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WKL0 and Stone's separation theorem for convex sets
ANNALS OF PURE AND APPLIED LOGIC ELSBVIIZR
Annals of Pure and Applied Logic 77 (1996) 245-249
WKLo and Stone’s separation theorem for convex sets Kostas Department
of Physics,
Hatzikiriakou*
University of Crete, Iraklion 71409, Greece
Received 25 November 1994; communicated by A. Nerode
Abstract The Stone’s Separation Theorem (for countable convex sets of countable vector spaces over countable ordered fields) is equivalent to Weak Kiinig’s Lemma (over RCA,,).
1. Introduction
This is a contribution to the program of “reverse mathematics”. We assume that the reader is familiar with this program as well as with RCA0 and WKLo, the two subsystems of second order arithmetic we are working with here. (If not, a good place to start is [2].) Our main result is that, over RCAo, Weak K&rig’s Lemma is equivalent to Stone’s Separation Theorem (for countable convex sets in countable vector spaces over countable ordered fields) whose statement is as follows: “Let V be a countable vector space over a countable ordered field F. Let C, D be disjoint countable convex subsets of V. Then, there exists a convex set H c V such that V/H is convex, C c H and D c V/H".
2. Definitions In RCA0 we give the following definitions. Let V be a countable vector space over a countable ordered field F.A subset C of V is called convex if it contains px + ry, for all p, z E F,p,z 2 0 and p + r = 1, whenever it contains x, y. A subset K of V is a cone with vertex 0 if K contains every point px, p > 0, p E F whenever it contains x. (We write K(0) to denote the cone with vertex 0.) Obviously, K *(0) = -K(O) is also a cone, the cone diametrically opposite to K(0). K(0)is a convex cone if it contains x + y whenever it contains x, y. A cone K(0)is a hypercone if K(O)uK*(O)= V/(0}and K(O), K*(O)are disjoint. (We follow the terminology of [4].)
* E-mail: [email protected]. 0168~0072/96/$15.00 0 1996-Elsevier Science B.V. All rights reserved SSDI 0168-0072(95)00038-O
246
Kostas Hatzikiriakou /Annals of Pure and Applied Logic 77 (1996) 245-249
3. The Separation Theorem is provable in WKL,
The proof that the Separation Theorem is provable in WKLo is inspired by the proof of the theorem given by Hammer (see [4]). 3.1. Lemma
(WKL,,). Let V be a countable vector space over a countable orderedjeld
F. Let C be a C,” convex subset of V such that 0 $ C. Then, there exists a hypercone K(0) of V such that C E K(0).
Let p0,p1,p2, . . . be an enumeration of F+ = {p: p E F be an and p > 01. Let ao,al ,a2, . . . be an enumeration of V/(O). Let b0,bl,b2,... enumeration of C. We define (by primitive recursion) a tree T of finite O-l sequences. Also, for each 0 E T we define a finite subset K, of V such that if r~is an initial segment of r, then K, E K,. At stage s, T, = (0 E T: lb(o) = s> is defined. For s = 0, To = (8) and Kfi = 0. Assume that T,_ 1 has been defined. The definition of T, splits into five cases. Let s = 5m + r, 0 < r < 5. Every natural number m encodes a triple of natural numbers (i,j, k). Case 1: r = 0. For each u E T,_ 1, put a0 in T, and let KoO = K,u {b,}. Case 2: r = 1. For each 0 E T,_ 1, put a0 in T, and let Kg0 = K,, unless m = (i, j, k) and ai, aj E K,, in which case let KaO = K,u {ai + aj}. Case 3: r = 2. For each r~E T,_ 1, put 00 in T, and let KbO = K,, unless m = (i, j, k) and aj E K,, in which case let K,,O = K, u { piaj}. Case 4: r = 3. For each cr E T,_ 1, put a0 in T, and let KaO = K,, unless m = (i, j, k) and ai + aj = 0, in which case put cr0, al in T, and let KoO = K,u {ai}, K,, = Proof. We argue in WKL,.
