List of Set Identities and Relations
Browse the List of Set Identities and Relations below. View Videos or join the discussion on this topic. Add List of Set Identities and Relations to your PopFlock.com topic list for future reference or share this resource on social media.
List of Set Identities and Relations

This article lists mathematical properties and laws of sets, involving the set-theoretic operations of union, intersection, and complementation and the relations of set equality and set inclusion. It also provides systematic procedures for evaluating expressions, and performing calculations, involving these operations and relations.

The binary operations of set union (${\displaystyle \cup }$) and intersection (${\displaystyle \cap }$) satisfy many identities. Several of these identities or "laws" have well established names.

## Notation

Throughout this article, capital letters such as ${\displaystyle A,B,C,L,M,R,S,}$ and ${\displaystyle X}$ will denote sets and ${\displaystyle \wp (X)}$ will denote the power set of ${\displaystyle X.}$ If it is needed then unless indicated otherwise, it should be assumed that ${\displaystyle X}$ denotes the universe set, which means that all sets that are used in the formula are subsets of ${\displaystyle X.}$ In particular, the complement of a set ${\displaystyle L}$ will be denoted by ${\displaystyle L^{C}}$ where unless indicated otherwise, it should be assumed that ${\displaystyle L^{C}}$ denotes the complement of ${\displaystyle L}$ in (the universe) ${\displaystyle X.}$

Typically, the set ${\displaystyle L}$ will denote the L eft most set, ${\displaystyle M}$ the M iddle set, and ${\displaystyle R}$ the R ight most set.

For sets ${\displaystyle L}$ and ${\displaystyle R,}$ define:

{\displaystyle {\begin{alignedat}{4}L\cup R&&~:=~\{~x~:~x\in L\;&&{\text{ or }}\;\,&&\;x\in R~\}\\L\cap R&&~:=~\{~x~:~x\in L\;&&{\text{ and }}&&\;x\in R~\}\\L\setminus R&&~:=~\{~x~:~x\in L\;&&{\text{ and }}&&\;x\notin R~\}\\\end{alignedat}}}
and
${\displaystyle L\triangle R~:=~\{~x~:~x{\text{ belongs to exactly one of }}L{\text{ and }}R~\}.}$

The symmetric difference of ${\displaystyle L}$ and ${\displaystyle R}$ is:[1][2]

{\displaystyle {\begin{alignedat}{4}L\;\triangle \;R~&=~(L~\setminus ~&&R)~\cup ~&&(R~\setminus ~&&L)\\~&=~(L~\cup ~&&R)~\setminus ~&&(L~\cap ~&&R).\end{alignedat}}}
If ${\displaystyle L}$ is a set that is understood (say from context, or because it is clearly stated) to be a subset of some other set ${\displaystyle X}$ then the complement of a set ${\displaystyle L}$ may be denoted by:
${\displaystyle L^{\operatorname {C} }~:=~X\setminus L.}$
The definition of ${\displaystyle L^{\operatorname {C} }=X\setminus L}$ may depend on context. For instance, had ${\displaystyle L}$ been declared as a subset of ${\displaystyle Y,}$ with the sets ${\displaystyle Y}$ and ${\displaystyle X}$ not necessarily related to each other in any way, then ${\displaystyle L^{\operatorname {C} }}$ would likely mean ${\displaystyle Y\setminus L}$ instead of ${\displaystyle X\setminus L.}$

## Finitely many sets and the algebra of sets

A family ${\displaystyle \Phi }$ of subsets of a set ${\displaystyle X}$ is said to be an algebra of sets if ${\displaystyle \varnothing \in \Phi }$ and for all ${\displaystyle L,R\in \Phi ,}$ all three of the sets ${\displaystyle X\setminus R,\,L\cap R,}$ and ${\displaystyle L\cup R}$ are elements of ${\displaystyle \Phi .}$[3] The article on this topic lists set identities and other relationships these three operations.

Every algebra of sets is also a ring of sets[3] and a ?-system.

Algebra generated by a family of sets

Given any family ${\displaystyle {\mathcal {S}}}$ of subsets of ${\displaystyle X,}$ there is a unique smallest[note 1] algebra of sets in ${\displaystyle X}$ containing ${\displaystyle {\mathcal {S}}.}$[3] It is called the algebra generated by ${\displaystyle {\mathcal {S}}}$ and it will be denote it by ${\displaystyle \Phi _{\mathcal {S}}.}$ This algebra can be constructed as follows:[3]

1. If ${\displaystyle {\mathcal {S}}=\varnothing }$ then ${\displaystyle \Phi _{\mathcal {S}}=\{\varnothing ,X\}}$ and we are done. Alternatively, if ${\displaystyle {\mathcal {S}}}$ is empty then ${\displaystyle {\mathcal {S}}}$ may be replaced with ${\displaystyle \{\varnothing \},\{X\},{\text{ or }}\{\varnothing ,X\}}$ and continue with the construction.
2. Let ${\displaystyle {\mathcal {S}}_{0}}$ be the family of all sets in ${\displaystyle {\mathcal {S}}}$ together with their complements (taken in ${\displaystyle X}$).
3. Let ${\displaystyle {\mathcal {S}}_{1}}$ be the family of all possible finite intersections of sets in ${\displaystyle {\mathcal {S}}_{0}.}$[note 2]
4. Then the algebra generated by ${\displaystyle {\mathcal {S}}}$ is the set ${\displaystyle \Phi _{\mathcal {S}}}$ consisting of all possible finite unions of sets in ${\displaystyle {\mathcal {S}}_{1}.}$

### One subset involved

Assume ${\displaystyle L\subseteq X.}$

Identity:[4]

{\displaystyle {\begin{alignedat}{10}L\cap X&\;=\;&&L~~~~{\text{ where }}L\subseteq X\\[1.4ex]L\cup \varnothing &\;=\;&&L\\[1.4ex]L\,\triangle \varnothing &\;=\;&&L\\[1.4ex]L\setminus \varnothing &\;=\;&&L\\[1.4ex]X\cap L&\;=\;&&L~~~~{\text{ where }}L\subseteq X\\[1.4ex]\varnothing \cup L&\;=\;&&L\\[1.4ex]\varnothing \,\triangle L&\;=\;&&L\\[1.4ex]\end{alignedat}}}
but
${\displaystyle \varnothing \setminus L=\varnothing }$
so ${\textstyle \varnothing \setminus L=L{\text{ if and only if }}L=\varnothing .}$
{\displaystyle {\begin{alignedat}{10}L\cup L&\;=\;&&L&&\quad {\text{ (Idempotence)}}\\[1.4ex]L\cap L&\;=\;&&L&&\quad {\text{ (Idempotence)}}\\[1.4ex]L\,\triangle \,L&\;=\;&&\varnothing &&\quad {\text{ (Nilpotence of index 2)}}\\[1.4ex]L\setminus L&\;=\;&&\varnothing &&\quad {\text{ (Nilpotence of index 2)}}\\[1.4ex]\end{alignedat}}}

Domination:[4]

{\displaystyle {\begin{alignedat}{10}X\cup L&\;=\;&&X~~~~{\text{ where }}L\subseteq X\\[1.4ex]\varnothing \cap L&\;=\;&&\varnothing \\[1.4ex]\varnothing \times L&\;=\;&&\varnothing \\[1.4ex]\varnothing \setminus L&\;=\;&&\varnothing \\[1.4ex]L\cup X&\;=\;&&X~~~~{\text{ where }}L\subseteq X\\[1.4ex]L\cap \varnothing &\;=\;&&\varnothing \\[1.4ex]L\times \varnothing &\;=\;&&\varnothing \\[1.4ex]\end{alignedat}}}
but
${\displaystyle L\setminus \varnothing =L}$
so ${\textstyle L\setminus \varnothing =\varnothing {\text{ if and only if }}L=\varnothing .}$

Double complement or involution law:

{\displaystyle {\begin{alignedat}{10}X\setminus (X\setminus L)&=L&&\qquad {\text{ Also written }}\quad &&\left(L^{C}\right)^{C}=L&&\quad &&{\text{ where }}L\subseteq X\quad {\text{ (Double complement/Involution law)}}\\[1.4ex]\end{alignedat}}}
${\displaystyle L\setminus \varnothing =L}$
{\displaystyle {\begin{alignedat}{4}\varnothing &=L&&\setminus L\\&=\varnothing &&\setminus L\\&=L&&\setminus X~~~~{\text{ where }}L\subseteq X\\\end{alignedat}}}
[4]
${\displaystyle L^{C}=X\setminus L\quad {\text{ (definition of notation)}}}$
{\displaystyle {\begin{alignedat}{10}L\,\cup (X\setminus L)&=X&&\qquad {\text{ Also written }}\quad &&L\cup L^{C}=X&&\quad &&{\text{ where }}L\subseteq X\\[1.4ex]L\,\triangle (X\setminus L)&=X&&\qquad {\text{ Also written }}\quad &&L\,\triangle L^{C}=X&&\quad &&{\text{ where }}L\subseteq X\\[1.4ex]L\,\cap (X\setminus L)&=\varnothing &&\qquad {\text{ Also written }}\quad &&L\cap L^{C}=\varnothing &&\quad &&\\[1.4ex]\end{alignedat}}}
[4]
{\displaystyle {\begin{alignedat}{10}X\setminus \varnothing &=X&&\qquad {\text{ Also written }}\quad &&\varnothing ^{C}=X&&\quad &&{\text{ (Complement laws for the empty set))}}\\[1.4ex]X\setminus X&=\varnothing &&\qquad {\text{ Also written }}\quad &&X^{C}=\varnothing &&\quad &&{\text{ (Complement laws for the universe set)}}\\[1.4ex]\end{alignedat}}}

Other properties:

If ${\displaystyle L}$ is any set then the following are equivalent:

1. ${\displaystyle L}$ is not empty (${\displaystyle L\neq \varnothing }$), meaning: ${\displaystyle \lnot [\forall x(x\not \in L)]}$
2. (In classical mathematics) ${\displaystyle L}$ is inhabited, meaning: ${\displaystyle \exists x(x\in L)}$
• In constructive mathematics, "not empty" and "inhabited" are not equivalent: every inhabited set is not empty but the converse is not always guaranteed; that is, in constructive mathematics, a set ${\displaystyle L}$ that is not empty (where by definition, "${\displaystyle L}$ is empty" means that the statement ${\displaystyle \forall x(x\not \in L)}$ is true) might not have an inhabitant (which is an ${\displaystyle x}$ such that ${\displaystyle x\in L}$).
3. ${\displaystyle L\not \subseteq R}$ for some set ${\displaystyle R}$

