Equivalence Relations and Partitions
1
Partitions.
Recall the definition of a partition of a set (text, §1.5, p.7):
Definition 1 Let S be a nonempty set. A partition P of S is a collection P = {Aα } of nonempty
subsets1 of S with the property that each element x of of S is contained in exactly one of the subsets Aα
In other words:
(i): S =
[
Aα ; and
α
(ii): If Aα1 6= Aα2 , then Aα1
\
Aα2 = ∅.
In mathematics, it is frequently essential to partition a set and to work with the classes of the partition;
this handout will introduce the method by which, in almost all cases, one constructs the partition one
needs.
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Binary Relations.
Definition 2 Let S be a nonempty set. A binary relation on S is a two-input predicate R with
domain S.
The assertion R(s1 , s2 ) is often written “s1 Rs2 .” For example, the predicate
R(x, y) :=
“x is less than y”
(domain = N)
is the usually written2 “x < y” on N.
A binary relation R on S can also be thought of as—even identified with—the set of inputs (s1 , s2 ) that
make R true:
R := {(s1 , s2 ) ∈ (S × S) : s1 Rs2 }.
In this case, one sometimes writes “(s1 , s2 ) ∈ R” instead of “(s1 Rs2 ).”
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Equivalence Relations.
There are several kinds of binary relations that have been singled out for their usefulness; the names of
some of the important kinds are partial orders, lattices, and well-orderings. This handout introduces
an important called equivalence relations, which are used to construct partitions. Roughly speaking:
if R is an equivalence relation on S, then s1 Rs2 means that s1 is “equivalent” to s2 in some sense or
other. (The simplest example is the binary relation s1 =s2 , in which the equivalence is simply equals.)
This “roughly speaking” description of equivalence relations is hoplessly vague; to be of any value, this
vague idea must to be made precise. Here is how this is done.
1
2
usually called classes
with “<” replacing the “R”
1
Definition 3 Let R be a binary relation on set S. R is an equivalence relation if it has the following
three properties:
(a)[Reflexivity]: sRs for all elements s ∈ S.
(b)[Symmetry]: Whenever sRt then tRs.
(c)[Transitivity]: Whenever sRt and tRw then sRw.
An important equivalence relation on Z is that of “congruence:”
Definition 4 Fix an integer m ≥ 2. For all (x, y) ∈ (Z × Z), say that
x is congruent to y (mod m)
m
m | (x − y).
I will sometimes indicate that x is congruent to y (mod m) by writing x
≡
(m)
y.
We will discuss this example in some detail in class; to begin, we will do Exercise 1 together.
Exercise 1 Show that
≡
(m)
is reflexive, symmetric, and transitive.
link
{Partitions} the←→
{Equivalence Relations}
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Equivalence relations are important because there is a one-to-one correspondence
{partitions} ←→ {equivalence relations}.
The ←− direction is simple: for any partition P = {Aα } of S, the associated equivalence relation on S
is given by:
xRP y ⇐⇒ x and y are in the same subset Aα ,
(1)
and it is easy to see that RP is an equivalence relation (Exercise 2). Theorem 1 describes the −→
direction of the correspondence.3
Theorem 1 Let R be any equivalence relation on (nonempty) set S. There is a partition P of S for
which R = RP .
Proof. We are given R (reflexive, symmetric, and transitive); we need to find the partition P for which
R = RP . Equation (1) shows the way here: for each a ∈ S, put
[a] := {x ∈ S: aRx};
then let
P := {[a]: a ∈ S} .
(The set [a] is called the equivalence class containing a.) I must establish two things: first, that P
is a partition; and second, that R = RP .
Showing that P is a partition. Since R is reflexive,
we have that a ∈ [a] for all a ∈ S; consequently,
[
each equivalence class [a] is nonempty and
[a] = S (so that P satisfies (i) of Definition 1). To
a∈S
establish that P satisfies (ii), of Definition 1, which in this context says
If [a] 6= [b], then [a]
\
[b] = ∅,
I will prove the logically equivalent assertion
If [a]
\
[b] 6= ∅, then [a] = [b].
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(2)
By so doing, Theorem (1) also establishes the interesting fact that these “RP ” relations are the only equivalence
relations on S—there are no others.
2
\
The argument is a little involved but easy to follow. Suppose that [a] [b] 6= ∅. Then there are elements
in the intersection; let z be any such element (so that z ∈ [a] and z ∈ [b]). Then
z ∈ [a] =⇒ aRz
=⇒
↑
zRa,
(symmetry of R)
and
z ∈ [b] =⇒ bRz
zRb.
=⇒
↑
(symmetry of R)
Next, I will apply transitivity:
bRz and zRa =⇒ bRa,
(3)
aRz and zRb =⇒ aRb.
(4)
and
Now, it is easy to show that Equation (3) implies that [b] ⊆ [a] and that Equation (4) implies that
[a] ⊆ [b] (see Exercise 3); hence [a] = [b]. This completes the proof that P is a partition.
Proof that R = RP . There are now two equivalence relations: besides the original equivalence relation
R, there is the equivalence relation RP given by Equation (1):
aRP b ⇐⇒ x and y are in the same P-class.
In order to show that these two equivalence relations match, I must show that for every element (a, b)
of (S × S),
aRb ⇐⇒ aRP b.
To do this, I will throw in a third condition—namely “[a] = [b]”—and I will show that each of these
three condition implies the other two, by verifying the implications shown below:
aRb
=⇒
|{z}
(i)
[a] = [b]
=⇒
|{z}
aRP b
(ii)
=⇒
|{z}
aRb.
(iii)
Verifying [=⇒
|{z}]: If aRb then (by symmetry) bRa, so (by Exercise 3), [a] and [b] are subsets of each
(i)
other.
Verifying [=⇒
|{z}]: By reflexivity, a ∈ [a] and b ∈ [b]. If [a] = [b], then both a and b are in the same
(ii)
P–class [a] = [b]. This is exactly what the statement “aRP b” says.
Verifying [=⇒
|{z}]: The statement aRP b means that a and b are in the same P–class; say that they’re
(iii)
both in the P–class
[c] := {x ∈ S: cRx}.
Then, since a ∈ [c], we have that cRa, so that by symmetry aRc; and since b ∈ [c], we also have that
cRb. Then, an application of transitivity gives aRb.
Since all three conditions are equivalent, we have that R = RP . The proof of the theorem is complete.
Exercise 2 Show that the binary relation RP defined in Equation (1) is an equivalence relation.
Exercise 3 Let R be an equivalence relation on S, let a and b be elements of S, and suppose that aRb.
Show that [a] ⊆ [b].
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