1.3 Extended Set Operations 1 Definition 1.3.7 o baxsbi m sd A Let {A,:a e A} be an indexed collection of sets. The union of the collec tion {A.:a e A} is the set U{A,:a e A} = {x:xe A, for some a A The intersection of the collection {A,:a e A} is the set {A:a e A} : {x:x e A, for every a e A}. CLET %3D Example 1.3.8 Let A = R* and for each a e A let A, = [0, 1 + a). Then {A.iaE A} [0, 1] and U{A.:a e A} = [0, +o0). %3D Example 1.3.9 Let A = Z+ and for each a e A let B, = (-a, a). Then {Ba:a E A} = (-1,1) and U{B,:a e A} = R. || %3D Example 1.3.10 Let A = Z and for each a e A let D, = [a, a + 1]. Then {Da:a e A} = Ø and U{D:a e A} = R. %3D ie product of sets also extends to indexed collections of sets. However. Process is somewhat complicated and will be developed later in the text. the * snghtly different notation is sometimes used for collections of sets. If ben

1.3 Extended Set Operations 1 Definition 1.3.7 o baxsbi m sd A Let {A,:a e A} be an indexed collection of sets. The union of the collec tion {A.:a e A} is the set U{A,:a e A} = {x:xe A, for some a A The intersection of the collection {A,:a e A} is the set {A:a e A} : {x:x e A, for every a e A}. CLET %3D Example 1.3.8 Let A = R* and for each a e A let A, = [0, 1 + a). Then {A.iaE A} [0, 1] and U{A.:a e A} = [0, +o0). %3D Example 1.3.9 Let A = Z+ and for each a e A let B, = (-a, a). Then {Ba:a E A} = (-1,1) and U{B,:a e A} = R. || %3D Example 1.3.10 Let A = Z and for each a e A let D, = [a, a + 1]. Then {Da:a e A} = Ø and U{D:a e A} = R. %3D ie product of sets also extends to indexed collections of sets. However. Process is somewhat complicated and will be developed later in the text. the * snghtly different notation is sometimes used for collections of sets. If ben

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter3: Groups

Section3.1: Definition Of A Group

Problem 43E: Write out the elements of P(A) for the set A={ a,b,c }, and construct an addition table for P(A)...

See similar textbooks

Similar questions

12. (See Exercise 10 and 11.) If each is identified with in prove that . (This means that the order relation defined in Exercise 10 coincides in with the original order relation in . We say that the ordering in is an extension of the ordering in .) 11. (See Exercise 10.) According to Definition 5.29, is defined in by if and only if . Show that if and only if . 10. An ordered field is an ordered integral domain that is also a field. In the quotient field of an ordered integral domain define by . Prove that is a set of positive elements for and hence, that is an ordered field. Definition 5.29 Greater than Let be an ordered integral domain with as the set of positive elements. The relation greater than, denoted by is defined on elements and of by if and only if . The symbol is read “greater than.” Similarly, is read “less than.” We define if and only if. As direct consequences of the definition, we have if and only if and if and only if . The three properties of in definition 5.28 translate at once into the following properties of in . If and then . If and then . For each one and only one of the following statements is true: . The other basic properties of are stated in the next theorem. We prove the first two and leave the proofs of the others as exercises.
Write out the elements of P(A) for the set A={ a,b,c }, and construct an addition table for P(A) using addition as defined in Exercise 42. (Sec. 1.1,7c) Sec. 1.1,7c 42. For an arbitrary set A, the power set P(A) was defined in Section 1.1 by P(A)={ XXA }, and addition in P(A) was defined by X+Y=(XY)(XY) =(XY)(YX) Prove that P(A) is a group with respect to this operation of addition. If A has n distinct elements, state the order of P(A).
(See Exercise 26) Let A be an infinite set, and let H be the set of all fS(A) such that f(x)=x for all but a finite number of elements x of A. Prove that H is a subgroup of S(A).
Let A be a given nonempty set. As noted in Example 2 of section 3.1, S(A) is a group with respect to mapping composition. For a fixed element a In A, let Ha denote the set of all fS(A) such that f(a)=a.Prove that Ha is a subgroup of S(A). From Example 2 of section 3.1: Set A is a one to one mapping from A onto A and S(A) denotes the set of all permutations on A. S(A) is closed with respect to binary operation of mapping composition. The identity mapping I(A) in S(A), fIA=f=IAf for all fS(A), and also that each fS(A) has an inverse in S(A). Thus we conclude that S(A) is a group with respect to composition of mapping.