If ${\displaystyle L}$ is any set then the following are equivalent:

1. ${\displaystyle L}$ is empty (${\displaystyle L=\varnothing }$), meaning: ${\displaystyle \forall x(x\not \in L)}$
2. ${\displaystyle L\cup R\subseteq R}$ for every set ${\displaystyle R}$
3. ${\displaystyle L\subseteq R}$ for every set ${\displaystyle R}$
4. ${\displaystyle L\subseteq R\setminus L}$ for some/every set ${\displaystyle R}$
5. ${\displaystyle \varnothing /L=L}$

### Two sets involved

In the left hand sides of the following identities, ${\displaystyle L}$ is the L eft most set and ${\displaystyle R}$ is the R ight most set. Assume both ${\displaystyle L{\text{ and }}R}$ are subsets of some universe set ${\displaystyle X.}$

{\displaystyle {\begin{alignedat}{10}L\cup R&\;=\;&&R\cup L&&\quad {\text{ (Commutativity)}}\\[1.4ex]L\cap R&\;=\;&&R\cap L&&\quad {\text{ (Commutativity)}}\\[1.4ex]L\,\triangle R&\;=\;&&R\,\triangle L&&\quad {\text{ (Commutativity)}}\\[1.4ex]\end{alignedat}}}
{\displaystyle {\begin{alignedat}{4}L\cup (L\cap R)&\;=\;&&L&&\quad {\text{ (Absorption)}}\\[1.4ex]L\cap (L\cup R)&\;=\;&&L&&\quad {\text{ (Absorption)}}\\[1.4ex]\end{alignedat}}}

For ${\displaystyle L,R\subseteq X:}$

{\displaystyle {\begin{alignedat}{10}X\setminus (L\cap R)&=(X\setminus L)\cup (X\setminus R)&&\qquad {\text{ Also written }}\quad &&(L\cap R)^{C}=L^{C}\cup R^{C}&&\quad &&{\text{ (De Morgan's law)}}\\[1.4ex]X\setminus (L\cup R)&=(X\setminus L)\cap (X\setminus R)&&\qquad {\text{ Also written }}\quad &&(L\cup R)^{C}=L^{C}\cap R^{C}&&\quad &&{\text{ (De Morgan's law)}}\\[1.4ex]\end{alignedat}}}

#### Subset inclusion

The following are equivalent for any ${\displaystyle L,R\subseteq X:}$[4]

1. ${\displaystyle L\subseteq R}$
2. ${\displaystyle L\cap R=L}$
3. ${\displaystyle L\cup R=R}$
4. ${\displaystyle L\,\triangle \,R=R\setminus L}$
5. ${\displaystyle L\,\triangle \,R\subseteq R\setminus L}$
6. ${\displaystyle L\setminus R=\varnothing }$
7. ${\displaystyle X\setminus R\subseteq X\setminus L\qquad }$ (that is, ${\displaystyle R^{C}\subseteq L^{C}}$)

Inclusion is a partial order: Explicitly, this means that inclusion ${\displaystyle \,\subseteq ,\,}$ which is a binary operation, has the following three properties:[4]

• Reflexivity: ${\textstyle L\subseteq L}$
• Antisymmetry: ${\textstyle (L\subseteq R{\text{ and }}R\subseteq L){\text{ if and only if }}L=R}$
• Transitivity: ${\textstyle {\text{If }}L\subseteq M{\text{ and }}M\subseteq R{\text{ then }}L\subseteq R}$
##### Meets, Joint, and lattice properties

The following proposition says that for any set ${\displaystyle S,}$ the power set of ${\displaystyle S,}$ ordered by inclusion, is a bounded lattice, and hence together with the distributive and complement laws above, show that it is a Boolean algebra.

Existence of a least element and a greatest element:

${\displaystyle \varnothing \subseteq L\subseteq X}$

Joins/supremums exist:[4]

${\displaystyle L\subseteq L\cup R}$

The union ${\displaystyle L\cup R}$ is the join/supremum of ${\displaystyle L{\text{ and }}R}$ with respect to ${\displaystyle \,\subseteq \,}$ because:

1. ${\displaystyle L\subseteq L\cup R{\text{ and }}R\subseteq L\cup R,}$ and
2. If ${\displaystyle Z}$ is a set such that ${\displaystyle L\subseteq Z{\text{ and }}R\subseteq Z}$ then ${\displaystyle L\cup R\subseteq Z.}$

The intersection ${\displaystyle L\cap R}$ is the join/supremum of ${\displaystyle L{\text{ and }}R}$ with respect to ${\displaystyle \,\supseteq .\,}$

Meets/infimums exist:[4]

${\displaystyle L\cap R\subseteq L}$

The intersection ${\displaystyle L\cap R}$ is the meet/infimum of ${\displaystyle L{\text{ and }}R}$ with respect to ${\displaystyle \,\subseteq \,}$ because:

1. If ${\displaystyle L\cap R\subseteq L{\text{ and }}L\cap R\subseteq R,}$ and
2. If ${\displaystyle Z}$ is a set such that ${\displaystyle Z\subseteq L{\text{ and }}Z\subseteq R}$ then ${\displaystyle Z\subseteq L\cap R.}$

The union ${\displaystyle L\cup R}$ is the meet/infimum of ${\displaystyle L{\text{ and }}R}$ with respect to ${\displaystyle \,\supseteq .\,}$

Other inclusion properties:

${\displaystyle L\setminus R\subseteq L}$
${\displaystyle (L\setminus R)\cap L=L\setminus R}$
${\displaystyle (L\setminus R)\cap R=\varnothing }$
• If ${\displaystyle L\subseteq X{\text{ and }}R\subseteq Y}$ then ${\displaystyle L\times R\subseteq X\times Y}$[4]
• If ${\displaystyle x\not \in L{\text{ then }}x\not \in L\setminus R}$

#### Elementary operations characterized

In the left hand sides of the following identities, ${\displaystyle L}$ is the L eft most set and ${\displaystyle R}$ is the R ight most set. Whenever necessary, both ${\displaystyle L{\text{ and }}R}$ should be assumed to be subsets of some universe set ${\displaystyle X,}$ so that ${\displaystyle L^{C}:=X\setminus L{\text{ and }}R^{C}:=X\setminus R.}$

{\displaystyle {\begin{alignedat}{9}L\cap R&=L&&\,\,\setminus \,&&(L&&\,\,\setminus &&R)\\&=R&&\,\,\setminus \,&&(R&&\,\,\setminus &&L)\\&=L&&\,\,\setminus \,&&(L&&\,\triangle \,&&R)\\&=L&&\,\triangle \,&&(L&&\,\,\setminus &&R)\\\end{alignedat}}}
{\displaystyle {\begin{alignedat}{9}L\cup R&=(&&L\,\triangle \,R)&&\,\,\cup &&&&L&&&&\\&=(&&L\,\triangle \,R)&&\,\triangle \,&&(&&L&&\cap \,&&R)\\&=(&&R\,\setminus \,L)&&\,\,\cup &&&&L&&&&~~~~~{\text{ (union is disjoint)}}\\\end{alignedat}}}
{\displaystyle {\begin{alignedat}{9}L\,\triangle \,R&=&&R\,\triangle \,L&&&&&&&&\\&=(&&L\,\cup \,R)&&\,\setminus \,&&(&&L\,\,\cap \,R)&&\\&=(&&L\,\setminus \,R)&&\cup \,&&(&&R\,\,\setminus \,L)&&~~~~~{\text{ (union is disjoint)}}\\&=(&&L\,\triangle \,M)&&\,\triangle \,&&(&&M\,\triangle \,R)&&~~~~~{\text{ where }}M{\text{ is an arbitrary set. }}\\&=(&&L^{C})&&\,\triangle \,&&(&&R^{C})&&\\\end{alignedat}}}
{\displaystyle {\begin{alignedat}{9}L\setminus R&=&&L&&\,\,\setminus &&(L&&\,\,\cap &&R)\\&=&&L&&\,\,\cap &&(L&&\,\triangle \,&&R)\\&=&&L&&\,\triangle \,&&(L&&\,\,\cap &&R)\\&=&&R&&\,\triangle \,&&(L&&\,\,\cup &&R)\\\end{alignedat}}}

#### Other properties

{\displaystyle {\begin{alignedat}{10}L\setminus R&=L\cap (X\setminus R)&&\qquad {\text{ Also written }}\quad &&L\setminus R=L\cap R^{C}&&\quad &&{\text{ where }}L,R\subseteq X\\[1.4ex]X\setminus (L\setminus R)&=(X\setminus L)\cup R&&\qquad {\text{ Also written }}\quad &&(L\setminus R)^{C}=L^{C}\cup R&&\quad &&{\text{ where }}R\subseteq X\\[1.4ex]L\setminus R&=(X\setminus R)\setminus (X\setminus L)&&\qquad {\text{ Also written }}\quad &&L\setminus R=R^{C}\setminus L^{C}&&\quad &&{\text{ where }}L,R\subseteq X\\[1.4ex]\end{alignedat}}}
${\displaystyle {\text{If }}L\cup R=X{\text{ and }}L\cap R=\varnothing \;{\text{ then }}\;R=X\setminus L\qquad {\text{ (Uniqueness of complements)}}}$
• If ${\displaystyle L\cap R=\varnothing }$ then ${\displaystyle L=R{\text{ if and only if }}L=\varnothing =R.}$
• Given any ${\displaystyle x,}$ ${\displaystyle \quad x\not \in L\setminus R\quad {\text{ if and only if }}\quad x\in L\cap R{\text{ or }}x\not \in L.}$
• If ${\displaystyle L\subseteq R}$ then:
1. ${\displaystyle L\,\triangle \,R=R\setminus L}$

The following statements are equivalent:

1. ${\displaystyle L=R}$
2. ${\displaystyle L\,\triangle \,R=\varnothing }$
3. ${\displaystyle L\,\setminus \,R=R\,\setminus \,L}$

### Three sets involved

In the left hand sides of the following identities, ${\displaystyle L}$ is the L eft most set, ${\displaystyle M}$ is the M iddle set, and ${\displaystyle R}$ is the R ight most set.

