6
Relations
Let R be a relation on a set A, i.e., a subset of AxA.
Notation: xRy iff (x, y) ∈ R ⊆ AxA.
Recall: A relation need not be a function.
Example: The relation R1 = {(x, y) ∈ RxR | x2 + y 2 = 1} is
not a function.
Some definitions
1. R is reflexive if xRx ∀x ∈ A.
2. R is symmetric if xRy ⇒ yRx ∀x, y ∈ A.
3. R is antisymmetric if xRy ∧ yRx ⇒ x = y ∀x, y ∈ A.
4. R is transitive if xRy ∧ yRz ⇒ xRz ∀x, y, z ∈ A.
We now give a number of examples.
1. R = “ < “ on Z.
R, S, A, T.
2. R = “ ≤ “ on Z.
R, S, A, T.
3. R = “ = “ on Z.
R, S, A, T.
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4.
R, S, A, T .
5.
R, S, A, T .
6.
R, S, A, T .
7.
R, S, A, T .
8.
R, S, A, T .
9.
R, S, A, T .
10.
R, S, A, T .
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Note:
1. In a reflexive relation there is a loop at each vertex.
2. In a symmetric relation, there are either 2 arcs or no arcs
between any two distinct nodes.
3. In an antisymmetric relation there is either 1 arc or no
arcs between any two distinct nodes.
Definition: If R is a relation from X to Y , then the inverse
of R is R−1 = {(y, x) | (x, y) ∈ R}.
Examples:
R:
R−1:
R:
R−1:
R:
R−1:
Theorem 6.1 R is symmetric iff R = R−1 .
Theorem 6.2 The reflexive, symmetric, antisymmetric and
transitive properties of relations are preserved by the inverse, i.e., if R has such a property then so does R−1.
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Equivalence Relations
Definition: A relation R ⊆ AxA is an equivalence relation if it is reflexive, symmetric and transitive.
Examples:
1.
00
=00 on Z.
2. The universal relation UA = AxA, i.e., the relation consisting of all elements of AxA.
3. Let A be the set of all triangles in the plane. Then T1 RT2
iff T1 and T2 are similar triangles.
4. Let A be the set of all points in the plane. Then p1 Rp2
iff the distance from p1 to the origin equals the distance
from p2 to the origin.
5. A = Z, m > 0. aRm b iff m | a − b, i.e., ∃c ∈ Z such that
m · c = a − b.
(a) Rm is reflexive since m | a − a.
(b) Rm is symmetric since m | a − b ⇒ m | b − a.
(c) Rm is transitive since if m | a − b and m | b − c, then
m | (a − b) + (b − c) or m | a − c.
Notation: If m | a − b we say a is congruent to b
mod m, or a ≡ b (mod m).
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6. Let f : A → B. Then Rf given by a1Rf a2 iff f (a1) =
f (a2) is an equivalence relation.
7.
Definition: Let R be an equivalence relation on A and b ∈ A.
Then [b] = {x ∈ A | xRb} is the equivalence class generated by b.
Example: Let A = Z and consider aR3 b. Thus a and b are
related iff 3 | a − b. Then
[0] = {. . . − 6, −3, 0, 3, 6, . . .}
[1] = {. . . − 5, −2, 1, 4, 7, . . .}
[2] = {. . . − 4, −1, 2, 5, 8, . . .}
Example: In (7) above, [1] = [2] = {1, 2} and
[3] = [4] = {3, 4}.
Definition: Let A = ∪α∈ΛAα, where each Aα 6= φ and the
Aα’s are pairwise disjoint. Then {Aα | α ∈ Λ} is a partition
of A.
Example: A1, A2, . . . , A7 is a partition of A.
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Note: A partition of a set defines an equivalence relation in a
very natural way.
Definition: Let P be a partition of a set A. Then the equivalence relation R(P) associated with P is given by: aR(P )b
iff a and b are in the same set in P .
Note: R(P ) is clearly an equivalence relation.
Example: The sets A1, A2, A3 partition Z.
A1 = {. . . − 6, −3, 0, 3, 6, . . .}
A2 = {. . . − 5, −2, 1, 4, 7, . . .}
A3 = {. . . − 4, −1, 2, 5, 8, . . .}
Thus 1R(P)7 and -4R(P)8.
We now wish to show that each equivalence relation on a set
A defines a partition in a natural way.
