Partially Ordered Set: A pair 〈P; ≤〉 is called a partially ordered set (poset) if P is a nonempty set

and ≤ is a partial order on P; that is ≤ is a binary relation on P which is reflexive, anti-symmetric and
transitive.
For all 𝑥, 𝑦, 𝑧 ∈ P
a) 𝑥 ≤ 𝑥 (reflexivity)
b) if 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑥, then 𝑥 = 𝑦 (anti-symmetry)
c) if 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑧, then 𝑥 ≤ 𝑧 (transitivity)
Example:
If X is a set and P(X) is the power set of X, then 〈P(X); ≤〉 is a poset.

Chain and Anti-Chain: A poset 〈P; ≤〉 that satisfies 𝑥 ≤ 𝑦 or 𝑦 ≤ 𝑥 for all 𝑥, 𝑦 ∈ P is called a chain
or linearly ordered set or totally ordered set. P is said to be anti-chain if 𝑥 ≤ 𝑦 in P iff 𝑥 = 𝑦.
Example:

Figure A and B are chain but C is not chain, D is anti-chain and E is not chain or anti-chain.

Covering: Let P be an ordered set and let 𝑥, 𝑦 ∈ P. We say 𝑥 is covered by 𝑦 (or 𝑦 covers 𝑥) and write
𝑥 ≺ 𝑦 or 𝑦 ≻ 𝑥 if 𝑥 < 𝑦 and 𝑥 < 𝑧 < 𝑦 for no 𝑧 ∈ P. We write 𝑥 ≼ 𝑦 for 𝑥 ≺ 𝑦 or 𝑥 = 𝑦 and 𝑥 ≽ 𝑦 for
𝑥 ≻ 𝑦 or 𝑥 = 𝑦.
If 𝑥 ≺ 𝑦, then a diagram P(𝑥) ≺ P(𝑦) i.e. P(𝑥) ≤ P(𝑦) by using a line segment from P(𝑥) to P(𝑦) in
a upward direction. The point P(𝑧) can’t be intersect in a line segment from P(𝑥) to P(𝑦).

Supremum and Infimum: Let 〈P; ≤〉 be a poset and let S ⊆ P. We call 𝑥 ∈ P an upper bound of S
if s ≤ 𝑥 for all s ∈ S. We denote the set of upper bounds of S by S 𝑢 .
Similarly, we call 𝑦 ∈ P a lower bound of S if s ≥ 𝑦 for all s ∈ S. We denote the set of lower bounds of
S by S 𝑙 . So,
S 𝑢 = {𝑥 ∈ P|(∀ s ∈ S) s ≤ 𝑥}
and S 𝑙 = {𝑦 ∈ P|(∀ s ∈ S) s ≥ 𝑦 }

We say that S ⊆ P possesses a least upper bound or supremum or join, if there exists a least element
in S 𝑢 . If S has least upper bound, then we denote it by Sup(S) or ∨ S.
Similarly, we say that S ⊆ P possesses a greatest lower bound or infimum or meet, if there exists a
greatest element in S 𝑙 . If S has greatest lower bound, then we denote it by Inf(S) or ∧ S.

An upper bound 𝑥 is a least upper bound or supremum of S if

1) 𝑥 is an upper bound of S and
2) 𝑥 ≤ 𝑦 for all upper bounds 𝑦 of S.
A lower bound 𝑥 is a greatest lower bound or infimum of S if
1) 𝑥 is an lower bound of S and
2) 𝑥 ≥ 𝑦 for all upper bounds 𝑦 of S

Top and bottom element: Let 〈P; ≤〉 be a poset. Whenever there exists 𝑥 ∈ P such that 𝑦 ≤ 𝑥 for
all 𝑦 ∈ P, we call 𝑥 the largest or top element of P and denote it by 1.
Similarly, whenever there exists 𝑥 ∈ P such that 𝑥 ≤ 𝑦 for all 𝑦 ∈ P, we call 𝑥 the infimum least or
bottom element of P and denote it by 0.

Down Set: Let P be an ordered set Q ⊆ P. Then Q is said to be a down set or decreasing set or order
ideal or ideal if 𝑥 ∈ Q, 𝑦 ∈ P with 𝑦 ≤ 𝑥 implies 𝑦 ∈ Q.
Example:

N = {𝑎, 𝑏, 𝑐, 𝑑}.

Down set is O(N) = {{𝑎}, {𝑏}, {𝑎, 𝑏}, {𝑏, 𝑑}, {𝑎, 𝑏, 𝑐}, {𝑎, 𝑏, 𝑑}, {𝑎, 𝑏, 𝑐, 𝑑}, ∅}.

