:x$xþ1, :x$3x, :x$2x5, :x$x1 be one-to-one mappings ofRonto itself. Since for any x2R
ððxịị ẳðxỵ1ị ẳ3ðxỵ1ị
while ððxịị ẳð3xị ẳ3xỵ1,
we see that
ðiị ððxịị 6ẳððxịịor simply6ẳ: However,
ððxịị ẳð2x5ị ẳ2x6
and đỡđđxỡỡ Ử đỡđxợ1ỡ Ử2đxợ1ỡ 6Ử2x4
while ððxịị ẳðxỵ1ị ẳ2x3
and đỡđxỡ Ửđ2x3ỡ Ử2x31Ử2x4
Thus điiỡ đỡỬđỡ
Now
ðxị ẳðxỵ1ị ẳx and ðxị ẳðx1ị ẳx
that is, followed by (also, followed by ) maps each x2Rinto itself. Denote byJ, theidentity mapping,
J :x$x
10 SETS [CHAP. 1
Then ðiiiị ẳẳ J
that is,undoes whateverdoes (also,undoes whateverdoes). In view of (iii),is called theinverse mapping of and we writeẳ1; also,is the inverse of and we writeẳ1.
See Problem 1.18.
In Problem 1.19, we prove
Theorem I. If is a one-to-one mapping of a set S onto a set T, then has a unique inverse, and conversely.
In Problem 1.20, we prove
Theorem II. If is a one-to-one mapping of a setSonto a setTand is a one-to-one mapping ofT onto a setU, thenđỡ1Ử11.
Solved Problems
1.1. Exhibit in tabular form: (a) Aẳ fa:a2N, 2<a<6g, (b) Bẳ fp:p2N,p<10, p is oddg, (c)Cẳ fx:x2Z, 2x2ỵx6ẳ0g.
(a) HereAconsists of all natural numbersða2Nịbetween 2 and 6; thus,Aẳ f3, 4, 5g.
(b) Bconsists of the odd natural numbers less than 10; thus,Bẳ f1, 3, 5, 7, 9g.
(c) The elements ofCare the integral roots of 2x2ỵx6ẳ ð2x3ịðxỵ2ị ẳ0; thus,Cẳ f2g.
1.2. LetAẳ fa,b,c,dg,Bẳ fa,c,gg,Cẳ fc,g,m,n,pg. ThenA[Bẳ fa,b,c,d,gg,A[Cẳ fa,b,c,d, g,m,n,pg,B[Cẳ fa,c,g,m,n,pg;
A\Bẳ fa,cg,A\Cẳ fcg,B\Cẳ fc,gg;A\ ðB[Cị ẳ fa,cg;
ðA\Bị [Cẳ fa,c,g,m,n,pg,ðA[Bị \Cẳ fc,gg, ðA\Bị [ ðA\Cị ẳA\ ðB[Cị ẳ fa,cg.
1.3. Consider the subsets Kẳ f2, 4, 6, 8g, Lẳ f1, 2, 3, 4g, Mẳ f3, 4, 5, 6, 8g of Uẳ f1, 2, 3,. . ., 10g.
(a) ExhibitK0,L0,M0 in tabular form. (b) Show thatðK[Lị0ẳK0\L0. (a) K0ẳ f1, 3, 5, 7, 9, 10g,L0ẳ f5, 6, 7, 8, 9, 10g,M0ẳ f1, 2, 7, 9, 10g.
(b) K[Lẳ f1, 2, 3, 4, 6, 8gso thatðK[Lị0ẳ f5, 7, 9, 10g. Then K0\L0ẳ f5, 7, 9, 10g ẳ ðK[Lị0:
1.4. For the sets of Problem 1.2, show: (a) ðA[Bị [CẳA[ ðB[Cị, (b) ðA\Bị \Cẳ A\ ðB\Cị.
(a) SinceA[Bẳ fa,b,c,d,ggandCẳ fc,g,m,n,pg, we have ðA[Bị [Cẳ fa,b,c,d,g,m,n,pg:
SinceAẳ fa,b,c,dgandB[Cẳ fa,c,g,m,n,pg, we have
A[ ðB[Cị ẳ fa,b,c,d,g,m,n,pg ẳ ðA[Bị [C:
(b) Since A\Bẳ fa,cg, we have ðA\Bị \Cẳ fcg. Since B\Cẳ fc,gg, we have A\ ðB\Cị ẳ fcg ẳ ðA\Bị \C.
