POWER SETS AND RELATIONS
L. MARIZZA A. BAILEY
1. The Power Set
Now that we have defined sets as best we can, we can consider a sets of sets.
If we were to assume nothing, except the existence of the empty set, we could
build the following sequence of sets:
∅
{∅}
{{∅}}
{{{∅}}}
and so on. This is one of the many methods of constructing the natural numbers.
We assign ∅ → 0 and {∅} → 1 and so forth.
Given a set, for example,
A = {1, 2, 3, 4, 5}
one can construct a set of subsets. For example, we can construct the set of
subsets of cardinality 2.
A2 = {{1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {3, 5}, {4, 5}}
Now consider the collection of all subsets of A. This is a well-defined set and
is called the Power set of A, denoted P(A). Therefore
{1, 2, 3} ∈ P(A)
A2 ( P(A)
1∈
/ P(A)
2. Relations
Definition 1 (relation). A relation, R on a set A is a nonempty subset
R ⊂ A × A.
If (a, b) ∈ R then we usually denote this by aRb.
Usually, the binary relation is defined by all the elements of A × A which satisfy
some property, P .
Here are some examples of relations.
Example 1 (A trivial example). {(5, 5)} ⊂ Z × Z is a relation on Z, but not
very interesting. So there isn’t much to say about it.
Date: September 29, 2006.
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2
Example 2 ( less than).
<⊂ N × N
is a relation defined by (n, m) ∈< if n is less than m.
Therefore, (1, 2) ∈< and (4, 8) ∈< but (5, 4) ∈<
/ and (3, 3) ∈<.
/
Example 3 ( Geometric). Let G be the set of regular polygons. Define a
relation, R, on G by
R = {(Pn , P2n ) | Pk is a k sided regular polygon }
2.1. Interesting Properties. Some relations have some interesting and useful
properties that one would like to identify.
(a) R is reflexive if ∀a ∈ A then (a, a) ∈ R
(b) R is symmetric if ∀a, b ∈ A then (a, b) ∈ R ⇒ (b, a) ∈ R
/R
(c) R is antisymmetric if ∀a, b ∈ A then (a, b) ∈ R and a 6= b then (b, a) ∈
(d) R is connected if ∀a, b ∈ A then a 6= b ⇒ (a, b) ∨ (b, a) ∈ R
(e) R is transitive if ∀a, b, c ∈ A then (a, b) ∧ (b, c) ∈ R ⇒ (a, c) ∈ R
Definition 2 (partial ordering). A partial ordering of a set A is a reflexive,
antisymmetric, and transitive relation.
In other words, if a relation on a set A satisfies the properties:
Every element in A is related to itself.
If A contains two distinct elements a, b ∈ A such that a is related to b, then b is
not related to a. If A contains three elements a, b, c ∈ A such that a is related
to b, and b is related to c, then a is related to c.
Note that order REALLY matters.
Example 4. Prove that the following are examples of partial orders.
(a) Define a relation L on R by (a, b) ∈ L if a is less than or equal to b.
Proof 1. We must show that each of the properties are satisfied by the
relation.
(r) [reflexive] Clearly, every real number is less than or equal to itself.
(a) [antisymmetric] If a, b ∈ R are distinct and a ≤ b then b a.
(t) [transitive] If a, b, c ∈ R and a ≤ b and b ≤ c then a ≤ c
(b) Define a relation D on N by (n, m) ∈ D if and only if m = nq for some
q. This relation is called divides and is denoted n | m.
(c) ⊂ on P(A) for a set A.
(d) Let F be the set of all real-valued functions differentiable on the real
line. Define a relation, R, by f Rg iff
df
|x=0
dx
is less than
dg
|x=0
dx
.
We denote partial orders by the symbol ≤.
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A total ordering on a set A is a connected, partial ordering.
This means that any two distinct elements in the set must be related in some
way.
We sometimes also call this a linear order.
Note that Example (a) is the only example from above that is a linear order.
Why?
Definition 3 (Equivalence Relation). An equivalence relation is a transitive,
reflexive, and symmetric relation.
We denote these relations by ∼.
Equivalence relations are the most interesting relations because they provide
the most structure for our sets.
Since these are the most useful in finding information about the set we are
studying, we will concentrate on these particular binary relations.
One can think about these particular relations as similar to equality or congruence.
Example 5. Here are some examples of equivalence relations.
(a) = is an equivalence relation on any of the number sets, N, Q, R, Z.
(b) Define a relation ∼ on N × N by (a, b) ∼ (c, d) if ad − bc = 0.
(c) Define a relation ∼ on the power set P(X) of an arbitrary finite set X
by ∀A, B ∈ P(X) A ∼ B if |A| = |B|.
(d) Let r ∈ N be a nonzero natural number.
Define a binary relation ∼k on Z by n ∼k m if and only if n = kq + m
for some q ∈ Z.
(e) Let FR be the set of all integrable functions on R.
Define a relation R on FR × FR by (f, g) ∈ R if
Z 1
Z 1
f=
g
.
