User:Mgkrupa/List of set identities and relations

This article lists mathematical properties and laws of sets, 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 ( $\cup$ ) and intersection ( $\cap$ ) satisfy many identities. Several of these identities or "laws" have well established names.

List of set identities and relations

Notation

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

For sets $A$ and $B,$ define:

{\begin{alignedat}{4}A\cup B&&~=~\{~x~:~x\in A\;&&{\text{ or }}\;\,&&\;x\in B~\}\\A\cap B&&~=~\{~x~:~x\in A\;&&{\text{ and }}&&\;x\in B~\}\\A\setminus B&&~=~\{~x~:~x\in A\;&&{\text{ and }}&&\;x\notin B~\}.\\\end{alignedat}}

The symmetric difference of $A$ and $B$ is:

A\;\triangle \;B~=~\left(A\setminus B\right)~\cup ~\left(B\setminus A\right)

and the complement of a set $B$ is:

B^{C}=X\setminus B

where

X

is the universe set.

Algebra of sets

A family $\Phi$ of subsets of a set $X$ is said to be an algebra of sets if $\varnothing \in \Phi$ and for all $A,B\in \Phi ,$ all three of the sets $X\setminus A,$ $A\cap B,$ and $A\cup B$ are elements of $\Phi .$ ^[1] The article on this topic lists set identities and other relationships these three operations.

Every algebra of sets is also a ring of sets^[1] and a π-system.

Algebra generated by a family of sets

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

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

Basic set relationships

Assume that $A,B,C\subseteq X.$

Commutativity:

$A\cup B=B\cup A$
$A\cap B=B\cap A$

Associativity:

$(A\cup B)\cup C=A\cup (B\cup C)$
$(A\cap B)\cap C=A\cap (B\cap C)$

Distributivity:

$A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$
$A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$

Identity:

$A\cup \varnothing =A$
$A\cap X=A$

Complement:

$A\cup A^{C}=X$
$A\cap A^{C}=\varnothing$

Idempotent:

$A\cup A=A$
$A\cap A=A$

Domination:

$A\cup X=X$
$A\cap \varnothing =\varnothing$

Absorption laws:

$A\cup (A\cap B)=A$
$A\cap (A\cup B)=A$

Intersection can be expressed in terms of set difference :

$A\cap B=A\setminus (A\setminus B)$

Algebra of inclusion

The following proposition says that the binary relation of inclusion is a partial order.

Reflexivity:

$A\subseteq A$

Antisymmetry:

$A\subseteq B$ and $B\subseteq A$ if and only if $A=B$

Transitivity:

If $A\subseteq B$ and $B\subseteq C,$ then $A\subseteq C$

The following proposition says that for any set $S,$ the power set of $S,$ 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:

$\varnothing \subseteq A\subseteq X$

Existence of joins:

$A\subseteq A\cup B$
If $A\subseteq C$ and $B\subseteq C,$ then $A\cup B\subseteq C$

Existence of meets:

$A\cap B\subseteq A$
If $C\subseteq A$ and $C\subseteq B,$ then $C\subseteq A\cap B$

The following are equivalent:

$A\subseteq B$
$A\cap B=A$
$A\cup B=B$
$A\setminus B=\varnothing$
$B^{C}\subseteq A^{C}$

Some relationships involving complements

Assume that $A,B,C\subseteq X.$

De Morgan's laws:

$(A\cup B)^{C}=A^{C}\cap B^{C}$
$(A\cap B)^{C}=A^{C}\cup B^{C}$

Double complement or involution law:

${(A^{C})}^{C}=A$

Complement laws for the universe set and the empty set:

$\varnothing ^{C}=X$
$X^{C}=\varnothing$

Uniqueness of complements:

If $A\cup B=X$ and $A\cap B=\varnothing$ then $B=A^{C}$

Algebra of relative complements

{\begin{alignedat}{2}\left(C\setminus B\right)\cup A&=\left(C\cup A\right)\setminus \left(B\setminus A\right)\\&=\left(C\setminus \left(B\cup A\right)\right)\cup A\;\;\;\;\;\;{\text{ (the outermost union is disjoint) }}\\\end{alignedat}}

{\begin{alignedat}{2}(C\setminus B)\cap A&=(C\cap A)\setminus B\\&=C\cap (A\setminus B)\\\end{alignedat}}

{\begin{alignedat}{2}(C\setminus B)\setminus A&=(C\setminus B)\cap (C\setminus A)\\&=C\setminus (B\cup A)\\\end{alignedat}}

C\setminus (B\cap A)=(C\setminus B)\cup (C\setminus A)

