**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 ? ≠ *S* ⊆ * L* be any subset of

*. Since*

__L__*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)**