Union and Intersection

Union

Given two sets

A, B \subseteq X

, then the union of

A

and

B

is defined as the set

A \cup B := {p \in X : p \in A \lor p \in B}

Intersection

Given two sets

A, B \subseteq X

, then the intersection of

A

and

B

is defined as the set

A \cap B := {p \in X : p \in A \land p \in B}

A set Intersects Another

Suppose that

A, B

are sets, we say that

A

intersects

B

when

A \cap B \neq \emptyset

Arbitrary Union

Suppose that

M

is a family of sets, then

⋃ M

is defined so that

x \in ⋃ M ⟺ \exists A \in M, x \in A

Arbitrary Intersection

Suppose that

M

is a family of sets, then

⋂ M

is defined so that

x \in ⋂ M ⟺ \forall A \in M, x \in A

Arbitrary Union Element of Notation

We define

⋃_{A \in M} A :=

⋃ M

Arbitrary Intersection Element of Notation

We define

⋂_{A \in M} A :=

⋂ M

Arbitrary Union Indexed Notation

Suppose that

I

is an index set for the collection

A = {A_{α} : α \in I}

where

A_{α}

is a set, then

⋃_{α \in I} A_{α} :=

⋃ A

. If the index set is known by context, then we may use the shorthand

⋃_{α} A_{α}

Arbitrary Intersection Indexed Notation

Suppose that

I

is an index set for the collection

A = {A_{α} : α \in I}

where

A_{α}

is a set, then

⋂_{α \in I} A_{α} :=

⋂ A

. If the index set is known by context, then we may use the shorthand

⋂_{α} A_{α}

Arbitrary Union Counting Notation

Suppose that

a, b \in Z

, with

a < b

, and

A := {A_{i} : i \in Z, a \leq i \leq b}

, then

⋃_{i = a}^{b} A_{i} =

⋃ A

Arbitrary Intersection Counting Notation

Suppose that

a, b \in Z

, with

a < b

, and

A := {A_{i} : i \in Z, a \leq i \leq b}

, then

⋂_{i = a}^{b} A_{i} =

⋂ A

disjoint sets

Given two sets

A, B

we say that

A

and

B

are disjoint when

A \cap B

\emptyset

pairwise disjoint sets

Suppose that

M

is a family of sets, then we say these sets are pairwise disjoint when given

A, B \in M

such that

A \neq B

, then

A \cap B = \emptyset

partition

Suppose that

X

is a set, then we say that a set

P

is a partition of

X

if and only if the following are true

$\emptyset \notin P$
$⋃ P$ $= X$
P is pairwise disjoint

partition of the integers

The family

{{p \in Z : p < 0}, {0}, {p \in Z : p > 0}}

is a partition of

Z

Note that the integers $- 1, 0, 1$ show that the empty set is not in this family. We'll now prove that it's union equals $Z$ , so let $p \in ⋃ P$ , therefore $p$ is in at least one of the sets included in $P$ , since each is a subset of $Z$ , then we know $p \in Z$ , for the reverse direction, we can assume that $p \in Z$ , therefore we know that $p$ is either positive, negative or zero, so that $p \in ⋃ P$ , this shows $⋃ P = Z$ .

To show this family is pairwise disjoint, note that $0 \notin {p \in Z : p < 0}$ and $0 \notin {p \in Z : p > 0}$ , which shows us that ${0} \cap {p \in Z : p < 0} = \emptyset$ and ${0} \cap {p \in Z : p > 0} = \emptyset$ .

Given any positive integer, we know it cannot be a negative integer, therefore ${p \in Z : p < 0} \cap {p \in Z : p > 0} = \emptyset$ , thus we know that $P$ is pairwise disjoint, which concludes the proof.

a set intersected with a superset is itself

Let

A, B

be sets such that

A \subseteq B

, then

A \cap B = A

We start by showing $A \cap B \subseteq A$ , so consider $x \in A \cap B$ , so we know that $x \in A$ and $x \in B$ , therefore we trivially know that $x \in A$ , as needed

Now consider the other direction, we need to show that $A \subseteq A \cap B$ , so consider that $x \in A$ , then we want to show that $x \in A$ and $x \in B$ , which really just simplifies to showing $x \in B$ , but we know that $A \subseteq B$ , therefore since $x \in A$ , we know that $x \in B$ finishing the proof

A Set Union a Subset is Itself

Let

A, B

be sets such that

B \subseteq A

, then

A \cup B = A

We start by showing $A \cup B \subseteq A$ , so consider $x \in A \cup B$ , so we know that $x \in A$ or $x \in B$ , if $x \in A$ we are done, on the other hand if $x \in B$ , then since $B \subseteq A$ , then $x \in A$ as needed.

