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Home / Lattices and Boolean Algebras / Complete lattices

Complete lattices

Definition: A lattice L is called a complete lattice if every non empty sub set of L has its Sup and Inf in L.

Theorem: Dual of a complete lattice is complete.

Proof: Let (L, ?) be a complete lattice and let (L, ?) be its dual. Then (L, ?) is a lattice.

Let ? ≠ SL be any subset of L. Since L is complete, Sup S and Inf S exist in L.

Let a = Inf S in L

Then a ? x ∀ x ∊ L

⟹ x ? a ∀ x ∊ L

⟹ a is upper bound of S in L.

Let b be any other upper bound of S in L.

Then x ? b ∀ x ∊ L

⟹ b ? x ∀ x ∊ L

⟹ b ? a as a = Inf S in L

⟹ a ? b or that a is L. u. b of S in L.

Similarly we can show that Sup S in L will be Inf S in L. Hence (L, ?) is complete. (Proved)

Theorem: If (P, ≤) is a poset with greatest element u such that every non empty subset S of P has Inf then P is a complete lattice.

Proof: Let S be any non empty subset of P. we need prove that Sup S exists.

Since u is greatest element of P, x ≤ u ∀ x ∊ P

and thus s ≤ u ∀ s ∊ S

⟹ u is an upper bound of S.

Let T = set of all upper bounds of S, then T is a non empty subset of P and, therefore, by given condition Inf T exists. Let k Inf T

Now s ∊ S ⟹ s ≤ x, ∀ x ∊ T

⟹ each element of S is a lower bound of T

⟹ s ≤ k ∀ s ∊ S

⟹ k is an upper bounds of S. But k being a lower bound of T means k ≤ x ∀ x ∊ T i.e., k ≤ x ∀ upper bounds of S ⟹ k = Sup S

Hence P is a poset in which every non empty subset has Sup and Inf and thus P is a complete lattice. (Proved)

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