math 103 - 10 relations
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Chapter 10
Relations
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Section 10.1
Relations on Sets
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Relations on a Set
A relation from a set A to a set B is a subsetof A × B.
A relation on a set A is a relation from A to
A, i.e., a subset of A × A.
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Examples: Relations on a Set
Define the relation on the integers asa b if a divides b.
Define the relation < on the real numbers as
x < y if x is less than y.
Define the relation ≤ on a Boolean algebra
as a ≤ b if a ⋅ b = a.
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Inverse Relations
Let R be a relation from A to B.
The inverse relation, denoted R–1, is defined
by (a, b) ∈ R–1 if and only if (b, a) ∈ R.
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Examples: Inverse Relations
The inverse of the “divides” relation onthe integers is the “is a multiple of” relation.
The inverse of the “less-than” relation < on
the real numbers is the “greater-than”relation >.
The inverse of the relation ≤ on a Boolean
algebra (defined by a ⋅ b = a) is the relation≥ (defined by a + b = a).
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Section 10.2
Reflexivity, Symmetry, and
Transitivity
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Reflexivity, Symmetry, andTransitivity
Let R be a relation on a set A.
R is reflexive if ( x, x) ∈ R for all x ∈ A.
R is symmetric if for all x, y ∈ A( x, y) ∈ R → ( y, x) ∈ R.
R is transitive if for all x, y, z ∈ A
( x, y) ∈ R and ( y, z) ∈ R → ( x, z) ∈ R.
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Examples
Which of the following relations arereflexive? symmetric? transitive?
a ⋅ b = a, on a Boolean algebra B.
R × R, on R.
∅, on R.
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Section 10.3
Equivalence Relations
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Equivalence Relations
An equivalence relation on a set A is arelation on A that is reflexive, symmetric,
and transitive.
We use the symbol ~ as a generic symbol
for an equivalence relation.
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Examples of EquivalenceRelations
Which of the following are equivalencerelations?
a ⋅ b = a, on a Boolean algebra B.
gcd(a, b) = 1, on Z+.
gcd(a, b) = a, on Z+.
A ∩ B = ∅, on℘(U ). A = B, on℘(U ).
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Examples of EquivalenceRelations
Which of the following are equivalencerelations?
R × R, on R.
∅, on R.
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Equivalence Classes
Let ~ be an equivalence relation on a set Aand let a ∈ A.
The equivalence class of a is
[a] = { x ∈ A x ~ a}.
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Examples: Equivalence Classes
Describe the equivalence classes of each of the following equivalence relations.
a ≡ b (mod 10), on Z.
A = B, on℘(U ).
p ↔ q, on a set of statements.
R
×R
, onR
.
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Equivalence Classes andPartitions
Theorem: Let ~ be an equivalence relationon a set A. The equivalence classes of ~
form a partition of A.
Proof:
We must show that
The equivalence classes are pairwise disjoint, The union of the equivalence classes equals A.
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Proof, continued
Proof that the equivalence classes arepairwise disjoint.
Let [a] and [b] be two distinct equivalence
classes.
Suppose [a] ∩ [b] ≠ ∅.
Let x ∈ [a] ∩ [b].
Then x ~ a and x ~ b.
Therefore, a ~ x and x ~ b.
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Proof, continued
By transitivity, a ~ b. Now let y ∈ [a].
Then y ~ a.
By transitivity, y ~ b.
So y ∈ [b].
Therefore, [a] ⊆ [b].
By a similar argument, [b] ⊆ [a].
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Proof, continued
Thus, [a] = [b], which is a contradiction Therefore, [a] ∩ [b] = ∅.
Thus, the equivalence classes are pairwise
disjoint.
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Proof, continued
Proof that the union of the equivalenceclasses is A.
Let a ∈ A.
Then a ∈ [a] since a ~ a.
Therefore, a is in the union of the equivalence
classes.
So, A is a subset of that union.
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Proof, concluded
On the other hand, every equivalence class is asubset of A.
Therefore, the union of the equivalence classes
is a subset of A. Therefore, the union of the equivalence classes
equals A.
