@ Applied General Topology c© Universidad Politécnica de Valencia Volume 4, No. 2, 2003 pp. 487–507 Some properties of the containing spaces and saturated classes of spaces Stavros Iliadis Dedicated to Professor S. Naimpally on the occasion of his 70th birthday. Abstract. Subjects of this paper are: (a) containing spaces con- structed in [2] for an indexed collection S of subsets, (b) classes con- sisting of ordered pairs (Q,X), where Q is a subset of a space X, which are called classes of subsets, and (c) the notion of universality in such classes. We show that if T is a containing space constructed for an indexed collection S of spaces and for every X ∈ S, QX is a subset of X, then the corresponding containing space T|Q constructed for the indexed collection Q ≡{QX : X ∈ S} of spaces, under a simple condition, can be considered as a specific subset of T. We prove some “commutative” properties of these specific subsets. For classes of subsets we introduce the notion of a (properly) universal element and define the notion of a (complete) saturated class of subsets. Such a class is “saturated” by (properly) universal elements. We prove that the intersection of (complete) saturated classes of subsets is also a (complete) saturated class. We consider the following classes of subsets: (a) IP(Cl), (b) IP(Op), and (c) IP(n.dense) consisting of all pairs (Q,X) such that: (a) Q is a closed subset of X, (b) Q is an open subset of X, and (c) Q is a never dense subset of X, respectively. We prove that the classes IP(Cl) and IP(Op) are complete saturated and the class IP(n.dense) is saturated. Saturated classes of subsets are convenient to use for the construction of new saturated classes by the given ones. 2000 AMS Classification: 54B99, 54C25 Keywords: containing space, universal space, saturated class of spaces, saturated class of subsets. 488 S. Iliadis 1. Introduction. Agreement concerning notations. We denote by τ a fixed infinite cardinal. The set of all finite subsets of τ is denoted by F. By a space we mean a T0-space of weight less than or equal to τ. In this paper we use all notions and notation introduced in [2]. In particular, if an indexed collection of spaces is denoted by the letter “S”, a co-mark of S is denoted by the letter “M”, and an M-admissible family of equivalence relations on S is denoted by the letter “R”, then we always denote by “T” the containing space T(M, R) and by “BT” the standard base for T. If moreover M = {{UXδ : δ ∈ τ} : X ∈ S}, then the elements of BT are denoted by UTδ (H), δ ∈ τ and H ∈ C ♦(R). As in [2] we shall be concerned with classes, sets, collections, and families. A class is not necessarily a set. A collection and a family are sets. Any equivalence relation on a set S (which is considered as a subset of S×S satisfying the well- known conditions) is denoted by the symbol ∼ supplied usually with one or more indices. Any ordinal α is identified with the set of all ordinals less than α. For every set X we denote by P(X) the set of all subsets of X. For every subset Q of a space X we denote by ClX(Q), IntX(Q), and BdX(Q) the closure, the interior, and the boundary of Q in X, respectively. We shall use the symbol “≡” in order to introduce new notations without mention to this fact. This will be done as follows. When we introduce an expression A as a notation of an object (a set, an indexed set, a mapping and so on) writing A ≡ B (or B ≡ A), where B is another new expression, or when we consider a known object with a known expression A as its notation writing A ≡ B (or B ≡ A), where B is a new expression, in both cases we mean that B is considered as another notation of the same object. In the second section we construct some specific subsets of the containing spaces given in [2]. Suppose that for every space X of an indexed collection S of spaces a subset QX of X is given. Then, the indexed collection Q ≡{QX : X ∈ S} is called a restriction of S. Such a restriction can be also treat as an indexed collection of spaces. Any co-mark M of S defines by a natural manner a co-mark M|Q of Q and any M-admissible family R of equivalence relations on S defines an admissible family R|Q of equivalence relations on Q. For “almost all” co-marks M and families R, R|Q is M|Q-admissible. In this case, in parallels with the containing space T, we can also consider the containing space T(M|Q, R|Q) ≡ T|Q for Q corresponding to the co-mark M|Q and the family R|Q. We show that there exists a natural embedding of T|Q into T, which gives us the possibility to consider T|Q as a subset of T. In the third section some “commutative” properties of subsets T|Q are given. A restriction Q ≡{QX : X ∈ S} is called closed (open) if QX is closed (open) in X. Also, the notion of a complete restriction of S is given. We show that Some properties of the containing spaces and saturated classes of spaces 489 closed and open restrictions are complete. The following restrictions are also considered: Cl(Q) ≡{ClX(QX) : X ∈ S}, Int(Q) ≡{IntX(QX) : X ∈ S}, Bd(Q) ≡{BdX(QX) : X ∈ S}, and Co(Q) ≡{X \QX : X ∈ S}. For “almost all” co-marks M and families R of equivalence relations on S the following “commutation” relations are proved: T|Cl(Q) = ClT(T|Q), T|Int(Q) = IntT(T|Q), T|Bd(Q) = BdT(T|Q), and T|Co(Q) = T \ T|Q. The first relation is true for any restriction Q and the others for complete restrictions. Classes consisting of ordered pairs (Q,X), where Q is a subset of a space X, and called classes of subsets, are considered in the fourth section. An element (QT ,T ) of a class IP of subsets is called universal (respectively, properly universal) if for every element (QX,X) of IP there exists a homeomorphism h of X into T such that h(QX) ⊂ QT (respectively, h−1(QT ) = QX). Using the above considered properties of subsets T|Q of T we define the so-called (complete) saturated classes of subsets. This definition is similar to that of classes of spaces (see [2]). As for the classes of spaces the (complete) saturated classes of subsets not only have (properly) universal elements but they are “saturated” by such elements. We prove that the intersection of (complete) saturated classes is also a (complete) saturated class. We also show that the classes of subsets IP(Cl) ≡{(Q,X) : Q is closed in X} and IP(Op) ≡{(Q,X) : Q is open in X} are complete saturated classes and the class IP(n.dense) ≡{(Q,X) : Q is never dense in X} is saturated. For classes IP(Cl) and IP(Op) this follows by the corresponding “commutation” relations. In the fifth section we introduce the notion of a “commutative operator”. The closure, the interior, and the boundary operators are such operators. Using 490 S. Iliadis these operators we construct new (complete) saturated classes of subsets by the given ones. Finally, in the last section we pose some problems. 2. Specific subspaces of the space T(M, R). Definition 2.1. Let {V Xδ : δ ∈ τ} be a mark of a space X. Then, for every subspace Q of X the indexed set {V Qδ ≡ Q∩V X δ : δ ∈ τ} is a mark of the space Q. This mark is called the trace on Q of the mark {V Xδ : δ ∈ τ} of X. Lemma 2.2. Let {V Xδ : δ ∈ τ} be the mark of a marked space X, Q a subspace of X, and {V Qδ : δ ∈ τ} the trace on Q of the mark {V X δ : δ ∈ τ}. Then, dXs (x) = d Q s (x) for every x ∈ Q and s ∈F \{∅}. Therefore, dXs (Q) = dQs (Q). (We note that the mapping dXs is constructed with respect to the mark {V Xδ : δ ∈ τ} of X and the mapping dQs is constructed with respect to the mark {V Q δ : δ ∈ τ} of Q). Proof. Let s ∈ F \ {∅}, x ∈ Q and dXs (x) = f ∈ 2s. We must prove that dQs (x) = f. By the definition of the mapping d X s we have x ∈ X(s,f) = ∩{X(δ,f(δ)) : δ ∈ s}. Therefore, x ∈ Q∩ (∩{X(δ,f(δ)) : δ ∈ s}) = ∩{Q∩X(δ,f(δ)) : δ ∈ s} = ∩{Q(δ,f(δ)) : δ ∈ s} = Q(s,f). This means that dQs (x) = f. � Definition 2.3. Suppose that for every element X of an indexed collection S of spaces a subspace QX of X is given. Then, the indexed collection Q ≡{QX : X ∈ S} is called a restriction of S. The element QX of Q will be also denoted by Q(X). Some properties of the containing spaces and saturated classes of spaces 491 Definition 2.4. Let S be an indexed collection of spaces, Q ≡{QX : X ∈ S} a restriction of S, M a co-mark of S, and let R ≡ {∼s: s ∈ F} be an indexed family of equivalence relations on S. The trace on Q of the co-mark M of S is the co-mark of Q denoted by M|Q and defined as follows: the mark (M|Q)(QX) of an element QX of Q is the trace on QX of the mark M(X) of X. The trace on Q of an equivalence relation ∼ on S is the equivalence relation on Q denoted by ∼|Q and defined as follows: two elements QX and QY of Q are ∼|Q-equivalent if and only if X ∼ Y . The indexed family R|Q ≡{∼s|Q : s ∈F} of equivalence relations on Q is called the trace on Q of the indexed family R. The trace on Q of an element H of C♦(R), denoted by H|Q, is the set of all elements QX of Q for which X ∈ H. It is easy to see that H|Q is an element of C♦(R|Q). Obviously, if H ∈ C(R), then H|Q ∈ C(R|Q). By definition, it follows that if ∼0 and ∼1 are two equivalence relations on S and ∼1 is contained in ∼0, then ∼1|Q is contained in ∼0|Q. Therefore, if R0 and R1 are two indexed families of equivalence relations on S and R1 is a final refinement of R0, then R1|Q is a final refinement of R0|Q. Agreement 1. In what follows in this section it is supposed that an indexed collection of spaces denoted by S, a restriction of S denoted by Q ≡{QX : X ∈ S}, a co-mark of S denoted by M ≡{{UXδ : δ ∈ τ} : X ∈ S}, and an M-admissible family of equivalence relations on S denoted by R ≡{∼s: s ∈F} are fixed. We note that, in general, the trace on Q of the M-standard family of equiv- alence relations on S is not the M|Q-standard family of equivalence relations on Q. This justifies the following definition. Definition 2.5. The M-admissible family R of equivalence relations on S is said to be (M, Q)-admissible if R|Q is an M|Q-admissible family of equivalence relations on Q. Lemma 2.6. The family R is (M, Q)-admissible if and only if for every el- ement s of F \ {∅} there exists an element t of F \ {∅} such that relation X ∼t Y implies relation dXs (QX) = dYs (QY ) for every X,Y ∈ S. 492 S. Iliadis Proof. Suppose that the condition of the lemma is satisfied. We must prove that the family R is (M, Q)-admissible. Since R is M-admissible it suffices to prove that the family R|Q = {∼s|Q : s ∈ F} is a final refinement of the M|Q-standard family of equivalence relations on Q. Denote the last family by RQ ≡{∼sQ: s ∈F}. Let s and t be elements of F \{∅} satisfying the condition of the lemma. We need to prove that the trace on Q of the equivalence relation ∼t on S, that is, the equivalence relation ∼t|Q on Q is contained in the equivalence relation ∼sQ on Q. Let Q X,QY ∈ Q and QX ∼t|QQY . This means that X ∼t Y . By the condition of the lemma, the last relation implies that dXs (Q X) = dYs (Q Y ). By Lemma 2.2, dQ X s (Q X) = dQ Y s (Q Y ) and by Lemma 1.4 of [2], QX ∼sQ Q Y . Thus, the equivalence relation ∼t |Q is contained in the equivalence relation ∼sQ. Conversely, suppose that the family R is (M, Q)-admissible. Let s ∈F\{∅}. Since the family R|Q is M|Q-admissible there exists