Up Set: Let P be an ordered set Q ⊆ P. Then Q is said to be a upset if 𝑥 ∈ Q, 𝑦 ∈ P with 𝑦 ≥ 𝑥 implies

𝑦 ∈ Q.
Example:

N = {𝑎, 𝑏, 𝑐, 𝑑}. Now up set is

O(N) = {{𝑐}, {𝑑}, {𝑎, 𝑐}, {𝑏, 𝑑}, {𝑎, 𝑏, 𝑐, 𝑑}, ∅}.

Direct product of two posets:

In the figure below, all connected posets are represented with four elements:

Lattice: A lattice is a partially ordered set in which every two elements have a supremum (also called
a least upper bound or join) and an infimum (also called a greatest lower bound or meet).
That is, a poset 〈P; ≤〉 is called a lattice if 𝑝 ∨ 𝑞 = Sup {𝑝, 𝑞} and 𝑝 ∧ 𝑞 = Inf {𝑝, 𝑞} for all 𝑝, 𝑞 ∈ P.
Examples:
(1) Here are Hasse diagrams of a couple of finite lattices –

(2) Any linearly ordered set is a lattice, where

(3) The following posets, however, are not lattices -

Theorem: Let L be a lattice. Then all nonempty finite subsets of L possess suprema and infima.
Proof: Let 𝑎1 , 𝑎2 , … , 𝑎𝑛 ∈ L. Then an easy induction gives:
∨ L = 𝑎1 ∨ 𝑎2 ∨ … ∨ 𝑎𝑛 , ∧ L = 𝑎1 ∧ 𝑎2 ∧ … ∧ 𝑎𝑛
where L = {𝑎1 , 𝑎2 , … , 𝑎𝑛 }
Therefore, ∨ {𝑎1 , 𝑎2 , … , 𝑎𝑛 } and ∧ {𝑎1 , 𝑎2 , … , 𝑎𝑛 } exists in L.

Complete Lattices: However, there exist lattices L in which not all subsets of L have suprema and
infima.
Example:
Let ℤ be the lattice of all integers with the usual (linear) ordering. Then the set of positive
integers has no supremum and the set of negative integers has no infimum in ℤ.

We call a poset 〈P; ≤〉 a complete lattice if every subset of P has both supremum and infimum.
Examples:
The interval [0, 1] with the usual (linear) ordering forms a complete lattice.
We call a lattice bounded if it has both top and bottom. Every complete lattice is bounded, but not vice
versa.

Lattices as Algebras: An algebra 〈L; ∨, ∧〉 is called a lattice if ∨ and ∧ satisfies the following
conditions–
 Commutative laws: 𝑎 ∨ 𝑏 = 𝑏 ∨ 𝑎, 𝑎 ∧ 𝑏 = 𝑏 ∧ 𝑎
 Associative laws : (𝑎 ∨ 𝑏) ∨ 𝑐 = 𝑎 ∨ (𝑏 ∨ 𝑐), (𝑎 ∧ 𝑏) ∧ 𝑐 = 𝑎 ∧ (𝑏 ∧ 𝑐)
 Absorption laws : 𝑎 ∨ (𝑎 ∧ 𝑏) = 𝑎, 𝑎 ∧ (𝑎 ∨ 𝑏) = 𝑎
 Idempotent laws : 𝑎 ∨ 𝑎 = 𝑎, 𝑎 ∧ 𝑎 = 𝑎
Moreover, 𝑎 ≤ 𝑏 iff 𝑎 ∧ 𝑏 = 𝑎, iff 𝑎 ∨ 𝑏 = 𝑏.
Conversely, suppose L is a nonempty set equipped with two binary operations ∧, ∨ : L × L → L
satisfying the identities above. Then we can define ≤ on L as follows: 𝑎 ≤ 𝑏 iff 𝑎 ∧ 𝑏 = 𝑎, iff 𝑎 ∨ 𝑏 = 𝑏.
We have that ≤ is a partial order on L, that Sup {𝑎, 𝑏} = 𝑎 ∨ 𝑏, and that Inf {𝑎, 𝑏} = 𝑎 ∧ 𝑏 for each
𝑎, 𝑏 ∈ L.
Thus, we can think of lattices as algebras 〈L; ∨, ∧〉 where ∧, ∨ : L × L → L are two binary operations
on L satisfying the commutativity, associativity, idempotency, and absorption laws.

Theorem: Any finite chain is a lattice.

Proof: Let L be a finite chain. We need to show that L is a lattice.
Let, 𝑎, 𝑏 ∈ L. As L is a lattice, so 𝑎 ≤ 𝑏 or 𝑏 ≤ 𝑎.
If 𝑎 ≤ 𝑏 ⇒ 𝑎 ∧ 𝑏 = 𝑎 ∈ L and 𝑎 ∨ 𝑏 = 𝑏 ∈ L
If 𝑏 ≤ 𝑎 ⇒ 𝑎 ∧ 𝑏 = 𝑏 ∈ L and 𝑎 ∨ 𝑏 = 𝑎 ∈ L
Hence L is a lattice.

