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1 Section 2.2 Families of Sets Section 2.2 2.2 Families of Sets Purpose of Section: To introduce the concept of classes or families of sets and show how operations on sets are applied to classes. Introduction Often in such areas of mathematics as point-set topology, measure theory, and abstract algebra one often works with sets of sets, generally called families of sets. Normally, we denote them with subscripts, such as A1 , A2 , A3 , , A10 for a collection of 10 sets. For a family of sets indexed by the natural numbers, we would denote the sets by A1 , A2 , A3 , ∞ { Ak }k =1 or { Ak }k∈N where N = {1, 2,3,...} . or maybe Indexing sets in this way has advantages. The reader will recall from calculus the notation for an infinite sums and products to be ∞ ∞ ∑a k = a1 + a2 + , k =1 ∏a k = a1a2 k =1 In a similar way, one defines the union and intersection of a family of A1 , A2 , A3 , as follows. Definition : Unions and Intersections of Families Families of Sets The union of an infinite number of sets A1 , A2 , in some universe U is defined to be n A1 ∪ A2 ∪ ∪ An = ∪ Ak = { x ∈U : x ∈ Ak for some k = 1, 2,..., n} k =1 The intersection of the sets A1 , A2 , is n A1 ∩ A2 ∩ A3 ∩ An = ∩ Ak = { x ∈U : x ∈ Ak for all k = 1, 2, , n } k =1 Sometimes one deals with a more general families of sets { Aα }α∈Λ where the index α on the set ranges over some index set Λ ( Λ could be a finite set, the natural numbers, or all values in an interval on the real line.). In this case the union of the family { Aα }α ∈Λ is ∪ Aα = { x ∈U : x ∈ Aα α ∈Λ for at least one α ∈ Λ } 2 Section 2.2 Families of Sets and the intersection of the family { Aα } is ∩ Aα = { x ∈U : x ∈ Aα for all α ∈ Λ } α ∈Λ The following examples illustrate these ideas. Example 1 Let Ak = x : 0 ≤ x ≤ k k = 1, 2,... k + 1 Draw these sets and the find the following intersections. a) c) Solution The sets 3 ∪ Ak k =1 3 ∩A k =1 k A1, A2 , A3 , b) d) ∞ ∪ Ak k =1 ∞ ∩A k =1 k are a family of monotomically increasing closed intervals A1 ⊆ A2 ⊆ , each member in the family a subset of the next. 3 Section 2.2 Families of Sets Figure 1 3 a) ∪ A = [ 0,3 / 4] k =1 k b) ∪ A = [ 0,1) k =1 k ∞ The right-endpoints of Ak get closer and closer to 1 but are always less than 1: hence 1 is not in the infinite union. 3 c) ∩ A = [ 0,1/ 2 ) k =1 k d) ∩ A = [ 0,1/ 2 ) k =1 k ∞ Example 2 Define the sequence of sets 1 Ak = x ∈ :0 < x ≤ 1 − , k = 1, 2,... k Find the union and intersection ∞ ∞ ∪ Ak and ∩A k =2 k =2 k . Solution Writing out a few sets, we see that by choosing k sufficiently large, every real number in the interval ( 0,1) will belong in at least one of the intervals Ak . Also, the only points that belong to all of the intervals Ak are 1 2 the numbers in the interval 0, . Hence ∞ 1 1 2 3 A = ∪ ∪ k 0,1 − = 0, ∪ 0, ∪ 0, ∪ = ( 0,1) k 2 3 4 k =2 k =2 ∞ ∞ 1 1 2 3 1 Ak = ∩ 0,1 − = 0, ∩ 0, ∩ 0, ∩ = 0, ∩ k 2 3 4 2 k =2 k =2 ∞ 4 Section 2.2 Families of Sets Margin Note: ▪ individual elements of sets are denoted by small letters x, y, a, b,... ▪ sets are denoted capital letters A, B, C ,... ▪ families of sets are denoted by script letters , , , … Example 3 (Indexed Family) If An = {n + 1, n + 2, , 2n} for each natural number n ∈ N , then ∞ ∪ A = {n ∈ N : n ≥ 2} k k =1 ∞ ∩A k =∅ k =1 Example 4 (Indexed (Indexed Family) Find the following infinite intersections of the open ( −α , α ) and closed intervals [ −α , α ] where the index α ranges over the real numbers . ∩ ( −α , α ) ∩ [ −α , α ] . α ∈ α ∈ Solution In both cases for any x ≠ 0 there exists a real number α ∈ sufficiently large so that x ∉ Aα . Thus any x ≠ 0 will not lie in either intersection. But x = 0 belongs to both ( −α , α ) and [ −α , α ] for any real number α ∈ . Hence, we conclude 0 belongs to both intersections, and so ∩ ( −α ,α ) = {0} ∩ [ −α , α ] = {0} α ∈ α ∈ Example 5 (Set (Set Projection) Let S ⊆ × = {( x, y ) : x, y ∈ } plane and define the infinite family be an arbitrary subset of the Cartesian { Ax }x∈ of sets Ax = { y ∈ : ( x, y ) ∈ S } ( x, y ) ∈ S . The of this family is the projection of the set S onto the y -axis. See which represents for arbitrary x ∈ the y values such that union ∪A x x∈ Figure 2. 