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Theorem 6.3 Let R be an equivalence relation on a set
A. Then
1. b ∈ [b] ∀b ∈ A.
2. ∀a, b ∈ A, [a] = [b] ⇔ aRb.
3. ∀a, b ∈ A, either [a] = [b] or [a] ∩ [b] = φ.
Proof: First recall that [b] = {x ∈ A | xRb}.
1. Since R is reflexive, bRb. Hence b ∈ [b].
2. (⇒) Suppose [a] = [b]. Since a ∈ [a], a ∈ [b]. Hence aRb.
(⇐) Suppose x ∈ [b]. Then xRb. Also, since aRb and R
is symmetric, bRa. Since R is also transitive, xRa, i.e.,
x ∈ [a]. Hence [b] ⊆ [a]. Similarly [a] ⊆ [b].
3. Suppose x ∈ [a] ∩ [b]. Then xRa and xRb. Since R is
symmetric, aRx. Now since R is transitive, aRb. Hence
by (2), [a] = [b]. 2
Let P (R) = {[a] | a ∈ A}. If R∗ is an equivalence relation on
a set A, the distinct sets P (R∗) partition A. Thus we have a
1-1 correspondence between the partitions of a set A and the
equivalence relations on A.
Note: P (R(P )) = P and R(P (R∗ )) = R∗.
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Posets
Let R be a relation on a set A. Then R is a partial ordering
on A if R is
1. reflexive
2. antisymmetric
3. transitive
Examples:
1. R : “ ≤ “ on Z.
2. R : “ ⊆ “ on P(A), the power set of A.
3. R : “divides” on Z +.
(a) a | a.
(b) a | b and b | a ⇒ a = b.
(c) a | b and b | c ⇒ a | c.
Definition: If R is a partial ordering on A we call (A, R) a
partially ordered set or poset.
Note: A subset of a partially ordered set is a partially ordered
set (with the same ordering).
Notation: When the relation is a partial ordering, we often
use a ≤ b instead of aRb.
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Definition: Suppose (A, R) is a poset. Elements a and b of
A are said to be comparable if, and only if, either aRb or
bRa. Otherwise they are noncomparable.
Definition: Let R be a partial order relation on a set A. If
any two elements a and b in A are comparable, then R is a
total order relation on A.
Examples:
1. R : “ ≤ “ on Z is a total order.
2. R : “ ⊆ “ on P (A) is not a total order if A has more than
1 element.
3. R : “divides” is not a total order on Z +, e.g., 3 does not
divide 5 and 5 does not divide 3.
Definition: Let (A, R) be a poset. A subset B of A is called a
chain if, and only if, each pair of elements in B is comparable.
The length of a chain is the number of elements in the chain.
Note: The book has a different definition of length.
Example: The set P ({a, b, c}) is partially ordered with respect
to subset inclusion. The set S = {φ, {a}, {a, b}, {a, b, c}} is
a chain of length 4 in P ({a, b, c}).
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Hasse Diagrams
Let A = {0, 1} and consider the poset (P (A), ⊆).
Hasse diagram
Properties of Hasse Diagrams
• arrows are omitted - edges are directed upward
• self loops are omitted
• edges implied by transitivity are omitted
More examples:
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Definition: A subset of a poset (A, R) is an antichain if no
two distinct elements of the subset are related.
Example: {c, f, e}
Definition: Let (A, ≤) be a poset. An element a ∈ A is a
maximal element if there does not exist b ∈ A such that
b 6= a and a ≤ b.
Note: minimal element is defined similarly.
In the examples above
• a is a maximal element
• g is a minimal element
• 1, 2, 7 are maximal elements
• 8, 9, 10 are minimal elements
Note: Any finite, nonempty partially ordered set has a minimal (and maximal) element.
Theorem 6.4 Let (A, ≤) be a poset. If n is the length of
a longest chain in (A, ≤), then A can be partitioned into n
disjoint antichains.
Proof: Later, by induction.
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Well Orderings
Let (A, ≤) be a poset and B ⊆ A. An element b ∈ B is a
0
0
greatest element of B if b ≤ b ∀ b ∈ B. It is a least
0
0
element of B if b ≤ b ∀ b ∈ B.
Example: Consider (N, ≤) and B = {3, 7, 12, 15}. Then 3 is
a least element of B and 15 is a greatest element of B.
