ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Seria I : PRACE MATEMATYCZNE X I I I (1969) ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I : COMMEN TATI ONES MATHEMATICAE X I I I (1969)
Je r z y Cz a j s n e r (Poznań)
Some equivalence relations in relational systems
§ 1. The relation of undistinguishability. Let (A,Ri)t<-T be a rela
tional system of an arbitrary order, each relation R t being of finite degree vt (see Tarski [9]). Thus, Rt is a subset of the Cartesian product A x ... XA (vt times). The word “element” will refer to elements of A, which will he denoted by x, y, z, a, b, c, ... (possibly with indices). Greek letters a, /3, £, 7], ... will stand for equivalence relations on 1 ; a a b will mean that (a , b)ea. All definitions and theorems will concern this fixed system
, Rt)uT •
De f i n i t io n 1. Elements a and b are undistinguishable (x) (in symbol:
a ^ .b) iff for any t in T and for any sequences <%>/^, the con
dition
Д [(oct = а л yi = b) v (Xi — b л yt = a) v a?* = y{\
implies
(1) (x11 .. ., Xvt)zR t о {y1, . . . , y4)eR t.
Eoughly speaking, elements are undistinguishable if replacing a by b, or b by a, in any relation Rt, does not affect the validity of this relation. The following proposition, which can be proved by induction, states that it is enough to assume that the elements are exchanged at one place only.
Pr o p o s it io n 1. Elements a and b are undistinguishable iff for any t in T, for any finite sequences (Xj}j<vt and ( У ) } ^ , and for any i (i < vt), the conditions xi = u, yi = 6, x^ = у к for ~k Ф i imply (1).
§ 2. r-congruences.
De f i n i t io n 2. An r-congruence (2) is an equivalence relation £ on the set A such that for any t in T and any sequences
(1) This relation (in a somewhat different formulation) was used by Scott and Suppes [8], p. 118.
(2) If Rt are functions, then this definition does not coincide with the standard definition of a congruence; cf. Grzegorczyk [3], p. 53.
Roczniki PTM — P ra ce M atem atyczne X III 6.
82 J. C z a js n e r
the conditions xt £yi (i = 1, ..., vt) imply (1). The set of all r-congruences will be denoted by Kng.
If the given system is an algebra, i.e., if each B t is an (vt—l)-ary operation on A, then each r-congruence is a congruence in the usual sense. The converse statement is not true; if the values of any of the operations fill up the set A, then there is only one r-congruence in the system, the identity relation = on A. Thus, the notion of an r-congruence may be useful only in certain relational systems. In order that there exist non-trivial r-congruences it is necessary and sufficient that ^ be different from = .
Theorem 1. The set Kng is a complete lattice with respect to inclu
sion <=. The zero of Kng is the identity relation = and the unit of Kng is the undistingmshability relation 4
Proof. Let {£s}s<r,s be a non-empty set of r-congruences. The infi- mum (3) of this set is the intersection £ of those relations; in other words, x£y iff x£sy for each s in S. The supremum rj of this set is the smallest equivalence relation generated by the union of all relations £s, s e S (i.e., xrjy iff there exist finite sequences x0 = x , xx, ..., xn_ i, xn = у and sx, . .., sn in S such that xi_l £SiXi for i = 1 , ..., n).
It is clear that ^ is an r-eongruence and contains each r-congruence;
it is also obvious that = is the smallest element in Kng.
In the sequel a and fi will denote two fixed r-congruences such that a c 0 ; [ a , /5] will denote the corresponding closed interval, i.e.,
[a, f i] = {£: £ eKng Д a c £ a /5}.
In particular, if a is the identity and /? is the undistinguishability relation, then [a, /3] = Kng.
Theorem 2. [a, /9] is a complementable lattice.
Proof. Let £ be an r-congruence in [a, /?]. We shall construct a (relative) complement of £, i.e., an r-congruence £' such that a is the infimum and /3 is the supremum of {£, £'}. By axiom of choice, there is a set В containing exactly one element of each equivalence class of £.
Let a£'b mean that (i) a(3b, and (ii) either aab or there exist a', b' in В such that aaa' and bob'. Let us note that if a',b 'eB and a'£b', then a' — b' and a a b ; hence a' £'b' is equivalent to a'fib' (4).
It is clear that £' belongs to [a, fi]. We shall show that the infimum y of £ and £' is just a. Since both £ and £' are in [a, fi], у is also in [a, fi];
thus, we have to show that ayb implies aab. Suppose that aab does not hold. Combining ayb with (ii) we infer that a£b and there exist a', b' in В
(3) Analogous constructions of supremum and infimum of algebraic congruences can be found in [1], p. 23.