an element t of F \{∅} such that the equivalence relation ∼t|Q on Q is contained in the equivalence relation ∼sQ on Q. Let X,Y ∈ S and X ∼t Y , that is, QX ∼t |QQY . Then, QX ∼sQ Q Y . By Lemma 1.4 of [2], dQ X s (Q X) = dQ Y s (Q Y ) and by Lemma 2.2, dXs (Q X) = dYs (Q Y ). � Remark 2.7. Lemma 2.6 implies the existence of (M, Q)-admissible families of equivalence relations on S. For example, such a family is the admissible family R0 ≡ {∼s0: s ∈ F} for which X ∼s0 Y if and only if X ∼sM Y and dXs (Q X) = dYs (Q Y ). (This family is admissible because the set P(2sX) = P(2 s Y ) is finite). Notation. Suppose that R is an (M, Q)-admissible family of equivalence re- lations on S. Then, besides of the space T we can also consider the containing space T(M|Q, R|Q) for the indexed collection Q corresponding to the co-mark M|Q = {{U QX δ = Q∩U X δ : δ ∈ τ} : Q X ∈ Q} and the M|Q-admissible family R|Q of equivalence relations on Q. This con- taining space is also denoted by T|Q. If H is an element of C♦(R), then we denote by T(H|Q) the subset of T consisting of all points a for which there exists an element (x,X) of a such that X ∈ H and x ∈ QX. It is easy to verify that T(H|Q) = T|Q ∩ T(H). For every δ ∈ τ and E ∈ C♦(R|Q) the set of all elements b of T|Q for which there exists an element (x,QX) of b such that x ∈ UQ X δ and Q X ∈ E is denoted by UT|Qδ (E). Also, we set Some properties of the containing spaces and saturated classes of spaces 493 BT|Q = {UT|Qδ (E) : δ ∈ τ, E ∈ C ♦(R|Q)}. Lemma 2.8. For every b ∈ T|Q there exists a unique element a of T such that for every x ∈ QX the pair (x,QX) belongs to b if and only if the pair (x,X) belongs to a. Proof. Let b ∈ T|Q. Consider an element (y,QY ) of b and denote by a the element of T containing the pair (y,Y ). Now, let (x,QX) ∈ b. Then, X ∼s|QY and dQ X s (x) = d QY s (y) (2.1) for every s ∈F \{∅}. By Lemma 2.2, dXs (x) = d QX s (x) and d Y s (y) = d QY s (y). (2.2) Therefore, X ∼s Y and dXs (x) = dYs (y) (2.3) for every s ∈ F \{∅}. This means that the pairs (x,X) and (y,Y ) belong to the same element of the set T. Thus, (x,X) ∈ a. Conversely, let x ∈ QX and (x,X) ∈ a. Then, relation (2.3) is true for every s ∈F\{∅}. Lemma 2.2 implies relation (2.2). For every s ∈F\{0} relations (2.2) and (2.3) imply relation (2.1), which means that the pairs (x,QX) and (y,QY ) belong to the same element of the set T|Q. Thus, (x,QX) ∈ b. Obviously, the element a ∈ T satisfying the condition of the lemma is uniquely determined. � Definition 2.9. Let b be an arbitrary element of T|Q and a the unique element of T satisfying the condition of Lemma 2.8. We define a mapping eT|QT of T|Q into T setting e T|Q T (b) = a. Below, we shall prove that this mapping is an embedding, which will be called the natural embedding of the space T|Q into the space T. Lemma 2.10. Let δ ∈ τ, H ∈ C♦(R) and let L be the trace on Q of H. Then, an element b of T|Q belongs to U T|Q δ (L) if and only if the element a ≡ e T|Q T (b) belongs to UTδ (H). Proof. For every X ∈ S the indexed set {UQ X δ = Q X ∩UXδ : δ ∈ τ} 494 S. Iliadis is the trace on QX of the mark M(X). Let b ∈ T|Q and e T|Q T (b) = a ∈ T. Suppose that b ∈ UT|Qδ (L) and let (x,Q X) ∈ b. By Lemma 2.8, (x,X) ∈ a. We have QX ∈ L and x ∈ UQ X δ = Q X ∩UXδ and, therefore, X ∈ H and x ∈ UXδ , which means that a ∈ UTδ (H). Similarly we prove that if a ∈ UTδ (H), then b ∈ U T|Q δ (L). � Proposition 2.11. The mapping eT|QT is an embedding of T|Q into T. Proof. First, we prove that the mapping eT|QT is one-to-one. Indeed, let b1 and b2 be two distinct elements of the space T|Q. Since T|Q is a T0-space (see Proposition 2.9 of [2]) there exists δ ∈ τ and L ∈ C♦(R|Q) such that one of the points b1 and b2 belongs to the open set U T|Q δ (L) and the other does not belong. Let H ∈ C♦(R) such that L is the trace on Q of H. By Lemma 2.10 one of the points a1 ≡ e T|Q T (b1) and a2 ≡ e T|Q T (b2) belongs to the set UTδ (H) and the other does not belong. This means that a1 6= a2, that is, e T|Q T is one-to-one. The continuity of the mappings eT|QT and (e T|Q T ) −1 follows by Lemma 2.10 and by the fact that every element of the base BT|Q of the space T|Q has the form UT|Qδ (L) and every element of the base B T of the space T has the form UTδ (H). Thus, the mapping e T|Q T is an embedding. � Agreement 2. In what follows in this paper we identify a point b of T(M|Q, R|Q) with the point e T|Q T (b) ≡ a of T(M, R) and consider the space T(M|Q, R|Q) as a subspace of the space T(M, R). The definitions of the natural embeddings eXT and e T|Q T imply the following consequence. Corollary 2.12. The following relations is true: T|Q = ∪{eXT (Q X) : X ∈ S}. Proposition 2.13. Let X ∈ S and let eQ X X be the identical embedding of Q X into X. Then, e T|Q T ◦e QX T|Q = eXT ◦e QX X . (2.4) Some properties of the containing spaces and saturated classes of spaces 495 Proof. Let x ∈ QX. Let also (eXT ◦e QX X )(x) = e X T (x) = a. Then, a is the point of T containing the pair (x,X). On the other hand, by the definition of the mapping eQ X T|Q , eQ X T|Q (x) ≡ b is the point of T|Q containing the pair (x,QX). By the construction of the embedding eT|QT we have e T|Q T (b) = a. This proves relation (2.4). � 3. Commutative properties of the subspaces T|Q. Agreement. In the present section it is supposed that S, M, Q, and R are the same as in the preceding section and they are fixed. Definition 3.1. The restriction Q of S is said to be closed (respectively, open) if for every X ∈ S, QX is a closed (respectively, an open) subset of X. The restriction Q is said to be an (M, R)-complete