Lattices and Boolean Algebra Questions and Answers

Q: Consider the following statements :(I) Every bounded lattice is complete.(II) Every Boolean lattice is a distributive lattice.(III) Every complete lattice is bounded.Then which of the following is true ?

Only (II) and (III) are true.(II) holds by definition: a Boolean lattice is a complemented distributive lattice. (III) holds because a complete lattice has a join and meet of every subset, in particular of the whole lattice, which are its greatest and least elements.(I) is false: the rationals in [0, 1] with the usual order form a bounded lattice (least element 0, greatest 1), but the set of rationals below \(1/\sqrt{2}\) has no least upper bound in it, so it is not complete. That rules out 'all are true' and the options leaving out (II) or (III).

Q: The number of non-isomorphic modular lattices on 5 elements is :

There are 4 non-isomorphic modular lattices on 5 elements.There are exactly five lattices with 5 elements: the chain; the diamond \(M_{3}\) (a top, a bottom and three incomparable atoms); the pentagon \(N_{5}\); and the two lattices formed from the four-element Boolean square by adding a new top or a new bottom.A lattice is modular unless it contains a copy of \(N_{5}\), so the pentagon is the only non-modular one. The chain and the two square-based lattices are even distributive, and \(M_{3}\) is modular but not distributive. That leaves 4 modular lattices.

Q: Consider the following statements :(I) Every distributive lattice is complemented and complementation is unique.(II) Every modular lattice is complemented and complementation is uniqueThen which of the following is true ?

Neither (I) nor (II) is true.Not every distributive lattice is complemented: the three-element chain \(0 \lt a \lt 1\) is distributive, but \(a\) has no complement (no \(b\) with \(a \wedge b = 0\) and \(a \vee b = 1\)). Distributivity only guarantees that complements are unique when they exist.The same chain is modular, so (II) fails for the same reason. Modularity does not even give uniqueness: in the diamond \(M_{3}\) each atom has two complements.

Q: Consider the following statements about a lattice L :(i) Every ideal of L is a prime ideal.(ii) L is a chain.Then which of the following is correct ?

The answer is (I) if and only if (II).If L is a chain and \(a \wedge b \in I\) for an ideal I, then \(a \wedge b\) is the smaller of \(a\) and \(b\), so one of them lies in I: every ideal is prime.Conversely, suppose every ideal is prime but \(a\) and \(b\) are incomparable. The principal ideal \((a \wedge b]\) contains \(a \wedge b\), so being prime it contains \(a\) or \(b\). If it contains \(a\), then \(a \le a \wedge b \le b\), contradicting incomparability; similarly for \(b\). So L is a chain. Each condition implies the other.

Consider the following statements :

(I) Every bounded lattice is complete.
(II) Every Boolean lattice is a distributive lattice.
(III) Every complete lattice is bounded.

Then which of the following is true ?

A.Only (III) is true
B.Only (II) and (III) are true
C.Only (II) is true
D.All are true

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

Only (II) and (III) are true.

(II) holds by definition: a Boolean lattice is a complemented distributive lattice. (III) holds because a complete lattice has a join and meet of every subset, in particular of the whole lattice, which are its greatest and least elements.

(I) is false: the rationals in [0, 1] with the usual order form a bounded lattice (least element 0, greatest 1), but the set of rationals below \(1/\sqrt{2}\) has no least upper bound in it, so it is not complete. That rules out 'all are true' and the options leaving out (II) or (III).

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The number of non-isomorphic modular lattices on 5 elements is :

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

There are 4 non-isomorphic modular lattices on 5 elements.

There are exactly five lattices with 5 elements: the chain; the diamond \(M_{3}\) (a top, a bottom and three incomparable atoms); the pentagon \(N_{5}\); and the two lattices formed from the four-element Boolean square by adding a new top or a new bottom.

A lattice is modular unless it contains a copy of \(N_{5}\), so the pentagon is the only non-modular one. The chain and the two square-based lattices are even distributive, and \(M_{3}\) is modular but not distributive. That leaves 4 modular lattices.

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Consider the following statements :

(I) Every distributive lattice is complemented and complementation is unique.
(II) Every modular lattice is complemented and complementation is unique

Then which of the following is true ?

A.(I) is true but (II) is not true
B.(II) is true but (I) is not true
C.Neither (I) is true nor (II) is true
D.Both statements are true

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

Neither (I) nor (II) is true.

Not every distributive lattice is complemented: the three-element chain \(0 \lt a \lt 1\) is distributive, but \(a\) has no complement (no \(b\) with \(a \wedge b = 0\) and \(a \vee b = 1\)). Distributivity only guarantees that complements are unique when they exist.

The same chain is modular, so (II) fails for the same reason. Modularity does not even give uniqueness: in the diamond \(M_{3}\) each atom has two complements.

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Consider the following statements about a lattice L :

(i) Every ideal of L is a prime ideal.
(ii) L is a chain.

Then which of the following is correct ?

A.(I) implies (II) but not conversely
B.(II) implies (I) but not conversely
C.(I) if and only if (II)
D.Neither (I) implies (II) nor (II) implies (I)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The answer is (I) if and only if (II).

If L is a chain and \(a \wedge b \in I\) for an ideal I, then \(a \wedge b\) is the smaller of \(a\) and \(b\), so one of them lies in I: every ideal is prime.

Conversely, suppose every ideal is prime but \(a\) and \(b\) are incomparable. The principal ideal \((a \wedge b]\) contains \(a \wedge b\), so being prime it contains \(a\) or \(b\). If it contains \(a\), then \(a \le a \wedge b \le b\), contradicting incomparability; similarly for \(b\). So L is a chain. Each condition implies the other.

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