Theorem 6.5 Let (A, R) be a poset and B ⊆ A. If x and
y are greatest elements of B, then x = y.
Proof: If x and y are greatest elements of B, then x ≤ y
and y ≤ x. Since R is antisymmetric, x = y. 2
Note: Least elements are also unique.
Definition: A relation R on A is a well ordering if R is
a total ordering and every nonempty subset of A has a least
element.
Examples:
• Any finite total ordering is a well ordering.
• (R+ ∪ {0}) is NOT a well ordering since R+ ⊆ R+ ∪ {0}
does not have a least element.
• (Z, ≤) is NOT a well ordering since Z has no least element.
• (N, ≤) is a well ordering - this is actually taken as an
axiom.
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Theorem 6.6 Every set can be well ordered.
Note: It is not always easy to find the ordering.
Example: The integers can be well ordered as follows:
Let f : Z → N be defined by



f (k) = 

2k
if k ≥ 0
-2k - 1 if k < 0.
Note: f (0) = 0, f (−1) = 1, f (1) = 2, f (−2) = 3, etc.
Note: Now a“ ≤ “b iff f (a) ≤ f (b) is a well ordering.
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Closure Operations on Relations
Example: Suppose we define a relation R on a set A of cities
as follows: aRb iff there is a direct communication link from
city a to city b for transmission of messages.
Problem: Find a relation that describes how messages can be
transmitted from one city to another, either through a direct
communication link, or through any number of intermediate
cities.
Definition: Let R be a relation on a set A. The transitive
closure of R is a relation Rt such that
1. Rt is transitive
2. R ⊆ Rt
3. If R1 is transitive and R ⊆ R1 , then Rt ⊆ R1 .
Note: The transitive closure is unique.
Note: The reflexive and symmetric closures are defined in an
analogous way.
Example:
Rt :
R:
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Theorem 6.7 Let {Sα | α ∈ Λ} be the set of all transitive
relations containing a relation R. Then Rt = ∩α∈Λ Sα.
Thus the transitive closure of a relation R is the “smallest”
transitive relation containing R. It is obtained by adding the
least number of ordered pairs to ensure transitivity.
Note: A similar theorem holds for reflexive and symmetric
closures.
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Composition of Relations
Definition: Let R1 be a relation from A to B and R2 be a
relation from B to C. The composition of R1 and R2 is a
relation from A to C given by
R1 R2 = {(a, c) | a ∈ A, c ∈ C ∧ ∃b ∈ B
such that [(a, b) ∈ R1 ∧ (b, c) ∈ R2 ]}.
Example:
Note: In general, R1 R2 6= R2 R1 .
In fact, if R1 is a relation from A to B and R2 is a relation
from B to C, then R2 R1 is not defined.
Example: Let A = {0, 1, 2, 3} and consider R1 and R2 on A.
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Theorem 6.8 Let R1 ⊆ AxB, R2 ⊆ BxC and R3 ⊆ CxD.
Then (R1 R2 )R3 = R1(R2 R3 ), i.e., the composition of relations is associative.
Proof: (⊆) Let (a, d) ∈ (R1 R2 )R3 . Then ∃c ∈ C such that
(a, c) ∈ R1 R2 and (c, d) ∈ R3 . Since (a, c) ∈ R1 R2 ∃b ∈ B
such that (a, b) ∈ R1 and (b, c) ∈ R2 . Now (b, c) ∈ R2
and (c, d) ∈ R3 ⇒ (b, d) ∈ R2 R3 . But now (a, b) ∈ R1 ⇒
(a, d) ∈ R1 (R2 R3 ).
(⊇) Similar.
2
Definition: Let R be a binary relation on a set A. Then Rn
is defined as follows:
1. R0 = {(x, x) | x ∈ A}.
2. Rn+1 = RnR.
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Example:
• R0 :
• R1 = R:
• R2 = R1R:
• R3 = R2R:
• R4 = R3R:
Note: In this example, R4 = R2.
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Theorem 6.9 Let |A| = n and R ⊆ AxA. Then ∃s, t,
2
0 ≤ s < t ≤ 2n , such that Rs = Rt .
Proof: First note that AxA has n2 elements. Hence there
2
are 2n distinct relations on A. By the pigeonhole principle, at
least two of them are equal. 2
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