CHAP. 1] SETS 11
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1.5. In Fig. 1-1(c), let CẳA\B, DẳA\B0, EẳB\A0 and F ẳ ðA[Bị0. Verify: (a) ðA[Bị0ẳ A0\B0, (b)ðA\Bị0ẳA0[B0.
(a) A0\B0ẳ ðE[Fị \ ðD[Fị ẳFẳ ðA[Bị0
(b) A0[B0ẳ ðE[Fị [ ðD[Fị ẳ ðE[Fị [DẳC0ẳ ðA\Bị0 1.6. Use the Venn diagram of Fig. 1-7 to verify:
ðaị Eẳ ðA\Bị \C0 ðcị A[B\Cis ambiguous
ðbị A[B[Cẳ ðA[Bị [CẳA[ ðB[Cị ðdị A0\C0ẳG[L
(a) A\BẳD[EandC0ẳE[F[G[L; then
ðA\Bị \C0ẳE
(b) A[B[CẳE[F[G[D[H[J[K. Now
A[BẳE[F[G[D[H[J
and CẳD[H[J[K
so that
ðA[Bị [CẳE[F[G[D[H[J[K
ẳA[B[C
Also,B[CẳE[G[D[H[J[K andAẳE[F[D[H so that A[ ðB[Cị ẳE[F[G[D[H[J[KẳA[B[C
(c) A[B\Ccould be interpreted either asðA[Bị \Cor asA[ ðB\Cị. NowðA[Bị \CẳD[H[J, whileA[ ðB\Cị ẳA[ ðD[Jị ẳA[J. Thus,A[B\Cis ambiguous.
(d) A0ẳG[J[K[LandC0ẳE[F[G[L; hence,A0\C0ẳG[L.
1.7. LetAandBbe subsets ofU. Use Venn diagrams to illustrate:A\B0ẳAif and only ifA\Bẳ ;.
SupposeA\Bẳ ;and refer to Fig. 1-1(b). NowAB0; henceA\B0ẳA.
SupposeA\B6ẳ ;and refer to Fig. 1-1(c). NowA6B0; henceA\B06ẳA.
Thus,A\B0ẳAif and only ifA\Bẳ ;.
Fig. 1-7
12 SETS [CHAP. 1
1.8. Prove: ðA[Bị [CẳA[ ðB[Cị.
Let x2 ðA[Bị [C. Then x2A[B or x2C, so that x2A or x2B or x2C. When x2A, then x2A[ ðB[Cị; when x2B or x2C, then x2B[C and hence x2A[ ðB[Cị.
Thus,ðA[Bị [CA[ ðB[Cị.
Letx2A[ ðB[Cị. Thenx2Aorx2B[C, so thatx2Aorx2Borx2C. Whenx2Aorx2B, then x2A[B and hence x2 ðA[Bị [C; when x2C, then x2 ðA[Bị [C. Thus, A[ ðB[Cị ðA[Bị [C.
Now ðA[Bị [CA[ ðB[Cị and A[ ðB[Cị ðA[Bị [C imply ðA[Bị [CẳA[ ðB[Cị as required. Thus,A[B[Cis unambiguous.
1.9. Prove: ðA\Bị \CẳA\ ðB\Cị.
Let x2 ðA\Bị \C. Thenx2A\Bandx2C, so thatx2Aandx2Bandx2C. Sincex2Band x2C, thenx2B\C; sincex2Aandx2B\C, thenx2A\ ðB\Cị. Thus,ðA\Bị \CA\ ðB\Cị.
Let x2A\ ðB\Cị. Thenx2Aandx2B\C, so thatx2Aandx2B andx2C. Since x2Aand x2B, thenx2A\B; sincex2A\Bandx2C, thenx2 ðA\Bị \C. Thus,A\ ðB\Cị ðA\Bị \C andðA\Bị \CẳA\ ðB\Cịas required. Thus,A\B\Cis unambiguous.