#### Distributivity and associativity

Definition: If ${\displaystyle \ast {\text{ and }}\bullet }$ are binary operators then ${\displaystyle \,\ast \,}$ left distributes over ${\displaystyle \,\bullet \,}$ if

${\displaystyle L\,\ast \,(M\,\bullet \,R)~=~(L\,\ast \,M)\,\bullet \,(L\,\ast \,R)\qquad \qquad {\text{ for all }}L,M,R}$
while ${\displaystyle \,\ast \,}$ right distributes over ${\displaystyle \,\bullet \,}$ if
${\displaystyle (L\,\bullet \,M)\,\ast \,R~=~(L\,\ast \,R)\,\bullet \,(M\,\ast \,R)\qquad \qquad {\text{ for all }}L,M,R.}$
The operator ${\displaystyle \,\ast \,}$ distributes over ${\displaystyle \,\bullet \,}$ if it both left distributes and right distributes over ${\displaystyle \,\bullet \,.\,}$ In the definitions above, to transform one side to the other, the innermost operator (the operator inside the parentheses) becomes the outermost operator and the outermost operator becomes the innermost operator.
{\displaystyle {\begin{alignedat}{9}(L\,\cap \,M)\,\cup \,R~&~~=~~&&(L\,\cup \,R)\,&&\cap \,&&(M\,\cup \,R)\qquad &&{\text{ (Right-distributivity of }}\,\cup \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex](L\,\cup \,M)\,\cup \,R~&~~=~~&&(L\,\cup \,R)\,&&\cup \,&&(M\,\cup \,R)\qquad &&{\text{ (Right-distributivity of }}\,\cup \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex](L\,\cup \,M)\,\cap \,R~&~~=~~&&(L\,\cap \,R)\,&&\cup \,&&(M\,\cap \,R)\qquad &&{\text{ (Right-distributivity of }}\,\cap \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex](L\,\cap \,M)\,\cap \,R~&~~=~~&&(L\,\cap \,R)\,&&\cap \,&&(M\,\cap \,R)\qquad &&{\text{ (Right-distributivity of }}\,\cap \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex](L\,\triangle \,M)\,\cap \,R~&~~=~~&&(L\,\cap \,R)\,&&\triangle \,&&(M\,\cap \,R)\qquad &&{\text{ (Right-distributivity of }}\,\cap \,{\text{ over }}\,\triangle \,{\text{)}}\\[1.4ex](L\,\cap \,M)\,\times \,R~&~~=~~&&(L\,\times \,R)\,&&\cap \,&&(M\,\times \,R)\qquad &&{\text{ (Right-distributivity of }}\,\times \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex](L\,\cup \,M)\,\times \,R~&~~=~~&&(L\,\times \,R)\,&&\cup \,&&(M\,\times \,R)\qquad &&{\text{ (Right-distributivity of }}\,\times \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex](L\,\setminus \,M)\,\times \,R~&~~=~~&&(L\,\times \,R)\,&&\setminus \,&&(M\,\times \,R)\qquad &&{\text{ (Right-distributivity of }}\,\times \,{\text{ over }}\,\setminus \,{\text{)}}\\[1.4ex](L\,\cup \,M)\,\setminus \,R~&~~=~~&&(L\,\setminus \,R)\,&&\cup \,&&(M\,\setminus \,R)\qquad &&{\text{ (Right-distributivity of }}\,\setminus \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex](L\,\cap \,M)\,\setminus \,R~&~~=~~&&(L\,\setminus \,R)\,&&\cap \,&&(M\,\setminus \,R)\qquad &&{\text{ (Right-distributivity of }}\,\setminus \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex](L\,\triangle \,M)\,\setminus \,R~&~~=~~&&(L\,\setminus \,R)&&\,\triangle \,&&(M\,\setminus \,R)\qquad &&{\text{ (Right-distributivity of }}\,\setminus \,{\text{ over }}\,\triangle \,{\text{)}}\\[1.4ex](L\,\setminus \,M)\,\setminus \,R~&~~=~~&&(L\,\setminus \,R)&&\,\setminus \,&&(M\,\setminus \,R)~~=~~L\,\setminus \,(M\cup R)\qquad &&{\text{ (Right-distributivity of }}\,\setminus \,{\text{ over }}\,\setminus \,{\text{)}}\\[1.4ex]\end{alignedat}}}
{\displaystyle {\begin{alignedat}{5}L\cup (M\cap R)&\;=\;\;&&(L\cup M)\cap (L\cup R)\qquad &&{\text{ (Left-distributivity of }}\,\cup \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex]L\cup (M\cup R)&\;=\;\;&&(L\cup M)\cup (L\cup R)&&{\text{ (Left-distributivity of }}\,\cup \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex]L\cap (M\cup R)&\;=\;\;&&(L\cap M)\cup (L\cap R)&&{\text{ (Left-distributivity of }}\,\cap \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex]L\cap (M\cap R)&\;=\;\;&&(L\cap M)\cap (L\cap R)&&{\text{ (Left-distributivity of }}\,\cap \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex]L\cap (M\,\triangle \,R)&\;=\;\;&&(L\cap M)\,\triangle \,(L\cap R)&&{\text{ (Left-distributivity of }}\,\cap \,{\text{ over }}\,\triangle \,{\text{)}}\\[1.4ex]L\times (M\cap R)&\;=\;\;&&(L\times M)\cap (L\times R)&&{\text{ (Left-distributivity of }}\,\times \,{\text{ over }}\,\cap \,{\text{)}}\\[1.4ex]L\times (M\cup R)&\;=\;\;&&(L\times M)\cup (L\times R)&&{\text{ (Left-distributivity of }}\,\times \,{\text{ over }}\,\cup \,{\text{)}}\\[1.4ex]L\times (M\,\setminus R)&\;=\;\;&&(L\times M)\,\setminus (L\times R)&&{\text{ (Left-distributivity of }}\,\times \,{\text{ over }}\,\setminus \,{\text{)}}\\[1.4ex]\end{alignedat}}}

Left distributivity and set subtraction:

Set subtraction is right distributive over itself. However, in general, set subtraction is not left distributive over itself:

{\displaystyle {\begin{alignedat}{5}L\,\setminus \,(M\,\setminus \,R)&~~{\color {red}{\supseteq }}~~&&\color {black}{\,}(L\,\setminus \,M)\,\setminus \,(L\,\setminus \,R)~~=~~L\cap R\,\setminus \,M\\[1.4ex]\end{alignedat}}}
where equality holds if and only if ${\displaystyle L\,\setminus \,M=L\,\cap \,R,}$ which happens if and only if ${\displaystyle L\cap M\cap R=\varnothing {\text{ and }}L\setminus M\subseteq R.}$

Set subtraction is not left distributive over unions or intersections in general:

{\displaystyle {\begin{alignedat}{5}L\,\setminus \,(M\,\cup \,R)~&~~{\color {red}{\subseteq }}~~&&\color {black}{\,}(L\,\setminus \,M)\,\cup \,(L\,\setminus \,R)~~=~~L\,\setminus \,(M\,\cap \,R)\\[1.4ex]\end{alignedat}}}
{\displaystyle {\begin{alignedat}{5}L\,\setminus \,(M\,\cap \,R)~&~~{\color {red}{\supseteq }}~~&&\color {black}{\,}(L\,\setminus \,M)\,\cap \,(L\,\setminus \,R)~~=~~L\,\setminus \,(M\,\cup \,R)\\[1.4ex]\end{alignedat}}}
where equality holds for one (or equivalently, for both) of the above two inclusion formulas if and only if ${\displaystyle L\,\setminus \,(M\,\cap \,R)\;\subseteq \;L\,\setminus \,(M\,\cup \,R),}$ which happens if and only if ${\displaystyle L\,\cap \,M=L\,\cap \,R.}$

In contrast, for symmetric difference, the sets ${\displaystyle L\,\setminus \,(M\,\triangle \,R)}$ and ${\displaystyle (L\,\setminus \,M)\,\triangle \,(L\,\setminus \,R)=L\,\cap \,(M\,\triangle \,R)}$ are always disjoint. So these two sets are equal if and only if they are both equal to ${\displaystyle \varnothing .}$ Moreover, ${\displaystyle L\,\setminus \,(M\,\triangle \,R)=\varnothing }$ if and only if ${\displaystyle L\cap M\cap R=\varnothing {\text{ and }}L\subseteq M\cup R.}$

Distributivity and symmetric difference:

Intersection distributes over symmetric difference:

{\displaystyle {\begin{alignedat}{5}L\,\cap \,(M\,\triangle \,R)~&~~=~~&&(L\,\cap \,M)\,\triangle \,(L\,\cap \,R)~&&~\\[1.4ex]\end{alignedat}}}
{\displaystyle {\begin{alignedat}{5}(L\,\triangle \,M)\,\cap \,R~&~~=~~&&(L\,\cap \,R)\,\triangle \,(M\,\cap \,R)~&&~\\[1.4ex]\end{alignedat}}}

Union does not distribute over symmetric difference because only the following is guaranteed in general

{\displaystyle {\begin{alignedat}{5}L\cup (M\,\triangle \,R)~~{\color {red}{\supseteq }}~~\color {black}{\,}(L\cup M)\,\triangle \,(L\cup R)~&~=~&&(M\,\triangle \,R)\,\setminus \,L&~=~&&(M\,\setminus \,L)\,\triangle \,(R\,\setminus \,L)\\[1.4ex]\end{alignedat}}}

Symmetric difference does not distribute over itself:

${\displaystyle L\,\triangle \,(M\,\triangle \,R)~~{\color {red}{\neq }}~~\color {black}{\,}(L\,\triangle \,M)\,\triangle \,(L\,\triangle \,R)~=~M\,\triangle \,R}$
and in general, for any sets ${\displaystyle L{\text{ and }}A}$ (where ${\displaystyle A}$ represents ${\displaystyle M\,\triangle \,R}$), ${\displaystyle L\,\triangle \,A}$ might not be a subset, nor a superset, of ${\displaystyle L}$ (ditto for ${\displaystyle A}$).
{\displaystyle {\begin{alignedat}{5}(L\cup M)\cup R&\;=\;\;&&L\cup (M\cup R)\\[1.4ex](L\cap M)\cap R&\;=\;\;&&L\cap (M\cap R)\\[1.4ex](L\,\triangle M)\,\triangle R&\;=\;\;&&L\,\triangle (M\,\triangle R)\\[1.4ex]\end{alignedat}}}

For set subtraction, instead of associativity, only the following is always guaranteed:

${\displaystyle (L\,\setminus \,M)\,\setminus \,R\;\subseteq \;L\,\setminus \,(M\,\setminus \,R)}$
where equality holds if and only if ${\displaystyle L\cap R=\varnothing }$ (this condition does not depend on ${\displaystyle M}$). Thus ${\textstyle \;(L\setminus M)\setminus R=L\setminus (M\setminus R)\quad {\text{ if and only if }}\quad (R\setminus M)\setminus L=R\setminus (M\setminus L),\;}$ where the only difference between the left and right hand side set equalities is that the locations of ${\displaystyle L{\text{ and }}R}$ have been swapped.