C\setminus (B\setminus A)=\left(C\setminus B\right)\cup \left(C\cap A\right)

So if $C\subseteq B$ then $C\setminus (B\setminus A)=C\cap A$

A^{C}=X\setminus A

(by definition of this notation)

B\setminus A=A^{C}\cap B

(B\setminus A)^{C}=A\cup B^{C}

A\setminus \varnothing =A

{\begin{alignedat}{2}\varnothing &=A\setminus A\\&=\varnothing \setminus A\\&=A\setminus X\\\end{alignedat}}

Intersection and unions of arbitrary families of sets

Let $\left(S_{i,j}\right)_{(i,j)\in I\times J}$ be a families of sets.

$\bigcup _{i\in I}\left(\bigcap _{j\in J}S_{i,j}\right)\subseteq \bigcap _{j\in J}\left(\bigcup _{i\in I}S_{i,j}\right)$

$\bigcup _{i\in I}\left(\bigcup _{j\in J}S_{i,j}\right)=\bigcup _{j\in J}\left(\bigcup _{i\in I}S_{i,j}\right)$

$\bigcap _{i\in I}\left(\bigcap _{j\in J}S_{i,j}\right)=\bigcap _{j\in J}\left(\bigcap _{i\in I}S_{i,j}\right)$

$\prod _{j\in J}\left(\bigcap _{i\in I}S_{i,j}\right)=\bigcap _{i\in I}\left(\prod _{j\in J}S_{i,j}\right)$

In particular, $\left(\prod _{i\in I}A_{i}\right)\cap \left(\prod _{i\in I}B_{i}\right)=\prod _{i\in I}\left(A_{i}\cap B_{i}\right)$

$\left(\bigcup _{i\in I}A_{i}\right)\cap B=\bigcup _{i\in I}\left(A_{i}\cap B\right)$

$\left(\bigcup _{i\in I}A_{i}\right)\cap \left(\bigcup _{j\in J}B_{j}\right)=\bigcup _{i\in I,j\in J}\left(A_{i}\cap B_{j}\right)$

More generally, suppose that for each $i\in I,$ $J_{i}$ is some non-empty index set and for each $j\in J_{i},$ $S_{i,j}$ is a set. Let ${\mathcal {F}}$ be the set of all functions $f$ on $I$ such that for each $i\in I,$ $f(i)\in J_{i},$ (note that if all $J_{i}$ are equal to some set, call it $J,$ then ${\mathcal {F}}=J^{I}$ ). Then

\bigcap _{i\in I}\left[\bigcup _{j\in J_{i}}S_{i,j}\right]=\bigcup _{f\in {\mathcal {F}}}\left[\bigcap _{i\in I}S_{i,f(i)}\right]

Sets and maps

Definitions

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

Many of the identities below do not actually require that the sets be somehow related to $f$ 's domain or codomain (i.e. to $X$ or $Y$ ) so when some kind of relationship is necessary then it will be clearly indicated. Because of this, in this article, if $S$ is declared to be "any set," and it is not indicated that $S$ must be somehow related to $X$ or $Y$ (say for instance, that it be a subset $X$ or $Y$ ) then it is meant that $S$ is truly arbitrary. So, for instance, it's even possible that $S\cap (X\cup Y)=\varnothing ,$ or that $S\cap X\neq \varnothing$ and $S\cap Y\neq \varnothing$ (which happens, for instance, if $X=Y$ ), etc.

If $S$ is any set then by definition,

f -1 (S) ≝ {x \in domain(f) : f (x) \in S}

and

f (S) ≝ {f (s) : s \in S \cap domain(f) }

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

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

The restriction of a $f:X\to Y$ to $S,$ denoted by $f{\big \vert }_{S},$ is defined to be the map

f{\big \vert }_{S}~:~S\cap \operatorname {domain} (f)~\to ~Y

with domain $S\cap \operatorname {domain} (f)$ that is defined by sending $x\in S\cap \operatorname {domain} (f)$ to $f(x);$ that is, $f{\big \vert }_{S}\left(x\right)=f(x).$

A set $S$ is said to be $f$ -saturated or simply saturated if $S=f^{-1}(f(S)),$ which is only possible if $S\subseteq \operatorname {domain} (f).$

Throughout, $f:X\to Y$ will be a map, $R,S\subseteq X,$ and $T,U\subseteq Y.$

Finitely many sets

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

Let $R,S,$ and $T$ be completely arbitrary sets. Assume $A\subseteq X$ and $C\subseteq Y.$