Theorem: Let the algebra 〈L; ∨, ∧〉 be a lattice. Set 𝑎 ≤ 𝑏 iff 𝑎 ∧ 𝑏 = 𝑎. Then 𝔏𝑝 = 〈L; ≤〉 is a
poset and the poset 𝔏𝑝 is a lattice.
Proof: To show 𝔏𝑝 is a poset, ≤ must be reflexive, anti-symmetric and transitive.

i. Since ∧ is idempotent, i.e. 𝑎 ∧ 𝑎 = 𝑎 and 𝑎 ∨ 𝑎 = 𝑎.

Therefore 𝑎 ≤ 𝑎.
i.e. ≤ is reflexive.

ii. Let 𝑎 ≤ 𝑏 and 𝑏 ≤ 𝑎

∴ 𝑎 ∧ 𝑏 = 𝑎, 𝑏 ∧ 𝑎 = 𝑏
By commutativity, 𝑎 ∧ 𝑏 = 𝑏 ∧ 𝑎
 Therefore 𝑎 = 𝑎 ∧ 𝑏 = 𝑏 ∧ 𝑎 = 𝑏.
i.e. ≤ is anti-symmetric.

iii. Let 𝑎 ≤ 𝑏 and 𝑏 ≤ 𝑐

∴ 𝑎 ∧ 𝑏 = 𝑎, 𝑏 ∧ 𝑐 = 𝑏
Now, 𝑎 ∧ 𝑐 = (𝑎 ∧ 𝑏) ∧ 𝑐 [∵𝑎 = 𝑎 ∧ 𝑏]
= 𝑎 ∧ (𝑏 ∧ 𝑐 ) [Associativity]
=𝑎∧𝑏 [∵𝑏 = 𝑏 ∧ 𝑐]
=𝑎
It shows that, 𝑎 ≤ 𝑐.
i.e. ≤ is transitive.

Hence, 𝔏𝑝 = 〈L; ≤〉 is a poset.

To show 𝔏𝑝 is a lattice, we need to show that Inf {𝑎, 𝑏} and Sup {𝑎, 𝑏} exists for all 𝑎, 𝑏 ∈ L. Now we have
to show that, Inf {𝑎, 𝑏} = 𝑎 ∧ 𝑏, Sup {𝑎, 𝑏} = 𝑎 ∨ 𝑏.
Indeed 𝑎 ∧ 𝑏 ≤ 𝑎, since (𝑎 ∧ 𝑏) ∧ 𝑎 = 𝑎 ∧ (𝑏 ∧ 𝑎) [Associativity]
(𝑎
= 𝑎 ∧ ∧ 𝑏) [Commutativity]
= (𝑎 ∧ 𝑎) ∧ 𝑏 [Associativity]
=𝑎∧𝑏
Similarly 𝑎 ∧ 𝑏 ≤ 𝑏 since (𝑎 ∧ 𝑏) ∧ 𝑏 = 𝑎 ∧ (𝑏 ∧ 𝑏) = 𝑎 ∧ 𝑏.

Therefore 𝑎 ∧ 𝑏 is a lower bound of 𝑎 and 𝑏. Let 𝑐 be another lower bound such that 𝑐 ≤ 𝑎 and 𝑐 ≤ 𝑏. That
means 𝑐 ∧ 𝑎 = 𝑐 and 𝑐 ∧ 𝑏 = 𝑐. Now,
𝑐 ∧ (𝑎 ∧ 𝑏) = (𝑐 ∧ 𝑎) ∧ 𝑏 = 𝑐 ∧ 𝑏 = 𝑐.

This shows that, 𝑐 ≤ 𝑎 ∧ 𝑏. Hence, Inf {𝑎, 𝑏} = 𝑎 ∧ 𝑏.

Now, 𝑎 ≤ 𝑎 ∨ 𝑏, 𝑏 ≤ 𝑎 ∨ 𝑏. Therefore 𝑎 ∨ 𝑏 is an upper bound of 𝑎 and 𝑏.

Let 𝑐 be another upper bound such that 𝑎 ≤ 𝑐 and 𝑏 ≤ 𝑐.