Set subtraction complexity: To manage the many identities involving set subtraction, this section is divided based on where the set subtraction operation and parentheses are located on the left hand side of the identity. The great variety and (relative) complexity of formulas involving set subtraction (compared to those without it) is in part due to the fact that unlike ${\displaystyle \,\cup ,\,\cap ,{\text{ and }}\triangle ,\,}$ set subtraction is neither associative nor commutative. The commutativity of set subtraction can be characterized: because ${\displaystyle (L\,\setminus \,R)\cap (R\,\setminus \,L)=\varnothing ,}$ it follows that:

${\displaystyle L\,\setminus \,R=R\,\setminus \,L\quad {\text{ if and only if }}\quad L=R.}$
Said differently, if distinct symbols always represented distinct sets, then the only true formulas of the form ${\displaystyle \,\cdot \,\,\setminus \,\,\cdot \,=\,\cdot \,\,\setminus \,\,\cdot \,}$ that could be written as those involving a single symbol; that is, those of the form: ${\displaystyle S\,\setminus \,S=S\,\setminus \,S.}$ But such formulas are necessarily true for every binary operation ${\displaystyle \,\ast \,}$ (because ${\displaystyle x\,\ast \,x=x\,\ast \,x}$ must hold by definition of equality), and so in this sense, set subtraction is as diametrically opposite to being commutative as is possible for a binary operation. Set subtraction is also neither left alternative nor right alternative; instead, ${\displaystyle (L\setminus L)\setminus R=L\setminus (L\setminus R)}$ if and only if ${\displaystyle L\cap R=\varnothing }$ if and only if ${\displaystyle (R\setminus L)\setminus L=R\setminus (L\setminus L).}$ Set subtraction is quasi-commutative and satisfies the Jordan identity.

#### Both operations are set subtraction

{\displaystyle {\begin{alignedat}{4}(L\setminus M)\setminus R&=&&L\setminus (M\cup R)\\[0.6ex]&=(&&L\setminus M)\cap (L\setminus R)\\[0.6ex]&=(&&L\setminus R)\setminus M\\[0.6ex]&=(&&L\,\setminus \,R)\,\setminus \,(M\,\setminus \,R)\\[1.4ex]\end{alignedat}}}
{\displaystyle {\begin{alignedat}{4}L\setminus (M\setminus R)&=(L\setminus M)\cup (L\cap R)\\[1.4ex]\end{alignedat}}}
• If ${\displaystyle L\subseteq M{\text{ then }}L\setminus (M\setminus R)=L\cap R}$
• ${\textstyle L\setminus (M\setminus R)\subseteq (L\setminus M)\cup R}$ with equality if and only if ${\displaystyle R\subseteq L.}$
${\displaystyle (L\setminus M)\setminus R~=~(L\setminus R)\setminus M\qquad {\text{ (Quasi-commutative)}}}$
but
${\displaystyle L\setminus (M\setminus R)~\subseteq ~L\setminus (R\setminus M)\qquad {\text{ if and only if }}\qquad L\cap R~\subseteq ~M}$
so that the following are equivalent:
1. ${\displaystyle L\setminus (M\setminus R)~=~L\setminus (R\setminus M)}$
2. ${\displaystyle L\cap R~\subseteq ~M\;{\text{ and }}\;L\cap M~\subseteq ~R}$
3. ${\displaystyle L\cap (M\cup R)~\subseteq ~M\cap R}$
4. ${\displaystyle L\cap (M\cup R)~=~L\cap M\cap R}$

#### Set subtraction on the left

Parentheses on the left

{\displaystyle {\begin{alignedat}{4}\left(L\setminus M\right)\cup R&=(L\cup R)\setminus (M\setminus R)\\&=(L\setminus (M\cup R))\cup R~~~~~{\text{ (the outermost union is disjoint) }}\\\end{alignedat}}}
{\displaystyle {\begin{alignedat}{4}(L\setminus M)\cap R&=(&&L\cap R)\setminus (M\cap R)~~~{\text{ (Distributive law of }}\cap {\text{ over }}\setminus {\text{ )}}\\&=(&&L\cap R)\setminus M\\&=&&L\cap (R\setminus M)\\\end{alignedat}}}[5]
{\displaystyle {\begin{alignedat}{4}(L\setminus M)~\triangle ~R&=(L\setminus (M\cup R))\cup (R\setminus L)\cup (L\cap M\cap R)~~~{\text{ (the three outermost sets are pairwise disjoint) }}\\\end{alignedat}}}
${\displaystyle (L\,\setminus M)\times R=(L\times R)\,\setminus (M\times R)~~~~~{\text{ (Distributivity)}}}$

Parentheses on the right

{\displaystyle {\begin{alignedat}{3}L\setminus (M\cup R)&=(L\setminus M)&&\,\cap \,(&&L\setminus R)~~~~{\text{ (De Morgan's law) }}\\&=(L\setminus M)&&\,\,\setminus &&R\\&=(L\setminus R)&&\,\,\setminus &&M\\\end{alignedat}}}
{\displaystyle {\begin{alignedat}{4}L\setminus (M\cap R)&=(L\setminus M)\cup (L\setminus R)~~~~{\text{ (De Morgan's law) }}\\\end{alignedat}}}
{\displaystyle {\begin{alignedat}{4}L\setminus (M~\triangle ~R)&=(L\setminus (M\cup R))\cup (L\cap M\cap R)~~~{\text{ (the outermost union is disjoint) }}\\\end{alignedat}}}

#### Set subtraction on the right

Parentheses on the left

{\displaystyle {\begin{alignedat}{4}(L\cup M)\setminus R&=(L\setminus R)\cup (M\setminus R)\\\end{alignedat}}}
{\displaystyle {\begin{alignedat}{4}(L\cap M)\setminus R&=(&&L\setminus R)&&\cap (M\setminus R)\\&=&&L&&\cap (M\setminus R)\\&=&&M&&\cap (L\setminus R)\\\end{alignedat}}}
{\displaystyle {\begin{alignedat}{4}(L\,\triangle \,M)\setminus R&=(L\setminus R)~&&\triangle ~(M\setminus R)\\&=(L\cup R)~&&\triangle ~(M\cup R)\\\end{alignedat}}}

Parentheses on the right

{\displaystyle {\begin{alignedat}{3}L\cup (M\setminus R)&=&&&&L&&\cup \;&&(M\setminus (R\cup L))&&~~~{\text{ (the outermost union is disjoint) }}\\&=[&&(&&L\setminus M)&&\cup \;&&(R\cap L)]\cup (M\setminus R)&&~~~{\text{ (the outermost union is disjoint) }}\\&=&&(&&L\setminus (M\cup R))\;&&\;\cup &&(R\cap L)\,\,\cup (M\setminus R)&&~~~{\text{ (the three outermost sets are pairwise disjoint) }}\\\end{alignedat}}}
{\displaystyle {\begin{alignedat}{4}L\cap (M\setminus R)&=(&&L\cap M)&&\setminus (L\cap R)~~~{\text{ (Distributive law of }}\cap {\text{ over }}\setminus {\text{ )}}\\&=(&&L\cap M)&&\setminus R\\&=&&M&&\cap (L\setminus R)\\&=(&&L\setminus R)&&\cap (M\setminus R)\\\end{alignedat}}}[5]
${\displaystyle L\times (M\,\setminus R)=(L\times M)\,\setminus (L\times R)~~~~~{\text{ (Distributivity)}}}$

#### Three operations

Operations of the form ${\displaystyle (L\bullet M)\ast (M\bullet R)}$:

{\displaystyle {\begin{alignedat}{9}(L\cup M)&\,\cup \,&&(&&M\cup R)&&&&\;=\;\;&&L\cup M\cup R\\[1.4ex](L\cup M)&\,\cap \,&&(&&M\cup R)&&&&\;=\;\;&&M\cap (L\cup R)\\[1.4ex](L\cup M)&\,\setminus \,&&(&&M\cup R)&&&&\;=\;\;&&L\,\setminus \,(M\cup R)\\[1.4ex](L\cup M)&\,\triangle \,&&(&&M\cup R)&&&&\;=\;\;&&(L\,\setminus \,(M\cup R))\,\cup \,(R\,\setminus \,(L\cup M))\\[1.4ex]&\,&&\,&&\,&&&&\;=\;\;&&(L\,\triangle \,R)\,\setminus \,M\\[1.4ex](L\cap M)&\,\cup \,&&(&&M\cap R)&&&&\;=\;\;&&M\cup (L\cap R)\\[1.4ex](L\cap M)&\,\cap \,&&(&&M\cap R)&&&&\;=\;\;&&L\cap M\cap R\\[1.4ex](L\cap M)&\,\setminus \,&&(&&M\cap R)&&&&\;=\;\;&&(L\cap M)\,\setminus \,R\\[1.4ex](L\cap M)&\,\triangle \,&&(&&M\cap R)&&&&\;=\;\;&&[(L\,\cap M)\cup (M\,\cap R)]\,\setminus \,(L\,\cap M\,\cap R)\\[1.4ex](L\,\setminus M)&\,\cup \,&&(&&M\,\setminus R)&&&&\;=\;\;&&(L\,\cup M)\,\setminus (M\,\cap \,R)\\[1.4ex](L\,\setminus M)&\,\cap \,&&(&&M\,\setminus R)&&&&\;=\;\;&&\varnothing \\[1.4ex](L\,\setminus M)&\,\setminus \,&&(&&M\,\setminus R)&&&&\;=\;\;&&L\,\setminus M\\[1.4ex](L\,\setminus M)&\,\triangle \,&&(&&M\,\setminus R)&&&&\;=\;\;&&(L\,\setminus M)\cup (M\,\setminus R)\\[1.4ex]&\,&&\,&&\,&&&&\;=\;\;&&(L\,\cup M)\setminus (M\,\cap R)\\[1.4ex](L\,\triangle \,M)&\,\cup \,&&(&&M\,\triangle \,R)&&&&\;=\;\;&&(L\,\cup \,M\,\cup \,R)\,\setminus \,(L\,\cap \,M\,\cap \,R)\\[1.4ex](L\,\triangle \,M)&\,\cap \,&&(&&M\,\triangle \,R)&&&&\;=\;\;&&((L\,\cap \,R)\,\setminus \,M)\,\cup \,(M\,\setminus \,(L\,\cup \,R))\\[1.4ex](L\,\triangle \,M)&\,\setminus \,&&(&&M\,\triangle \,R)&&&&\;=\;\;&&(L\,\setminus \,(M\,\cup \,R))\,\cup \,((M\,\cap \,R)\,\setminus \,L)\\[1.4ex](L\,\triangle \,M)&\,\triangle \,&&(&&M\,\triangle \,R)&&&&\;=\;\;&&L\,\triangle \,R\\[1.7ex]\end{alignedat}}}