restriction if for every point a ∈ T|Q and for every element (x,X) of a we have x ∈ QX. Notation. Below besides of the restriction Q we also consider the following restrictions connected with Q: Cl(Q) ≡{ClX(QX) : X ∈ S}, Bd(Q) ≡{BdX(QX) : X ∈ S}, Int(Q) ≡{IntX(QX) : X ∈ S}, and Co(Q) ≡{X \QX : X ∈ S}. Lemma 3.2. Suppose that Q is a closed restriction of S and R is an (M, Q)- admissible family of equivalence relations on S. Then, the following statements are true: (1) The subset T|Q of T is closed. (2) If, moreover, R is (M, Co(Q))-admissible, then T|Co(Q) = T \ T|Q. (3.1) (3) The restriction Q is an (M, R)-complete restriction. Proof. Let a be a point of T for which there exists an element (x,X) of a such that x /∈ QX. Since QX is closed in X there exists δ ∈ τ such that x ∈ UXδ and UXδ ∩Q X = ∅. Let s = {δ}. Since R is (M, Q)-admissible by Lemma 2.6 there exists an element t of F such that ∼t⊂∼sM and d X s (Q X) = dYs (Q Y ) for every Y ∈ S for which X ∼t Y . Let H be the ∼t-equivalence class of X. Then, UTδ (H) is an open neigh- bourhood of a in T. We prove that UTδ (H) ∩ T|Q = ∅. (3.2) Indeed, in the opposite case, there exists a point b belonging to the set UTδ (H)∩ T|Q. Let (y,QY ) be an element of b ∈ T|Q. By Lemma 2.8, b as a point of the 496 S. Iliadis space T contains the pair (y,Y ). Since b ∈ UTδ (H) we have Y ∈ H and y ∈ U Y δ . Therefore, X ∼t Y . By the choice of t, X ∼sM Y and d X s (Q X) = dYs (Q Y ). Since y ∈ QY there exists a point z ∈ QX such that dXs (z) = dYs (y). This equality and the relation y ∈ UYδ imply that z ∈ U X δ . Therefore, Q X ∩UXδ 6= ∅, which contradicts to the choice of δ. Thus, the relation (3.2) is proved. Now we prove the statements of the lemma. (1). By Corollary 2.12 as the above point a we can take any point of the set T \ T|Q. In this case, relation (3.2) implies that the set T|Q is closed. (2). As the point a we can take any point of the set T|Co(Q). In this case, relation (3.2) implies that a /∈ T|Q, that is, T|Q ∩T|Co(Q) = ∅. The Corollary 2.12 implies that T|Q ∪ T|Co(Q) = T. The last two relations are equivalent to the relation (3.1). (3). If Q is not an (M, R)-complete restriction, then there exists a point a of T|Q and an element (x,X) of a such that x /∈ QX. Then, relation (3.2) is true for some open neighbourhood UTδ (H) of a, which is a contradiction. Thus, Q is an (M, R)-complete restriction. � Lemma 3.3. Suppose that the restriction Q of S is open and the family R of equivalence relations on S is (M, Q)-admissible and (M, Co(Q))-admissible. Then: (1) The following relation is true: T|Co(Q) = T \ T|Q. (3.3) (2) The subset T|Q of T is open. (3) The restriction Q is an (M, R)-complete restriction. Proof. (1). Relation (3.3) follows immediately by the statement (2) of Lemma 3.2 if instead of the restriction Q of this lemma we consider the closed restriction Co(Q). (Note that Co(Co(Q)) = Q). (2). Since the restriction Co(Q) is closed by Lemma 3.2 the subset T|Co(Q) of T is closed. Therefore, by relation (3.3) the subset T|Q of T is open. (3). Let a ∈ T|Q and (x,X) ∈ a. If x /∈ QX, then x ∈ X\QX. By Corollary 2.12, a ∈ T|Co(Q), which contradicts to the relation (3.3). � Proposition 3.4. Suppose that the family R is (M, Q)-admissible and (M, Cl(Q))-admissible. Then, T|Cl(Q) = ClT(T|Q). Proof. Since the restriction Cl(Q) is closed by Lemma 3.2 the subset T|Cl(Q) of T is closed. Therefore, since T|Q ⊂ T|Cl(Q), it suffices to prove that T|Cl(Q) ⊂ ClT(T|Q). Let a ∈ T|Cl(Q). Then, by Corollary 2.12 there exists an element (x,X) of a such that x ∈ ClX(QX). Suppose that a /∈ ClT(T|Q). Then, there exists a neighbourhood UTδ (H) ∈ B T of a such that UTδ (H) ∩ T|Q = ∅. Corollary 2.12 implies that UYδ ∩ Q Y = ∅ for every Y ∈ H. Since X ∈ H, UXδ ∩ Q X = ∅. This means that x /∈ ClX(QX), which is a contradiction. Thus, a ∈ ClT(T|Q) and, therefore, T|Cl(Q) ⊂ ClT(T|Q). � Some properties of the containing spaces and saturated classes of spaces 497 Proposition 3.5. Suppose that Q is (M, R)-complete (in particular, by Lem- mas 3.2 and 3.3, Q may be a closed or an open restriction of S) and the family R of equivalence relations on S is (M, Q)-admissible, (M, Int(Q))-admissible, and (M, Co(Int(Q)))-admissible. Then, T|Int(Q) = IntT(T|Q). Proof. Obviously, the restriction Int(Q) is open. By assumptions of the lemma and Lemma 3.3, the subset T|Int(Q) is open in T. Therefore, since T|Int(Q) ⊂ T|Q, it suffices to prove that IntT(T|Q) ⊂ T|Int(Q). Let a ∈ IntT(T|Q). There exists an open neighbourhood UTδ (H) ∈ B T of a such that UTδ (H) ⊂ IntT(T|Q). Let (x,X) ∈ a. Then, x ∈ U X δ and X ∈ H. We prove that UXδ ⊂ Q X. Indeed, in the opposite case, there exists a point y belonging to the set UXδ \ Q X. Let b be the point of T containing the pair (y,X). Then, b ∈ UTδ (H). Since Q is an (M, R)-complete restriction, (y,X) ∈ b, and y /∈ QX we have b /∈ T|Q, which contradicts of the fact that UTδ (H) ⊂ IntT(T|Q) ⊂ T|Q. Thus, U X δ ⊂ Q X, which means that x ∈ IntX(QX) and, therefore, a ∈ T|Int(Q). � Proposition 3.6. Suppose that Q is (M, R)-complete and R is: (a) (M, Q)-admissible, (b) (M, Cl(Q))-admissible, (c) (M, Int(Q))-admissible, (d) (M, Co(Int(Q)))-admissible, and (e) (M, Bd(Q))-admissible. Then, T|Bd(Q) = BdT(T|Q). (3.4) Proof. Obviously, BdT(T|Q) = ClT(T|Q) \ IntT(T|Q). By Proposition 3.4, T|Cl(Q) = ClT(T|Q) and by Proposition 3.5, T|Int(Q) = IntT(T|Q). Therefore, it suffices to prove that T|Bd(Q) = T|Cl(Q) \ T|Int(Q). Let a ∈ T|Bd(Q). There exists an element (x,X) of a such that x ∈ BdX(QX). Then, x ∈ ClX(QX) and, therefore, a ∈ T|Cl(Q). On the other hand, since