KG u {aj>. Case 5: r = 4. For each 0 E T,_ 1, put a0 in T, and let Kg0 = K,, unless 0 E K,, in
which case do not put 00 nor al in T, and do not define Kg0 nor KG1. Claim (RCAo). T is infinite. Proof. Consider the n,” formula e(s) = 3a E T,(O$ C,), where C, is the cone with
vertex 0 generated by the convex closure of C u K, in V. (Note that C, and the convex closure of CuK, are definable by C,” formulas; we do not assume that they exist as sets. Also, note that C, is a convex cone.) Now, $(O) holds, since 0 $ C. Assume that $(s) holds. In cases 1,2,3,5, C, = Coo, thus there is 0’ E Ts such that 0 # C,,.. In case 4, assume, towards a contradiction, that 0 E Co0 and 0 E C,r. Then, /zici + Ila) = p’(C, ~ t pidi + ,u( -a)), where p, p’, 1, p E F+, Ai,pi E F+ U 0 = P(C,,, (O}, xi up /li + I = 1, xi of pi + p = 1, a E V/(O), -a E V/(O), and ci, di are elements of the convex closure of C UK,. Note that xi 4 p li # 0 and Ci ~ f ,Ui# 0. Then, = p(CiS,lidi)~ C,. Hence, p@‘paE C, and p’pa = ~‘(1, GP p
Kostas Hatzikiriakou/Annals of Pure and Applied Logic 77 (1996) 245-249
247
So, by Weak KSnig’s Lemma, T has an infinite path g. Let X = u, K,, where d ranges over all finite segments of g. Note that X is defined by a CF formula, but for a # 0, a E X if and only if -a 6 X. So X exists by d 1”comprehension (included in RCA,) and, clearly, is the required hypercone IT(O). q 3.2. Theorem (WKLO). Let V be a countable vector space over a countable ordered field F. Let C, D be disjoint countable convex subsets of V. Then, there exists a convex set A c V such that V/A is convex, C c A and D E V/A.
Proof. We argue in WKLe. Let ao,al, a2, . . . be an enumeration of V. Let Po,Pl,Pz,**~ be an enumeration of F +. Let co, cl, c2, . . . be an enumeration of C. Let do, d,, d2, . . . be an enumeration of D. Let C - D = M. Then, M is a EF convex subset of V such that 04 M. Hence, by the previous theorem, there exists hypercone K(O) of V such that M G K(0). Let H = nxeD (x + K(0)). Note that Note V/H = UXED V/(x + K(0)) = Uxc,, (x + K*(O)u(O)) = 1) -I-(~*(O)u~O~). that H is a IZ: convex subset of V and H* = V/H is a Cy convex subset of V such that C c: H and D E V/H. We consider the following tree T: CTE T iff for all i, j, k, m, f < lb(o): (1) ai = dj * o(i) = 0. (2) ai = Cj =j o(i) = 1. (3) o(i) = 0 and g(j) = 0 and pkai + play = a, and ,& + pt = 1 *a(m) = 0. (4) o(i) = 1 and c(j) = 1 and pkai + ptaj = a, and Pk + pr = 1 *e(m) = 1. Claim. T is infinite. Proof. Given s E N, let X = (i < s: ai E H*). X exists by bounded Cp comprehension (included in RCA,; see Lemma 1.6 in [2]), Define r~E 2” by a(i) = 0 if i E X and a(i) = 1 if i$X. Then, CFexists and belongs to T (given the properties of H, H *). Hence, T is infinite. f-J By Weak Kiinig’s Lemma, letf be an infinite path through T. Let B = (ai:f(i) = 01. Then, B exists by d: comprehension. Let A = V/B. Clearly, D s B, C & A, and A,B convex. 0
4. The main theorem 4.1. Theorem (RCAo). The foElowing are equivalent. (1) Weak K&zig’s Lemma. (2) Let V be a countable vector space over a countable orderedfield F. Let C, D be disjoint countable convex subsets of V. Then, there exists a convex set H c V such that V/H is convex, C G H and D E V/H.