Operations of the form ${\displaystyle (L\bullet M)\ast (R\,\setminus \,M)}$:

{\displaystyle {\begin{alignedat}{9}(L\cup M)&\,\cup \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&L\cup M\cup R\\[1.4ex](L\cup M)&\,\cap \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&(L\cap R)\,\setminus \,M\\[1.4ex](L\cup M)&\,\setminus \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&M\cup (L\,\setminus \,R)\\[1.4ex](L\cup M)&\,\triangle \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&M\cup (L\,\triangle \,R)\\[1.4ex](L\cap M)&\,\cup \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&[L\cap (M\cup R)]\cup [R\,\setminus \,(L\cup M)]\qquad {\text{ (disjoint union)}}\\[1.4ex]&\,&&\,&&\,&&&&\;=\;\;&&(L\cap M)\,\triangle \,(R\,\setminus \,M)\\[1.4ex](L\cap M)&\,\cap \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&\varnothing \\[1.4ex](L\cap M)&\,\setminus \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&L\cap M\\[1.4ex](L\cap M)&\,\triangle \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&(L\cap M)\cup (R\,\setminus \,M)\qquad {\text{ (disjoint union)}}\\[1.4ex](L\,\setminus \,M)&\,\cup \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&L\cup R\,\setminus \,M\\[1.4ex](L\,\setminus \,M)&\,\cap \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&(L\cap R)\,\setminus \,M\\[1.4ex](L\,\setminus \,M)&\,\setminus \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&L\,\setminus \,(M\cup R)\\[1.4ex](L\,\setminus \,M)&\,\triangle \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&(L\,\triangle \,R)\,\setminus \,M\\[1.4ex](L\,\triangle \,M)&\,\cup \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&(L\cup M\cup R)\,\setminus \,(L\cap M)\\[1.4ex](L\,\triangle \,M)&\,\cap \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&(L\cap R)\,\setminus \,M\\[1.4ex](L\,\triangle \,M)&\,\setminus \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&[L\,\setminus \,(M\cup R)]\cup (M\,\setminus \,L)\qquad {\text{ (disjoint union)}}\\[1.4ex]&\,&&\,&&\,&&&&\;=\;\;&&(L\,\triangle \,M)\setminus (L\,\cap R)\\[1.4ex](L\,\triangle \,M)&\,\triangle \,&&(&&R\,\setminus \,M)&&&&\;=\;\;&&L\,\triangle \,(M\cup R)\\[1.7ex]\end{alignedat}}}

Operations of the form ${\displaystyle (L\,\setminus \,M)\ast (L\,\setminus \,R)}$:

{\displaystyle {\begin{alignedat}{9}(L\,\setminus M)&\,\cup \,&&(&&L\,\setminus R)&&\;=\;&&L\,\setminus \,(M\,\cap \,R)\\[1.4ex](L\,\setminus M)&\,\cap \,&&(&&L\,\setminus R)&&\;=\;&&L\,\setminus \,(M\,\cup \,R)\\[1.4ex](L\,\setminus M)&\,\setminus \,&&(&&L\,\setminus R)&&\;=\;&&(L\,\cap \,R)\,\setminus \,M\\[1.4ex](L\,\setminus M)&\,\triangle \,&&(&&L\,\setminus R)&&\;=\;&&L\,\cap \,(M\,\triangle \,R)\\[1.4ex]&\,&&\,&&\,&&\;=\;&&(L\cap M)\,\triangle \,(L\cap R)\\[1.4ex]\end{alignedat}}}

Other properties:

${\displaystyle L\cap M=R\;{\text{ and }}\;L\cap R=M\qquad {\text{ if and only if }}\qquad M=R\subseteq L.}$
• If ${\displaystyle L\subseteq M}$ then ${\displaystyle L\setminus R=L\cap (M\setminus R).}$[5]
• ${\displaystyle L\times (M\,\setminus R)=(L\times M)\,\setminus (L\times R)}$
• If ${\displaystyle L\subseteq R}$ then ${\displaystyle M\setminus R\subseteq M\setminus L.}$
• ${\displaystyle L\cap M\cap R=\varnothing }$ if and only if for any ${\displaystyle x\in L\cup M\cup R,}$ ${\displaystyle x}$ belongs to at most two of the sets ${\displaystyle L,M,{\text{ and }}R.}$

## Arbitrary families of sets

Let ${\displaystyle \left(L_{i}\right)_{i\in I},}$ ${\displaystyle \left(R_{j}\right)_{j\in J},}$ and ${\displaystyle \left(S_{i,j}\right)_{(i,j)\in I\times J}}$ be families of sets. Whenever the assumption is needed, then all indexing sets, such as ${\displaystyle I}$ and ${\displaystyle J,}$ are assumed to be non-empty.

### Definitions

Arbitrary unions defined

If ${\displaystyle I=\varnothing }$ then ${\displaystyle \bigcup _{i\in \varnothing }L_{i}=\{x~:~{\text{ there exists }}i\in \varnothing {\text{ such that }}x\in L_{i}\}=\varnothing ,}$ which is somethings called the nullary union convention (despite being called a convention, this equality follows from the definition).

Arbitrary intersections defined

If ${\displaystyle I\neq \varnothing }$ then[4]

Nullary intersections

If ${\displaystyle I=\varnothing }$ then

${\displaystyle \bigcap _{i\in \varnothing }L_{i}=\{x~:~{\text{ for all }}i,{\text{ if }}i\in \varnothing {\text{ then }}x\in L_{i}\}}$
where every possible thing ${\displaystyle x}$ in the universe vacuously satisfied the condition: "if ${\displaystyle i\in \varnothing }$ then ${\displaystyle x\in L_{i}}$". Consequently, ${\displaystyle \bigcap _{i\in \varnothing }L_{i}=\{x~:~{\text{ for all }}i,{\text{ if }}i\in \varnothing {\text{ then }}x\in L_{i}\}=\{x:{\text{ for all }}i,{\text{ true }}\}}$ consists of everything in the universe.

So if ${\displaystyle I=\varnothing }$ and:

1. if you are working in a model in which there exists some universe set ${\displaystyle X}$ then ${\displaystyle \bigcap _{i\in \varnothing }L_{i}=\{x~:~x\in L_{i}{\text{ for every }}i\in \varnothing \}~=~X.}$
2. otherwise, if you are working in a model in which "the class of all things ${\displaystyle x}$" is not a set (by far the most common situation) then ${\displaystyle \bigcap _{i\in \varnothing }L_{i}}$ is undefined. This is because ${\displaystyle \bigcap _{i\in \varnothing }L_{i}=\{x~:~{\text{ for all }}i,{\text{ if }}i\in \varnothing {\text{ then }}x\in L_{i}\}}$ consists of everything, which makes ${\displaystyle \bigcap _{i\in \varnothing }L_{i}}$ a proper class and not a set.
Assumption: Henceforth, whenever a formula requires some indexing set to be non-empty in order for an arbitrary intersection to be well-defined, then this will automatically be assumed without mention.

A consequence of this is the following assumption/definition:

A finite intersection of sets or an intersection of finitely many sets refers to the intersection of a finite collection of one or more sets.

Some authors adopt the so called nullary intersection convention, which is the convention that an empty intersection of sets is equal to some canonical set. In particular, if all sets are subsets of some set ${\displaystyle X}$ then some author may declare that the empty intersection of these sets be equal to ${\displaystyle X.}$ However, the nullary intersection convention is not as commonly accepted as the nullary union convention and this article will not adopt it (this is due to the fact that unlike the empty union, the value of the empty intersection depends on ${\displaystyle X}$ so if there are multiple sets under consideration, which is commonly the case, then the value of the empty intersection risks becoming ambiguous).

Multiple index sets

${\displaystyle \bigcup _{\stackrel {i\in I,}{j\in J}}S_{i,j}~~\colon =~\bigcup _{(i,j)\in I\times J}S_{i,j}}$
${\displaystyle \bigcap _{\stackrel {i\in I,}{j\in J}}S_{i,j}~~\colon =~\bigcap _{(i,j)\in I\times J}S_{i,j}}$

### Commutativity and associativity

Commutativity:[4]

${\displaystyle \bigcup _{\stackrel {i\in I,}{j\in J}}S_{i,j}~~\colon =~\bigcup _{(i,j)\in I\times J}S_{i,j}~=~\bigcup _{i\in I}\left(\bigcup _{j\in J}S_{i,j}\right)~=~\bigcup _{j\in J}\left(\bigcup _{i\in I}S_{i,j}\right)}$
${\displaystyle \bigcap _{\stackrel {i\in I,}{j\in J}}S_{i,j}~~\colon =~\bigcap _{(i,j)\in I\times J}S_{i,j}~=~\bigcap _{i\in I}\left(\bigcap _{j\in J}S_{i,j}\right)~=~\bigcap _{j\in J}\left(\bigcap _{i\in I}S_{i,j}\right)}$

Unions of unions and intersections of intersections:[4]

${\displaystyle \left(\bigcup _{i\in I}L_{i}\right)\cup R~=~\bigcup _{i\in I}\left(L_{i}\cup R\right)}$
${\displaystyle \left(\bigcap _{i\in I}L_{i}\right)\cap R~=~\bigcap _{i\in I}\left(L_{i}\cap R\right)}$

and if ${\displaystyle I=J}$ then also:[note 3]

### Distributing unions and intersections

#### Binary intersection of arbitrary unions

• If all ${\displaystyle \left(L_{i}\right)_{i\in I}}$ are pairwise disjoint and all ${\displaystyle \left(R_{j}\right)_{j\in J}}$ are also pairwise disjoint, then so are all ${\displaystyle \left(L_{i}\cap R_{j}\right)_{(i,j)\in I\times J}}$ (that is, if ${\displaystyle (i,j)\neq \left(i_{2},j_{2}\right)}$ then ${\displaystyle \left(L_{i}\cap R_{j}\right)\cap \left(L_{i_{2}}\cap R_{j_{2}}\right)=\varnothing }$).