Bd(Q) is a closed restriction, by Lemma 3.2, Bd(Q) is an (M, R)- complete restriction. This means that for every (y,Y ) ∈ a we have y ∈ 498 S. Iliadis BdY (QY ), that is, y /∈ IntY (QY ). By Corollary 2.12, a /∈ T|Int(Q), that is, a ∈ T|Cl(Q) \ T|Int(Q). Conversely, let a ∈ T|Cl(Q) \T|Int(Q). Since Cl(Q) and Int(Q) are (M, R)- complete restrictions, for every (x,X) ∈ a we have x ∈ ClX(QX) \ IntX(QX), that is, x ∈ BdX(QX). This means that a ∈ T|Bd(Q). Thus, relation (3.4) is proved. � Proposition 3.7. Suppose that the restriction Q of S is (M, R)-complete and the family R of equivalence relations on S is (M, Q)-admissible and (M, Co(Q))- admissible. Then, Co(Q) is also (M, R)-complete restriction and T|Co(Q) = T \ T|Q. (3.5) Proof. Let a ∈ T|Co(Q). There exists an element (x,X) of a such that x ∈ X \QX. Let (y,Y ) ∈ a. If y /∈ Y \QY , then a ∈ T|Q and since Q is a (M, R)- complete restriction we have x ∈ QX, which is a contradiction. Therefore, y ∈ Y \QY , which means that Co(Q) is a (M, R)-complete restriction. This also means that a /∈ T|Q. By the above T|Co(Q) ⊂ T \ T|Q. On the other hand T|Co(Q) ∪ T|Q = T. The last two relations imply (3.5). � Definition 3.8. Suppose that for every element λ of a set Λ, F(λ) ≡{FX(λ) : X ∈ S} is a restriction of S. The union (respectively, the intersection) of the restrictions F(λ) is the restriction {FX : X ∈ S} of S for which FX = ∪{FX(λ) : λ ∈ Λ} (respectively, FX = ∩{FX(λ) : λ ∈ Λ}) for every X ∈ S. These restrictions are also denoted by ∨{F(λ) : λ ∈ Λ} and ∧{F(λ) : λ ∈ Λ}, respectively. Proposition 3.9. Suppose that for every element λ of a set Λ of cardinality ≤ τ an (M, R)-complete restriction F(λ) of S is given and let F be either the restriction ∧{F(λ) : λ ∈ Λ} or the restriction ∨{F(λ) : λ ∈ Λ}. If the family R is (M, F)-admissible and (M, F(λ))-admissible for every λ ∈ Λ, then F is an (M, R)-complete restriction. Some properties of the containing spaces and saturated classes of spaces 499 Proof. Suppose that F(λ) = {FX(λ) : X ∈ S}. Let F = ∧{F(λ) : λ ∈ Λ}≡{FX : X ∈ S}, a ∈ T|F ≡ T(M|F, R|F) and (x,X) ∈ a. There exists a pair (y,Y ) ∈ a such that y ∈ FY . Therefore, y ∈ FY (λ) for every λ ∈ Λ. By Corollary 2.12, a ∈ T|F(λ). Since by assumption F(λ) is an (M, R)-complete restriction, x ∈ FX(λ) for every λ ∈ Λ, which means that x ∈ FX. Thus, F is also an (M, R)-complete restriction. The case where F = ∨{F(λ) : λ ∈ Λ} is proved similarly. � 4. Saturated classes of subsets. Definition 4.1. In our considerations by a class of subsets we mean a class IP consisting of ordered pairs (Q,X), where Q is a subset of a space X. Such a class is said to be topological if for every homeomorphism h of a space X onto a space Y the condition (Q,X) ∈ IP implies that (h(Q),Y ) ∈ IP. In what follows all considered classes of subsets are assumed to be topological. Definition 4.2. Let IP be a class of subsets. A restriction Q of an indexed collection S of spaces is said to be a IP-restriction if (Q(X),X) ∈ IP for every X ∈ S. (We recall that Q = {Q(X) : X ∈ S}). Definition 4.3. A class IP of subsets is said to be saturated if for every indexed collection S of spaces and for every IP-restriction Q of S there exists a co-mark M+ of S satisfying the following condition: for every co-extension M of M+ there exists an (M, Q)-admissible family R+ of equivalence relations on S such that for every admissible family R of equivalence relations on S, which is a final refinement of R+, and for every elements H and L of C♦(R) for which H ⊂ L, we have (T(HQ), T(L)) ∈ IP. The considered co-mark M+ is said to be an initial co-mark of S (corre- sponding to IP -restriction Q) and the family R+ is said to be an initial family of S (corresponding to the co-mark M and IP -restriction Q). The proof of the next proposition is similar to the proof of Proposition 3.3 of [2]. (About the “intersection of classes” see the Note to Proposition 3.3 of [2]). Proposition 4.4. The intersection of no more than τ many saturated classes of subsets is also a saturated class of subsets. 500 S. Iliadis Notation. We shall denote by: (a) IP(Cl), (b) IP(Op), and (c) IP(n.dense) the classes of subsets consisting of all ordered pairs (Q,X) such that: (a) Q is closed, (b) Q is open, and (c) Q is nowhere dense in X, respectively. Proposition 4.5. The classes IP(Cl), IP(Op), and IP(n.dence) are saturated classes of subsets. Proof. By Lemma 3.2 it follows immediately that IP(Cl) is a saturated class. (For every IP(Cl)-restriction Q of an indexed collection S of spaces as an initial co-mark M+ we can take any co-mark of S and for every co-extension M of M+ as an initial family we can take any (M, Q)-admissible family of equivalence relations on S). Similarly, Lemma 3.3 implies that IP(Op) is a saturated class. (In this case, for every IP(Op)-restriction Q of an indexed collection S of spaces, as an initial co-mark M+ we can take any co-mark of S and for every co-extension M of M+ as an initial family we can consider any (M, Q)-admissible and (M, Co(Q))- admissible family of equivalence relations on S). Now, we prove that IP(n.dense) is a saturated class. Consider an indexed collection S of spaces and let Q ≡{QX : X ∈ S} be a IP(n.dense)-restriction of S. Therefore, QX is a nowhere dense subset of X ∈ S. Denote by M+ an arbitrary co-mark of S. We prove that M+ is an initial co-mark of S corresponding to the IP(n.dense)-restriction Q. For this purpose we consider an arbitrary co-mark M ≡{{UXδ : δ ∈ τ} : X ∈ S} of S, which is a co-extension of M+, and let R+ be any (M, Q)-admissible family of equivalence relations on S. We show