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Unformatted text preview: SECTION 2.2 Set Operations 2. 4. 10. 12. a) AﬂB b) Aﬂﬁnvhich is the same as AEB c) AUB d) EU?
Note that A f; B. a) {a,b,c,d,e,f,g, h} : B b) {a,b,(:,d,e} : A c) There are no elements in A that are not in B, so the answer is (25. d) {f,g, h} . a) AU®:{;El;L‘EAVxEG}:{:r$EAVF}:{m3:EA}:A b) AﬂU:{$l:cEAAmEU}:{;It3:EAAT}:{3:I$EA}:A .a) AUA:{J:].1:EAVmEA}:{:r:l:J:EA}r—A b) AﬂA={:rmEAAJJEA}={$CFEA}=A a) AEE')={2;]:1:EAAngQ}:{mmEAAT}:{:1:$EA}:A
b) (ii—A:{IIJEEOA$¢A}={3:IFA$§£A}={$F}=® We will show that these two sets are equal by showing that each is a subset of the other. Suppose a: E
A U (A H B). Then .1: E A or :1: E A ﬂ B by the deﬁnition of union. 1n the former case, we have .7? E A, and
in the latter case we have rt E A and :c E B by the deﬁnition of intersection; thus in any event, 3 E A, so
we have proved that the left—hand side is a subset of the rightshand side. Conversely, let :1: E A. Then by the
deﬁnition of union, (L' E A U (A n B) as well. Thus we have shown that the righthand side is a subset of the
left—hand side. Section 2.2 14. 16. 18. 20. 22. 24. Set Operations 37
Since A = (A — B) U (A 0 B), we conclude that A : {1,5,7,8} U {3,6,9} : {1,3,5,6,7,8,9}. Similarly
B : (B — A) u (A n s) : {2,10} u {3,6,9} : {2,3,s,9,10}. a) If (r is in A O B, then perforce it is in A (by deﬁnition of intersection). b) If :1; is in A, then perforce it is in A U B (by deﬁnition of union). e) If .1: is in A E B, then perforce it is in A (by deﬁnition of difference). (:1) If :r E A then :i; gt B  A. Therefore there can be no elements in A 0 (B ~ A), so A F) (B E A) : 0. e) The lefthand side consists precisely of those things that are either elements of A or else elements of B but not. A, in other words, things that are elements of either A or B (or, of course, both). This is precisely
the definition of the righthand side. a) Suppose that .1: E A U B. Then either a: E A or .r E B. In either case, certainly :1; E A U B U C. This
establishes the desired inelusion. b) Suppose that a: E A (‘I B F] C. Then :1: is in all three of these sets. In partieular, it is in both A and B
and therefore in A F] B, as desired. 0) Suppose that 3: E (A — B) — C. Then .7: is in A — B but not in C‘. Sinee .i: E A — B, we know that a: E A
(we also know that Lt ¢ B, but that won’t be used here). Since we have established that a; E A but :i: C,
we have proved that a: E A — C. d) To show that the set given on the lefthand side is empty, it sufﬁees to assume that .1: is some element in that.
set and derive a contradiction, thereby showing that no such .1: exists. So suppose that a: E (A E C) m (C — B).
Then :r E A — C and x E C — B. The ﬁrst of these statements implies by deﬁnition that 3: ¢ C, while the
second implies that a: E C. This is impossible, so our proof by coutradietion is complete. e) To establish the equality, we need to prove inclusion in both directions. To prove that (B E A) U (C — A) g
(B UC) E A, suppose that a: E (B E A) U (C E A). Then either I E (B — A) or :r E (C — A). Without loss of
generality, assume the former (the proof in the latter case is exactly parallel.) Then a: E B and it A. From
the first. of these assertions, it follows that :1: E BUC. Thus we can eonelude that :i: E (BUC)  A, as desired.
For the converse, that is, to show that (B U C) — A Q (B — A) U (C — A), suppose that .1: E (B U C) E A.
This means that I E (B U C) and .1: at A. The ﬁrst of these assertions tells us that either .1”? E B or a: E C.
Thus either .7: E B — A or 3; E C E A. In either case, .1: E (B — A) U (C — A). (An alternative proof could be
given by using Venn diagrams, showing that both sides represent the same region.) That A Q (A F] B) U (A F] B) follows from the fact that every element .r E A is an element of either A n B
(if :L' E B) or A FIB (if a; g! B). On the other hand, if :i: E (A H B) U (A FIB), then either a: E Am B or
w E A D B. in either case, :r E A by the deﬁnition of intersection. First we show that every element of the leftnhand side must be in the righthand side as well. If it E Aﬂ(BFiC),
then a: must be in A and also in B DC. Hence I must be in A and also in B and in C. Since .1' is in both
A and B, we conclude that a: E AﬂB. This, together with the fact. that x E C tells us that .r E (ADB) DC,
as desired. The argument in the other direction (if m E (A n B) 00 then rt must be in Am (B m (7)) is nearly identical. First suppose :1: is in the lefthand side. Then 9: must be in A but in neither B nor C. Thus :1: E A — C,
but .1: B E C, so .7: is in the righthand side. Next suppose that :r is in the righthand side. Thus (r must
be in A — C' and not in B — C. The ﬁrst of these implies that :L' E A and .1: C. But now it must also be
the case that a; B, since otherwise we would have :t' E B — C. Thus we have shown that rr is in A but in
neither B nor 0, which implies that a: is in the lefthand side. 38 Chapter 2 Basic Structures: Sets, FLu‘ictions, Sequences, and Sums 26. The set is shaded in each case. (b) 28. Here is a Venn diagram that can be. used for four sets. Notice that sets A and B are not convex in this picture.