• Importantly, if ${\displaystyle I=J}$ then in general,
${\displaystyle ~\left(\bigcup _{i\in I}L_{i}\right)\cap \left(\bigcup _{i\in I}R_{i}\right)~~\color {Red}{\neq }\color {Black}{}~~\bigcup _{i\in I}\left(L_{i}\cap R_{i}\right)~}$
(see this[note 4] footnote for an example). The single union on the right hand side must be over all pairs ${\displaystyle (i,j)\in I\times I:}$
${\displaystyle ~\left(\bigcup _{i\in I}L_{i}\right)\cap \left(\bigcup _{i\in I}R_{i}\right)~~=~~\bigcup _{\stackrel {i\in I,}{j\in I}}\left(L_{i}\cap R_{j}\right).~}$
The same is usually true for other similar non-trivial set equalities and relations that depend on two (potentially unrelated) indexing sets ${\displaystyle I}$ and ${\displaystyle J}$ (such as Eq. 4b or Eq. 7g[5]). Two exceptions are Eq. 2c (unions of unions) and Eq. 2d (intersections of intersections), but both of these are among the most trivial of set equalities and moreover, even for these equalities there is still something that must be proven.[note 3]

#### Arbitrary intersections and arbitrary unions

Naively swapping ${\displaystyle \;\bigcup _{i\in I}\;}$ and ${\displaystyle \;\bigcap _{j\in J}\;}$ may produce a different set

The following inclusion always holds:

In general, equality need not hold and moreover, the right hand side depends on how, for each fixed ${\displaystyle i\in I,}$ the sets ${\displaystyle \left(S_{i,j}\right)_{j\in J}}$ are labelled (see this footnote[note 5] for an example) and the analogous statement is also true of the left hand side as well. Equality can hold under certain circumstances, such as in 7e and 7f, which are respectively the special cases where ${\displaystyle S_{i,j}\colon =L_{i}\setminus R_{j}}$ and ${\displaystyle \left({\hat {S}}_{j,i}\right)_{(j,i)\in J\times I}\colon =\left(L_{i}\setminus R_{j}\right)_{(j,i)\in J\times I}}$ (for 7f, ${\displaystyle I}$ and ${\displaystyle J}$ are swapped).

Formula for equality

For an equality of sets that extends the distributive laws, an approach other than just switching ${\displaystyle \cup }$ and ${\displaystyle \cap }$ is needed. Suppose that for each ${\displaystyle i\in I,}$ there is some non-empty index set ${\displaystyle J_{i}}$ and for each ${\displaystyle j\in J_{i},}$ let ${\displaystyle T_{i,j}}$ be any set (for example, with ${\displaystyle \left(S_{i,j}\right)_{(i,j)\in I\times J}}$ use ${\displaystyle J_{i}\colon =J}$ for all ${\displaystyle i\in I}$ and use ${\displaystyle T_{i,j}\colon =S_{i,j}}$ for all ${\displaystyle i\in I}$ and all ${\displaystyle j\in J_{i}=J}$). Let

${\displaystyle {\mathcal {F}}~\colon =~\prod _{i\in I}J_{i}}$
be the Cartesian product, which can be interpreted as the set of all functions ${\displaystyle f~:~I~\to ~\bigcup _{i\in I}J_{i}}$ such that ${\displaystyle f(i)\in J_{i}}$ for every ${\displaystyle i\in I.}$ Then

where ${\displaystyle {\mathcal {F}}~\colon =~\prod _{i\in I}J_{i}.}$

Example application: In the particular case where all ${\displaystyle J_{i}}$ are equal (that is, ${\displaystyle J_{i}=J_{i_{2}}}$ for all ${\displaystyle i,i_{2}\in I,}$ which is the case with the family ${\displaystyle \left(S_{i,j}\right)_{(i,j)\in I\times J},}$ for example), then letting ${\displaystyle J}$ denote this common set, the Cartesian product ${\displaystyle {\mathcal {F}}~\colon =~\prod _{i\in I}J_{i}}$ will be ${\displaystyle {\mathcal {F}}=J^{I};}$ that is, ${\displaystyle {\mathcal {F}}}$ will be the set of all functions of the form ${\displaystyle f~:~I~\to ~J.}$ The above set equalities Eq. 5 to and Eq. 6 to , respectively become:

• ${\displaystyle \bigcap _{i\in I}\left[\;\bigcup _{j\in J}S_{i,j}\right]=\bigcup _{f\in J^{I}}\left[\;\bigcap _{i\in I}S_{i,f(i)}\right]}$[4]
• ${\displaystyle \bigcup _{i\in I}\left[\;\bigcap _{j\in J}S_{i,j}\right]=\bigcap _{f\in J^{I}}\left[\;\bigcup _{i\in I}S_{i,f(i)}\right]}$[4]

which when combined with Inclusion 1 is a subset of implies:

${\displaystyle \bigcup _{i\in I}\left[\bigcap _{j\in J}S_{i,j}\right]~=~\bigcap _{f\in J^{I}}\left[\;\bigcup _{i\in I}S_{i,f(i)}\right]~~\color {Red}{\subseteq }\color {Black}{}~~\bigcup _{g\in I^{J}}\left[\;\bigcap _{j\in J}S_{g(j),j}\right]~=~\bigcap _{j\in J}\left[\bigcup _{i\in I}S_{i,j}\right]}$
where
• on the  left   hand side, the indices ${\displaystyle f{\text{ and }}i}$ range over ${\displaystyle f\in J^{I}{\text{ and }}i\in I}$ (so the subscripts of ${\displaystyle S_{i,f(i)}}$ range over ${\displaystyle i\in I{\text{ and }}f(i)\in f(I)\subseteq J}$)
• on the right hand side, the indices ${\displaystyle g{\text{ and }}j}$ range over ${\displaystyle g\in I^{J}{\text{ and }}j\in J}$ (so the subscripts of ${\displaystyle S_{g(j),j}}$ range over ${\displaystyle j\in J{\text{ and }}g(j)\in g(J)\subseteq I}$).

Example application: To apply the general formula to the case of ${\displaystyle \left(C_{k}\right)_{k\in K}}$ and ${\displaystyle \left(D_{l}\right)_{l\in L},}$ use ${\displaystyle I\colon =\{1,2\},}$ ${\displaystyle J_{1}\colon =K,}$ ${\displaystyle J_{2}\colon =L,}$ and let ${\displaystyle T_{1,k}\colon =C_{k}}$ for all ${\displaystyle k\in J_{1}}$ and let ${\displaystyle T_{2,l}\colon =D_{l}}$ for all ${\displaystyle l\in J_{2}.}$ Every map ${\displaystyle f\in {\mathcal {F}}~\colon =~\prod _{i\in I}J_{i}=J_{1}\times J_{2}=K\times L}$ can be bijectively identified with the pair ${\displaystyle \left(f(1),f(2)\right)\in K\times L}$ (the inverse sends ${\displaystyle (k,l)\in K\times L}$ to the map ${\displaystyle f_{(k,l)}\in {\mathcal {F}}}$ defined by ${\displaystyle 1\mapsto k}$ and ${\displaystyle 2\mapsto l;}$ this is technically just a change of notation). Expanding and simplifying the left hand side of Eq. 5 to , which recall was

${\displaystyle ~\bigcap _{i\in I}\left[\;\bigcup _{j\in J_{i}}T_{i,j}\right]=\bigcup _{f\in {\mathcal {F}}}\left[\;\bigcap _{i\in I}T_{i,f(i)}\right]~}$
gives
${\displaystyle \bigcap _{i\in I}\left[\;\bigcup _{j\in J_{i}}T_{i,j}\right]=\left(\bigcup _{j\in J_{1}}T_{1,j}\right)\cap \left(\;\bigcup _{j\in J_{2}}T_{2,j}\right)=\left(\bigcup _{k\in K}T_{1,k}\right)\cap \left(\;\bigcup _{l\in L}T_{2,l}\right)=\left(\bigcup _{k\in K}C_{k}\right)\cap \left(\;\bigcup _{l\in L}D_{l}\right)}$
and doing the same to the right hand side gives:
${\displaystyle \bigcup _{f\in {\mathcal {F}}}\left[\;\bigcap _{i\in I}T_{i,f(i)}\right]=\bigcup _{f\in {\mathcal {F}}}\left(T_{1,f(1)}\cap T_{2,f(2)}\right)=\bigcup _{f\in {\mathcal {F}}}\left(C_{f(1)}\cap D_{f(2)}\right)=\bigcup _{(k,l)\in K\times L}\left(C_{k}\cap D_{l}\right)=\bigcup _{\stackrel {k\in K,}{l\in L}}\left(C_{k}\cap D_{l}\right).}$

Thus the general identity Eq. 5 to reduces down to the previously given set equality Eq. 3b:

${\displaystyle \left(\bigcup _{k\in K}C_{k}\right)\cap \left(\;\bigcup _{l\in L}D_{l}\right)=\bigcup _{\stackrel {k\in K,}{l\in L}}\left(C_{k}\cap D_{l}\right).}$

### Distributing subtraction

The following set equalities can be deduced from the equalities 7a - 7d above (see this note on why the following equalties are atypical):