K&as Ha~i~r~ak~u~A~aIs@Pure and AppliedLogic 77 (1996) 245-249
248
Proof. (1) 3 (2). Because of Theorem 3.2, (2) * (1). We work in RCA0 and we assume (2). Given functions f, g : I%+ N which are l-l and have disjoint ranges, we want to find a set X such that Vn(f(n) E X and g(n)$X). (This is equivalent, over RCAo, to Weak K&rig’s Lemma, see Lemma 3.2 in [2].) Let V a vector space over Q with basis co,er,.... Let C be the convex closure of (u,: k E N] = (uk:k E N) and D the convex closure of (uk: k E N} = (uk : k E N). The elements uk are defined by primitive recursion as follows: u. = efto) and for k > 0, uk = (1 - c(k)- leg(O) - ak(1 - ak)-‘e&k) SO that O(~~&P,O<~~<1andfora~~~~kifu&(~o,~~,...,~k-~)then~~(~O,~~,...,Z(k).TO see that such ak exists we prove the following claim. Claim. For any v $ ( uo, ul, . . , , uk) there is at most one a” such that a” E Q$,0 < a” < 1 - a,)-leg(o) - a,(1 - a,)-‘egckI),for k > 0.
and VE (~~,u~,.~.,u~_~,(l
Proof. Let 21= Ci
xi
ck
liUi+
/A%
+
with &,J.,/Q,~,~~,E,E Ci,,pi+p=l.Then,
(JO-
p0)ef(0)
- xoIick
A((1
-
aJ'e,(0)
-
&(l
-
a,f-'e,d
AU
-
C
-
f%U
-
&de
‘e,(0)
Q, 0
legtkdp
< 1, 0 < 6, < 1, Ai,gi 2 0, 2,~ > 0, Ci
+ Ml - aJ1 - P(l - G-l + CO
(ni - flj) ai(l - &)-leg(j) + (p&(1 -
k
(Ai
-
PiIt
-
ai)-l)egcO~
&)-’ - ila”(l - a,)-‘)eg(k)= 0.
Hence, Lo = ,uo, li = pi, 0 < i < k, pcts,(l -e,)-’
A(1 -01,)-r Then, ,I(1 - a,)-1
= Aa,(l - a,)-‘,
- ~((1 - E,)-~ + CO
SO, a, = 6,.
- 4-l
= 0.
Cl
Thus, in the definition of uk, k > 0, we need only choose a, E Q, 0 < ~1,< 1 outside the finite set (a,: u E V and ZI< k). Hence, by d p comprehension, C exists, since u E C iffv85(z40,z41,..., u,). The elements ak are defined by primitive recursion as follows: u. = ego) and for so that /& o Q, 0 < /& < 1 and for all k > 0, uk = (1 - #de’e/(O) pk(l - Pk)-le.f(kJ u < k if u~(uo,ul,..., r&l) then v~(v*,u1, . ..) ok). Then, similarly, D exists and clearly CnD = 8. So, using (2) let H be a convex subset of V whose complement is convex and C E H and D c V/H. Claim. Qn ef(,,) E H and ‘v’n e,(,) E V/H.
Kmtas ~a~i~iria~~~~nnals
ofPure
and Applied Logic 77 (1996) 245-249
249
Proof. ef(oj E H. Let n > 0 and eftnj E V/N. Then, e/(oj = Pnefc,,, + (1 - B,Jvn = Pnescn, + (1 - Pn)(U - Pn)- ’ efco, - P,,(1 - P,,)- ’ efcn,) E V/H. Contradiction. Hence, Vn efcn) E H. Similarly, Vdne,(,, E V/H. IJ Thus X = (m : e, E H > is the required set.
0
Since Weak KGnig’s Lemma does not hold in the ar-model of RCA0 consisting of the recursive sets, the above theorem gives us a well-known result proved independently by Kalantari and Downey; see [l] and [3]. 4.2. Corollary. There is a recursive vector space V (over a recursive ordered$eld)
and two recursive disjoint convex subsets of it that cannot be separated by a recursive convex set, whose complement is also convex.
References [l} R. Downey, Some remarks on a theorem of Iraj Kalantari concerning convexity and recursion theory, Z. Math. Logik Grundlag. Math. 30 (1984) 295-302. [2] I-I. Friedman, S. Simpson and R. Smith, Countable algebra and set existence axioms, Ann. Pure Appl. Logic 25 (1983) 141-181; addendum 27 (1983) 319-320. [S] I. Kalantari, Effective content of a theorem of MN. Stone, in: J.N. Crossley, ed., Aspects of Effective Algebra (U.D.A. Book Co., 1981) 209-219. [4] G. K&he, Topological Vector Spaces I (Springer, New York, 1969).