1.10. Prove:A\ ðB[Cị ẳ ðA\Bị [ ðA\Cị.
Letx2A\ ðB[Cị. Thenx2Aandx2B[C(x2Borx2C), so thatx2Aandx2Borx2Aand x2C. When x2Aand x2B, thenx2A\B and so x2 ðA\Bị [ ðA\Cị; similarly, whenx2Aand x2C, thenx2A\Cand sox2 ðA\Bị [ ðA\Cị. Thus,A\ ðB[Cị ðA\Bị [ ðA\Cị.
Letx2 ðA\Bị [ ðA\Cị, so thatx2A\Borx2A\C. Whenx2A\B, thenx2Aandx2Bso that x2Aandx2B[C; similarly, whenx2A\C, thenx2Aandx2Cso thatx2Aandx2B[C. Thus, x2A\ ðB[CịandðA\Bị [ ðA\Cị A\ ðB[Cị. Finally,A\ ðB[Cị ẳ ðA\Bị [ ðA\Cịas required.
1.11. Prove:ðA[Bị0ẳA0\B0.
Let x2 ðA[Bị0. Nowx2=A\B, so thatx2=Aandx2=B. Thenx2A0 andx2B0, that is,x2A0\B0; henceðA[Bị0A0\B0.
Let x2A0\B0. Nowx2A0 andx2B0, so thatx2=Aandx2=B. Thenx2=A[B, so thatx2 ðA[Bị0; henceA0\B0 ðA[Bị0. Thus,ðA[Bị0ẳA0\B0 as required.
1.12. Prove:ðA\Bị [Cẳ ðA[Cị \ ðB[Cị.
C[ ðA\Bị ẳ ðC[Aị \ ðC[Bị byð1:10ị
Then ðA\Bị [Cẳ ðA[Cị \ ðB[Cị byð1:9ị
1.13. Prove:A ðB[Cị ẳ ðABị \ ðACị.
Letx2A ðB[Cị. Nowx2Aandx2=B[C, that is,x2Abutx2=Bandx2=C. Thenx2ABand x2AC, so thatx2 ðABị \ ðACịandA ðB[Cị ðABị \ ðACị.
Letx2 ðABị \ ðACị. Nowx2ABandx2AC, that is,x2Abutx2=Bandx2=C. Thenx2A but x2=B[C, so that x2A ðB[Cị and ðABị \ ðACị A ðB[Cị. Thus, A ðB[Cị ẳ ðABị \ ðACịas required.
1.14. Prove:ðA[Bị \B0ẳAif and only ifA\Bẳ ;.
Using (1.100) and (1.70), we find
ðA[Bị \B0ẳ ðA\B0ị [ ðB\B0ị ẳA\B0
We are then to prove:A\B0ẳAif and only ifA\Bẳ ;.
CHAP. 1] SETS 13
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(a) SupposeA\Bẳ ;. ThenAB0 andA\B0ẳA.
(b) SupposeA\B0ẳA. ThenAB0andA\Bẳ ;.
Thus,ðA[Bị \B0ẳAif (by (a)) and only if (by (b))A\Bẳ ;.
1.15. Prove: XY if and only ifY0X0.
(i) SupposeXY. Lety02Y0. Theny02=Xsincey02=Y; hence,y02X andY0X0. (ii) Conversely, supposeY0X0. Now, by (i),ðX0ị0 ðY0ị0; hence,XY as required.