(4) The construction of £' is similar to a construction of Ore [6].
E qu ivalen ce relations 83
such that aaa' and bob'. Consequently, a£a', b£b', and a£&. Thns, a'£b' and a' = b’ (since a ',b ’ ^B) and aab, a contradiction.
Now, let <5 be the supremum of £ and £'. We claim that 6 = (3. The inclusion 6 c /5 is obvious. Suppose that a(3b. Let x and у be elements of В such that xtja and y£b. Then x(3a and у(3a, and hence x(3y. Therefore xgy. We have shown that aijx, xgy, and у £6 ; hence ad(3.
§ 3. Boolean r-congruences. In general, an r-congruence may have several complements in [a, (3]. It is well known that if Ut is a complemen- table and distributive lattice, then the complements are unique, i.e., is a Boolean algebra (cf. e.g., Hermes [4], p. 49). This suggests the follow
ing definition: ♦
De f i n i t io n 3 . An r-congruence £ is Boolean in [a, (3] iff £e[a,^]
and for every a,
[Д(ж£а) => (xaa)] v [Д (ура) => (?/£«)].
X у
In general, this notion depends on the given interval [a, /?].
Let E (a , a), E {a , (3), E {a , £) denote the equivalence class of the element a with respect to the relation a, /?, £, respectively, and let £ be in [a, /?]. I t is clear that £ is Boolean in [a, /?] iff for every element a either E (a , £) = E (a , a) or E (a , £) = E (a , (3).
Pr o p o s it io n 2. Let a c a ' c £ c ( ? ' c |5 be r-congruences. I f £ is Boolean in [a', /?'] and a', j3' are Boolean in [a, /?], then £ is Boolean in
[ a , 0 ] .
We omit the proof.
Ex a m p l e 1. Let A = {1, 2, 3 , '4, 5, 6} and let {Bt}tiT consist of one relation В = 4 x 4 . Then ^ is also A x A. Let a, (3, £,y be the equi
valence relations determined by the following decompositions of A, respectively:
a: A = {1, 2} ^ {3} w {4} ^ {5} w {6}, (3: A = {1, 2, 3 , 4 } ^ {5,6},
f: 4 = {1,2, 3 , 4 } w { 5 } ^ { 6 } , V- A = { 1 , 2 , 3 } ^ { 4 } ^ { 5 , 6}.
It is clear that a, (3, £, у are r-congruences, а c £ <= [3, a а у <z j3, £ is Boolean in [a, /?], but у is not (though у is Boolean in [a, y]). The relation £' determined by the decomposition
A = {1, 2} w {3} w {4} w {5, 6}
84 J . C z a js n e r
is the unique complement of £ in [a, /?], and the relations щ and r{2 deter
mined by the decompositions
A = { l , 2 } w { 3 , 4 } w { 5 } w { 6 } , rj'2: A = {1, 2, 4} w {3} w {5} w {6}, are two different complements of rj in [«, /3].
Theorem 3. An r-congruence £ is Boolean in [a, /3] i f f it has a unique complement in [a, /3].
Proof. Let rjx, rj2 be complements of a Boolean r-congruence £ in [a, /3], i.e., inf(£, щ) = a and sup(£, щ) = /3 for i = 1, 2. We claim that yx c: r\2. Let « ^ 6 ; hence a(3b and there exists a finite sequence a = x x, x2, . . . , x n = b such that for each i either a?i£#i+i or Xiy2Xi+ l.
Consequently, Xi(3xi+1. Suppose that for a certain к the relation хку2хк + 1 does not hold. Then xk lxk+1) in virtue of a c rj2, xk axk + 1 does not hold either. Since £ is Boolean in [a, /3], for every у the condition y/3xk + 1 im
plies у lx k+l. Applying this for у = a and у = Ъ, we infer that a lxk + 1
and blxk+1\ hence alb. On the other hand, ayxb and inf (£, yx) — a; there
fore aab and arj2b. If хкд2хк + 1 holds for each k, then arj2b as well. We have thus shown that rjx <= r\2. The proof of the inclusion у2 <= у1 is ana
logous. Hence yx = у2, i.e., £ has a unique complement in [a, /3].