Therefore, the equivalence classes form apartition of A.
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Example: A Partition Induced byan Equivalence Relation
Let F be the set of all functions f : R→ R. For f , g ∈ F , define f ~ g to mean that f is
O(g) and g is O( f ).
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Example: A Partition Induced byan Equivalence Relation
Theorem: ~ is an equivalence relation on F . Proof:
Reflexivity Obviously, f ~ f for all f ∈ F .
Symmetry
By the symmetry of the definition of ~, it isclear that f ~ g implies that g ~ f .
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Example: A Partition Induced byan Equivalence Relation
Transitivity Let f , g, h ∈ F and suppose that f ~ g and g ~ h.
Then there exist M 1 and x1 and M 2 and x2 such
that
f ( x) ≤ M 1g( x)for all x ≥ x
1
and
g( x) ≤ M 2h( x)for all x ≥ x2.
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Example: A Partition Induced byan Equivalence Relation
Let M = M 1 ⋅ M 2 and let x0 = max( x1, x2). Then for all x ≥ x0,
f ( x)≤ M 1g( x)
≤ M 1 ⋅ M 2h( x)= M h( x).
Therefore, f is O(h).
Similarly, we can show that h is O( f ).
Therefore, f ~ h.
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Example: A Partition Induced byan Equivalence Relation
Therefore, ~ is an equivalence relation on F . The equivalence class of f is the set [ f ] of all
functions with the same growth rate as f .
The most important equivalence classes are
[ xa], a ∈ R.
[b x], b ∈ R. [ xa log x], a ∈ R.
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The Equivalence RelationInduced by a Partition
Let A be a set and let { Ai}i∈ I be a partitionof A.
Define a relation ~ on A as
x ~ y ↔ x, y ∈ Ai for some i ∈ I .
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The Equivalence RelationInduced by a Partition
Theorem: The relation ~ defined above isan equivalence relation on A.
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Proof
Proof: We must prove that ~ is reflexive,
symmetric, and transitive.
Proof that ~ is reflexive.
Let a ∈ A.
Then a is in Ai for some i ∈ I . So a ~ a.
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Proof, continued
Proof that ~ is symmetric. Let a, b ∈ A and suppose that a ~ b.
Then a, b ∈ Ai for some i ∈ I .
So b, a ∈ Ai for some i ∈ I .
Therefore b ~ a.
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Proof, concluded
Proof that ~ is transitive. Let a, b, c ∈ A and suppose a ~ b and b ~ c.
Then a, b ∈ Ai for some i ∈ I and b, c ∈ A j for
some j ∈ I .
That means that b ∈ Ai ∩ A j.
This is possible only if Ai = A j.
Therefore, a, c ∈ Ai.
So, a ~ c.
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Section 10.4
Partial Order Relations
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Antisymmetry
A relation R on a set A is antisymmetric if for all a, b ∈ A,
[(a, b) ∈ R and (b, a) ∈ R] → a = b.
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Examples: Antisymmetry
The following relations are antisymmetric. a b, on Z+.
A ⊆ B, on ℘(U ).
p → q, on the set of all statements (= means ≡).
x ≤ y, on R.
a ⋅ b = a, on a Boolean algebra B.
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Partial Order Relations
A relation R on a set A is a partial order relation if
R is reflexive.
R is antisymmetric.
R is transitive.
We use ≤ as the generic symbol for a partialorder relation.
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Examples: Partial OrderRelations
The following relations are partial orderrelations.
a b, on Z+.
A ⊆ B, on ℘(U ).
p → q, on the set of all statements (= means ≡).
x ≤ y, on R.
a ⋅ b = a, on a Boolean algebra B.
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Total Order Relations
A relation < on a set A is a total order relation if it has the following properties.
< is transitive.
Trichotomy – For all a, b ∈ A, exactly one of the following holds:
a < b.
b < a.
a = b.
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Partial Orders and Total Orders
It is important to realize that not everypartial order is a total order.
