366 ORDERED SETS AND LATTICES [CHAP. 14
14.50. (a)e,f,g; (b) none; (c) sup(B)=e; (d) none.
14.51. (a) 1, 2, 3; (b) 8; (c) sup(A)=3; (d) inf(A)=8.
14.52. (a) None; (b) 8; (c) none; (d) inf(B)=8.
14.53. (a) Both; (b) sup(A)=3; inf(A)does not exist.
14.54. (a) Both; (b) sup(A)=3; inf(A)=^3
√
5
14.55. Four: (1)a,b,c; (2)a,b/c; (3)a/b,a/c.
(4)a/b/c.
14.56. Four: See Fig. 14-24.
Fig. 14-24
14.57. (a) One: Identity mapping; (b) one; (c) two.
14.59. (b)b 1 ,c 1 ; (c)Nhas no limit points.
14.60. (a) Definef:S→Nbyf(a)=1,f(b)=2,
f( 3 )=3,f (n)=n+3.
(b) The elementais a limit point ofS′, butNhas no
limit points.
14.61. Ais a finite linearly ordered set.
14.65. (a) Six: 0abdI,0acdI,0adeI,0bceI,0aceI,0cdeI;
(b) (i)a,b,c, 0; (ii)a,b,c. (c)candeare comple-
ments ofa.bhas no complement. (d) No. No.
14.66. (a)a,b,c,g,0.(b)a,b,c. (c)ahasg;bhas none.
(d)I=a∨g,f=a∨b,e=b∨c,d=a∨c.
Other elements are join-irreducible. (e) No. No.
14.67. (a)ehas none;fhasbandc. (b)I =c∨f =
b∨f=b∨d∨f.(c)No, since decompositions are
not unique. (d) Two: 0,d,e,f,Imust be mapped
into themselves. ThenF= (^1) L, identity map onL,
orF={(b, c), (c, b)}.
14.68. (a)ahasc;chasaandb.(b)I=a∨c=b∨c. (c)No.
(d) Two: 0,c,d, I must be mapped into themselves.
Thenf= (^1) Lorf={(a, b), (b, a)}.
14.69. (a)ahase,chasbande. (b)I=a∨e=b∨c=c∨e.
(c) No. (d) Two: 0,d,Iare mapped into themselves.
Thenf= (^1) Lorf={(a, b), (b, a), (c, d), (d, c)}.
14.70. (a) See Fig. 14-25. (b) 1, 2, 3, 4, 5. The atoms are 2, 3
and 5. (c) 2 has none, 10 has none. (d) 60= 4 ∨ 3 ∨ 5 ;
30 = 2 ∨ 3 ∨ 5 ; 20 = 4 ∨ 5 ; 15 = 3 ∨ 5 ; 12 = 3 ∨ 4 ;
10 = 2 ∨ 5 ; 6 = 2 ∨3.
Fig. 14-25
14.73. (a) [1, 2, 3, 4], [14, 2, 3], [13, 24], [14, 23], [123, 4],
[124, 3], [134, 2], [234, 1], [1234], [23, 1, 4] [24, 1, 3],
[34, 1, 2]. (b) See Fig. 14-26.
Fig. 14-26