Annals of Pure and Applied Logic 77 (1996) 245-249
WKLo and Stone’s separation theorem for convex sets Kostas Department
of Physics,
Hatzikiriakou*
University of Crete, Iraklion 71409, Greece
Received 25 November 1994; communicated by A. Nerode
Abstract The Stone’s Separation Theorem (for countable convex sets of countable vector spaces over countable ordered fields) is equivalent to Weak Kiinig’s Lemma (over RCA,,).
1. Introduction
This is a contribution to the program of “reverse mathematics”. We assume that the reader is familiar with this program as well as with RCA0 and WKLo, the two subsystems of second order arithmetic we are working with here. (If not, a good place to start is [2].) Our main result is that, over RCAo, Weak K&rig’s Lemma is equivalent to Stone’s Separation Theorem (for countable convex sets in countable vector spaces over countable ordered fields) whose statement is as follows: “Let V be a countable vector space over a countable ordered field F. Let C, D be disjoint countable convex subsets of V. Then, there exists a convex set H c V such that V/H is convex, C c H and D c V/H".
2. Definitions In RCA0 we give the following definitions. Let V be a countable vector space over a countable ordered field F.A subset C of V is called convex if it contains px + ry, for all p, z E F,p,z 2 0 and p + r = 1, whenever it contains x, y. A subset K of V is a cone with vertex 0 if K contains every point px, p > 0, p E F whenever it contains x. (We write K(0) to denote the cone with vertex 0.) Obviously, K *(0) = -K(O) is also a cone, the cone diametrically opposite to K(0). K(0)is a convex cone if it contains x + y whenever it contains x, y. A cone K(0)is a hypercone if K(O)uK*(O)= V/(0}and K(O), K*(O)are disjoint. (We follow the terminology of [4].)
* E-mail: [email protected]. 0168~0072/96/$15.00 0 1996-Elsevier Science B.V. All rights reserved SSDI 0168-0072(95)00038-O
246
Kostas Hatzikiriakou /Annals of Pure and Applied Logic 77 (1996) 245-249
3. The Separation Theorem is provable in WKL,
The proof that the Separation Theorem is provable in WKLo is inspired by the proof of the theorem given by Hammer (see [4]). 3.1. Lemma
(WKL,,). Let V be a countable vector space over a countable orderedjeld
F. Let C be a C,” convex subset of V such that 0 $ C. Then, there exists a hypercone K(0) of V such that C E K(0).
Let p0,p1,p2, . . . be an enumeration of F+ = {p: p E F be an and p > 01. Let ao,al ,a2, . . . be an enumeration of V/(O). Let b0,bl,b2,... enumeration of C. We define (by primitive recursion) a tree T of finite O-l sequences. Also, for each 0 E T we define a finite subset K, of V such that if r~is an initial segment of r, then K, E K,. At stage s, T, = (0 E T: lb(o) = s> is defined. For s = 0, To = (8) and Kfi = 0. Assume that T,_ 1 has been defined. The definition of T, splits into five cases. Let s = 5m + r, 0 < r < 5. Every natural number m encodes a triple of natural numbers (i,j, k). Case 1: r = 0. For each u E T,_ 1, put a0 in T, and let KoO = K,u {b,}. Case 2: r = 1. For each 0 E T,_ 1, put a0 in T, and let Kg0 = K,, unless m = (i, j, k) and ai, aj E K,, in which case let KaO = K,u {ai + aj}. Case 3: r = 2. For each r~E T,_ 1, put 00 in T, and let KbO = K,, unless m = (i, j, k) and aj E K,, in which case let K,,O = K, u { piaj}. Case 4: r = 3. For each cr E T,_ 1, put a0 in T, and let KaO = K,, unless m = (i, j, k) and ai + aj = 0, in which case put cr0, al in T, and let KoO = K,u {ai}, K,, = Proof. We argue in WKL,.