That means 𝑎 ∧ 𝑐 = 𝑎 and 𝑏 ∧ 𝑐 = 𝑏, 𝑎 ∨ 𝑐 = 𝑐 and 𝑏 ∨ 𝑐 = 𝑐. [Absorption law]

Now, (𝑎 ∨ 𝑏) ∧ 𝑐 = (𝑎 ∨ 𝑏) ∧ (𝑎 ∨ 𝑐)
= (𝑎 ∨ 𝑏) ∧ (𝑎 ∨ 𝑏 ∨ 𝑐)
= (𝑎 ∨ 𝑏) ∧ ((𝑎 ∨ 𝑏) ∨ 𝑐)
=𝑎∨𝑏
This shows that, 𝑎 ∨ 𝑏 ≤ 𝑐. i.e. 𝑎 ∨ 𝑏 is the least upper bound of {𝑎, 𝑏}. Hence, Sup {𝑎, 𝑏} = {𝑎 ∨ 𝑏}.

∴ 𝔏𝑝 = 〈L; ≤〉 is a lattice.

Distributive Laws: Let L be a lattice and let 𝑎, 𝑏, 𝑐 ∈ L. It is easy to see that –

(𝑎 ∧ 𝑏) ∨ (𝑎 ∧ 𝑐) ≤ 𝑎 ∧ (𝑏 ∨ 𝑐) and 𝑎 ∨ (𝑏 ∧ 𝑐) ≤ (𝑎 ∨ 𝑏) ∧ (𝑎 ∨ 𝑐)

We say that in L meet distributes over join if for each 𝑎, 𝑏, 𝑐 ∈ L we have 𝑎 ∧ (𝑏 ∨ 𝑐) = (𝑎 ∧ 𝑏) ∨ (𝑎 ∧ 𝑐).
Dually, we say that join distributes over meet if 𝑎 ∨ (𝑏 ∧ 𝑐) = (𝑎 ∨ 𝑏) ∧ (𝑎 ∨ 𝑐).
These laws are called the distributive laws.
In fact, in every lattice the two distributive laws are equivalent to each other.
We show that 𝑎 ∧ (𝑏 ∨ 𝑐) = (𝑎 ∧ 𝑏) ∨ (𝑎 ∧ 𝑐) implies 𝑎 ∨ (𝑏 ∧ 𝑐) = (𝑎 ∨ 𝑏) ∧ (𝑎 ∨ 𝑐). The converse is
proved similarly.
Suppose 𝑎 ∧ (𝑏 ∨ 𝑐) = (𝑎 ∧ 𝑏) ∨ (𝑎 ∧ 𝑐) holds in a lattice L. For 𝑝, 𝑞, 𝑟 ∈ L we have:
(𝑝 ∨ 𝑞) ∧ (𝑝 ∨ 𝑟) = [(𝑝 ∨ 𝑞) ∧ 𝑝] ∨ [(𝑝 ∨ 𝑞) ∧ 𝑟]
= 𝑝 ∨ [(𝑝 ∨ 𝑞) ∧ 𝑟]
= 𝑝 ∨ [𝑟 ∧ (𝑝 ∨ 𝑞)]
= 𝑝 ∨ [(𝑟 ∧ 𝑝) ∨ (𝑟 ∧ 𝑞)]
= [𝑝 ∨ (𝑟 ∧ 𝑝)] ∨ (𝑟 ∧ 𝑞)
= 𝑝 ∨ (𝑟 ∧ 𝑞)
= 𝑝 ∨ (𝑞 ∧ 𝑟)
A lattice L is called distributive if the above distributive laws hold in it.
Example:
Each linearly ordered set is a distributive lattice.

Example:
(1) The lattice depicted below, and called the diamond, is not distributive.

(2) Another non-distributive lattice, called the pentagon, is shown below –

Sub-lattice: A non-empty subset S of a lattice L is called a sub-lattice if S itself a lattice.

Let L be a lattice and S ⊆ L. If for each 𝑎, 𝑏 ∈ S we have 𝑎 ∧ 𝑏, 𝑎 ∨ 𝑏 ∈ S, then we call S a sub-lattice
of L.
If in addition L is bounded and 0,1 ∈ S, then we call S a bounded sub-lattice of L. We say that a lattice K
is isomorphic to a (bounded) sub-lattice S of L if there exists a (bounded) lattice isomorphism from K
onto S.
Birkhoff’s Characterization Theorem: A lattice L is distributive if and only if neither the diamond nor
the pentagon is isomorphic to a sub-lattice of L.
Examples:

{0, 𝑎, 𝑏, 1}, {0, 𝑎}, {0, 𝑏} are sub-lattice but {0, 𝑎, 𝑏} is not a sub-lattice. Because 𝑎 ∨ 𝑏 = 1 ∉ L. {𝑎, 1},
{𝑏, 1} are sub-lattice of L.
Convex Sub-lattices: A non-empty sub-set K of a lattice L is said to be a convex sub-lattice if 𝑎, 𝑏 ∈
K and 𝑐 ∈ L with 𝑎 ≤ 𝑐 ≤ 𝑏, implies that 𝑐 ∈ K.