that R+ is an initial family corresponding to the co-mark M and the IP(n.dense)-restriction Q. Indeed, let R ≡{∼s: s ∈F} be an admissible family of equivalence relations on S, which is a final refinement of R+, H, L ∈ C♦(R), and H ⊂ L. Then, R is also (M, Q)-admissible. In order to prove that R+ is an initial family (and, therefore, M+ is an initial co-mark) we need to prove that the subset T(H|Q) of T(L) is nowhere dense. For this purpose it suffices to prove that T|Q is a nowhere dense subset of T. Let U be an open subset of T. Without loss of generality, we can suppose that U has the form UTδ (H) for some δ ∈ τ and some ∼ t-equivalence class H, t ∈ F. Let X ∈ H. Since QX is nowhere dense in X there exists an element Some properties of the containing spaces and saturated classes of spaces 501 ε ∈ τ such that UXε ⊂ UXδ and U X ε ∩QX = ∅. Let {δ,ε}∪ t = s. By Lemma 2.6 there exists an element q of F such that ∼q⊂∼sM and d X s (Q X) = dYs (Q Y ) if X ∼q Y . Let E be the ∼q-equivalence class of X. In order to prove that the subset T|Q of T is nowhere dense it suffices to prove that UTε (E) ⊂ UTδ (H) and U T ε (E) ∩ T|Q = ∅. (4.1) Let Y be an arbitrary element of E. By the choice of q, X ∼sM Y . By Lemma 1.1 of [2], UYε ⊂ UYδ . Therefore, U T ε (E) ⊂ UTδ (E) ⊂ U T δ (H). We now prove that the set UTε (E) ∩ T|Q is empty. Indeed, in the opposite case, there exists a point a ∈ T|Q belonging to the set UTε (E). By Corollary 2.12 there exists a pair (y,Y ) ∈ a such that y ∈ QY . Then, y ∈ UYε and Y ∈ E. By the choice of q, X ∼sM Y and d X s (Q X) = dYs (Q Y ). Therefore, there exists a point x ∈ QX such that dXs (x) = dYs (y). This means that x ∈ UXε , that is, UXε ∩QX 6= ∅, which is a contradiction proving relation (3.1). Thus, the set T|Q is a nowhere dense subsets of T and, therefore, IP(n.dense) is a saturated class. � Corollary 4.6. If IP is a saturated class of subsets, then the classes IP(Cl) ∩ IP , IP(Op) ∩ IP , and IP(n.dense) ∩ IP are also saturated classes. Definition 4.7. A class IP of subsets is said to be closed (respectively, open) if for every element (Q,X) ∈ IP the subset Q of X is closed (respectively, open). A restriction Q of an indexed collection S of spaces is said to be complete if there exists a co-mark M of S and an (M, Q)-admissible family R of equivalence relations on S such that Q is an (M, R)-complete restriction. A class IP of subsets is said to be complete if for every indexed collection S of spaces any IP-restriction Q of S is complete. Lemmas 3.2 and 3.3 imply that any closed or any open restriction of any indexed collection of spaces is complete. Therefore, any closed or open class of subsets is complete. In particular, the classes IP(Cl) and IP(Op) are complete. Lemma 4.8. Let Q be a restriction of a collection S of spaces, M0 a co-mark of S and R0 a family of equivalence relations on S such that Q is (M0, R0)- complete. Then, for every co-extension M of M0 and (M, Q)-admissible fam- ily R of equivalence relations on S, which is a final refinement of R0, Q is an (M, R)-complete restriction. Proof. Let M be a co-extension of M0 and θ an indicial mapping of this co- extension. Let also R ≡ {∼s: s ∈ F} be an (M, Q)-admissible family of equivalence relations on S, which is a final refinement of R0 ≡ {∼s0: s ∈ F}. Consider a point a ∈ T|Q and let (x,X) ∈ a. We must prove that x ∈ Q(X). There exists an element (y,Y ) of a such that y ∈ Q(Y ). Let b be a point of T(M0, R0) such that (y,Y ) ∈ b. Since y ∈ Q(Y ), b ∈ T(M0|Q, R0|Q). We 502 S. Iliadis prove that (x,X) ∈ b. Indeed, since R is a final refinement of R0 and since X ∼s Y for every s ∈F we have X ∼t0 Y for every t ∈F. Let t be an arbitrary element of F \{∅} and s = θ(t). For every Z ∈ S denote by ÃZt the t-algebra of Z related to the mark M0(Z) and by A Z s the s-algebra of Z related to the mark M(Z). Also denote by d̃Zt the mapping of Z into 2t constructed for the algebra ÃZt and by d Z s the corresponding mapping constructed for the algebra AZs . It is easy to verify that if for some point z ∈ Z, dZs (z) = f ∈ 2s, then d̃Zt (z) = f ◦ θ|t ∈ 2t, where θ|t is the restriction of θ to t ⊂ τ. Since (x,X), (y,Y ) ∈ a we have dXs (x) = dYs (y) for every s ∈ F. By the above, d̃Xt (x) = d̃ Y t (y) for every t ∈ F. Therefore, the pairs (x,X) and (y,Y ) belong to the same point of T(M0, R0). Since (y,Y ) ∈ b we have (x,X) ∈ b. Since Q is an (M0, R0)-complete restriction and y ∈ Q(Y ) we have x ∈ Q(X), which proves the lemma. � Definition 4.9. Let IP be a class of subsets. An element (QT ,T ) ∈ IP is said to be universal (respectively, properly universal) in IP if for every element (QZ,Z) of IP there exists a homeomorphism h of Z into T such that h(QZ) ⊂ QT (respectively, h−1(QT ) = QZ). Proposition 4.10. In any non-empty complete saturated class of subsets there exist properly universal elements. Proof. Let IP be a non-empty complete saturated class of subsets. Since IP is a topological class there exists an indexed collection S of spaces and a IP- restriction Q ≡ {QX : X ∈ S} of S such that for every element (QZ,Z) of IP there exists an element X of S and a homeomorphism f of Z onto X for which f(QZ) = QX. Since IP is complete, Q is a complete restriction of S. Therefore, there exists a co-mark M0 of S and an (M0, Q)-admissible family R0 of equivalence relations on S such that Q is an (M0, R0)-complete restriction. On the other hand, since IP is a saturated class there exists an initial co- mark M+ of S corresponding to the IP-restriction Q. Let M be a co-mark of S, which is simultaneously a co-extension of M+ and M0. There