We now suppose that £e[a, /3] and £ is not Boolean in [a, /?]. Thus there exists an a such that
(2) \ JxeE (a, £) ^ E (a , a) and y e E (a , (3) ^ Е (а , £).
x,y
We shall now construct two complements r\x and цг using the method shown in the proof of Theorem 2. By (2), there exists a set B x containing one element of each equivalence class of £, such that x e B x and y e B x, and there exists a set B2 containing one element of each equivalence class of £, such that a e B2 and y e B 2. Let щ (i = 1, 2) be defined by the requirement Ьгцс iff (i) &/3c and (ii) either bac or there exist b', & in Bi such that bab' and cac'. By (2), xpa and y($a\ hence x($y and хуху. Suppose, if possible, that xr\2y. Since x ly does not hold, xay does not hold either.
Therefore there exist x', y' in B2 such that xax' and у ay'. By (2) again, x'la. The set B2 contains one element of E (a , £) and a e B 2-, consequently, x' = a. Thus, xaa\ this contradicts (2). We have shown that the comple
ments rjx and rj2 are different.
Proposition 3. The set of Boolean congruences in [a, /3] is a complete sublattice of [a, /3]; in other words, i f { l s}seS is a fam ily of Boolean r-con- gruences, then sup£s and inf£s are also Boolean. Moreover, i f £ is Boolean in [a, /3] and £' is the unique complement of £ in [a, /3], then £' is also Boolean.
E qu iv alen ce relations 85
Proof. Let aeA . Suppose that for each s the set E ( a , |3) is either E (a , a) or E (a , (3). Then the sets
and
E (a , inf |s) = П E (a , |a)
SeS SeS
E (a , sup Is) = U
SeS SeS
also have this property. Moreover, if | is a Boolean r-congruence, then the equivalence relation |' corresponding to the decomposition A
= U E (a , I'), where A E (a , I')
CteA
E (a , a) if E (a , |) = E (a , (3), E ( a, /3) if E ( a , |) = E (a , a)
is also a Boolean r-congruence and is the unique complement of | in
§ 4 . Decomposition of [a, $] into ideals. Let AI (3 be the quotient set of A with respect to /3 (i.e., the set of all equivalence classes of /9).
If | e [ a , j 8 ] and S€A[{3, let
x%sy iff xay v [x, у es a xtjy].
It can easily be verified that |se[a, [3]. Moreover, let Is = {y*[a, 0]: У c &}•
I t is clear that |el3 iff I = |e.
Pr o p o s it io n 4. iw r-congruence | is Boolean in [a, /3] iff for every s in A 1(3 the congruence |s is either the zero or the unit of the corresponding lattice Is.
Pr o p o s it io n 5. Let s , reA](3 and г Ф s. Т/ш г I s ^ I r == •
BTow,if |€[a,/?],let/(|) product
= (ls)se^ * Thus,/(I) belongs to the cardinal p = p
SeAjfi
(the Cartesian product, ordered pointwise). Obviously, if |,?/e[a,/3]
and I c ^ then |3 c: and hence /(I) ^ f(*l)-
Proposition 6. The map f : {a, /3] - » P is a lattice isomorphism. In particular, i f z = (C(s))S€^/(3 is awy element of P, then z = /(|), where | is defined as follows: x£y if f there exists an s in Aj(3 such that x£(s)y.
We omit the proofs.
§ 5. Symmetries. By a permutation we mean a one-to-one transfor
mation from M onto M. If a?!, ..., then will denote the
86 J . C z a js n e r
permutation that sends xx to x2, x2 to x3, ..., xn_ x to xn, and xn to xx, leaving other points fixed. The image of x under a permutation <p will be denoted by xcp and the set of all permutations by P.
By a symmetry we mean a permutation 90 such that for every t in T the condition (x1} ..., xV() eRt is equivalent to (xx<p, . .., xvpp) elłt. The symmetries form a group of permutations (cf. Mostowski [5], p. 371), which will be denoted by Gs .
If £ is a subgroup of Gs, then the set p K of all subgroups of К is a complete lattice with respect to inclusion (cf. Hermes [4], p. 34).
The purpose of this section is to establish an embedding from Kng into pGs.
Pr o p o s it io n 7. Let <peGs. Then x and у are undistinguishable if f xq?
and yep are undistinguishable.
We omit the proof.
De f i n i t io n 4. If £eKng, then the set g (i) = {< peP : Д хЦхер)},
XeA
is conjugate with £. The set g (^ ) will be denoted by IV; this set can also be defined in terms of r-congruences, without referring to
Pr o p o s it io n 8. IV is a normal subgroup of Gs. I f leK n g , then g(l) is a subgroup o f Ж.
Pr o p o s it io n 9. The correspondence I -> g (l) is one-to-one and order
preserving, i.e., £ c: r\ if f </(£) c gig).
We omit the proofs.
