Union
Given two sets $A,B\subseteq X$, then the union of $A$ and $B$ is defined as the set $A\cup B:=\left\{p\in X:p\in A\vee p\in B\right\}$
Intersection
Given two sets $A,B\subseteq X$, then the intersection of $A$ and $B$ is defined as the set $A\cap B:=\left\{p\in X:p\in A\wedge p\in B\right\}$
A set Intersects Another
Suppose that $A,B$ are sets, we say that $A$ intersects $B$ when $A\cap B\ne \mathrm{\varnothing }$
Arbitrary Union
Suppose that $M$ is a family of sets, then $\bigcup M$ is defined so that $x\in \bigcup M\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mathrm{\exists }A\in M,x\in A$
Arbitrary Intersection
Suppose that $M$ is a family of sets, then $\bigcap M$ is defined so that $x\in \bigcap M\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }A\in M,x\in A$
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 $\mathcal{A}=\left\{{A}_{\alpha }:\alpha \in I\right\}$ 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 $\mathcal{A}=\left\{{A}_{\alpha }:\alpha \in I\right\}$ 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, and $\mathcal{A}:=\left\{{A}_{i}:i\in \mathbb{Z},a\le i\le b\right\}$, then $\bigcup _{i=a}^{b}{A}_{i}=$ $\bigcup \mathcal{A}$
Arbitrary Intersection Counting Notation
Suppose that $a,b\in \mathbb{Z}$, with $a, and $\mathcal{A}:=\left\{{A}_{i}:i\in \mathbb{Z},a\le i\le b\right\}$, 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 $A\cap B$ = $\mathrm{\varnothing }$
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 $A\cap B=\mathrm{\varnothing }$
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
partition of the integers
The family $\left\{\left\{p\in \mathbb{Z}:p<0\right\},\left\{0\right\},\left\{p\in \mathbb{Z}:p>0\right\}\right\}$ 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 $0\notin \left\{p\in \mathbb{Z}:p<0\right\}$ and $0\notin \left\{p\in \mathbb{Z}:p>0\right\}$, which shows us that $\left\{0\right\}\cap \left\{p\in \mathbb{Z}:p<0\right\}=\mathrm{\varnothing }$ and $\left\{0\right\}\cap \left\{p\in \mathbb{Z}:p>0\right\}=\mathrm{\varnothing }$.

Given any positive integer, we know it cannot be a negative integer, therefore $\left\{p\in \mathbb{Z}:p<0\right\}\cap \left\{p\in \mathbb{Z}:p>0\right\}=\mathrm{\varnothing }$, 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 $\mathcal{C}=\left\{{B}_{a}:a\in A\right\}$ 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 $A=\bigcup _{a\in A}{B}_{a}$.