Union

Given two sets \( A , B \subseteq X \), then the union of \( A \) and \( B \) is defined as the set \( A \cup B := \left \lbrace p \in X : p \in A \lor p \in B \right \rbrace \)

Intersection

Given two sets \( A , B \subseteq X \), then the intersection of \( A \) and \( B \) is defined as the set \( A \cap B := \left \lbrace p \in X : p \in A \land p \in B \right \rbrace \)

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 \( \bigcup M \) is defined so that \( x \in \bigcup M \iff \exists A \in M , x \in A \)

Arbitrary Intersection

Suppose that \( M \) is a family of sets, then \( \bigcap M \) is defined so that \( x \in \bigcap M \iff \forall A \in M , x \in A \)

Arbitrary Union Element of Notation

We define \( \bigcup_{A \in M} A := \) \( \bigcup M \)

Arbitrary Intersection Element of Notation

We define \( \bigcap_{A \in M} A := \) \( \bigcap M \)

Arbitrary Union Indexed Notation

Suppose that \( I \) is an index set for the collection \( \mathcal{A} = \left \lbrace A_{\alpha} : \alpha \in I \right \rbrace \) where \( A_{\alpha} \) is a set, then \( \bigcup_{\alpha \in I} A_{\alpha} := \) \( \bigcup \mathcal{A} \). If the index set is known by context, then we may use the shorthand \( \bigcup_{\alpha} A_{\alpha} \)

Arbitrary Intersection Indexed Notation

Suppose that \( I \) is an index set for the collection \( \mathcal{A} = \left \lbrace A_{\alpha} : \alpha \in I \right \rbrace \) where \( A_{\alpha} \) is a set, then \( \bigcap_{\alpha \in I} A_{\alpha} := \) \( \bigcap \mathcal{A} \). If the index set is known by context, then we may use the shorthand \( \bigcap_{\alpha} A_{\alpha} \)

Arbitrary Union Counting Notation

Suppose that \( a , b \in \mathbb{Z} \), with \( a < b \), and \( \mathcal{A} := \left \lbrace A_{i} : i \in \mathbb{Z} , a \le i \le b \right \rbrace \), then \( \bigcup_{i = a}^{b} A_{i} = \) \( \bigcup \mathcal{A} \)

Arbitrary Intersection Counting Notation

Suppose that \( a , b \in \mathbb{Z} \), with \( a < b \), and \( \mathcal{A} := \left \lbrace A_{i} : i \in \mathbb{Z} , a \le i \le b \right \rbrace \), then \( \bigcap_{i = a}^{b} A_{i} = \) \( \bigcap \mathcal{A} \)

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 \( A \ne B \), 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

- \( \emptyset \notin P \)
- \( \bigcup P \) \( = X \)
- P is pairwise disjoint

partition of the integers

The family \( \left \lbrace \left \lbrace p \in \mathbb{Z} : p < 0 \right \rbrace , \left \lbrace 0 \right \rbrace , \left \lbrace p \in \mathbb{Z} : p > 0 \right \rbrace \right \rbrace \) is a partition of \( \mathbb{Z} \)

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 \( \mathbb{Z} \), so let \( p \in \bigcup P \), therefore \( p \) is in at least one of the sets included in \( P \), since each is a subset of \( \mathbb{Z} \), then we know \( p \in \mathbb{Z} \), for the reverse direction, we can assume that \( p \in \mathbb{Z} \), therefore we know that \( p \) is either positive, negative or zero, so that \( p \in \bigcup P \), this shows \( \bigcup P = \mathbb{Z} \).

To show this family is pairwise disjoint, note that \( 0 \notin \left \lbrace p \in \mathbb{Z} : p < 0 \right \rbrace \) and \( 0 \notin \left \lbrace p \in \mathbb{Z} : p > 0 \right \rbrace \), which shows us that \( \left \lbrace 0 \right \rbrace \cap \left \lbrace p \in \mathbb{Z} : p < 0 \right \rbrace = \emptyset \) and \( \left \lbrace 0 \right \rbrace \cap \left \lbrace p \in \mathbb{Z} : p > 0 \right \rbrace = \emptyset \).

