Section 3 : Partially Ordered Sets and Semilattices
Let X be a set. A binary relation ρ on X is a subset of
X × X = {(x, y)|x, y ∈ X}.
For ρ a binary relation on X, we normally write “x ρ y” rather than “(x, y) ∈ ρ”.
A binary relation “≤” is a partial order on X if the following hold for all
a, b, c ∈ X
• P(i) :
• P(ii) :
a≤a
a ≤ b, b ≤ a ⇒ a = b
• P(iii) : a ≤ b, b ≤ c ⇒ a ≤ c
Then we say (X, ≤) is a partially ordered set or a poset. If it is also the case that
for all a, b ∈ X, either a ≤ b or b ≤ a, we say (X, ≤) is a totally ordered set
and “≤” is a total order .
Examples:
• (i) (R, ≤) where “≤” is the usual ordering of reals, is a totally ordered set.
Similarly, (Q, ≤), (Z, ≤), (N, ≤) are totally ordered.
• (ii) On N, define m|n if m divides n; i.e. n = km for some k ∈ N. e.g. 2|4, 3
does not divide 5, etc. Then (N, |) is a poset, but NOT totally ordered.
• (iii) Let S be a set and let X = 2S , the power set of S = the set of all subsets
of S. Then if A ⊆ B has its usual meaning, (X, ⊆) is a poset (not totally
ordered). Try the proof yourself as an exercise. This last example is the most
important for what lies ahead of us.
Any subset Y of a poset (X, ≤) gives a poset (Y, ≤Y ), defined by y1 ≤Y y2 iff
y1 ≤ y2 , ∀y1 , y2 ∈ Y . We normally just write (Y, ≤), and call it a sub-poset of (X, ≤).
For two posets (X1 , ≤1 ) and (X2 , ≤2 ), we say they are isomorphic if there is a
bijection f : X1 → X2 such that ∀a, b ∈ X1 , a ≤1 b iff f (a) ≤2 f (b).
just requiring a ≤1 b ⇒ f (a) ≤2 f (b) is NOT enough. (WHY?)
Note:
Theorem 3.1: Every poset (X, ≤) is isomorphic to a poset of subsets of X ordered
by “⊆” (that is, isomorphic to a sub-poset of (2X , ⊆)).
1
Proof: Let (X, ≤) be a poset. For each x ∈ X, define Sx = {a ∈ X|a ≤ x},
so Sx ⊆ X. Let Y = {Sx |x ∈ X}, then (Y, ⊆) is a sub-poset of (2X , ⊆). Define
f : X → Y by setting f (x) = Sx , ∀x ∈ X. We shall show that f is an isomorphism.
If f (x1 ) = f (x2 ) then Sx1 = Sx2 . So a ≤ x1 ⇔ a ≤ x2 for all a ∈ X. But
x1 ≤ x1 , so x1 ≤ x2 , and similarly, x2 ≤ x1 . So by P(ii), x1 = x2 . So f is 1:1.
For Sx ∈ Y, f (x) = Sx , so f is onto.
If a, b ∈ X are such that a ≤ b, then need to show f (a) ⊆ f (b), i.e. Sa ⊆ Sb .
If x ∈ Sa , then x ≤ a, but a ≤ b, so x ≤ b, by P(iii). That is, x ∈ Sb , so
Sa ⊆ Sb by definition.
Also must show f (a) ⊆ f (b) ⇒ a ≤ b, i.e. Sa ⊆ Sb ⇒ a ≤ b. If Sa ⊆ Sb , then
since a ≤ a by P(i), a ∈ Sa , so a ∈ Sb , so a ≤ b. It is possible to draw finite posets as follows. Put bigger elements higher than
those they are larger than and draw a line between a and b if a > b (i.e. a ≥ b AND
a 6= b) and there is no c such that a > c > b. The result is a Hasse Diagram.
X = {1, 2, 3, 4, 6, 8, 12} is a poset under the partial order | (it is a sub-poset
e.g.
of (N, |)).
