Section 2: Classes of Sets

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Section 2: Classes of Sets Notation: If A, B are subsets of X, then A \ B denotes the set difference, A \ B = {x A : x B}. A B denotes the symmetric difference. A B = (A \ B) (B \ A) = (A B) \ (A B). Remarks : (i) A A =. If A B =, then A B = A B. (iii) If B A then A B = A \ B. (iv) In fact A \ B = A (A B), A B = (A B) (A B). Let X be a set. Then (X) denotes the set of all subsets of X. If we write we mean that A 1,..., A n are pairwise disjoint and A =. A = A i A i. Definition 2.1. Let X be a set. Then a collection of sets (X) is a semi-ring if (i), if A, B then A B, (iii) if A, B then there is an n and there are sets A 1, A 2,..., A n such that A i are pairwise disjoint and A \ B = A i. Example 2.2. Set P = {(a, b]: a, b, a b}. Then P is a semi-ring of subsets of.

- 2 - [It is not hard to check this. For example, if a < c < d < b, then (a, b] \ (c, d] = (a, c] (d, b].] Similarly in 2 or n we can consider P n, the collection of all subsets of n of the form (a 1, b 1 ] (a 2, b 2 ]... (a n, b n ], e.g. P 2 = {(a, b] (c, d] : a, b, c, d, a b and c d}. Then P n is a semi-ring of subsets of n. This is not as obvious as in the case n = 1. For example, for P 2 note that the set difference of two half-open rectangles is the disjoint union of (at most) four half-open rectangles. Other examples of semi-rings of subsets of X: (i) (X) is a semi-ring; { } is a semi-ring; (iii) { } {{x}: x X} = collection of all subsets of X containing 1 point. Definition 2.3. Let X be a set, let R (i) Remarks R; (X). Then R is a ring of subsets of X if if A, B R then A B, A B and A \ B are all in R. (i) Every ring is a semi-ring. Rings are closed under finite intersection and union: if A 1, A 2,..., A n R, then n A i R and A i R. Examples (i) (X), { } are both rings of subsets of X. R = {A X: A is finite}. (iii) R = {A : A is bounded }. In this course, our main example of a ring will be the following. Example 2.4. Set = {A : A is a finite union of half open intervals in, each of the form (a,b]}. is called the collection of elementary figures in. ( ] a 1 b 1 ( ] a 2 b 2 ( ] a 3 b 3 ( ]... ( ] a n b n contains all sets of form (a i, b i ].

- 3 - The fact that is a ring follows from Lemma 2.6 below. First we give a definition. Definition 2.5. Let be a semi-ring of subsets of a set X. Then R( ) is defined to be the collection of all finite disjoint unions of sets in. Lemma 2.6. Let be a semi-ring of subsets of a set X. Then R( ) is a ring, and for any ring R satisfying R, we have R( ) R also. Remarks. It will follow that R( ) is also the collection of all finite unions of sets in smallest ring containing.. R( ) is the Proof. First note R( ). Next note that if A, B, then A \ B is a finite disjoint union of sets in semi-rings). Thus A \ B R( ). Suppose now that A, B R( ) and A B =. (by property (iii) of Then A =. A i, B = m. B j with all A i, B j in. Then A B =. A i. m. B j, a finite disjoint union of sets in. Thus A B R( ). Hence if A 1, A 2,..., A n R( ) and A i are pairwise disjoint then A i R( ). Now suppose that A, B R( ) with A =. A i, B = m. B j, A i, A j. Set C ij = A i B j. Then C ij (1 i n, 1 j m), and tions. A B = A ( i) m B ( j) = (A i B j ) = i, j 1 i n 1 j m C ij. i, j The sets C ij are pairwise disjoint, so A B R( ). Hence R( ) is closed under finite intersec - Suppose A, B are as above in R( ). Then A \ B = (. =. A i) ( \ m. B j) ( A i \ m B j) m (A i \ B j )) ( =. R( ).

- 4 - Finally, if A, B R( ), then A B = (A \ B). B R( ). Hence R( ) is a ring. The rest of the result is obvious, since any ring containing of sets in. must also contain all finite unions In particular, with P as in Example 2.2, we see that R(P) =, and so is a ring. Similarly in n, the ring generated by P n is the set of elementary figures in n, n consisting of all finite (disjoint) unions of sets in P n. There is an alternative definition of ring, equivalent to ours, R is a ring if (i) R, for A, B R, A B is in R, and A B R. With Operations as multiplication, as addition, R really is a ring in the algebraic sense. This is NOT true for fields: fields of sets are not usually really fields in the algebraic sense. Definition 2.7. Let X be a set. A collection of sets (X) is a field of subsets of X if is a ring and X. Examples (i) {, X} is the smallest possible field of subsets of X. (X) is a field of subsets of X. (iii) Let A X. Then {, A, X \ A, X} is a field of subsets of X. (iv) Set = {A X: either A or X \ A are finite (or both)}. Exercise: Check this is a field. Fields are also called algebras of sets. The next type of collection of sets is the σ -field (also known as σ -algebra or Borel algebra or Borel family). σ-fields A σ -field of subsets of X is a field of subsets of X which is closed under countable unions.