### Distributing products

• If ${\displaystyle I=J}$ then
${\displaystyle \left(\prod _{i\in I}L_{i}\right)\cap \left(\prod _{i\in I}R_{i}\right)~=~\prod _{i\in I}\left(L_{i}\cap R_{i}\right)}$
• If ${\displaystyle I\neq J}$ then ${\displaystyle \left(\prod _{i\in I}L_{i}\right)\cap \left(\prod _{j\in J}R_{j}\right)=\varnothing }$ in general (meaning, unless ${\displaystyle I}$ and ${\displaystyle J}$ as somehow identified as the same set through some bijection or one of these products is identified as a subset of the other via some injective map), so only the case ${\displaystyle I=J}$ is useful.
• For example, if ${\displaystyle I:=\{1,2\}}$ and ${\displaystyle J:=\{1,2,3\}}$ with all sets equal to ${\displaystyle \mathbb {R} }$ then ${\displaystyle \prod _{i\in I}L_{i}=\prod _{i\in \{1,2\}}\mathbb {R} =\mathbb {R} ^{2}}$ and ${\displaystyle \prod _{j\in J}R_{j}=\prod _{j\in \{1,2,3\}}\mathbb {R} =\mathbb {R} ^{3}}$ where ${\displaystyle \mathbb {R} ^{2}\cap \mathbb {R} ^{3}=\varnothing }$ unless, for example, ${\displaystyle \prod _{i\in \{1,2\}}\mathbb {R} =\mathbb {R} ^{2}}$ is identified as a subset of ${\displaystyle \prod _{j\in \{1,2,3\}}\mathbb {R} =\mathbb {R} ^{3}}$ through some injection, such as maybe ${\displaystyle (x,y)\mapsto (x,y,0)}$ for instance; however, in this particular case the product ${\displaystyle \prod _{i\in I=\{1,2\}}L_{i}}$ actually represents the ${\displaystyle J}$-indexed product ${\displaystyle \prod _{j\in J=\{1,2,3\}}L_{i}}$ where ${\displaystyle L_{3}:=\{0\}.}$
• For another example, take ${\displaystyle I:=\{1,2\}}$ and ${\displaystyle J:=\{1,2,3\}}$ with ${\displaystyle L_{1}:=\mathbb {R} ^{2}}$ and ${\displaystyle L_{2},R_{1},R_{2},{\text{ and }}R_{3}}$ all equal to ${\displaystyle \mathbb {R} .}$ Then ${\displaystyle \prod _{i\in I}L_{i}=\mathbb {R} ^{2}\times \mathbb {R} }$ and ${\displaystyle \prod _{j\in J}R_{j}=\mathbb {R} \times \mathbb {R} \times \mathbb {R} ,}$ which can both be identified as the same set via the bijection that sends ${\displaystyle ((x,y),z)\in \mathbb {R} ^{2}\times \mathbb {R} }$ to ${\displaystyle (x,y,z)\in \mathbb {R} \times \mathbb {R} \times \mathbb {R} .}$ Under this identification, ${\displaystyle \left(\prod _{i\in I}L_{i}\right)\cap \left(\prod _{j\in J}R_{j}\right)~=~\mathbb {R} ^{3}.}$

More generally, if ${\displaystyle \left(S_{i,j}\right)_{(i,j)\in I\times J}}$ is a family of sets then

• Moreover, a tuple ${\displaystyle \left(x_{j}\right)_{j\in J}}$ belongs to the set in equation 8 if and only if ${\displaystyle x_{j}\in S_{i,j}}$ for all ${\displaystyle i\in I}$ and all ${\displaystyle j\in J.}$

But for unions, it is in general only guaranteed that:

${\displaystyle \bigcup _{i\in I}\left(\prod _{j\in J}S_{i,j}\right)~~\color {Red}{\subseteq }\color {Black}{}~~\prod _{j\in J}\left(\bigcup _{i\in I}S_{i,j}\right)}$

## Functions and sets

### Definitions

Let ${\displaystyle f:X\to Y}$ be any function, where we denote its domain ${\displaystyle X}$ by ${\displaystyle \operatorname {domain} f}$ and denote its codomain ${\displaystyle Y}$ by ${\displaystyle \operatorname {codomain} f.}$

Many of the identities below do not actually require that the sets be somehow related to ${\displaystyle f}$'s domain or codomain (that is, to ${\displaystyle X}$ or ${\displaystyle Y}$) so when some kind of relationship is necessary then it will be clearly indicated. Because of this, in this article, if ${\displaystyle L}$ is declared to be "any set," and it is not indicated that ${\displaystyle L}$ must be somehow related to ${\displaystyle X}$ or ${\displaystyle Y}$ (say for instance, that it be a subset ${\displaystyle X}$ or ${\displaystyle Y}$) then it is meant that ${\displaystyle L}$ is truly arbitrary.[note 6] This generality is useful in situations where ${\displaystyle f:X\to Y}$ is a map between two subsets ${\displaystyle X\subseteq U}$ and ${\displaystyle Y\subseteq V}$ of some larger sets ${\displaystyle U}$ and ${\displaystyle V,}$ and where the set ${\displaystyle L}$ might not be entirely contained in ${\displaystyle X=\operatorname {domain} f}$ and/or ${\displaystyle Y=\operatorname {codomain} f}$ (e.g. if all that is known about ${\displaystyle L}$ is that ${\displaystyle L\subseteq U}$); in such a situation it may be useful to know what can and cannot be said about ${\displaystyle f(L)}$ and/or ${\displaystyle f^{-1}(L)}$ without having to introduce a (potentially unnecessary) intersection such as: ${\displaystyle f(L\cap X)}$ and/or ${\displaystyle f^{-1}(L\cap Y).}$

Images and preimages of sets

If ${\displaystyle L}$ is any set then the image of ${\displaystyle L}$ under ${\displaystyle f}$ is defined to be the set:

${\displaystyle f(L)~:=~\{\,f(s)~:~s\in L\cap \operatorname {domain} f\,\}}$
while the preimage of ${\displaystyle L}$ under ${\displaystyle f}$ is:
${\displaystyle f^{-1}(L)~:=~\{\,x\in \operatorname {domain} f~:~f(x)\in L\,\}}$
where if ${\displaystyle L=\{s\}}$ is a singleton set then the preimage of ${\displaystyle s}$ under ${\displaystyle f}$ is
${\displaystyle f^{-1}(s)~:=~f^{-1}(\{s\})~=~\{\,x\in \operatorname {domain} f~:~f(x)=s\,\}.}$

Denote by ${\displaystyle \operatorname {Im} f}$ or ${\displaystyle \operatorname {image} f}$ the image or range of ${\displaystyle f:X\to Y,}$ which is the set:

${\displaystyle \operatorname {Im} f~=~f(X)~:=~f(\operatorname {domain} f)~=~\{f(x)~:~x\in \operatorname {domain} f\}.}$

Saturated sets

A set ${\displaystyle L}$ is said to be ${\displaystyle f}$-saturated or a saturated set if any of the following equivalent conditions are satisfied:

1. There exists a set ${\displaystyle R}$ such that ${\displaystyle L=f^{-1}(R).}$
• Any such set ${\displaystyle R}$ necessarily contains ${\displaystyle f(L)}$ as a subset.
2. ${\displaystyle L=f^{-1}(f(L)).}$

For a set ${\displaystyle L}$ to be ${\displaystyle f}$-saturated, it is necessary that ${\displaystyle L\subseteq \operatorname {domain} f.}$

Compositions and restrictions of functions

If ${\displaystyle f}$ and ${\displaystyle g}$ are maps then ${\displaystyle g\circ f}$ denotes the composition map

${\displaystyle g\circ f~:~\{\,x\in \operatorname {domain} f~:~f(x)\in \operatorname {domain} g\,\}~\to ~\operatorname {codomain} g}$
with domain and codomain
{\displaystyle {\begin{alignedat}{4}\operatorname {domain} (g\circ f)&=\{\,x\in \operatorname {domain} f~:~f(x)\in \operatorname {domain} g\,\}\\[0.4ex]\operatorname {codomain} (g\circ f)&=\operatorname {codomain} g\\[0.7ex]\end{alignedat}}}
defined by
${\displaystyle (g\circ f)(x):=g(f(x)).}$

The restriction of ${\displaystyle f:X\to Y}$ to ${\displaystyle L,}$ denoted by ${\displaystyle f{\big \vert }_{L},}$ is the map

${\displaystyle f{\big \vert }_{L}~:~L\cap \operatorname {domain} f~\to ~Y}$
with ${\displaystyle \operatorname {domain} f{\big \vert }_{L}~=~L\cap \operatorname {domain} f}$ defined by sending ${\displaystyle x\in L\cap \operatorname {domain} f}$ to ${\displaystyle f(x);}$ that is,
${\displaystyle f{\big \vert }_{L}(x)~:=~f(x).}$
Alternatively, ${\displaystyle ~f{\big \vert }_{L}~=~f\circ \operatorname {In} ~}$ where ${\displaystyle ~\operatorname {In} ~:~L\cap X\to X~}$ denotes the inclusion map, which is defined by ${\displaystyle \operatorname {In} (s):=s.}$

### Finitely many sets

Let ${\displaystyle f:X\to Y}$ be any function.