(Of course there are other ways to draw it.)
A Hasse Diagram completely determines a partial order on a finite poset.
e.g.
1-element posets:
e.g.
2-element posets:
x ≤ x, y ≤ y, x ≤ y
x ≤ x, y ≤ y
2
e.g.
(2{a,b,c} , ⊆)
Let (X, ≤) be a poset. Two elements x, y have z as a lower bound if z ≤ x
and z ≤ y. Suppose every x and y have a greatest lower bound , glb(x, y), a lower
bound such that if z ≤ x and z ≤ y, then z ≤ glb(x, y). Then the following are true
for all x, y, z ∈ X:
• G(i):
glb(x, x) = x
• G(ii): glb(x, y) = glb(y, x)
• G(iii): glb(glb(x, y), z) = glb(x, glb(y, z)).
The proofs of G(i) and G(ii) are clear. The proof of G(iii) is left as an exercise.
So glb defines a binary operation on any such a poset for which glb(x, y) always
exists (and we call the poset a semilattice in this case). So (X, glb) is a semigroup
if (X, ≤) is a semilattice. Indeed it is a commutative semigroup also satisfying
x2 = x.
We can instead start with such semigroups. Let (X, ∧) be a semigroup satisfying
also: x ∧ y = y ∧ x and x ∧ x = x, ∀x, y ∈ X. We call (X, ∧) a semilattice
algebra. So if (X, ≤) is a semilattice, then (X, glb) is a semilattice algebra. In
fact all semilattice algebras arise in this way.
Theorem 3.2:
Let (S, ∧) be a semilattice algebra. Define a ≤ b to mean a∧b = a.
Then ≤ is a partial order and (S, ≤) is a semilattice with glb(a, b) = a∧b, ∀a, b ∈ S.
Proof: We show ≤, as defined, is a partial order on S.
P(i):
∀a ∈ S, a ∧ a = a, so a ≤ a.
P(ii):
If a ≤ b and b ≤ a, then a = a ∧ b, b = b ∧ a = a ∧ b = a.
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If a ≤ b and b ≤ c, then a = a ∧ b, b = b ∧ c; so
P(iii):
a =
=
=
=
a∧b
a ∧ (b ∧ c)
(a ∧ b) ∧ c
a ∧ c,
and so a ≤ c.
We show (S, ≤) is a semilattice for which glb(a, b) = a ∧ b, ∀a, b ∈ S. Now
(a ∧ b) ∧ a =
=
=
=
(b ∧ a) ∧ a
b ∧ (a ∧ a)
b∧a
a ∧ b.
So by definition, a∧b ≤ a. Interchanging a and b gives b∧a ≤ b. But b∧a = a∧b. So
a ∧ b ≤ a, b, so a ∧ b is a lower bound of a, b. If also c ≤ a, c ≤ b, then c = c ∧ a = c ∧ b.
So
c ∧ (a ∧ b) = (c ∧ a) ∧ b
= c∧b
= c.
S c ≤ a ∧ b. Hence glb(a, b) = a ∧ b.
Note: when one starts with a semilattice (S, ≤), forms a semilattice algebra (S, glb),
and defines a “new” partial order as in Theorem 3.2 proof, we find the new partial
order equals the original one. (The proof is left as an exercise!)
Overall then, these ways of converting between semilattices and semilattice algebras are mutually inverse: a ≤ b iff a = a ∧ b with glb(a, b) = a ∧ b. So semilattices
and semilattice algebras are “really” the same things. (This is an example of a cryptomorphism).
From now on, we use the term “semilattice” interchangably for both. We call ∧
“meet” in a semilattice (S, ∧) (since one of the main examples is intersection of sets
– see below).
Two examples of semilattices:
(i)
(2X , ∩) where X is a set.
under intersection.
Note:
(ii)
H1 , H2 ≤ G ; H1 ∪ H2 ≤ G.
Exercise. Show these claims.
4
the subuniverses of algebras