- 5 - In full, the definition of a σ -field is: Definition 2.8. Let X be a set and let (X). Then is a σ - field of subsets of X if satis fies Since (i), X, for all A, B, A \ B, (iii) whenever A 1, A 2, A 3,..., then is closed under infinite intersections. in. Examples A n. A n = X (X \ A n )) \(n, =1 (i) {, X}, (X) are both σ -fields of subsets of X. Finite unions and intersections then follow, since φ and X are If is a field and has only finitely many elements, then is always a σ -field. This is because only finitely many sets are involved in the countable union. (iii) Set X =. Set = {A : A or \ A is countable } (here countable means finite or countably infinite). Exercise. Show that is a σ -field. The following lemma remains true if σ -field is replaced throughout by field, ring but NOT semi-ring. Lemma 2.9. Let { γ : } be a set of σ -fields on a set X, where Γ is a non-empty indexing set. Then γ is also a σ -field on X. Proof (i) For each γ, and X are in γ. Thus γ and X γ. Let A, B γ. Then, for all, A and B are in γ. Since γ is a σ -field, A \ B γ for all γ, and so A \ B γ. (iii) Let A 1, A 2, A 3,... be a sequence of sets in and so A n γ. γ. Then, just as before, A n γ for every γ

- 6 - Definition 2.10. Suppose that X is a set and on X containing, namely, (X). Now define is any set of subsets of X. There is at least one σ -field ( ) = { a σ-field on X: } the intersection of all the σ -fields on X containing. By Lemma 2.9, ( ) is a σ -field on X. Any σ -field on X which contains must contain ( ) also. ( ) is called the σ -field generated by. Definition 2.11. Let (X, d) be a metric space. Let be the collection of all open subsets of X. Then the Borel subsets of X are the sets in ( ). Thus the collection of Borel sets on X is the σ -field generated by the set of open subsets of X. We are interested mainly in and. Let denote the collection of Borel subsets of. So is a σ -field which includes all open sets. Since fields are closed under complement, all closed subsets of are also in. Since is closed under countable unions and intersections, we see that every countable subset of is in, in particular. There are very many sets in but we shall see later that ( ). We have, for example, the Cantor middle thirds set is in. Example 2.12 (the Cantor Middle Thirds Set). Start with X 0 = [0, 1]. Delete the middle third ( 1 3, 2 3 ) to form X 1 = [0, 1 3 ] [ 2 3, 1]. X 1 consists of two closed intervals. Form X 2 by deleting the middle third of both intervals to leave four closed intervals. e.g. X 0 X 1 X 2 X n consists of 2 n closed intervals, each with length 1 3 n obtained by deleting the middle third of all the intervals forming X n 1. Set C = n =0 X n = those points in none of the deleted open intervals, but in [0, 1]. Then C is a closed subset of [0, 1], called the Cantor middle thirds set. In fact C consists of all x in [0, 1] which have a base 3 expansion of the form where all a i are 0 or 2. 0. a 1 a 2 a 3...

- 7 - C is an example of a metric space with no isolated points but such that the only connected subsets are single points. Although C contains no intervals of positive length, C has the same cardinality as. Every half-open interval (a, b] is a Borel set. This is because [ (a, b] = a + (b a), n b]. Thus (a, b] is a countable union of closed sets and hence (a, b]. We have P. Since is a ring we have, i.e. every elementary figure is a Borel set. Also, since is a σ -field containing P, we have (P). However, every open interval (a, b) is a countable union of sets in P. ( (a, b) = a, b (b a) n ]. Thus (a, b) (P) for all a < b in. But any open set U is a countable union of open intervals, e.g. U = {(p, q): p, q and (p, q) U}. Thus (P) is a σ -field containing all open subsets of. Since is the smallest σ -field containing all the open sets, it follows that (P). We already had (P). Thus (P) =. We have thus proven the following. Proposition 2.13. The σ -field generated by P is precisely the set of Borel subsets of.

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