Union

Given two sets $$ A,B\subseteq X$$, then the union of $$ A$$ and $$ B$$ is defined as the set $$ A\cup B:=\{p\in X:p\in A\vee p\in B\}$$

Intersection

Given two sets $$ A,B\subseteq X$$, then the intersection of $$ A$$ and $$ B$$ is defined as the set $$ A\cap B:=\{p\in X:p\in A\wedge p\in B\}$$

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 $$ x\in \bigcup M\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\exists}A\in M,x\in A$$

Arbitrary Intersection

Suppose that $$ M$$ is a family of sets, then $$ \bigcap M$$ is defined so that $$ x\in \bigcap M\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}A\in M,x\in A$$

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 $$ \mathcal{A}=\{{A}_{\alpha}:\alpha \in I\}$$ 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 $$ \mathcal{A}=\{{A}_{\alpha}:\alpha \in I\}$$ 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 $$ \mathcal{A}:=\{{A}_{i}:i\in \mathbb{Z},a\le i\le b\}$$, 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 $$ \mathcal{A}:=\{{A}_{i}:i\in \mathbb{Z},a\le i\le b\}$$, 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

- $$ \mathrm{\varnothing}\notin P$$
- $$ \bigcup P$$ $$ =X$$
- P is pairwise disjoint

partition of the integers

The family $$ \{\{p\in \mathbb{Z}:p<0\},\left\{0\right\},\{p\in \mathbb{Z}:p>0\}\}$$ 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 $$ 0\notin \{p\in \mathbb{Z}:p<0\}$$ and $$ 0\notin \{p\in \mathbb{Z}:p>0\}$$, which shows us that $$ \left\{0\right\}\cap \{p\in \mathbb{Z}:p<0\}=\mathrm{\varnothing}$$ and $$ \left\{0\right\}\cap \{p\in \mathbb{Z}:p>0\}=\mathrm{\varnothing}$$.

Given any positive integer, we know it cannot be a negative integer, therefore $$ \{p\in \mathbb{Z}:p<0\}\cap \{p\in \mathbb{Z}:p>0\}=\mathrm{\varnothing}$$, 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 $$ A\subseteq (B\cap C)$$ 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 $$ {U}_{\alpha}$$ is an indexed family of sets, and $$ Y$$ is any set, then

$$ \bigcup _{\alpha \in I}({U}_{\alpha}\cap Y)=\left(\bigcup _{\alpha \in I}{U}_{\alpha}\right)\cap Y$$

Suppose that $$ x\in \bigcup _{\alpha \in I}({U}_{\alpha}\cap Y)$$, 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 $$ x\in \bigcup _{\alpha \in I}({U}_{\alpha}\cap Y)$$

intersection factors from intersection

Suppose that $$ {U}_{\alpha}$$ is an indexed family of sets, and $$ Y$$ is any set, then

$$ \bigcap _{\alpha \in I}({U}_{\alpha}\cap Y)=\left(\bigcap _{\alpha \in I}{U}_{\alpha}\right)\cap Y$$

Suppose that $$ x\in \bigcap _{\alpha \in I}({U}_{\alpha}\cap Y)$$, 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 $$ x\in \bigcap _{\alpha \in I}({U}_{\alpha}\cap Y)$$

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 $$ \mathcal{C}=\{{B}_{a}:a\in A\}$$ is a collection of subsets of $$ A$$, so since a union of a subsets is still a subset we have $$ \bigcup _{a\in A}{B}_{a}\subseteq A$$.

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 $$ A=\bigcup _{a\in A}{B}_{a}$$.