Examples:

(1) {𝑎, 𝑐, 1} is a sub-lattice and convex sub-lattice because 𝑎 ≤ 𝑐 ≤ 1 and 𝑎, 1 ∈ {𝑎, 𝑐, 1} implies 𝑐 ∈
{𝑎, 𝑐, 1}.

Semi-lattices: A poset 〈P; ≤〉 is a join semi-lattice if Sup{𝑎, 𝑏} exists for any 𝑎, 𝑏 ∈ P and the poset
is a meet semi-lattice if Inf{𝑎, 𝑏} exists for any 𝑎, 𝑏 ∈ P.
An algebra 〈A; 0〉 with one binary operation ‘0’ is a semi-lattice if ′0′ is idempotent, commutative and
associative.

Theorem: Any chain of a lattice L is a sub-lattice of L.

OR Let 〈L; ∨, ∧〉 be a lattice. Then ∀𝑎, 𝑏 ∈ L we have 𝑎 ∨ 𝑏 = 𝑏 iff 𝑎 ∧ 𝑏 = 𝑎.
Proof: Assume that, 𝑎 ∧ 𝑏 = 𝑎.
Now, 𝑏 = 𝑏 ∨ (𝑏 ∧ 𝑎) [absorption law]
= 𝑏 ∨ (𝑎 ∧ 𝑏)
=𝑏∨𝑎
=𝑎∨𝑏
∴𝑏 =𝑎∨𝑏
Conversely, let 𝑎 ∨ 𝑏 = 𝑏
Now, 𝑎 = 𝑎 ∧ (𝑎 ∨ 𝑏) [absorption law]
=𝑎∧𝑏
∴𝑎 =𝑎∧𝑏

Example:

(1) For any set A, the collection of all subsets of A (called the power set of A) can be ordered via subset
inclusion to obtain a lattice bounded by A itself and the null set. Set intersection and union interpret
meet and join, respectively (pic.1).
(2) For any set A, the collection of all finite subsets of A, ordered by inclusion, is also a lattice, and will
be bounded if and only if A is finite.
(3) For any set A, the collection of all partitions of A, ordered by refinement, is a lattice (pic.3).
(4) The positive integers in their usual order form a lattice, under the operations of "min" and "max".
1 is bottom; there is no top (pic.4).
(5) The Cartesian square of the natural numbers, ordered so that (𝑎, 𝑏) ≤ (𝑐, 𝑑) if 𝑎 ≤ 𝑐 and 𝑏 ≤ 𝑑.
The pair (0,0) is the bottom element; there is no top (pic.5).
(6) The natural numbers also form a lattice under the operations of taking the greatest common divisor
and least common multiple, with divisibility as the order relation: 𝑎 ≤ 𝑏 if 𝑎 divides 𝑏. 1 is bottom;
0 is top. Pic.2 shows a finite sub lattice.
(7) Every complete lattice (also see below) is a (rather specific) bounded lattice. This class gives rise to
a broad range of practical examples.
(8) The set of compact elements of an arithmetic complete lattice is a lattice with a least element,
where the lattice operations are given by restricting the respective operations of the arithmetic
lattice. This is the specific property which distinguishes arithmetic lattices from algebraic lattices,
for which the compacts do only form a join-semi lattice.
Counter Examples: partial ordered sets are not lattices, including the following-

(1) A discrete poset, meaning a poset such that 𝑥 ≤ 𝑦 implies 𝑥 = 𝑦, is a lattice if and only if it has at
most one element. In particular the two-element discrete poset is not a lattice.
(2) Although the set {1,2,3,6} partially ordered by divisibility is a lattice, the set {1,2,3} so ordered is
not a lattice because the pair 2,3 lacks a join, and it lacks a meet in {2,3,6}.
(3) The set {1,2,3,12,18,36} partially ordered by divisibility is not a lattice. Every pair of elements has
an upper bound and a lower bound, but the pair 2,3 has three upper bounds, namely 12, 18, and
36, none of which is the least of those three under divisibility (12 and 18 do not divide each other).
Likewise the pair 12,18 has three lower bounds, namely 1, 2, and 3, none of which is the greatest
of those three under divisibility (2 and 3 do not divide each other).
Theorem: A poset 〈L; ≤〉 is a lattice iff it is meet and join semi-lattice.
Proof:
As algebra
Let, 〈L; ≤〉 be a lattice as algebra. Then the binary operations ∨ and ∧ satisfies the idempotency,
commutativity, associativity and absorption. By the definition of semi-lattice, we say that 〈L; ≤〉 is meet and
join semi-lattice.