Let ${\displaystyle L{\text{ and }}R}$ be completely arbitrary sets. Assume ${\displaystyle A\subseteq X{\text{ and }}C\subseteq Y.}$

#### (Pre)Image of a single set

{\displaystyle {\begin{alignedat}{4}f(L)&=f(L\cap \operatorname {domain} f)\\&=f(L\cap X)\\&=Y~~~~\,\setminus \left\{y\in Y~~~~\,:f^{-1}(y)\subseteq X\setminus L\right\}\\&=\operatorname {Im} f\setminus \left\{y\in \operatorname {Im} f:f^{-1}(y)\subseteq X\setminus L\right\}\\\end{alignedat}}} {\displaystyle {\begin{alignedat}{4}f^{-1}(L)&=f^{-1}(L\cap \operatorname {Im} f)\\&=f^{-1}(L\cap Y)\end{alignedat}}} None
${\displaystyle f(X)=\operatorname {Im} f\subseteq Y}$ {\displaystyle {\begin{alignedat}{4}f^{-1}(Y)&=X\\f^{-1}(\operatorname {Im} f)&=X\end{alignedat}}} None
{\displaystyle {\begin{alignedat}{4}f(L)&=f(L\cap R~&&\cup ~&&(&&L\setminus R))\\&=f(L\cap R)~&&\cup ~f&&(&&L\setminus R)\end{alignedat}}} {\displaystyle {\begin{alignedat}{4}f^{-1}(L)&=f^{-1}(L\cap R&&\cup &&(&&L&&\setminus &&R))\\&=f^{-1}(L\cap R)&&\cup f^{-1}&&(&&L&&\setminus &&R)\\&=f^{-1}(L\cap R)&&\cup f^{-1}&&(&&L&&\setminus [&&R\cap \operatorname {Im} f])\\&=f^{-1}(L\cap R)&&\cup f^{-1}&&([&&L\cap \operatorname {Im} f]&&\setminus &&R)\\&=f^{-1}(L\cap R)&&\cup f^{-1}&&([&&L\cap \operatorname {Im} f]&&\setminus [&&R\cap \operatorname {Im} f])\end{alignedat}}} None
${\displaystyle \operatorname {Im} f=f(X)~=~f(L)\cup f(X\setminus L)}$ {\displaystyle {\begin{alignedat}{4}X&=f^{-1}(L)\cup f^{-1}(Y&&\setminus L)\\&=f^{-1}(L)\cup f^{-1}(\operatorname {Im} f&&\setminus L)\end{alignedat}}} None
${\displaystyle f{\big \vert }_{L}(R)=f(L\cap R)}$ ${\displaystyle \left(f{\big \vert }_{L}\right)^{-1}(R)=L\cap f^{-1}(R)}$ None
${\displaystyle (g\circ f)(L)~=~g(f(L))}$ ${\displaystyle (g\circ f)^{-1}(L)~=~f^{-1}\left(g^{-1}(L)\right)}$ None (${\displaystyle f}$ and ${\displaystyle g}$ are arbitrary functions).

Equivalences and implications of images and preimages

• ${\displaystyle f(L)\cap R=\varnothing \quad {\text{ if and only if }}\quad L\cap f^{-1}(R)=\varnothing .}$[6]
• So for any ${\displaystyle t,}$
${\displaystyle t\not \in f(L)\quad {\text{ if and only if }}\quad L\cap f^{-1}(t)=\varnothing .}$
[6]
Image Preimage Additional assumptions on sets
${\displaystyle L\subseteq R}$ implies ${\displaystyle f(L)\subseteq f(R)}$[6] ${\displaystyle L\subseteq R}$ implies ${\displaystyle f^{-1}(L)\subseteq f^{-1}(R)}$[6] None
${\displaystyle f(L)\subseteq \operatorname {Im} f\subseteq Y}$ ${\displaystyle f^{-1}(L)=X}$ if and only if ${\displaystyle \operatorname {Im} f\subseteq L}$ None
${\displaystyle f(L)=\varnothing }$ if and only if ${\displaystyle L\cap \operatorname {domain} f=\varnothing }$ ${\displaystyle f^{-1}(L)=\varnothing }$ if and only if ${\displaystyle L\cap \operatorname {Im} f=\varnothing }$ None
${\displaystyle f(A)=\varnothing }$ if and only if ${\displaystyle A=\varnothing }$ ${\displaystyle f^{-1}(C)=\varnothing }$ if and only if ${\displaystyle C\subseteq Y\setminus \operatorname {Im} f}$ ${\displaystyle A\subseteq X}$ and ${\displaystyle C\subseteq Y}$
The following are equivalent:
1. ${\displaystyle f(L)\subseteq f(R)}$
2. ${\displaystyle f(L\cup R)=f(R)}$
3. ${\displaystyle L\cap \operatorname {domain} f~\subseteq ~f^{-1}(f(R))}$
The following are equivalent:
1. ${\displaystyle f^{-1}(L)\subseteq f^{-1}(R)}$
2. ${\displaystyle f^{-1}(L\cup R)=f^{-1}(R)}$
3. ${\displaystyle L\cap \operatorname {Im} f\subseteq R}$

If ${\displaystyle C~\subseteq ~\operatorname {Im} f}$ then ${\displaystyle f^{-1}(C)~\subseteq ~f^{-1}(R)}$ if and only if ${\displaystyle C~\subseteq ~R.}$

${\displaystyle C\subseteq \operatorname {Im} f}$
The following are equivalent when ${\displaystyle C\subseteq Y:}$
1. ${\displaystyle C\subseteq f(R)}$
2. ${\displaystyle f(B)=C}$ for some ${\displaystyle B\subseteq R}$
The following are equivalent:
1. ${\displaystyle L\subseteq f^{-1}(R)}$
2. ${\displaystyle f(L)\subseteq R}$ and ${\displaystyle L\subseteq \operatorname {domain} f}$

The following are equivalent when ${\displaystyle A\subseteq X:}$

1. ${\displaystyle A\subseteq f^{-1}(R)}$
2. ${\displaystyle f(A)\subseteq R}$
${\displaystyle A\subseteq X}$ and ${\displaystyle C\subseteq Y}$
The following are equivalent:
1. ${\displaystyle f(A)~\supseteq ~f(X\setminus A)}$
2. ${\displaystyle f(A)=\operatorname {Im} f}$
The following are equivalent:
1. ${\displaystyle f^{-1}(C)~\supseteq ~f^{-1}(Y\setminus C)}$
2. ${\displaystyle f^{-1}(C)=X}$
${\displaystyle A\subseteq X}$ and ${\displaystyle C\subseteq Y}$

#### (Pre)Images of set operations

Throughout, let ${\displaystyle L{\text{ and }}R}$ be any sets and let ${\displaystyle f:X\to Y}$ be any function.

Summary

As the table below shows, set equality is not guaranteed only for images of: intersections, set subtractions, and symmetric differences.

Image Preimage
${\displaystyle \,~~~~f(L\cup R)~=~f(L)\cup f(R)}$[7] ${\displaystyle f^{-1}(L\cup R)~=~f^{-1}(L)\cup f^{-1}(R)}$[4] None
${\displaystyle f(L\cap R)~\subseteq ~f(L)\cap f(R)}$
${\displaystyle f^{-1}(L\cap R)~=~f^{-1}(L)\cap f^{-1}(R)}$[4] None
${\displaystyle f(L\setminus R)~\supseteq ~f(L)\setminus f(R)}$
{\displaystyle {\begin{alignedat}{4}f^{-1}(L)\setminus f^{-1}(R)&=f^{-1}&&(&&L&&\setminus &&R)\\&=f^{-1}&&(&&L&&\setminus [&&R\cap \operatorname {Im} f])\\&=f^{-1}&&([&&L\cap \operatorname {Im} f]&&\setminus &&R)\\&=f^{-1}&&([&&L\cap \operatorname {Im} f]&&\setminus [&&R\cap \operatorname {Im} f])\end{alignedat}}}[6][4] None
${\displaystyle f(X\setminus R)~\supseteq ~f(X)\setminus f(R)}$
{\displaystyle {\begin{alignedat}{4}X\setminus f^{-1}(R)&=f^{-1}(&&Y&&\setminus &&R)\\&=f^{-1}(&&Y&&\setminus [&&R\cap \operatorname {Im} f])\\&=f^{-1}(&&\operatorname {Im} f&&\setminus &&R)\\&=f^{-1}(&&\operatorname {Im} f&&\setminus [&&R\cap \operatorname {Im} f])\end{alignedat}}}[note 7] None
${\displaystyle f\left(L~\triangle ~R\right)~\supseteq ~f(L)~\triangle ~f(R)}$
${\displaystyle f^{-1}\left(L~\triangle ~R\right)~=~f^{-1}(L)~\triangle ~f^{-1}(R)}$ None

Preimages preserve set operations

Preimages of sets are well-behaved with respect to all basic set operations:

{\displaystyle {\begin{alignedat}{4}f^{-1}(L\cup R)~&=~f^{-1}(L)\cup f^{-1}(R)\\f^{-1}(L\cap R)~&=~f^{-1}(L)\cap f^{-1}(R)\\f^{-1}(L\setminus R)~&=~f^{-1}(L)\setminus f^{-1}(R)\\f^{-1}(L\triangle R)~&=~f^{-1}(L)\triangle f^{-1}(R)\\\end{alignedat}}}

In words, preimages distribute over unions, intersections, set subtraction, and symmetric difference.

Images only preserve unions

Images of unions are well-behaved:

{\displaystyle {\begin{alignedat}{4}f(L\cup R)~&=~f(L)\cup f(R)\\\end{alignedat}}}

but images of the other basic set operations are not since only the following are guaranteed in general:

{\displaystyle {\begin{alignedat}{4}f(L\cap R)~&\subseteq ~f(L)\cap f(R)\\f(L\setminus R)~&\supseteq ~f(L)\setminus f(R)\\f(L\triangle R)~&\supseteq ~f(L)\triangle f(R)\\\end{alignedat}}}

In words, images distribute over unions but not necessarily over intersections, set subtraction, or symmetric difference.

In general, equality is not guaranteed for images of set subtraction ${\displaystyle L\setminus R}$ nor for images of the other two elementary set operators that can be defined as the difference of two sets:

${\displaystyle L\cap R=L\setminus (L\setminus R)\quad {\text{ and }}\quad L\triangle R=(L\cup R)\setminus (L\cap R).}$

If ${\displaystyle L=X}$ then ${\displaystyle f(X\setminus R)\supseteq f(X)\setminus f(R)}$ where as in the more general case, equality is not guaranteed. If ${\displaystyle f}$ is surjective then ${\displaystyle f(X\setminus R)~\supseteq ~Y\setminus f(R),}$ which can be rewritten as: ${\displaystyle f\left(R^{\operatorname {C} }\right)~\supseteq ~f(R)^{\operatorname {C} }}$ if ${\displaystyle R^{\operatorname {C} }:=X\setminus R}$ and ${\displaystyle f(R)^{\operatorname {C} }:=Y\setminus f(R).}$

##### Counter-examples to set equality
Picture showing ${\displaystyle f}$ failing to distribute over set intersection:
${\displaystyle f\left(A_{1}\cap A_{2}\right)\subsetneq f\left(A_{1}\right)\cap f\left(A_{2}\right).}$
The map ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ is defined by ${\displaystyle x\mapsto x^{2},}$ where ${\displaystyle \mathbb {R} }$ denotes the real numbers. The sets ${\displaystyle A_{1}=[-4,2]}$ and ${\displaystyle A_{2}=[-2,4]}$ are shown in blue immediately below the ${\displaystyle x}$-axis while their intersection ${\displaystyle A_{3}=[-2,2]}$ is shown in green.

Examples will be given demonstrating that the set containments