Image	Preimage	Additional assumptions on sets
${\begin{alignedat}{4}f\left(S\right)&=f\left(S\cap \operatorname {domain} f\right)\\&=f\left(S\cap X\right)\end{alignedat}}$	${\begin{alignedat}{4}f^{-1}\left(S\right)&=f^{-1}\left(S\cap \operatorname {Im} f\right)\\&=f^{-1}\left(S\cap Y\right)\end{alignedat}}$	None
$f\left(S\right)\subseteq Y$	$f^{-1}\left(S\right)=X$ if and only if $\operatorname {Im} f\subseteq S$	None
$f(X)=\operatorname {Im} f\subseteq Y$	${\begin{alignedat}{4}f^{-1}(Y)&=X\\f^{-1}\left(\operatorname {Im} f\right)&=X\end{alignedat}}$	None
$f(S)=\varnothing$ if and only if $S\cap \operatorname {domain} f=\varnothing$	$f^{-1}(S)=\varnothing$ if and only if $S\cap \operatorname {Im} f=\varnothing$	None
$f(A)=\varnothing$ if and only if $A=\varnothing$	$f^{-1}(C)=\varnothing$ if and only if $C\subseteq Y\setminus \operatorname {Im} f$	$A\subseteq X$ and $C\subseteq Y$
$\operatorname {Im} f\setminus f(S)=f(X)\setminus f(S)~\subseteq ~f\left(X\setminus S\right)$	${\begin{alignedat}{4}X\setminus f^{-1}(S)&=f^{-1}\left(Y\setminus S\right)\\&=f^{-1}\left(Y\setminus \left[S\cap \operatorname {Im} f\right]\right)\end{alignedat}}$ ^{[note 3]}	None
$\operatorname {Im} f=f(X)~=~f(S)\cup f\left(X\setminus S\right)$	${\begin{alignedat}{4}X&=f^{-1}\left(S\right)\cup f^{-1}\left(Y\setminus S\right)\\&=f^{-1}\left(S\right)\cup f^{-1}\left(\operatorname {Im} f\setminus S\right)\end{alignedat}}$	None
$f(T)~=~f(S)\cup f\left(T\setminus S\right)$	${\begin{alignedat}{4}f^{-1}\left(T\right)&=f^{-1}\left(S\right)\cup f^{-1}\left(T\setminus S\right)\\&=f^{-1}\left(S\right)\cup f^{-1}\left(\left(\operatorname {Im} f\cap T\right)\setminus S\right)\end{alignedat}}$	None
$f(S)\supseteq C$ if and only if there exists $B\subseteq S$ such that $f(B)=C$	$f^{-1}(S)\supseteq A$ if and only if $f(A)\subseteq S$	$A\subseteq X$ and $C\subseteq Y$
$f(A)~\supseteq ~f\left(X\setminus A\right)$ if and only if $f(A)=\operatorname {Im} f$	$f^{-1}(C)~\supseteq ~f^{-1}\left(Y\setminus C\right)$ if and only if $f^{-1}(C)=X$	$A\subseteq X$ and $C\subseteq Y$
$(g\circ f)(S)~=~g(f(S))$	$\left(g\circ f\right)^{-1}\left(S\right)~=~f^{-1}\left(g^{-1}(S)\right)$	$f$ and $g$ are functions such that $\operatorname {Im} f\subseteq \operatorname {domain} g$ .

Also:

$f(S)\cap T=\varnothing$ if and only if $S\cap f^{-1}\left(T\right)=\varnothing .$
When preimages preserve $\subseteq$ : If $C~\subseteq ~\operatorname {Im} f$ then $f^{-1}(C)~\subseteq ~f^{-1}(T)$ if and only if $C~\subseteq ~T.$

Image	Preimage	Additional assumptions on sets
$S\subseteq T$ implies $f(S)\subseteq f(T)$	$S\subseteq T$ implies $f^{-1}(S)\subseteq f^{-1}(T)$	None
$f\left(S\cup T\right)~=~f(S)\cup f(T)$ ^[2]	$f^{-1}\left(S\cup T\right)~=~f^{-1}(S)\cup f^{-1}(T)$	None
$f\left(S\cap T\right)~\subseteq ~f(S)\cap f(T)$ If $f$ is injective then equality holds.^[3]	$f^{-1}\left(S\cap T\right)~=~f^{-1}(S)\cap f^{-1}(T)$	None
$f\left(S\setminus T\right)~\supseteq ~f(S)\setminus f(T)$ If $f$ is injective then equality holds. If $T~=~f^{-1}(f(T))$ then equality holds.^{[note 4]} If $f$ is surjective then $f\left(X\setminus S\right)~\supseteq ~Y\setminus f(S).$ ^{[note 5]}	$f^{-1}\left(S\setminus T\right)~=~f^{-1}(S)\setminus f^{-1}(T)$	None
$f\left(S\triangle T\right)~\supseteq ~f(S)\triangle f(T)$ If $f$ is injective then equality holds.	$f^{-1}\left(S\triangle T\right)~=~f^{-1}(S)\triangle f^{-1}(T)$	None