5 Section 2.2 Families of Sets Projection of a Set Figure 2 Extended Laws for Sets Many laws of intersection and union of sets can be extended to families of sets. We leave most of these proofs left to the reader. Laws for Families of Sets a) A ∩ ∪ Bα = ∪ ( A ∩ Bα ) α ∈Λ α ∈Λ b) A ∪ ∩ Bα = ∩ ( A ∪ Bα ) α ∈Λ α ∈Λ c) ∩ Aα = α∪ Aα α ∈Λ d) (DeMorgan’s Law) ∈Λ ∪ Aα = α∩ Aα α ∈Λ (DeMorgan’s Law) ∈Λ Proof of d) Let x ∈ ∪ Aα α ∈Λ 6 Section 2.2 Hence x ∉ ∪ Aα Families of Sets and so x ∉ Aα for any α ∈ Λ . Thus x ∈ Aα for all α ∈ Λ and α ∈Λ thus x ∈ ∩ Aα . Hence α ∈Λ ∪ Aα ⊆ α∩ Aα . α ∈Λ ∈Λ The proof the other way follows along the same lines and is left to the reader. Defining Topologies on a Set A topology on a set is a family of subsets of a set that that allows for the study of convergence in the set. The study of topology forms much of the basis of many areas of modern mathematics, such as real and complex analysis. The area of mathematics that studies topologies and topological spaces is called topology. The idea is to introduce a family J of subsets of a given universal set U (the universe might be the real numbers, complex numbers, or even a family sets These open of functions), where the sets in the family J are called open sets. sets act as neighborhoods of points in U , thus allowing in a general way for the discussion of convergence of elements in U . This family of open sets J is called a topology on (or for) the universal set U . So is any collection of subsets of U called a topology? The answer is no. There are three restrictions on a family of subsets in order that they form a topology on U . They are: Definition: A topology J on a set U is any collection of subsets of U that satisfies the following conditions: (1) The empty set ∅ and U belong to the family J . (2) The union of any collection of sets in J also belongs to J . (3) The intersection of any two (or equivalently a finite number) sets in J is also a member of J . The sets in the topology J are called the open sets in the topology (or just open sets). sets Properties (2) and (3) say that the family J is closed under unions and finite intersections. intersections As an illustration consider the universal set U of only three points U = {a, b, c} 7 Section 2.2 Families of Sets So what subsets of U can we choose as our open sets? (i.e. sets in J ) We know U has a total of 23 = 8 subsets. Below we list five families of subsets of U , each of which satisfies the three given conditions for being a topology on U . The reader can verify (See Problem 7) each family J1 , J 2 , J 3 , J 4 , J 5 satisfies the conditions for being a topology on U . The topology J1 contains two open sets, the empty set ∅ and the universe U , and is called the indiscrete topology for U . At the other extreme the topology J 5 contains all subsets of U is the discrete topology, topology which means that every subset of U is an open set. The other topologies J 2 , J 3 , J 4 are in between the two extreme topologies. J1 = {∅,U } indiscrete topology J 2 = {∅, {a} , U } J 3 = {∅, {a} , {b, c} ,U } J 4 = {∅, {a, b} ,U } J 5 = {∅, {a} , {b} , {c} , {a, b} , {a, c} , {b, c} ,U } discrete topology Theorem 1 The family J = {∅, {a} , {b, c} ,U } is a topology on the set U = {a, b, c} . Proof: We verify the three conditions required for a topology. The first condition is verified since the topology J contains both the empty