Th e o r e m 4. Let leK n g . Then g (l) is a normal subgroup of Ж i f f £ is Boolean in Kng.
Proof. Suppose that £ is not Boolean in Kng. Thus, there is an a in A such that E (a , = ) Ф E (a , £) 7^ E (a , ^ ), i.e., there are x, у such that x Ф a, xtja, у ^ a, but у la does not hold; hence x ly does not hold either. Define ер = Tx>a, гр = Тх<Уьа. Then epeg(l), греЖ, and w W = Tx>y.
Therefore 1р~1ргр^д(1) and g (l) is not a normal subgroup of Ж.
How, let £ be Boolean in Kng. Let <peg{l) and y«JV. We have to show that for every ж in 4 the relation x l i x ^ ' e p i p ) holds. By Proposi
tion 8, x ^ x i p- 1 and xyr'ep ^ {xy~l (p)ip) moreover, {xip~l)l{x y T l(p). Let us distinguish two cases.
a) If xip~l = then x — xxp~lep\p.
b) If хуГ1 Фху)~г(р, then хгр~ 1 €E{xip~l(p, l)'s-E (xf~1<p, = ).
Since £ is Boolean in Kng, this implies E(xip~1<p, £) = Е(хгр~1(р, ^ ) .
E qu ivalen ce relations 87
We know that xeE{xip~lep, ^ ) ; hence х^{хгр~1ср). Similarly, in virtue of xip~lepip ^. хгр~1(р, we get (хгр-1 ep'ip) ^{хцГ1 cp). Consequently, x£(xip~lpip).
This concludes the proof.
§ 6 . Closed groups of permutations. Let X be a set. An operation M M which assigns to each subset M of X a subset M of X will be called a quasi-closure if the following conditions are satisfied: (i) I c I ,
(ii) M — M, (iii) Mx c- M 2 implies M 1 a M2. (Mx w M2 — M x w J f 2 is not postulated). A set will be called closed if M = M.
De f i n i t io n 5. If К is a subset of P, let К = {среР: Д V xcp = xip}.
X t A y>eK
The following proposition can easily be proved:
Pr o p o s it io n 10. The operation К -* К is a quasi-closure. Moreover, i f К is a group o f permutations, then К is also a group o f permutations.
Pr o p o s it io n 11. The group N is closed.
Ex a m p l e 2. Let A — {<a, Ъ, c} and let {BtjuT consist of the single relation P — A x A. Denote p0 — identity, yx = Tap , <p2 — <Рз
= Ta>C)b, щ = Tb>c, (p5 = Ta>c. Then {p0,(p4} is a closed group while Vo, <Pz, <Pz) is a non-closed group (its closure is {%, ..., <p6}). Furthermore, the sets {99,,} and {(ряуфз} are closed and their union is not.
Ex a m p l e 3. Let A = Z (the set of integers). Define f : Z ->Z as f(n ) = \n if n is even and f(n ) = %{n—1) if n is odd. Let В be the set of all pairs (n, m) in Z x Z such that f(n ) < f(m ). It is clear that the undis- tinguishability relation ^ consists of all pairs (n, m) such that either n is even and m = n -\-1 or else m is even and n = m +1. The permutation 99 defined as 9o(n) = n-\- 2 belongs to Gs.
How, let yj = P8>10oP 9)11 (thus, ip interchanges 8 with 10 and 9 with 11). For each n the number ip(n) is one of the three numbers n, 9)(n),
<P~1(n)-, consequently, ip belongs to the closure of the group { e ,9? ,9?-1}, where e: Z -> Z is the identity. Hence, ipeGs and yet ipiGs (indeed, (8, 10)eP whereas (y>(&), ip(1 0))4B).
Thus, G need not be closed and N may be different from Gs (we recall that N <=. Gs). The operation К К acts from pN to pN. Therefore the closed subgroups of N form a complete lattice (cf. Hermes [4], p. 34, Birkhoff [1], p. 60, Ward [10]).
The symbol p K will denote the set of closed subgroups of K .
Th e o r e m 5. The correspondence g: Kng -> pN is a lattice isomorphism.
Proof. In virtue of Propositions 8 and 9, we have only to show that each group g(tj) is closed and, conversely, each closed group in N is conjugate to an r-congruence (cf. Basiowa and Sikorski [7], p. 37).
88 J . C z a js n e r
Let £ be an r-eongruenee. We claim that g(£) cz g(£). Let cpeg(£).
We are to show that x£(xcp) for any x. By assumption, there exists a ip in </(£) such that xip = x<p. Since ipeg(i), x£(xip). Thus, 9oeg(£).