Given any positive integer, we know it cannot be a negative integer, therefore \( \left \lbrace p \in \mathbb{Z} : p < 0 \right \rbrace \cap \left \lbrace p \in \mathbb{Z} : p > 0 \right \rbrace = \emptyset \), 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 \( A \subseteq \left( B \cap C \right) \) iff \( A \subseteq B \) and \( A \subseteq C \)

Suppose that \( A \subseteq \left( B \cap C \right) \), 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 \( U_{\alpha} \) is an indexed family of sets, and \( Y \) is any set, then

\( \bigcup_{\alpha \in I} \left ( U_{\alpha} \cap Y \right ) = \left ( \bigcup_{\alpha \in I} U_{\alpha} \right ) \cap Y \)

Suppose that \( x \in \bigcup_{\alpha \in I} \left ( U_{\alpha} \cap Y \right ) \), therefore there is some \( \beta \in I \) such that \( x \in U_{\beta} \cap Y \), so that \( x \in U_{\beta} \) and \( x \in Y \). Since there is some \( \beta \in I \) such that \( x \in U_{\beta} \) then \( x \in \bigcup_{\alpha \in I} U_{\alpha} \), additionally we had that \( x \in Y \) so that \( x \in \left ( \bigcup_{\alpha \in I} U_{\alpha} \right ) \cap Y \) as needed

Now suppose that \( x \in \left ( \bigcup_{\alpha \in I} U_{\alpha} \right ) \cap Y \), therefore \( x \in \left ( \bigcup_{\alpha \in I} U_{\alpha} \right ) \) and \( x \in Y \), due to this we have some \( \beta \in I \), such that \( x \in U_{\beta} \) and thus \( x \in U_{\beta} \cap Y \) which by definition shows that \( x \in \bigcup_{\alpha \in I} \left ( U_{\alpha} \cap Y \right ) \)

intersection factors from intersection

Suppose that \( U_{\alpha} \) is an indexed family of sets, and \( Y \) is any set, then

\( \bigcap_{\alpha \in I} \left ( U_{\alpha} \cap Y \right ) = \left ( \bigcap_{\alpha \in I} U_{\alpha} \right ) \cap Y \)

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

Now suppose that \( x \in \left ( \bigcap_{\alpha \in I} U_{\alpha} \right ) \cap Y \), therefore \( x \in \left ( \bigcap_{\alpha \in I} U_{\alpha} \right ) \) and \( x \in Y \), due to this, for every \( \alpha \in I \), we know \( x \in U_{\alpha} \) and thus \( x \in U_{\alpha} \cap Y \) which by definition shows that \( x \in \bigcap_{\alpha \in I} \left ( U_{\alpha} \cap Y \right ) \)

Union of Subsets is Still a Subset

Suppose that \( \mathcal{C} \) is a collection of subsets of \( X \), then
\[
\bigcup \mathcal{ C } \subseteq X
\]

Let \( x \in \bigcup_{C \in \mathcal{C}} C \), then \( x \in U \) for some \( U \in \mathcal{C} \), 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 \( \mathcal{C} \) is a collection of supersets of \( X \), then
\[
\bigcap \mathcal{ C } \supseteq X
\]

Let \( x \in X \), suppose that \( C \in \mathcal{ C } \), thus by definition \( C \supseteq X \), and since \( x \in X \) we must deduce \( x \in C \), therefore \( x \in \bigcap \mathcal{ C } \) as needed.

Intersection Decreases as It Intersects More Things

Suppose that \( N, M \) are families of sets such that \( N \subseteq M \), then
\[
\bigcap M \subseteq \bigcap N
\]

Let \( x \in \bigcap M \), 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 \( x \in \bigcap N \) 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 \( A = \bigcup_{a \in A} B_{a} \)

We can see that \( \mathcal{C} = \left \lbrace B_{a} : a \in A \right \rbrace \) is a collection of subsets of \( A \), so since a union of a subsets is still a subset we have \( \bigcup_{a \in A} B_{a} \subseteq A \).

But also given \( p \in A \) we know that \( p \in B_{p} \) so \( p \in \bigcup_{a \in A} B_{a} \) which shows \( A \subseteq \bigcup_{a \in A} B_{a} \), so we can conclude \( A = \bigcup_{a \in A} B_{a} \).