0
0
3. Partitions and Equivalence Classes
The most useful consequence of finding a nontrivial equivalence relation on a
set is that it provides structure on the set.
Definition 4 (partition). A partition on a set X is a subset of P ⊂ P(X) such
that
(m) [mutually disjoint] If A, B ∈ P such that A 6= B then A ∩ B = ∅.
(c) [cover] The union of all the elements of P is the set X.
[
A=X
A∈P
(d) [non-empty] All elements of P are non-empty. A collection of sets with
this property is said to cover X.
Properties [m] and [c] are equivalent to the property that every element of X
must be contained in exactly one member of the partition of X.
We will find that equivalence relations yield a partition on a set which, depending on the relation, exposes information about said set.
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3.1. Equivalence Classes. Let (X, ∼) be a set with an equivalence relation.
Let x ∈ X be an element of X.
Define the set x = {a ∈ X | a ∼ x} to be the set of all elements in X which are
equivalent to x.
This is called the equivalence class of x. Now, let y ∈ X such that y x is not
equivalent to x.
Define y = {a ∈ X | a ∼ y} to be all of the elements in X equivalent to y.
Lemma 1. Suppose x, y ∈ X such that y x. Then x ∩ y = ∅
Proof 2. Suppose a ∈ y ∩ x.
Then a ∼ y and a ∼ x.
By symmetry of equivalence relation, a ∼ y ⇒ y ∼ a.
Thus, y ∼ a and a ∼ x.
By transitivity of an equivalence relation y ∼ x.
But this contradicts the claim that y x.
Hence, there does not exist a ∈ x ∩ y.
The lemma above show distinct equivalence classes are mutually disjoint. Now
we need only show that they are a cover for X.
Lemma 2. The union of all equivalence classes is the whole set.
[
x=X
x∈X
Proof 3. EXERCISE!
These Lemmas show that the set of equivalence classes does indeed form a
partition of X.
Let us look at some examples of equivalence classes.
Example 6. Let ∼ be an equivalence relation on R2 defined by
(x1 , y1 ) ∼ (x2 , y2 )
if and only if there exists k ∈ R r {0} such that
x1 = kx2
y1 = ky2
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We write this as (x1 , y1 ) = k(x2 , y2 ), and call it scalar multiplication. Geometrically, two points in R2 are equivalent if they lay on the same line through
the origin. If you consider the points in R2 to be vectors, then the equivalence is
defined by direction. In other words, two vectors are equivalent if they ( or their
negatives) point in the same direction. We call this projective 2-dimensional real
space.
First we should prove, that this actually does define an equivalence relation.
Reflexive Let (x, y) ∈ R2 be an arbitrary point.
Then , if k = 1, then (x, y) = 1(x, y).
Hence (x, y) ∼ (x, y).
Symmetric If (x, y) ∼ (a, b), then ∃k ∈ R nonzero, such that (x, y) = k(a, b).
By dividing both sides by k, k1 (x, y) = (a, b) implies (a, b) ∼ (x, y).
Transitive Let (x1 , y1 ) ∼ (x2 , y2 ) and (x2 , y2 ) ∼ (x3 , y3 ).
Then (x1 , y1 ) = k1 (x2 , y2 ) and (x2 , y2 ) = k2 (x3 , y3 ).
By substitution, we have
(x1 , y1 ) = k1 k2 (x3 , y3 )
Allowing k = k1 k2 yields (x1 , y1 ) ∼ (x3 , y3 )
Since every line through the origin intersects the unit circle twice, there will be
exactly two points on the unit circle in each equivalence class. If restrict ourselves
to only the portion of the unit circle which lies on the upper half plane, than the
intersection would be unique.
Hence, we can represent each equivalence class by a point on the unit circle that
lies on the upper half plane.
However, the point (1, 0) and (−1, 0) are also equivalent, and are, therefore,
identified as one object. So the set of equivalence classes becomes, geometrically,
like a circle.
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4. Problems
The first few problems involve examples and Lemmas which were not proven
in the previous sections.
Problem 1. Prove that the relations defined in Example 4, (b), (c), (d) are
partial orders.
Problem 2. Show that the partial orders in Example 4, (a) is a linear order.
Problem 3. Show that the relations in Example 5 are equivalence relations
and describe the set of equivalence classes. You must state how many equivalence classes there are, describe the elements in each equivalence class and give
representatives for each.
Problem 4. Show that the set of equivalence classes covers the set.
Problem 5. For the following problems, either give example of a relation which
has the following properties, or prove that it cannot be done.
(a) reflexive and symmetric but not transitive.
(b) symmetric and transitive but not reflexive.
(c) reflexive and transitive but not symmetric.
Problem 6 (Discussion). Give an example of a partial order on R2 .
Do you think it’s possible to find a total order on R2 ?
Does this extend to all dimensions?
Department of Mathematics, Arkansas School of Mathematics, Sciences and the
Arts
E-mail address: [email protected]