exists a family R+ of equivalence relations on S, which is an initial family corresponding to the co-mark M and the IP-restriction Q. Denote by R an admissible family of equivalence relations on S, which is simultaneously a final refinement of R+ and R0. By Lemma 4.8, Q is an (M, R)-complete restriction. Now, we consider the containing space T and its subset T|Q. By construc- tion, the ordered pair (T|Q, T) is an element of IP. We prove that this element is properly universal in IP. Indeed, let (QZ,Z) be an element of IP. There exists an element X ∈ S and a homeomorphism f of Z onto X such that f(QZ) = QX. Let eXT be the natural embedding of X into T and h = eXT ◦f. Then, h is a homeomorphism of Z into T. We prove that h−1(T|Q) = QZ. It suffices to prove that Some properties of the containing spaces and saturated classes of spaces 503 (eXT ) −1(T|Q) ⊂ QX. (4.2) Let a ∈ T|Q and x ∈ (eXT ) −1(a). By the definition of the mapping eXT , (x,X) ∈ a. Since Q is an (M, R)-complete restriction we have x ∈ QX proving relation (3.2). This completes the proof of the proposition. � Similarly we can prove the following proposition. Proposition 4.11. In any non-empty saturated class of subsets there exist universal elements. Corollary 4.12. In the classes IP(Cl), IP(Op), and IP(Cl) ∩ IP(n.dense) there exist properly universal elements. 5. Commutative operators. Definition 5.1. Suppose that for every space X a mapping OX of the set P(X) into itself is given. Then, the class of all such mappings is said to be an operator. Such an operator is said to be topological if for every homeomorphism h of a space X onto a space Y we have h(OX(Q)) = OY (h(Q)) for every Q ∈P(X). In what follows, all considered operators are assumed to be topological. Notation. The class of all spaces will be denoted by S. Let O ≡{OX : X ∈S} be an operator. For every indexed collection S of spaces and for every restriction Q ≡{QX : X ∈ S} of S we set O(Q) = {OX(QX) : X ∈ S}. Obviously, O(Q) is a restriction of S. Let IP be a class of subsets. Denote by O(IP) and O−1(IP) the classes of subsets, which are defined as follows: O(IP) = {(OX(Q),X) : (Q,X) ∈ IP} and O−1(IP) = {(Q,X) : (OX(Q),X) ∈ IP}. It is clear that O(IP) and O−1(IP) are topological classes of subsets. Below, the following operators will be considered: Bd = {BdX : X ∈S}, 504 S. Iliadis Cl = {ClX : X ∈S}, and Int = {IntX : X ∈S}, where BdX, ClX, and IntX are the boundary, the closure, and the interior operators in a space X, respectively. Definition 5.2. Let O ≡ {OX : X ∈ S} be an operator. This operator is said to be commutative with respect to a restriction Q of an indexed collection S of spaces if there exists a co-mark M+ of S satisfying the following condi- tion: for every co-extension M of M+ there exists an (M, Q)-admissible and (M, O(Q))-admissible family R+ of equivalence relations on S such that for every admissible family R of equivalence relations on S, which is a final refine- ment of R+, and for every elements H and L of C♦(R) for which H ⊂ L, we have OT(L)(T(H|Q)) = T(H|O(Q)). The considered co-mark M+ is said to be an O-commutative co-mark (corre- sponding to the restriction Q) and the family R+ is said to be an O-commutative family (corresponding to the co-mark M and the restriction Q). (We note that any co-extension of M+ is also an O-commutative co-mark and any admissible family of equivalence relation on S, which is a final refinement of R+, is also an O-commutative family). The operator O is said to be commutative with respect to a class IP of subsets if it is commutative with respect to any IP-restriction of any indexed collection of spaces. The operator O is said to be (completely) commutative if it is commutative with respect to any (complete) restriction of any indexed collection of spaces. It is clear that a (completely) commutative operator is commutative with respect to any (complete) class of subsets. Lemmas 3.2 and 3.3 and Propositions [3.4-3.6] imply the following two con- sequences. Corollary 5.3. The operator Cl is commutative. Corollary 5.4. The operators Bd and Int are completely commutative. The proofs of the following two proposition are similar. We prove only the second. Proposition 5.5. Let IP be a saturated class of subsets and O an operator, which is commutative with respect to IP . Then, O(IP) is a saturated class of subsets. Some properties of the containing spaces and saturated classes of spaces 505 Proposition 5.6. Let IP and IF be saturated classes of subsets and O an operator, which is commutative with respect to IP . Then, IP ∩ O−1(IF) is a saturated class of subsets. Proof. Suppose that O = {OX : X ∈ S}. Let S be an indexed collection of spaces and Q a (IP ∩O−1(IF))-restriction of S. Therefore, the restriction Q is a IP-restriction, the restriction G ≡ O(Q) of S is an IF-restriction, and O is commutative with respect to Q. Since IP and IF are saturated classes and O is commutative with respect to IP there exists a co-mark M+ of S, which is simultaneously an initial co- mark corresponding to the IP-restriction Q, an initial co-mark corresponding to the IF-restriction G, and an O-commutative co-mark corresponding to the restriction Q. We prove that M+ is also an initial co-mark of S corresponding to the (IP ∩ O−1(IF))-restriction Q. Consider a co-extension M of the co-mark M+. There exists a family R+ of equivalence relations on S, which is simultaneously an initial family corresponding to the co-mark M and the IP-restriction Q, an initial family corresponding to the co-mark M and the IF-restriction G, and an O-commutative family