KG u {aj>. Case 5: r = 4. For each 0 E T,_ 1, put a0 in T, and let Kg0 = K,, unless 0 E K,, in
which case do not put 00 nor al in T, and do not define Kg0 nor KG1. Claim (RCAo). T is infinite. Proof. Consider the n,” formula e(s) = 3a E T,(O$ C,), where C, is the cone with
vertex 0 generated by the convex closure of C u K, in V. (Note that C, and the convex closure of CuK, are definable by C,” formulas; we do not assume that they exist as sets. Also, note that C, is a convex cone.) Now, $(O) holds, since 0 $ C. Assume that $(s) holds. In cases 1,2,3,5, C, = Coo, thus there is 0’ E Ts such that 0 # C,,.. In case 4, assume, towards a contradiction, that 0 E Co0 and 0 E C,r. Then, /zici + Ila) = p’(C, ~ t pidi + ,u( -a)), where p, p’, 1, p E F+, Ai,pi E F+ U 0 = P(C,,, (O}, xi up /li + I = 1, xi of pi + p = 1, a E V/(O), -a E V/(O), and ci, di are elements of the convex closure of C UK,. Note that xi 4 p li # 0 and Ci ~ f ,Ui# 0. Then, = p(CiS,lidi)~ C,. Hence, p@‘paE C, and p’pa = ~‘(1, GP p
Kostas Hatzikiriakou/Annals of Pure and Applied Logic 77 (1996) 245-249
247
So, by Weak KSnig’s Lemma, T has an infinite path g. Let X = u, K,, where d ranges over all finite segments of g. Note that X is defined by a CF formula, but for a # 0, a E X if and only if -a 6 X. So X exists by d 1”comprehension (included in RCA,) and, clearly, is the required hypercone IT(O). q 3.2. Theorem (WKLO). Let V be a countable vector space over a countable ordered field F. Let C, D be disjoint countable convex subsets of V. Then, there exists a convex set A c V such that V/A is convex, C c A and D E V/A.
Proof. We argue in WKLe. Let ao,al, a2, . . . be an enumeration of V. Let Po,Pl,Pz,**~ be an enumeration of F +. Let co, cl, c2, . . . be an enumeration of C. Let do, d,, d2, . . . be an enumeration of D. Let C - D = M. Then, M is a EF convex subset of V such that 04 M. Hence, by the previous theorem, there exists hypercone K(O) of V such that M G K(0). Let H = nxeD (x + K(0)). Note that Note V/H = UXED V/(x + K(0)) = Uxc,, (x + K*(O)u(O)) = 1) -I-(~*(O)u~O~). that H is a IZ: convex subset of V and H* = V/H is a Cy convex subset of V such that C c: H and D E V/H. We consider the following tree T: CTE T iff for all i, j, k, m, f < lb(o): (1) ai = dj * o(i) = 0. (2) ai = Cj =j o(i) = 1. (3) o(i) = 0 and g(j) = 0 and pkai + play = a, and ,& + pt = 1 *a(m) = 0. (4) o(i) = 1 and c(j) = 1 and pkai + ptaj = a, and Pk + pr = 1 *e(m) = 1. Claim. T is infinite. Proof. Given s E N, let X = (i < s: ai E H*). X exists by bounded Cp comprehension (included in RCA,; see Lemma 1.6 in [2]), Define r~E 2” by a(i) = 0 if i E X and a(i) = 1 if i$X. Then, CFexists and belongs to T (given the properties of H, H *). Hence, T is infinite. f-J By Weak Kiinig’s Lemma, letf be an infinite path through T. Let B = (ai:f(i) = 01. Then, B exists by d: comprehension. Let A = V/B. Clearly, D s B, C & A, and A,B convex. 0
4. The main theorem 4.1. Theorem (RCAo). The foElowing are equivalent. (1) Weak K&zig’s Lemma. (2) Let V be a countable vector space over a countable orderedfield F. Let C, D be disjoint countable convex subsets of V. Then, there exists a convex set H c V such that V/H is convex, C G H and D E V/H.