1.16. Prove the identity ðABị [ ðBAị ẳ ðA[Bị ðA\Bị of Example 10 using the identity ABẳA\B0 of Example 9.
We have
ðABị [ ðBAị ẳ ðA\B0ị [ ðB\A0ị
ẳ ẵðA\B0ị [B \ ẵðA\B0ị [A0 byð1:10ị
ẳ ẵðA[Bị \ ðB[B0ị \ ẵðA[A0ị \ ðB0[A0ị byð1:10ị
ẳ ẵðA[Bị \U \ ẵU\ ðB0[A0ị byð1:7ị
ẳ ðA[Bị \ ðB0[A0ị byð1:40ị
ẳ ðA[Bị \ ðA0[B0ị byð1:9ị
ẳ ðA[Bị \ ðA\Bị0 byð1:110ị
ẳ ðA[Bị ðA\Bị
1.17. In Fig. 1-8, show that any two line segments have the same number of points.
Let the line segments be AB and A0B0 of Fig. 1-8. We are to show that it is always possible to establish a one-to-one correspondence between the points of the two line segments. Denote the intersection of AB0 and BA0 by P. On AB take any point C and denote the intersection of CP and A0B0 by C0. The mapping
C!C0
is the required correspondence, since each point ofABhas a unique image onA0B0and each point ofA0B0is the image of a unique point onAB.
1.18. Prove: (a) x!xþ2 is a mapping of N into, but not onto, N. (b) x!3x2 is a one-to-one mapping of Qonto Q, (c)x!x33x2x is a mapping ofR onto R but is not one-to-one.
(a) Clearlyxþ22Nwhenx2N. The mapping is not onto since 2 is not an image.
(b) Clearly 3x22Qwhenx2Q. Also, eachr2Qis the image ofxẳ ðrỵ2ị=32Q.
(c) Clearly x33x2x2R when x2R. Also, when r2R,x33x2xẳr always has a real rootx whose image isr. Whenrẳ 3,x33x2xẳrhas 3 real rootsxẳ 1, 1, 3. Since each hasrẳ 3 as its image, the mapping is not one-to-one.
1.19. Prove: If is a one-to-one mapping of a setS onto a set T, then has a unique inverse and conversely.
Supposeis a one-to-one mapping ofSontoT; then for anys2S, we have ðsị ẳt2T
14 SETS [CHAP. 1
Sincetis unique, it follows thatinduces a one-to-one mapping ðtị ẳs
Nowđỡđsỡ Ửđđsỡỡ Ửđtỡ Ửs; hence,Ử J andis an inverse of. Suppose this inverse is not unqiue;
in particular, supposeandare inverses of. Since
ẳẳ J and ẳẳ J
it follows that
Ửđỡ Ử J Ử
and Ử đỡỬ J Ử
Thus,ẳ; the inverse ofis unique.
Conversely, let the mappingofSintoThave a unique inverse1. Suppose fors1,s22S, withs16ẳs2, we have ðs1ị ẳðs2ị. Then 1ððs1ịị ẳ1ððs2ịị, so that ð1ịðs1ị ẳ ð1ịðs2ị and s1ẳs2, a contradiction. Thus, is a one-to-one mapping. Now, for anyt2T, we have ð1ðtịị ẳ ð1ịðtị ẳ t J ẳt; hence,tis the image ofsẳ1ðtị 2Sand the mapping is onto.
1.20. Prove: Ifis a one-to-one mapping of a setSonto a setTandis a one-to-one mapping ofT onto a setU, thenđỡ1Ử11.
Sinceđỡđ11ỡ Ửđ1ỡ1Ử1Ử J,11is an inverse of. By Problem 1.19 such an inverse is unique; hence,đỡ1Ử11.
Supplementary Problems
1.21. Exhibit each of the following in tabular form:
(a) the set of negative integers greater than6, (b) the set of integers between3 and 4,
(c) the set of integers whose squares are less than 20, (d) the set of all positive factors of 18,
(e) the set of all common factors of 16 and 24, (f) fp:p2N,p2<10g
(g) fb:b2N, 3b8g (h) fx:x2Z, 3x2ỵ7xỵ2ẳ0g
Fig. 1-8
CHAP. 1] SETS 15
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(i) fx:x2Q, 2x2ỵ5xỵ3ẳ0g
Partial Answer: (a) f5,4,3,2,1g, (d) f1, 2, 3, 6, 9, 18g, (f) f1, 2, 3g, (h) f2g 1.22. Verify: (a)fx:x2N,x<1g ẳ ;, (b)fx:x2Z, 6x2ỵ5x4ẳ0g ẳ ;
1.23. Exhibit the 15 proper subsets ofSẳ fa,b,c,dg.
1.24. Show that the number of proper subsets ofSẳ fa1,a2,. . .,angis 2n1.
1.25. Using the sets of Problem 1.2, verify: (a) ðA[Bị [CẳA[ ðB[Cị, (b) ðA\Bị \CẳA\ ðB\Cị, (c)ðA[Bị \C6ẳA[ ðB\Cị.
1.26. Using the sets of Problem 1.3, verify: (a)ðK0ị0ẳK, (b)ðK\Lị0ẳK0[L0, (c)ðK[L[Mị0ẳK0\L0\M0, (d)K\ ðL[Mị ẳ ðK\Lị [ ðK\Mị.