Conversely let 〈L; ≤〉 be a meet and join semi lattice as an algebra. So ∨ and ∧ satisfies idempotency,
commutativity and associativity.

To prove 〈L; ≤〉 is a lattice we have to only check that, 𝑎 ∧ (𝑎 ∨ 𝑏) = 𝑎 and 𝑎 ∨ (𝑎 ∧ 𝑏) = 𝑎.

Define 𝑎 ∧ 𝑏 = inf {𝑎, 𝑏} and 𝑎 ∨ 𝑏 = sup {𝑎, 𝑏}

Now, 𝑎 ≤ 𝑎 ∨ 𝑏
∴ 𝑎 ∧ 𝑎 ≤ 𝑎 ∧ (𝑎 ∨ 𝑏)
⇒ 𝑎 = 𝑎 ∧ 𝑎 ≤ 𝑎 ∧ (𝑎 ∨ 𝑏) ≤ 𝑎
Hence, 𝑎 = 𝑎 ∧ (𝑎 ∨ 𝑏)

Again, 𝑎 ∧ 𝑏 ≤ 𝑎
∴ 𝑎 ∨ (𝑎 ∧ 𝑏) ≤ 𝑎 ∨ 𝑎
⇒ 𝑎 ≤ 𝑎 ∨ (𝑎 ∧ 𝑏) ≤ 𝑎 ∨ 𝑎 = 𝑎
Hence, 𝑎 = 𝑎 ∨ (𝑎 ∧ 𝑏)

So, 〈L; ≤〉 is a lattice as algebra iff it is meet and join semi-lattice.

As poset
Let, 〈L; ≤〉 be a lattice as a poset. Then 𝑥 ∨ 𝑦 and 𝑥 ∧ 𝑦 exists for all 𝑥, 𝑦 ∈ L.

By the definition of semi-lattice, we can say that 〈L; ≤〉 is a meet and join semi-lattice.

Conversely, let 〈L; ≤〉 be a meet and join semi lattice as a poset. So inf {𝑎, 𝑏} and sup {𝑎, 𝑏} exists for all
𝑎, 𝑏 ∈ L.

Define 𝑎 ∧ 𝑏 = inf {𝑎, 𝑏} and 𝑎 ∨ 𝑏 = sup {𝑎, 𝑏}

Then 𝑎 ∧ 𝑏 and 𝑎 ∨ 𝑏 exists for all 𝑎, 𝑏 ∈ L.

So, 〈L; ≤〉 is a lattice as poset.

Sub-lattice generated by a subset: Let H be a subset of L. Then sub-lattice generated by H is the

intersection of all sub-lattice of L containing H. It is denoted by [H].
i.e. [H] = ⋂𝑖{H𝑖 are the sublattice of L containing H}.
Example:

Let H = {𝑎, 𝑐}, then 𝑎 ∨ 𝑐 = 𝑐 ∈ H and 𝑎 ∧ 𝑐 = 𝑎 ∈ H. Then H is a sub-lattice.

Then {𝑎, 𝑏, 𝑐, 1}, {0, 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 1} contains H.

∴ [H] = {𝑎, 𝑏, 𝑐, 1} ∩ {0, 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 1} = {𝑎, 𝑏, 𝑐, 1}.

Prime elements: Let L be a lattice. We call an element 𝑗 ≠ 0 of L join-prime, if 𝑗 ≤ 𝑎 ∨ 𝑏 implies

𝑗 ≤ 𝑎 or 𝑗 ≤ 𝑏 for all 𝑎, 𝑏 ∈ L. Let ℑ(L) denote the set of join-prime elements of L.
Dually, we call an element 𝑚 ≠ 1 of L meet-prime if 𝑎 ∧ 𝑏 ≤ 𝑚 implies 𝑎 ≤ 𝑚 or 𝑏 ≤ 𝑚 for all 𝑎, 𝑏 ∈
L. Let ℳ(L) denote the set of meet-prime elements of L.
Examples: In the lattice U(P) of upsets of a poset P, the up-sets ↑ 𝑝 = {𝑥 ∈ P|𝑥 ≥ 𝑝} are join-prime
elements for any 𝑝 ∈ P.
Similarly, in the lattice D(P) of down-sets of P, the down-sets ↓ 𝑝 = {𝑥 ∈ P|𝑥 ≤ 𝑝}are join-prime for
all 𝑝 ∈ P.
Note that instead of working with ℑ(𝐿) and U(P), we could alternatively work with M(L) and D(P).
The result would be the same!
One of the consequences of the Birkhoff duality is the following representation theorem for finite
distributive lattices:
Representation Theorem for Finite Distributive Lattices: Every finite distributive lattice can be
represented as the lattice of upsets (down-sets) of some poset. It is our goal to extend the Birkhoff
duality to all distributive lattices.

