Equivalence relations and partitions.

1
Equivalence relations and partitions.
Consider the following relation on a set of
all people B (x, y) x has the same birthday
as y
B is reflexive, symmetric and transitive. We
can think about this relation as splitting all
people into 366 categories, one for each
possible day.
Definition. A relation R?A?A is called an
equivalence relation on A if it is symmetric,
reflexive and transitive.
An equivalence relation on a set A represents
some partition of this set.
2

• Definition. For any set A subsets Ai?A
  partition set A if
• A A1? A2 ? ?An
• Ai?Aj ?, for any i?j.
• Ai ? ? for any i

Example. A1, 2, 3, 4, ? 2, 1, 3, 4 is
a partition of A.
3
Definition. Suppose R is equivalence relation on
a set A, and x?A. Then the equivalence class of
x with respect to R is the set xRy?A yRx
In the case of the same birthday relation B, if p
is any person, then the equivalence class of
p pBq?P pBq q?P q has the same
birthday as p
For example, if John was born on Aug. 10,
JohnB q ?P q was born on Aug.10
The set of all equivalence classes of elements of
A is called A modulo R and is denoted
A/R A/RxR x ?A
4
We are going to prove that any equivalence
relation R on A induces a partition of A and any
partition of A gives rise to an equivalence
relation.
Theorem 1. Suppose R is an equivalence relation
on a set A. Then A/R xR x ?A is a
partition of A.
Proof. To prove that A/R defines a partition, we
must prove three properties of a partition. 1)
The union of all equivalence classes xR equals
A, i. e. Since any equivalence class is a
subset of A, their union is also a subset of A.
So, all we need to show is To prove this,
suppose x?A. Then x?xR because xR y ?A
yRx and xRx due to reflexive property of R.
5
x?xR implies
2) To prove that A/R is pairwise disjoint we need
to show, that for any x, y ?A if x?y then
x?y?. It is easier to prove the
contrapositive if x?y ? ?, then xy.
So, assume x?y ? ?, then we can find an
element z ? x?y. It implies that xRz and zRy
by definition of equivalence classes. Then xRz
and zRy implies xRy due to transitive property
of R. We now claim that x?y. Take any a
?x, it implies aRx, which implies aRy because
of xRy and transitive property of R. aRy implies
a ?y, i. e. x?y. Similarly y?x, or
xy.
3) None of equivalence classes is empty, because
for any x?A, x?x.
6
Theorem 2. Suppose A is a set and ? is a
partition of A, i. e. ? is a set of disjoint
nonempty subsets, such that any element of A
belongs to exactly one subset. Then the relation
R on A defined as R (x, y) x, y ?A and x
and y belong to the same subset in ? is an
equivalence relation on A.

• Proof. We need to prove that R is reflexive,
  symmetric and transitive.
• R is reflexive because for any x ?A, x and
  itself belong to the same
• subset in ?.
• R is obviously symmetric
• Suppose xRy and yRz, that is x and y belong to
  the same subset and
• y and z belong to the same subset. This implies
  that x and z belong
• to the same subset in ?, thus xRz.

7
Definition. A partition P2 is called a refinement
of P1 if every set in P2 is a subset of one of
sets in P1.
Example. Consider the partition P1.1, 3, 5,
7, 9, 2, 4, 6, 8, 10, induced by the
relation R1 (x, y) x y mod(2) on a set A
1, 2, 10. Then the relation R2 (x, y) x
y mod(4) induces a partition P2.1, 5, 9,
3, 7, 2, 6, 10, 4, 8, which is a
refinement of P1.
Theorem. Suppose R1 and R2 are equivalence
relations on a set A. Let P1 and P2 are
partitions that correspond to R1 and R2
respectively. Then R1 ? R2 iff P1 is a
refinement of P2.
8
Partial Orders.
A particular type of binary relation on a set.
Definition. A binary relation R ? A?A is a
partial order, if it is reflexive, transitive
and anti-symmetric.
The set A together with the partial order
relation is called a poset.
Examples less or equal greater or
equal subset relation divides
9
Functions.
Function is a special type of relation.
Recall that in a relation R ?A ?B an element a
?A may be related to more then one element of B,
or it may be not related to any.
Definition. A function f A ? B is a binary
relation from A to B such that, for every a?A
there exists a unique (i. e. exactly one)
element b ?B such that (a, b)? f .
10
relation f is not a function
relation f is not a function
11
f (A) b1, b3
The set of all images f (A) f (a) a?A is
the range of f .
In general f (A) is a proper subset of B.
12
Assume An, Bm. How many different
functions f A ? B exist?
Recall that there are 2n?m different relations R
?A ?B, but not all of them are functions.
For relations we counted all subsets of A ?B. A
function must include exactly n pairs, one for
each element in A. Each of elements from the
domain can be related to any element from
co-domain. So, we have m choices for each pair,
the total number of functions is mn.
where
13
Properties of functions (surjective, injective,
bijective )
Definition. Surjective (onto) range
co-domain.
A function f A?B is called surjective, if for
any b?B there exists a?A, such that f (a) b.
? b?B, ?a?A (f (a)b)
This function is not surjective because b2 does
not have any pre-image.
14
If we have two finite sets A lt B
a surjective function f A?B is impossible.
The necessary condition for a surjective function
f A?B is A ? B.
15
Definition. Injective (one-to-one) ?a1,
a2?A, a1 ? a2 ? f (a1) ? f (a2)
A function f A?B is called injective, if for
all a1, a2?A, a1 ? a2 ? f (a1) ? f (a2).
Equivalently, it means f (a1) f (a2) ? a1 a2
This function is not injective.
16
An injective function f A?B is possible if
only A? B.
A function that is both injective and surjective
is called bijective. For this we need A? B
and A ? B, i. e. A B.
17
Pigeonhole Principle If n pigeons are put into m
holes, n gt m, then at least one hole contains
two or more pigeons.
Actually it is not about pigeons and holes...
Restate in terms of functions Theorem. Let A and
B be finite sets, A gt B . If f A ? B
then there exist distinct elements a1 and a2 in A
such that f (a1) f (a2) (that is f is not
injective)
A pigeons B holes f says where each pigeon
gets put
Examples. Birthdays.
18

• Examples
• f (x)x2 R ? R
• is not injective, since both x and -x have the
  same image.
• f (x) x R? R,
• is not surjective (but is surjective if regarded
  R ? R)

19
Composition of functions
(Note, that in the context of function
composition, the order in which the functions
appear is backward, i. e. the rightmost function
is applied first)
Theorem 4. Let f A?B and g B?C be two
functions. The composition of f and g as
relations defines a function g?f A ?C, such
that g?f (a) g(f(a)).
20
Proof. We need to prove that composite relation
g?f is a function. For this we need to
prove two things 1) the composition g?f
relates each element a?A to some c ?C. 2) an
element c ?C assigned to a by g?f is unique, so
we can denote it c g(f (a)).
1) existence Let bf (a) ?B. Let cg(b) ?C. So,
by the definition of composition of relations,
(a, c)?g?f . Thus, ?c?C (a, c)?g?f
2) uniqueness Suppose (a, c1)?g?f and (a,
c2)?g?f . Then by the definition of composition,
?b1?B, such that (a, b1)?f and (b1, c1)?g, and
?b2?B, such that (a, b2)?f and (b2, c2)?g.
Since f is a function, there may be only one b
?B, such that (a, b)?f, so b1 b2. Since g is
a function, (b, c1)?g and (b, c2)?g imply that
c1 c2.
Thus, (g?f )(a) c g (b) g (f (a)).