Kow, let GepN. We define £ by the requirement that x£y iff there exists a <p in G such that xcp — y. I t will be shown that £ is an r-congruence.
Since the identity belongs to G, x£x. If xijy, then xcp = у for some 9? in G\
hence x = ycp~1 and y£x. If x£y and y£z, then xcp = у and yip — z for some <p, ip in G ; hence xcpip — z and xt-z. Thus, £ is an equivalence rela
tion. Let xlf . .. , xV(, 2/1, y4 be elements such that Xi£iji, i.e., Xi<pi = yi for some cpi in G (i = 1, . .. , vf). Since G cz N, <px, ...,cp4 are elements of N. Consequently, xx ^ Xicpi and hence (1) holds.
I t remains to be shown that G = </(£). Let 9>eG. Since G is closed, for every x there is a ip in G such that xcp = xip; hence x£(x<p), i.e., 99 e#(£). Thus, G cz </(£). Arguing backwards we get #(£) c: G. This concludes the proof.
Combining Theorems 4 and 5 with Proposition 3 we get the follow
ing corollary: the set of all closed normal subgroups of N is a complete Boolean algebra (in general, the set of normal subgroups of a group is a complete modular lattice, not necessarily distributive, cf. Birkhoff, MacLane [2], p. 367).
§ 7. Characterization of r-congniences in terms of symmetries. The following proposition characterizes r-congruences in terms of symmetries and the undistinguishability relation
Pr o p o s i t i o n 12. A binary relation £ on A is an r-congruence if f there exists a closed subgroup G of N such that
x£y о [V xcp = у].
<ptQ
Pr o p o s i t i o n 13. Elements x and у are undistinguishable i f f у = xcp for some <p in N.
We omit the proofs.
All the remaining propositions will be shown under the following hypothesis:
(H) At most one equivalence class in A /^. is a singleton.
Pr o p o s i t i o n 14. Assume (H). Then pN = pGs.
Proof. I t is enough to show that pGs pN. Suppose that GepGs^pN.
Then G ip N and there is a cp in G^N. Thus there is an x such that x ^ x ę does not hold. By (H), there is an x' such that x' Ф x and x' ^ x or else x' Ф xcp and x' ^ xcp. Suppose that the former case occurs, i.e., x' Ф x and x' ^ x (in the latter case the argument is similar). Denote ip = T x>X(p.
I t is clear that ipeG; hence ip is a symmetry. Combining x ^ x ' with Proposition 7 we infer that xip ^ x' ip. Consequently, xcp ^ x' ^ x, a con
tradiction.
E q u iv alen ce relations 8»
Co r o l l a r y. N is the greatest dosed subgroup of Gs .
Combining Theorem 5 with Propositions 12, 13, and 11 we get the following propositions:
Pr o p o s it io n 15. Assume (H). The Kng is lattice-isomorphic to pGs.
Moreover, the set o f closed subgroups o f Gs which are normal subgroups o f the greatest closed subgroup of Gs is a complete Boolean algebra.
Pr o p o s it io n 16. Assume (H). Then a binary relation | is an r-con- gruence if f there is a closed subgroup G of Gs such that
x ly о [V Щ = у].
фев
Pr o p o s it io n 17. Assume (H). Then x ^ у i f f there exists a у in a closed subgroup G of Gs such that xcp = y.
Ex a m p l e 4. Let A = {1, 3, 4} and let xRy mean that either x — у or \x—y\ > 1. Then ^ is the identity = (there are no other r-congruences) and (H) is not fulfilled. Let e be the identity permutation and у = Тгл.
It can easily be verified that Gs — {e, ip}, N = {<?}, and the only sub
groups of Gs are trivial ones (they are closed). Thus, pN Ф pGs (cf. Pro
position 14) and N ф U {&: GepGs}. Moreover, Kng is not isomorphic to pGs (cf. Proposition 15). The equivalence relation
x £ y о V 0C(p — у
<рев3
(which identificies 3 with 4) is not an r-congruence (cf. Proposition 16). Finally, 4 is the image of 3 under a member of the closed subgroup {e, yj] but these numbers are not undistinguishable (cf. Proposition 17).
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[5] A. M o stow sk i, Zarys teorii Galois (an appendix to the book by W. Sierpiński, Zasady algebry wyższej).
[6] O. Ore, Theory o f equivalence relations, Duke Math. 9 (1942), pp. 573-627.
[7] H. R a sio w a and R. S ik o rsk i, The mathematics o f metamathematics, Warszawa, 1963.
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UNIWERSYTET IM. A. MICKIEWICZA, KATEDRA LOGIKI
