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Sublattice and Convex sublattice

Sublattice: Let (L, ˄, ˅) be a lattice. A non empty subset S of L is called a Sublattice of L if S itself is a lattice under same operations ˄ and ˅ in L.

Or, A non empty subset S of a lattice L is called a Sublattice of L if a, b ∊ S      ⟹ a ˄ b , a ˅ b ∊ S.

Example: Let L = { 0, a, b, 1} be a lattice.

Sublattice of L are: {0, a, b, 1}, { 0}, {a}, {b}, {1}, {0,a}, {0, b}, {a, 1}.

Example: If L is any lattice and a ∊ L be any element then {a} is a sublattice of L.

Again any lattice L is a sublattice of itself.

Problem: Show that union of sublattices may not be a sublattice.

Solution: Consider the lattice L = {1, 2, 3, 4, 6, 12} of factors of 12 under divisibility.

Then S = {1, 2} and T ={ 1, 3} are sublattices of L. But S ∪ T = {1, 2, 3} is not a sublattice as 2, 3 ∊ S ∪ T, But 2 ˅ 3 = 6 ∉ S ∪ T.

Hence union of sublattices may not be a sublattice.(Proved)

Problem: Show that a lattice L is a chain if and only if every non empty subset of it is a sublattice.

Solution: If the lattice is a chain then we have every non empty subset of it is a sublattice.

Conversely, let L be a lattice such that every non empty subset of L is a sublattice. We show L is a chain.

Let a, b ∊ L be any elements.

Then {a, b} being a non empty subset of L will be a sublattice of L. Thus by definition of sublattice a ˄ b ∊ {a, b}

⟹ a ˄ b = a or, a ˄ b = b

⟹ a ≤ b or b ≤ a

i.e., a, b are comparable. Hence L is a chain. (Proved)

Convex sublattice: A sublattice S of a lattice L is called a convex sublattice of L, if for all a, b ∊ S, [a ˄ b, a ˅ b ] ⊆ S.

Example: {0, a, b, c, 1} be a lattice.

Here {0, a, b, c} is convex sublattice.

 

Or, A sublattice S of a lattice L is a convex sublattice if ∀ a, b ∊ S, [a ˄ b, a ˅ b ] ⊆ S.

Theorem: A sublattice S of a lattice L is a convex sublattice if and only if ∀ a, b ∊ S, [a ˄ b, a ˅ b ] ⊆ S.

Proof: Let S be a convex sublattice of L.

Let a, b ∊ S, (a ≤ b),be any elements, then by definition [a ˄ b, a ˅ b ] ⊆ S

⟹ [a, b] ⊆ S as a ≤ b ⟹ a ˄ b = a, a ˅ b = b.

Conversely, let [a, b] ⊆ S ∀ a, b, (a ≤ b)

Let a, b ∊ S be any elements.

Then a ˄ b, a ˅ b ∊ S is a sublattice.

Also these are comparable. Thus by given condition [a ˄ b, a ˅ b ] ⊆ S.

Problem: Is the set of all convex sublattice of a lattice under set inclusion a lattice?

Solution: Suppose L is a lattice. Let us consider C = { A: [x, y] ⊆ A, ∀ x, y ∊ A}

We have to show that (C, ⊆) is a lattice. ∀ P, Q ∊ C

P ˄ Q = P ∩ Q ; ∀ x, y ∊ P ∩ Q.

x, y ∊ P and x, y ∊ Q

∴ [x, y] ⊆ P and [x, y] ⊆ Q

∴ [x, y] ⊆ P ∩ Q

So P ˄ Q ∊ C.

Again suppose P ˅ Q = {x:x ≤ p ˅ q; p ∊ P, q ∊ Q}

Let ?, ? ∊ P ˅ Q, then there exists p1, p2 ∊ P and q1, q2 ∊ Q such that

? ≤ p1 ˅ q1, ? ≤ p2 ˅ q2

Now for all γ ∊ p2 ˅ q2

So ? ≤ γ ≤ ?

⟹ p1 ˅ q1 ≤ γ ≤ p2 ˅ q2

⟹ γ ≤ p2 ˅ q2

∴ γ ∊ P ˅ Q

∴ [?, ?] ⊆ P ˅ Q

∴ P ˅ Q ∊ C

Hence C is a lattice.

i.e., Set of all convex sublattice of a lattice under set inclusion is a lattice.

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