K&as Ha~i~r~ak~u~A~aIs@Pure and AppliedLogic 77 (1996) 245-249
248
Proof. (1) 3 (2). Because of Theorem 3.2, (2) * (1). We work in RCA0 and we assume (2). Given functions f, g : I%+ N which are l-l and have disjoint ranges, we want to find a set X such that Vn(f(n) E X and g(n)$X). (This is equivalent, over RCAo, to Weak K&rig’s Lemma, see Lemma 3.2 in [2].) Let V a vector space over Q with basis co,er,.... Let C be the convex closure of (u,: k E N] = (uk:k E N) and D the convex closure of (uk: k E N} = (uk : k E N). The elements uk are defined by primitive recursion as follows: u. = efto) and for k > 0, uk = (1 - c(k)- leg(O) - ak(1 - ak)-‘e&k) SO that O(~~&P,O<~~<1andfora~~~~kifu&(~o,~~,...,~k-~)then~~(~O,~~,...,Z(k).TO see that such ak exists we prove the following claim. Claim. For any v $ ( uo, ul, . . , , uk) there is at most one a” such that a” E Q$,0 < a” < 1 - a,)-leg(o) - a,(1 - a,)-‘egckI),for k > 0.
and VE (~~,u~,.~.,u~_~,(l
Proof. Let 21= Ci
xi
ck
liUi+
/A%
+
with &,J.,/Q,~,~~,E,E Ci,,pi+p=l.Then,
(JO-
p0)ef(0)
- xoIick
A((1
-
aJ'e,(0)
-
&(l
-
a,f-'e,d
AU
-
C
-
f%U
-
&de
‘e,(0)
Q, 0
legtkdp
< 1, 0 < 6, < 1, Ai,gi 2 0, 2,~ > 0, Ci
+ Ml - aJ1 - P(l - G-l + CO
(ni - flj) ai(l - &)-leg(j) + (p&(1 -
k
(Ai
-
PiIt
-
ai)-l)egcO~
&)-’ - ila”(l - a,)-‘)eg(k)= 0.
Hence, Lo = ,uo, li = pi, 0 < i < k, pcts,(l -e,)-’
A(1 -01,)-r Then, ,I(1 - a,)-1
= Aa,(l - a,)-‘,
- ~((1 - E,)-~ + CO
SO, a, = 6,.
- 4-l
= 0.
Cl
Thus, in the definition of uk, k > 0, we need only choose a, E Q, 0 < ~1,< 1 outside the finite set (a,: u E V and ZI< k). Hence, by d p comprehension, C exists, since u E C iffv85(z40,z41,..., u,). The elements ak are defined by primitive recursion as follows: u. = ego) and for so that /& o Q, 0 < /& < 1 and for all k > 0, uk = (1 - #de’e/(O) pk(l - Pk)-le.f(kJ u < k if u~(uo,ul,..., r&l) then v~(v*,u1, . ..) ok). Then, similarly, D exists and clearly CnD = 8. So, using (2) let H be a convex subset of V whose complement is convex and C E H and D c V/H. Claim. Qn ef(,,) E H and ‘v’n e,(,) E V/H.
Kmtas ~a~i~iria~~~~nnals
ofPure
and Applied Logic 77 (1996) 245-249
249
Proof. ef(oj E H. Let n > 0 and eftnj E V/N. Then, e/(oj = Pnefc,,, + (1 - B,Jvn = Pnescn, + (1 - Pn)(U - Pn)- ’ efco, - P,,(1 - P,,)- ’ efcn,) E V/H. Contradiction. Hence, Vn efcn) E H. Similarly, Vdne,(,, E V/H. IJ Thus X = (m : e, E H > is the required set.
0
Since Weak KGnig’s Lemma does not hold in the ar-model of RCA0 consisting of the recursive sets, the above theorem gives us a well-known result proved independently by Kalantari and Downey; see [l] and [3]. 4.2. Corollary. There is a recursive vector space V (over a recursive ordered$eld)
and two recursive disjoint convex subsets of it that cannot be separated by a recursive convex set, whose complement is also convex.
References [l} R. Downey, Some remarks on a theorem of Iraj Kalantari concerning convexity and recursion theory, Z. Math. Logik Grundlag. Math. 30 (1984) 295-302. [2] I-I. Friedman, S. Simpson and R. Smith, Countable algebra and set existence axioms, Ann. Pure Appl. Logic 25 (1983) 141-181; addendum 27 (1983) 319-320. [S] I. Kalantari, Effective content of a theorem of MN. Stone, in: J.N. Crossley, ed., Aspects of Effective Algebra (U.D.A. Book Co., 1981) 209-219. [4] G. K&he, Topological Vector Spaces I (Springer, New York, 1969).