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CS 173:

Discrete Mathematical Structures

Partially Ordered Sets (POSets)

Let R be a relation then R is a Partially

Ordered Set (POSet) if it is

Reflexive - aRa, ∀a

Transitive - aRb ∧ bRc → aRc, ∀a,b,c

Antisymmetric - aRb ∧ bRa → a=b,

∀a,b

Ex. ( R ,≤), the relation “≤” on the real numbers, is a

partial order.

How do you check?

Reflexive? a ≤ a for any real

Transitive?

Antisymmetric?

If a ≤ b, b ≤ c then a ≤ c

If a ≤ b, b ≤ a then a = b

order

Actually totalorder. <

Partially Ordered Sets (POSets)

Ex. (Z, | ), the relation “divides” on integers.

Reflexive? Yes, x|x since x=1x (k=1)

Transitive?

Antisymmetric?

a|b means b=ak, b|c means c=bj. Does c=am for some m?

c = bj = akj (m=kj)

3|-3, and -3|3, but 3 ≠ -3.

Not a poset.

Yes, or No?

Partially Ordered Sets (POSets)

Ex. (2 S^ , ⊆ ), the relation “subset” on set of all subsets of S.

Reflexive? Yes, A ⊆ A, ∀A∈ 2 S

Transitive?

Antisymmetric?

A ⊆ B, B ⊆ C. Does that mean A ⊆ C? A ⊆ B means x ∈A → x ∈B

A ⊆ B, B ⊆ A → A=B

A poset.

B ⊆ C means x ∈B → x ∈C

Now take an x, and suppose it’s in A. Must it also be in C? (^) Yes, by… A. Modus Ponens B. Modus Tollens C. DeMorgan’s D. Transitivity

Yes, by MP

Partially Ordered Sets (POSets)

Ex. A common partial order on bit

strings of length n, {0,1} n, is defined

as:

a 1 a 2 …a n ≤ b 1 b 2 …bn

If and only if a0110 and 1000 are “incomparable” … We can’t tell which is i ≤ bi , ∀ i.

“bigger.”

As a bit of an aside, this relation is exactly the same as the last example, (2 S^ , ⊆ ).

A. 0110 ≤ 1000

B. 0110 ≤ 0000

C. 0110 ≤ 1110

D. 0110 ≤ 10111

Huh?

Set S, on which we build 2 S, has a size. That’s n.

Suppose S is {a,b}. Then 2 S^ = { {}, {a}, {b}, {a,b} }

Think of bit strings as membership indicators for the elts of S

Then 2 S^ can be represented by {00,10,01,11}

Partially Ordered Sets (POSets)

0110 and 1000 are “incomparable” … We can’t tell which is “bigger.”

As a bit of an aside, this relation is exactly the same as the last example, (2 S^ , ⊆ ).

In the string relation, we said 00≤01 because every bit

in 00 is less than or = the corresp bit in 01. String on the right has at least all the 1 bits of the left, maybe more. If each 1 represents an element in S, then right side has all elts of the left, maybe more.

Set S, on which we build 2 S, has a size. That’s n.

Suppose S is {a,b}. Then 2 S^ = { {}, {a}, {b}, {a,b} }

Think of bit strings as membership indicators for the elts of S

Then 2 S^ can be represented by {00,10,01,11}

Hasse Diagrams

Hasse diagrams are a special kind of

graphs used to describe posets.

Ex. In poset ({1,2,3,4}, ≤), we can draw the following picture

to describe the relation.

1. Draw edge (a,b) if a ≤b

2. Don’t draw up arrows
3. Don’t draw self loops
4. Don’t draw transitive edges

4

3

2

1

Hasse Diagrams

Have you seen this one before?

{a,b,c} or 111

{a,b} or 110 {a,c} or 101 {b,c} or 011

{a} or 100 {b} or 010 {c} or 001

{} or 000

Hasse Diagrams

Definition: In a poset S, an element z is

a minimum element if ∀b∈S, z≤b.

Write the defn

of maximum!

Did you get it right?

Intuition: If a is maxiMAL, then no one beats a. If a is maxiMUM, a beats everything.

Must minimum and maximum exist? A. Only if set is finite. B. No. C. Only if set is transitive. D. Yes.

Hasse Diagrams

Theorem: In every poset, if the maximum

element exists, it is unique. Similarly for

minimum.

Proof: Suppose there are two maximum elements, a 1 and a 2 , with a 1 ≠a 2. Then a 1 ≤ a 2 , and a 2 ≤a 1 , by defn of maximum. So a 1 =a 2 , a contradiction. Thus, our supposition was incorrect, and the maximum element, if it exists, is unique. Similar proof for minimum.

Upper and Lower Bounds

Defn: Let (S, ≤) be a partial order. If

A⊆S, then an upper bound for A is any

element x ∈ S (perhaps in A also) such

that ∀ a ∈ A, a ≤ x.

Ex. The upper bound of {g,j} is a. Why not b?

A lower bound for A is any x ∈ S such that

∀ a ∈ A, a ≥ x.

a b

d

f j

h i

e

c

g

Ex. The upper bounds of {g,i} is/are… A. I have no clue. B. c and e C. a D. a, c, and e

{a, b} has no UB.

Upper and Lower Bounds

Defn: Let (S, ≤) be a partial order. If

A⊆S, then an upper bound for A is any

element x ∈ S (perhaps in A also) such

that ∀ a ∈ A, a ≤ x.

Ex. The lower bounds of {a,b} are d, f, i, and j.

A lower bound for A is any x ∈ S such that

∀ a ∈ A, a ≥ x.

a b

d

f j

h i

e

c

g

Ex. The lower bounds of {c,d} is/are… A. I have no clue. B. f, i C. j, i, g, h D. e, f, j

{g, h, i, j} has no LB.

Upper and Lower Bounds

Ex. In the following poset, c and d are

lower bounds for {a,b}, but there is no

GLB.

Similarly, a and b are upper bounds for {c,d},

but there is no LUB.

a b

c d

This is because c and d are incomparable.

Total Orders

Consider the problem of getting dressed.

Precedence constraints are modeled by a poset in

which a ≤ b iff you must put on a before b.

In what order will you get dressed while respecting constraints?

shoes belt jacket

socks jeans swter

uwear shirt

jwlry

Let (S, ≤) be a poset (S finite). We will extend ≤ to a total order on S, so we can decide for all incomparable pairs whether to make a ≤ b, or vice versa w/o violating T,R,A.