Supposing that $x \in A$ , then we know that $x \in A \cup B$ is true.

Subset of an Intersection

Suppose that

A, B, C

are sets then

A \subseteq (B \cap C)

iff

A \subseteq B

and

A \subseteq C

Suppose that $A \subseteq (B \cap C)$ , suppose $x \in A$ , therefore $x \in B$ and $x \in C$ showing $A \subseteq B$ and $A \subseteq C$

Now suppose that $A \subseteq B$ and $A \subseteq C$ , therefore given $x \in A$ , we can see that $x \in B \cap C$ as needed.

intersection factors from union

Suppose that

U_{α}

is an indexed family of sets, and

Y

is any set, then

⋃_{α \in I} (U_{α} \cap Y) = (⋃_{α \in I} U_{α}) \cap Y

Suppose that $x \in ⋃_{α \in I} (U_{α} \cap Y)$ , therefore there is some $β \in I$ such that $x \in U_{β} \cap Y$ , so that $x \in U_{β}$ and $x \in Y$ . Since there is some $β \in I$ such that $x \in U_{β}$ then $x \in ⋃_{α \in I} U_{α}$ , additionally we had that $x \in Y$ so that $x \in (⋃_{α \in I} U_{α}) \cap Y$ as needed

Now suppose that $x \in (⋃_{α \in I} U_{α}) \cap Y$ , therefore $x \in (⋃_{α \in I} U_{α})$ and $x \in Y$ , due to this we have some $β \in I$ , such that $x \in U_{β}$ and thus $x \in U_{β} \cap Y$ which by definition shows that $x \in ⋃_{α \in I} (U_{α} \cap Y)$

intersection factors from intersection

Suppose that

U_{α}

is an indexed family of sets, and

Y

is any set, then

⋂_{α \in I} (U_{α} \cap Y) = (⋂_{α \in I} U_{α}) \cap Y

Suppose that $x \in ⋂_{α \in I} (U_{α} \cap Y)$ , therefore for every $α \in I$ we have: $x \in U_{α} \cap Y$ , so that $x \in U_{α}$ and $x \in Y$ . Since it's true for every $α \in I$ then $x \in ⋂_{α \in I} U_{α}$ , additionally we had that $x \in Y$ so that $x \in (⋂_{α \in I} U_{α}) \cap Y$ as needed

Now suppose that $x \in (⋂_{α \in I} U_{α}) \cap Y$ , therefore $x \in (⋂_{α \in I} U_{α})$ and $x \in Y$ , due to this, for every $α \in I$ , we know $x \in U_{α}$ and thus $x \in U_{α} \cap Y$ which by definition shows that $x \in ⋂_{α \in I} (U_{α} \cap Y)$

Union of Subsets is Still a Subset

Suppose that

C

is a collection of subsets of

X

, then

⋃ C \subseteq X

Let

x \in ⋃_{C \in C} C

, then

x \in U

for some

U \in C

, since

U

must be a subset of

X

then x is also in

X

. As needed

An Intersection of Supersets is still a Superset

Suppose that

C

is a collection of supersets of

X

, then

⋂ C \supseteq X

Let

x \in X

, suppose that

C \in C

, thus by definition

C \supseteq X

, and since

x \in X

we must deduce

x \in C

, therefore

x \in ⋂ C

as needed.

Intersection Decreases as It Intersects More Things

Suppose that

N, M

are families of sets such that

N \subseteq M

, then

⋂ M \subseteq ⋂ N

Let

x \in ⋂ M

, then for every

S \in M, x \in S

. Now consider any

K \in N

, then

K \in M

therefore

x \in K

which shows that

x \in ⋂ N

as needed.

A set Covered in Subsets is a Union

Suppose

A

is a set and that for each

a \in A

, there is a

B_{a}

such that

a \in B_{a} \subseteq A

, then

A = ⋃_{a \in A} B_{a}

We can see that $C = {B_{a} : a \in A}$ is a collection of subsets of $A$ , so since a union of a subsets is still a subset we have $⋃_{a \in A} B_{a} \subseteq A$ .

But also given $p \in A$ we know that $p \in B_{p}$ so $p \in ⋃_{a \in A} B_{a}$ which shows $A \subseteq ⋃_{a \in A} B_{a}$ , so we can conclude $A = ⋃_{a \in A} B_{a}$ .