set ∅ and the universe U . The second condition (closure under unions) also holds since one can check that the union any combination of sets in J is also a member of J For example {a} ∪ {b, c} = {a, b, c} ∈ J ∅ ∪ {a} = {a} ∈ J {a} ∪ {a, b, c} = {a, b, c} ∈ J Thirdly, it is also easy to see that the intersection of any two sets in J is also a member of J . For example 8 Section 2.2 . Families of Sets {a} ∩ {b, c} = ∅ ∈ J {a} ∩ {b, c} = {a} ∈ J {b, c} ∩ {a, b, c} = {b, c} ∈ J In other words, J is closed under finite intersections. So what do these topologies on U have to do with convergence? Wait till we get to the study of analysis in Chapter 5 where we introduce the concepts of neighborhoods. 9 Section 2.2 Families of Sets Problems A1 = {1, 2 } , A2 = {2,3 } , A3 = {3, 4} 1. Let and in general explicitly the following sets. 5 a) ∪A k k =1 b) ∪A k k∈ c) ∪A k k ≥5 d) ∪ Ak 1≤ k ≤ 4 5 e) ∩A k k =1 f) ∩A k k∈ g) ∩ Ak 1≤ k ≤5 5 h) ∩A k k =1 i) ∩A k k∈ 2. For the following indexed families of sets find ∞ ∪ Ak ∞ and k =1 a) Ak = [ 0,1 − 1/ k ] b) Ak = [ −1/ k ,1/ k ] c) Ak = ( 0,1/ k ) d) Ak = {k} ∪ [1/ k , 2k ] e) Ak = [ k , k + 1] f) Ak = [ 0,1 + 1/ k ] g) Ak = {( x, y ) : x 2 + y 2 ≤ k} ∩A k k =1 . Ak = {k , k + 1 } . Write 10 Section 2.2 Families of Sets (Algebra (Algebra of Sets) A family Y of subsets of a universe U is called an algebra1 of sets if 3. (i ) ( ii ) A ∪ B is in Y whenever A and B are in Y A is in Y whenever A is in Y If this happens we say the family Y is closed under unions and complementation complementation. mplementation Which of the following families of subsets of U = {a, b, c} are an algebra of subsets on U . a) P (U ) b) c) d) {∅,U } {∅, {a} ,U } {∅, {a} , {b, c} ,U } 4. (Sets Indexed by Two Indices) Let Ai , j be a family of sets where i, j belong to the index sets I , J respectively. A simple example might be the collection of rectangles Ai , j = {( x, y ) : 0 ≤ x ≤ i, 0 ≤ y ≤ j} in the plane. a) Prove ∪∩ A i, j i∈I j∈J ⊆ ∩∪ Ai , j j∈J i∈I b) Find a counter example showing that the two sets in a) are not equal. 5. (Sets of Length Zero) A subset A of the real numbers is said to have length zero (called measure zero) zero if ∀ε > 0 there exists a sequence Ak = ( ak , bk ) of intervals such that they “cover” A : i.e. ∞ A ⊆ ∪ ( ak , bk ) k =1 1 Algebras of sets and sigma algebras (families of sets closed under countable unions) are fundamental to the study of measure theory. Note the difference between an algebra of subsets and a topology of subsets on a universe; just a minor difference makes for vastly different structures on the universe. 11 Section 2.2 Families of Sets ∞ and their total length is less than ε ; that is ∑b k − ak < ε . Show that any k =1 sequence of real numbers A = {ck } has measure zero. Hint: Cover each element ck in the sequence by an interval ( ak , bk ) of length bk − ak = ε / 2k . 6. (Compact Sets) A subset A of the real numbers is said to be compact if for every collection ℑ = ( a, b ) of open intervals that contains (or covers) A ; { } i.e. A⊆ ∪ ( a, b ) ( a ,b )∈ℑ there exists a finite subcollection of ℑ that also contains (covers) A . Show the set A = ( 0,1) is not compact by showing the following. ∞ a) A is covered by ∪ ( 0,1 − 1/ k ) k =1 b) There does not exist a finite subcollection of ℑ whose union contains A . 7. (Topologies) Verify that the following families are topologies for {a, b, c} . J1 = {∅,U } indiscrete topology J 2 = {∅, {a} , U } J 3 = {∅, {a} , {b, c} ,U } J 4 = {∅, {a, b} ,U } J 5 = {∅, {a} , {b} , {c} , {a, b} , {a, c} , {b, c} ,U } discrete topology 8.. (Topologies) Which of the following families of {a, b, c} are topologies on {a, b, c} ? { } J = {∅, {a, b} , {b, c} , {a, b, c}} J = {∅, {a} , {b, c}} J = {∅, {b} , {b, c} , {a, b, c}} a) J = ∅, {b} , {c} , {a, b, c} b) c) d)