corresponding to the co-mark M and the restriction Q. We show that this family is also an initial family corresponding to the co-mark M and the (IP ∩ O−1(IF))-restriction Q. Indeed, let R be an admissible family of equivalence relations on S, which is a final refinement of R+, and H and L two elements of C♦(R) such that H ⊂ L. Then, we can consider the space T(L) and its subsets T(H|G) and T(H|Q). By construction, (T(H|Q), T(L)) ∈ IP and (T(H|G), T(L)) ∈ IF. Since O is commutative with respect to IP we have T(H|G) = T(H|O(Q)) = OT(L)(T(H|Q)). This relation shows that (T(H|Q), T(L)) ∈ IP ∩ O−1(IF), which means that the family R+ is an initial family corresponding to the co-mark M and the (IP ∩ O−1(IF))-restriction Q and, therefore, the co-mark M+ is an initial co- mark of S corresponding to the (IP∩O−1(IF))-restriction Q. Thus, IP∩O−1(IF) is a saturated class. � Corollary 5.7. If IF is a saturated class of subsets, then Cl(IF) and Cl−1(IF) are also saturated classes of subsets. Corollary 5.8. If IP is a complete saturated class of subsets, then Bd(IP) and Int(IP) are also complete saturated classes of subsets. Corollary 5.9. If IF is a saturated class of subsets and IP is a complete satu- rated class of subsets, then IP ∩Bd−1(IF) and IP ∩Int−1(IF) are also complete saturated classes of subsets. In the next proposition, which is easy to prove, we use the following definition of a saturated class of spaces. 506 S. Iliadis Definition 5.10. A class IP of spaces is said to be saturated if for every indexed collection S of elements of IP there exists a co-mark M+ of S satisfying the fol- lowing condition: for every co-extension M of M+ there exists an M-admissible family R+ of equivalence relations on S such that for every admissible family R of equivalence relations on S, which is a final refinement of R+, and for every element L of C♦(R), the space T(L) belongs to IP. Remark 5.11. The above definition of a saturated class of spaces is slightly different from that of [2]. In [2] instead of the space T(L) we consider only the space T. However, for the new notion of a saturated class of spaces all results of [2] concerning saturated classes of spaces are hold. Proposition 5.12. The following statements are true: (1) If IP is a (complete) saturated class of subsets and IE is a saturated class of spaces, then the classes {(Q,X) ∈ IP : X ∈ IE} and {(Q,X) ∈ IP : Q ∈ IE} are (complete) saturated classes of subsets. (2) If IF is a saturated class of subsets, then the classes {X ∈S : (Q,X) ∈ IF for some subset Q of X} and {Q ∈S : (Q,X) ∈ IF for some space X ∈S} are saturated classes of spaces. (3) If IP and IE are saturated classes of spaces, then the class {(Q,X) : Q ∈ IP , X ∈ IE, and Q ⊂ X} is a saturated class of subsets. 6. Concluding remarks and some problems. 1. In [3] (see also [1]) for a given countable ordinal α “very simple examples of Borel sets Mα and Aα (lying in the Hilbert cube H) which are exactly of the multiplicative and additive class α respectively” are constructed. These sets satisfy the following property: “If X is a metric space and B ⊂ X is a Borel set of the multiplicative (additive) class α in X, then there exists a continuous mapping ϕ of X into H such that ϕ−1(Mα) = B (such that ϕ−1(Aα) = B).” Actually in [3] the class of subsets consisting of all pairs (B,X), where B is a Borel set of multiplicative (additive) class α in a metric space X, is considered. The proving property of the elements (Mα,H) and (Aα,H) shows that these elements in some sense can be considered as a “properly universal element” in this class. By our method we can prove the following result: For a given countable ordinal α the class IPm (respectively, the class IPa) of all pairs (Q,X), where Q is a Borel set of the multiplicative (respectively, Some properties of the containing spaces and saturated classes of spaces 507 additive) class α in a separable metrizable space X, is a complete saturated class of subsets. Therefore, in these classes there exist properly universal elements. In connection with the above we put the following problem. (1) Are the elements (Mα,H) and (Aα,H) properly universal elements in the classes IPm and IPa, respectively? 2. The problem of the existence of (properly) universal elements for different classes of subsets is arisen. Below, we set two questions concerning the class of all subsets, that is, the class of subsets consisting of all pairs (Q,X), where Q is a subset of a space X, whose answers it seems to be negative. (2) Is there a properly universal element in the class of all subsets? (3) Is the class of all subsets complete saturated? The second question is equivalent to the following: (3a) Is any saturated class of subsets complete saturated? (Obviously, the class of all subsets is saturated and, therefore, in this class there are universal elements). Another general problem is the following: (4) Construct (completely) commutative operators, which are distinct from that of Section 5 and its compositions. (The composition of two operators O1 ≡{O1X : X ∈S} and O 2 ≡{O2X : X ∈S} is the operator O ≡{OX ≡ O1X ◦ O 2 X : X ∈S}). References [1] R. Engelking, W. Holsztyński and R. Sikorski, Some examples of Borel sets, Colloq. Math. 15 (1966), 271–274. [2] S. D. Iliadis, A construction of containing spaces, Topology Appl. 107 (2000), 97–116. [3] R. Sikorski, Some examples of Borel sets, Colloq. Math. 5 (1958), 170–171. Received January 2002 Revised September 2002 Stavros Iliadis Department of Mathematics, University of Patras, Patras, Greece. E-mail address : iliadis@math.upatras.gr