1.27. Let ‘‘njm’’ mean ‘‘nis a factor ofm.’’ GivenAẳ fx:x2N, 3jxgandBẳ fx:x2N, 5jxg, list 4 elements of each of the setsA0,B0,A[B,A\B,A[B0,A\B0,A0[B0whereA0andB0are the respective complements ofAandBinN.
1.28. Prove the laws of (1.8)–(1.120), which were not treated in Problems 1.8–1.13.
1.29. LetAandBbe subsets of a universal setU. Prove:
(a) A[BẳA\Bif and only ifAẳB, (b) A\BẳAif and only ifAB,
(c) ðA\B0ị [ ðA0\Bị ẳA[Bif and only ifA\Bẳ ;.
1.30. Given nðUị ẳ692, nðAị ẳ300, nðBị ẳ230, nðCị ẳ370, nðA\Bị ẳ150, nðA\Cị ẳ180, nðB\Cị ẳ90, nðA\B0\C0ị ẳ10 wherenðSịis the number of distinct elements in the setS, find:
ðaị nðA\B\Cị ẳ40 ðcị nðA0\B0\C0ị ẳ172
ðbị nðA0\B\C0ị ẳ30 ðdị nððA\Bị [ ðA\Cị [ ðB\Cịị ẳ340
1.31. Given the mappings :n!n2ỵ1 and :n!3nỵ2 of N into N, find: ẳn4ỵ2n2ỵ2, , ẳ3n2ỵ5, and.
1.32. Which of the following mappings ofZintoZ:
ðaị x!xỵ2 ðdị x!4x ðbị x!3x ðeị x!x3 ðcị x!x2 ðf ị x!x2x
are (i) mappings ofZontoZ, (ii) one-to-one mappings ofZontoZ? Ans. (i), (ii); (a), (d)
1.33. Same as Problem 32 withZreplaced byQ. Ans. (i), (ii); (a), (b), (d) 1.34. Same as Problem 32 withZreplaced byR. Ans. (i), (ii); (a), (b), (d), (e) 1.35. (a) IfEis the set of all even positive integers, show thatx!xþ1,x2Eis not a mapping ofEonto the set
Fof all odd positive integers.
(b) IfEis the set consisting of zero and all even positive integers (i.e., the non-negative integers), show that x!xþ1,x2Eis a mapping ofEontoF.
16 SETS [CHAP. 1
1.36. Given the one-to-one mappings
J : J ð1ị ẳ1, J ð2ị ẳ2, J ð3ị ẳ3, J ð4ị ẳ4 : ð1ị ẳ2, ð2ị ẳ3, ð3ị ẳ4, ð4ị ẳ1 : ð1ị ẳ4, ð2ị ẳ1, ð3ị ẳ2, ð4ị ẳ3 : ð1ị ẳ3, ð2ị ẳ4, ð3ị ẳ1, ð4ị ẳ2 : ð1ị ẳ1, ð2ị ẳ4, ð3ị ẳ3, ð4ị ẳ2 ofSẳ f1, 2, 3, 4gonto itself, verify:
(a) ẳẳ J, hence, ẳ1; (b) ẳẳ; (c) 6ẳ;
(d) 2ẳẳ; (e) 2ẳ J, hence, 1ẳ; (f) 4ẳ J, hence, 3ẳ1; (g) ð2ị1ẳ ð1ị2.
CHAP. 1] SETS 17
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Relations and Operations
INTRODUCTION
The focus of this chapter is on relations that exist between the elements of a set and between sets. Many of the properties of sets and operations on sets that we will need for future reference are introduced at this time.