Ideals: Let L be a lattice. A nonempty subset I of L is called an ideal of L if the following two conditions
are satisfied:
1) If 𝑎, 𝑏 ∈ I, then 𝑎 ∨ 𝑏 ∈ I
2) 𝑎 ∈ L, 𝑏 ∈ I and 𝑎 ≤ 𝑏 implies that 𝑎 ∈ I.

Equivalently, I ≠ ∅ is an ideal of L if for each 𝑎, 𝑏 ∈ L we have 𝑎, 𝑏 ∈ I iff 𝑎 ∨ 𝑏 ∈ I.

Example:

In fig-1, {0, 𝑏, 𝑐, 𝑓} is an ideal.

 In fig-2, {𝑏, 𝑐, 𝑓} is not an ideal, because 0 ∈ L, 𝑏 ∈ {𝑏, 𝑐, 𝑓} doesn’t implies that 0 ∈ {𝑏, 𝑐, 𝑓},
(0 ≤ 𝑏)
In fig-3, {0, 𝑎, 𝑏, 𝑐, 𝑓} is not an ideal, because 𝑎 ∨ 𝑏 = 𝑑 ∉ {𝑏, 𝑐, 𝑓}.

Principal Ideals: An ideal I, generated by a single element 𝑎 is called a principal ideal which is
denoted by (𝑎] and defined as (𝑎] = {𝑥 ∈ L|𝑥 ≤ 𝑎}, where L is a lattice.
Example:

In the figure, principle ideals are (0] = {0}; (𝑎] = {0, 𝑎}; (𝑏] = {0, 𝑏}; (𝑐] = {0, 𝑎, 𝑏, 𝑐}; (1] =
{0, 𝑎, 𝑏, 𝑐, 1} = L

Proper Ideal and Improper Ideal: An ideal I of a lattice L is proper if I ≠ L and improper if I = L.

Prime Ideal: A 𝑝𝑟𝑜𝑝𝑒𝑟 𝑖𝑑𝑒𝑎𝑙 P of a lattice L is said to be a prime ideal if for any 𝑥, 𝑦 ∈ L with 𝑥 ∧ 𝑦 ∈
P implies either 𝑥 ∈ P or 𝑦 ∈ P.

Example:

Here {0, 𝑎} is a proper ideal and (1] = {0, 𝑎, 𝑏, 1} is an improper ideal.

{0} is a proper ideal but not a prime ideal because 𝑎 ∧ 𝑏 = 0 ∈ {0}, but 𝑎 ∉ {0} and 𝑏 ∉ {0}.
{0, 𝑎} is a prime ideal because 𝑎 ∧ 𝑏 = 0 ∈ {0, 𝑎} and 𝑎 ∈ {0, 𝑎}.
{0, 𝑏} is a prime ideal because 𝑎 ∧ 𝑏 = 0 ∈ {0, 𝑏} and 𝑏 ∈ {0, 𝑏}.

Filter: Let L be a lattice. A nonempty subset F of L is called a filter of L if the following two conditions
are satisfied:
1) If 𝑎, 𝑏 ∈ F, then 𝑎 ∧ 𝑏 ∈ F.
2) 𝑎 ∈ F, 𝑏 ∈ L and 𝑎 ≤ 𝑏 implies that 𝑏 ∈ F.

Prime Filter: Let L be a lattice and F ≠ L be a filter of L. We call F a prime filter of L if for all 𝑎, 𝑏 ∈ L
we have: 𝑎 ∨ 𝑏 ∈ F implies that 𝑎 ∈ F or 𝑏 ∈ F.
Let X(L) denote the set of prime filters of L and Y(L) denote the set of prime ideals of L. Thus, a filter
F is prime iff its complement I = L − F is an ideal, which is then a prime ideal. Similarly, an ideal I is
prime iff its complement F = L − I is a filter, which is then a prime filter.
Example:
In a linear order, every upset is a prime filter, and every down-set is a prime ideal.

Principal Dual Ideal (Principal Filter): A dual ideal generated by a single element 𝑎 is called a
principal filter, which is denoted by [𝑎) and defined as [𝑎) = {𝑥 ∈ L|𝑥 ≤ 𝑎}, where L is a lattice.
Example:

Principle filters are

[1) = {1};
[𝑎) = {𝑎, 𝑐, 1};
[𝑏) = {𝑏, 𝑐, 1};
[𝑐) = {𝑐, 1};
[0) = {0, 𝑎, 𝑏, 𝑐, 1} = L.