We have shaded set A. Notice thateach of the 16 different combinations are represented by a region. 30. a) We cannot conclude that A = B. For instance, if A and B are both subsets of C, then this equation will
always hold, and A need not equal B.
b) We cannot conclude that A : B; let. G = Q), for example.
0) By putting the two conditions together, we can now conclude that A : B. By symmetry, it sufﬁces to
prove that A Q B. Suppose that a: E A. There are two cases. ’If x E C, then .1: E A D C = 13 ﬂ 0, which
forces a; E B. On the other hand, if a: C, then because :33 E A U C : B U C, we must have :1: E B. Section 2.2 32. 34. 36. 38. 40. 42. 44. 46. 48. Set Operations 39 This is the set of elements in exactly one of these sets, namely {2, 5}. The ﬁgure is as shown; we shade that portion of A that is not in B and that portion of B that is not in A. There are precisely two ways that. an item can be in either A or B but not both. It can be in A but not B
(which is equivalent to saying that it is in A E B), or it can be in B but not A (which is equivalent to saying
that it is in B E A). Thus an element is in A (9 B if and only if it is in (A E B) LJ (B — A). a) This is clear from the symmetry (between A and B) in. the deﬁnition of symmetric difference. b) We prove two things. To show that A g (A EB B) (9 B, suppose a; E A. If :1: E B, then :i: 59 A EBB, so
a: is an element of the right—hand side. On the other hand if a: 9! B, then :i’: E A 6) B, so again a: is in the
righthand side. Conversely, suppose .1: is an element of the right—hand side. There are two cases. If .1: gt B,
then necessarily :L‘ E A 69 B, whence .‘L‘ E A. If a: E B, then necessarily :1: E A (P. B, and the only way for that
to happen (since I!) E B) is for a: to be in A. This is an identity; each side consists of those things that are in an odd number of the sets A, B, and C.
This is an identity; each side consists of those things that are in an odd number of the sets A, B, C, and D. To count the elements of A U B U C we proceed as follows. First we count the elements in each of the sets and
add. This certainly gives us all the elements in the union, but we'have overcounted. Each element in A 0 B,
A 0 C, and B D C has been counted twice. Therefore we subtract the cardinalities of these intersections to
make up for the overcount. Finally, we have compensated a bit too much, since the elements of A ﬂ B F1 0 have now been counted three times and subtracted three times. We adjust by adding back the cardinality of
A 0 B m C. \Ve note that these sets are increasing, that is, Al c; A2 g A3 g Therefore, the union of any collection
of these sets is just the one with the largest subscript, and the intersection is just the one with the smallest
subscript. a) A, 2 {...,—2,E1,0,1,...,n} b) A, 2 {...,—2,71,0,1}   C .443 C A2 C A] . All the sets are subsets of A1, which is the set
of positive integers, 21+. It follows that. A, : Z+. Every positive integer is excluded from at least one a) As 1' increases, the sets get smaller: of the sets (in fact from inﬁnitely many), so A5 : (D. b) All the sets are subsets of the set of natural numbers N (the nonnegative integers). The number 0 is in
each of the sets, and every positive integer is in exactly one of the sets, so A, : N and A, = c) As 27 increases, the sets get larger: A1 C Ag C A3  All the sets are subsets of the set of positive real
numbers R+ , and every positive real number is included eventually, so U2); A, : R+ . Because A1 is a subset
of each of the others, A, : A1 2 (0,1) (the interval of all real numbers between 0 and l, exclusive).  C A3 C A2 C A]. Because
(1,00) (all real numbers greater than 1). Every number eventually d) This time, as in part (a), the sets are getting smaller as 1' increases:
A] includes all the others, U2, A1 gets excluded as 7' increases, so A; = (3. Notice that 00 is not a real number, so we cannot write A. = {co}. 40 50. 52. 54. 56. 58. 60. 62. Chapter 2 Basic Structures: Sets, Functions, Sequences, and Sums a) 00 11100000 b) 10 1001 0001 c) 01 1100 1110 a) No elements are included, so this is the empty set. b) All elements are included, so this is the universal set. The bit string for the symmetric difference is obtained by taking the bitwise exclusive OR of the two bit
strings for the two sets, since we want to include those elements that are in one set or the other but not both. We can take the bitwise OR (for union) or AND (for intersection) of all the bit strings for these sets. The successor set has one more element than the original set. namely the original set itself. Therefore the answer is n + l. a) If the departments share the equipment, then the maximum number of each type is all that is required, so
we want to take the union of the multisets, A U B. b) Both departments will use the minimum number of each type, so we want to take the intersection of the
multisets, A F] B. c) This will be the difference B — A of the multisets.
d) if no sharing is allowed, then the university needs to purchase a quantity of each type of equipment that
is the sum of the quantities used by the departments; this is the sum of the multisets, A + B. Taking the maximum for each person, we have 5' U T = {0.6 Alice,0.9 Brian,0.4 Fred,0.9 Oscar, 0.? Rita}. ...
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This note was uploaded on 02/18/2012 for the course MATH 640:244 taught by Professor Ming during the Spring '09 term at Rutgers.
 Spring '09
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