Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

CHAP. 14] ORDERED SETS AND LATTICES 359

Hence

bi=bi∧(c 1 ∨c 2 ∨···∨cs)=(bi∧c 1 )∨(bi∧c 2 )∨···∨(bi∧cs)

Sincebiis join irreducible, there exists ajsuch thatbi=bi∧cj, and sobi
cj. By a similar argument, forcj

there exists abksuch thatcj
bk. Therefore

bi
cj
bk

which givesbi=cj=bksince theb’s are irredundant. Accordingly, theb’s andc’s may be paired off. Thus the

representation forais unique except for order.

14.29. ProveTheorem 14.10: LetLbe a complemented lattice with unique complements.Then the join irreducible
elements ofL, other than 0, are its atoms.

Supposeais join irreducible andais not an atom. Thenahas a unique immediate predecessorb=0. Letb′be

the complement ofb. Sinceb=0 we haveb′=I.Ifaprecedesb′, thenb
a
b′, and sob∧b′=b′, which

is impossible sinceb∧b′=I. Thus a does not precedeb′, and soa∧b′must strictly precedea. Sincebis the

unique immediate predecessor ofa, we also have thata∧b′precedesbas in Fig. 14-16(a). Buta∧b′precedesb′.

Hence

a∧b′
inf(b, b′)=b∧b′= 0

Thusa∧b′=0. Sincea∨b=a, we also have that

a∨b′=(a∨b)∨b′=a∨(b∨b′)=a∨I=I

Thereforeb′is a complement ofa. Since complements are unique,a=b. This contradicts the assumption thatbis

an immediate predecessor ofa. Thus the only join irreducible elements ofLare its atoms.

Fig. 14-16

14.30. Give an example of an infinite latticeLwith finite length.

LetL={ 0 , 1 ,a 1 ,a 2 ,a 3 ,...}and letLbe ordered as in Fig. 14-16(b). Accordingly, for eachn∈N, we have

0 <an<1. ThenLhas finite length sinceLhas no infinite linearly ordered subset.