Theorem: Every ideal is a convex sub-lattice and every filter (dual ideal) is also a convex sub-lattice
but converse is not true.
Proof:
For ideal: Let 𝑎, 𝑏 ∈ I where I is an ideal with 𝑎 ≤ 𝑐 ≤ 𝑏.
Since I is an ideal and 𝑐 ≤ 𝑏 implies that 𝑐 ∈ I.
Therefore I is a convex sub-lattice.

For filter: Let, Let 𝑎, 𝑏 ∈ F where F is a filter with 𝑎 ≤ 𝑐 ≤ 𝑏.

Since F is a filter and 𝑎 ≤ 𝑐 implies that 𝑐 ∈ F.
Therefore F is a convex sub-lattice.

Converse: Let us consider the following diagram-

Here {0, 𝑎, 1} is a convex sub-lattice of L, but not an ideal. Because 𝑏 ∈ L and 1 ∈ I with 𝑏 ≤ 1 but 𝑏 ∉ I.

Here {0, 𝑎, 1} is a convex sub-lattice of L, but not a filter. Because 𝑏 ∈ L and 0 ∈ F with 0 ≤ 𝑏 but 𝑏 ∉ F.
Theorem: Intersection of two ideal is an ideal but the intersection of two prime ideal need not to be
prime.
Proof: Let A, B be two ideals of a lattice L. Since A, B are non-empty, there exists some 𝑎 ∈ A, 𝑏 ∈ B.
Now 𝑎 ∈ A, 𝑏 ∈ B ≤ L ⇒ 𝑎 ∧ 𝑏 ∈ A. Similarly 𝑎 ∧ 𝑏 ∈ B.
Which implies 𝑎 ∧ 𝑏 ∈ A ∩ B
Thus, A ∩ B ≠ ∅
Let, 𝑥, 𝑦 ∈ A ∩ B be any elements.
⇒ 𝑥, 𝑦 ∈ A and 𝑥, 𝑦 ∈ B
⇒ 𝑥 ∨ 𝑦 ∈ A and 𝑥 ∨ 𝑦 ∈ B [since A, B both are ideals of lattice L]
⇒ 𝑥∨𝑦 ∈A∩B
Again, if 𝑥 ∈ A ∩ B and 𝑙 ∈ L be any elements then
𝑥 ∈ A, 𝑥 ∈ B, 𝑙 ∈ L
⇒ 𝑥 ∧ 𝑙 ∈ A and 𝑥 ∧ 𝑙 ∈ B
⇒𝑥∩𝑦 ∈A∩B

Hence, A ∩ B is an ideal.

Let us consider the following diagram-

Here I1 = {0, 𝑎}, I2 = {0, 𝑏} are two prime ideal.

∴ I1 ∩ I2 = {0, 𝑎} ∩ {0, 𝑏} = {0}
∴ {0} is a proper ideal but not prime. Because 𝑎 ∧ 𝑏 ∈ {0}, but 𝑎 ∉ {0}, 𝑏 ∉ {0}

Hence the intersection of two prime ideal need not to be prime.

Diamond Lattice: The lattice L = {0, 𝑎, 𝑏, 𝑐, 1} as shown in the figure is called the diamond lattice,
which is denoted by 𝐌𝟑 .

Pentagonal Lattice: The lattice L = {0, 𝑎, 𝑏, 𝑐, 1} shown in the figure is called the pentagonal lattice,
which is denoted by 𝐍𝟓 or 𝐑 𝟓 .
Distributive Lattice: A lattice L is called distributive lattice iff it doesn’t contains diamond or
pentagonal lattice. Let L be a lattice. Then L is said to be a distributive lattice if it satisfies the
distributive law. That is ∀𝑎, 𝑏, 𝑐 ∈ L
𝑎 ∧ (𝑏 ∨ 𝑐) = (𝑎 ∧ 𝑏) ∨ (𝑎 ∧ 𝑐) and 𝑎 ∨ (𝑏 ∧ 𝑐) = (𝑎 ∨ 𝑏) ∧ (𝑎 ∨ 𝑐)

Modular Lattice: A lattice L is said to be a modular if ∀𝑥, 𝑦, 𝑧 ∈ L with 𝑧 ≤ 𝑥 implies 𝑥 ∧ (𝑦 ∨ 𝑧) =

(𝑥 ∧ 𝑦) ∨ 𝑧.

Example: Diamond lattice i.e. M3 is modular but pentagonal lattice N5 or R 5 is not modular.

Question: Find ideals of diamond lattice M3 and pentagonal lattice N5 or R 5 .

Solution:

{0}; {0, 𝑎}; {0, 𝑏}; {0, 𝑐}; {0, 𝑎, 𝑏, 𝑐, 1} are ideals of M3 .
{0}; {0, 𝑎}; {0, 𝑏}; {0, 𝑎, 𝑐}; {0, 𝑎, 𝑏, 𝑐, 1} are ideals of N5 .