 Ng and yn 2 An . T We show T next that yn 2 An . We have: fyn g = f[pT nm ; qnm ] \ An : m 2 Ng = f[pnm ; qnm ] \ (An nAn [ An ) : m 2 Ng = f([pnm ; qnm ] \ An nAn ) [ ([pnm ; qnm ] \ An ) : m 2 Ng. Since the family f[pnm ; qnm ] : m 2 Ng T is descending, it follows easily that T the latter intersection is equal to f[pnm ; qnm ] \ An nAn : m 2 Ng [ f[pnm ; qT nm ] \ An : m 2 Ng. Since, for every m 2 N; M \ [p ; q ] = ? and f[pnm ; qnm ] \ An nAn : m 2 mn S nm nm T Ng f[pnm ; qnm ] \ ( fMmn : m 2 Ng : m 2 Ng = ?; it follows that T f[pnm ; qnm ] \ An nAn : m 2 Ng = ?. Thus, yn 2 An as required. Hence, fyn : n 2 Ng is a choice set of A and x 2 A. Contradiction! Thus AnA 6= A as required. (viii) ! (i) We work as in the proof of (vii) ! (viii). Fix A a sequentially closed subset of R. Assume, aiming for a contradiction, that A is not closed and …x x 2 AnA. For every n 2 N let An be as in the proof of (vii) ! (viii). Since R is a Loeb space, it follows that there exists a choice function f of the family of all non-empty closed subsets of R. For every n 2 N let Dn = fdpq = f ([p; q] \ An ) : p; q 2 Q and [p; q] \ An 6= ?g. It is easy to see that for every n 2 N, Dn with the obvious order it inherits from Q Q is a well-ordered subset of An such that Dn = An . We claim that for every n 2 N, Dn \ An 6= ?. Indeed, if for some n 2 N, Dn \ An = ? then, Dn An nAn and consequently, An nAn = An contradicting our hypothesis. Let xn be the …rst element of Dn which belongs to An . Clearly, (xn )n2N is a sequence of points of A converging to x. Thus, x 2 A. Contradiction! Theorem 12 The following are equivalent: (i) R is sequential. (ii) Every sequentially closed subspace of R is Cantor complete. 12
(iii) Every complete subspace of R is Cantor complete. (iv) Every complete subspace of R is closed. Proof. (i) ! (ii) Fix a sequentially closed subset A of R. By our hypothesis, A is closed and the conclusion follows from Proposition 6. (ii) ! (iii) Fix a complete subset A of R. Clearly, A is sequentially closed i¤ it is complete. Thus, A is sequentially closed. Hence, by our hypothesis, A is Cantor complete as required. (iii) ! (iv) This, in view of Proposition 6, is straigthforward. (iv) ! (i) This has been established in [1].
References [1] G. Gutierres, Sequential topological conditions in R in the absence of the axiom of choice, Math. Log. Quart., 49, No. 3, (2003) 293-298. [2] G. Gutierres, On countable choice and sequential spaces, Math. Log. Quart., 54, No. 2, (2008) 145-152. [3] H. Herrlich, Axiom of Choice, Lecture Notes Math., 1876, Springer, New York, 2006. [4] P. Howard, K. Keremedis, J. Rubin, A. Stanley & E. Tachtsis, Nonconstructive properties of the real line, Math. Logic Quart. 47 (2001) 423-431. [5] P. Howard and J. E. Rubin, Consequences of the Axiom of Choice, Math. Surveys and Monographs, Amer. Math. Soc., Vol. 59, Providence (RI), 1998. [6] K. Keremedis and E. Tachtsis, On Loeb and weakly Loeb Hausdor¤ spaces, Scientiae Mathematicae Japonicae 53 (2001), 247-251. Kyriakos Keremedis Department of Mathematics University of the Aegean Karlovassi, Samos 83200, Greece E-mail:
[email protected] 13