Example
Let ≤ be the subset relation ⊆ and let A = ℘({a, b, c}).
Then {a, b}, {a, c} ∈℘( A),
But {a, b} ⊄ {a, c}, {a, c} ⊄ {a, b}, and{a, b} ≠ {a, c}.
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Partial Orders and Total Orders
Theorem: If < is a total ordering of a set Aand we define a ≤ b to mean that a < b or
a = b, then ≤ is a partial ordering of A.
Proof:
Reflexivity.
a ≤ a because a = a.
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Proof, continued
Antisymmetry. Suppose a ≤ b and b ≤ a.
Then
a < b or a = b, and
b < a or b = a.
Suppose a < b.
Then a ≠ b, and b < a is false.
This contradicts b ≤ a.
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Proof, continued
Therefore, a < b is false. So, a = b, since a ≤ b.
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Proof, continued
Transitivity. Suppose a ≤ b and b ≤ c.
Then
a < b or a = b, and
b < c or b = c.
Case 1: a < b and b < c.
Then a < c.
Therefore, a ≤ c.
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Proof, concluded
Case 2: a < b and b = c. Then a < c.
Therefore, a ≤ c.
Cases 3 and 4 are similar.
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Hasse Diagrams
A Hasse diagram is a drawing thatrepresents a partial order relation.
Draw a diagram in which
a ≤ b is represented by a b.
a is drawn below b.
If there exists c such that a ≤ c and c ≤ b, thenwe represent only a ≤ c and c ≤ b; a ≤ b is
implied by transitivity.
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Example: Hasse Diagram
Let the relation be ⊆ on ℘({a, b, c}).
{a , b , c }
{a , c } {b , c }{a , b }
{a } {b } {c }
{}
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Example: Hasse Diagram
Let the relation be on {1, 2, 3, 4, 6, 12}12
6
3
4
2
1
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Example: Partial Order Relation
Let F be the set of all functions f : R→ R. Define a relation ≤ on the equivalence
classes [ f ] of F , under the equivalence
relation f ~ g if f is O(g) and g is O( f ), by
[ f ] ≤ [g] if f is O(g).
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[ f ] ≤ [g] is Well Defined
First, we must show that ≤ is well defined on F . Suppose that f is O(g) and let f ′ in [ f ] and g′ in [g].
Then f ′ ~ f and g′ ~ g.
Then f ′ is O( f ) and f is O(g), so f ′ is O(g).
Furthermore, f ′ is O(g) and g is O(g′ ), so f ′ is
O(g′ ).
Therefore, [ f ′ ] <= [g′ ].
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Example: Partial Order Relation
Theorem: ≤ is a partial order relation on F . Proof:
Reflexivity
It is clear that [ f ] ≤ [ f ] . (Let M = 1, x0 = 0.)
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Example: Partial Order Relation
Antisymmetry Suppose that [ f ] ≤ [g] and [g] ≤ [ f ] .
Then f is O(g) and g is O( f ).
This is the definition of f ~ g.
Therefore, [ f ] = [g].
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Example: Partial Order Relation
Transitivity Suppose that [ f ] ≤ [g] and [g] ≤ [h].
Then f is O(g) and g is O(h).
We have already shown that this implies that f is O(h).
Therefore, [ f ] ≤ [h].
Thus, ≤ is a partial order relation.
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Total Orderings and Sorting
In C++, if we define an order relation < on aclass, then we may sort the class, provided <
is a total ordering of the members of the
class.
E l T l O d i d
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Example: Total Ordering and
Sorting
Define a Point object to consist of twodoubles.
class Point
{
double x;
double y;
};
E l T l O d i d
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Example: Total Ordering and
Sorting
Define operator<() on the Point class asfollows.
bool operator<(const Point& p, const Point& q)
{
if (p.x != q.x)
return p.x < q.x;
else
return p.y < q.y;}
E l T l O d i d
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Example: Total Ordering and
Sorting
Sort the list{(2, 3), (3, 3), (3, 2), (2, 2)}.
The problem is that < is not a total ordering
of the Point class.
Is < a partial ordering?