Images of preimages and preimages of images

Let $S$ and $T$ be arbitrary sets, $f:X\rightarrow Y$ be any map, and let $A\subseteq X$ and $C\subseteq Y$ .

Image of preimage	Preimage of image	Additional assumptions on sets
$f\left(f^{-1}(T)\cap S\right)=T\cap f(S)$	$f^{-1}\left(f(T)\cap S\right)~\supseteq ~T\cap f^{-1}(S)$	None
$f\left(f^{-1}(T)\right)=T\cap \operatorname {Im} f$ $f\left(f^{-1}(T)\right)=T$ if and only if $T\subseteq \operatorname {Im} f$	$f^{-1}\left(f(T)\right)~\supseteq ~T\cap f^{-1}\left(\operatorname {Im} f\right)=T\cap f^{-1}(Y)$	None
$f\left(f^{-1}(T)\cup S\right)~\subseteq ~T\cup f(S)$	$f^{-1}\left(f(A)\cup S\right)~\supseteq ~A\cup f^{-1}(S)$ ^[4]	$A\subseteq X$
$f\left(f^{-1}(Y)\right)=f(X)$	$f^{-1}(f(X))=X$	None
$f\left(f^{-1}(S)\right)=S\cap f(X)$	$\left(f{\big \vert }_{T}\right)^{-1}(S)=T\cap f^{-1}(S)$	None
$f\left(f^{-1}(f(S))\right)=f(S)$	$f^{-1}\left(f\left(f^{-1}(S)\right)\right)=f^{-1}(S)$	None
$f\left(f^{-1}(S)\right)\subseteq S$ If $S\subseteq \operatorname {Im} f$ then equality holds.^[5]^[6] If $S\subseteq Y$ and $f:X\to Y$ is surjective then equality holds.	$f^{-1}\left(f(A)\right)\supseteq A$ If $f$ is injective then equality holds.^[5]^[6] If $A$ is $f$ -saturated then equality holds.	$A\subseteq X$

Infinitely many sets

Images and preimages of unions and intersections

Images and preimages of unions are always preserved. Inverse images preserve both unions and intersections. It is only images of intersections that are not always preserved.

If $\left(S_{i}\right)_{i\in I}$ is a family of arbitrary sets indexed by $I\neq \varnothing$ then:

{\begin{alignedat}{2}f^{-1}\left(\bigcap _{i\in I}S_{i}\right)&~=~\bigcap _{i\in I}f^{-1}\left(S_{i}\right)\\f^{-1}\left(\bigcup _{i\in I}S_{i}\right)&~=~\bigcup _{i\in I}f^{-1}\left(S_{i}\right)\\f\left(\bigcup _{i\in I}S_{i}\right)&~=~\bigcup _{i\in I}f\left(S_{i}\right)\\f\left(\bigcap _{i\in I}S_{i}\right)&~\subseteq ~\bigcap _{i\in I}f\left(S_{i}\right)\\\end{alignedat}}

If all $S_{i}$ are $f$ -saturated then $\bigcap _{i\in I}S_{i}$ be will be $f$ -saturated and equality will hold in the last relation below. Explicitly, this means:

f\left(\bigcap _{i\in I}S_{i}\right)~=~\bigcap _{i\in I}f\left(S_{i}\right)

IF

X\cap S_{i}=f^{-1}(f(S_{i}))

for all

i\in I.

If $\left(A_{i}\right)_{i\in I}$ is a family of arbitrary subsets of $X=\operatorname {domain} f,$ which means that $A_{i}\subseteq X$ for all $i,$ then this becomes:

f\left(\bigcap _{i\in I}A_{i}\right)~=~\bigcap _{i\in I}f\left(A_{i}\right)

IF

A_{i}=f^{-1}(f(A_{i}))

for all

i\in I.

Preimage from a Cartesian product

This subsection will describe the preimage of a subset $B\subseteq \prod _{j\in J}Y_{j}$ under a map of the form $F~:~X~\to ~\prod _{j\in J}Y_{j}.$ For every $k\in J,$

let $\pi _{k}~:~\prod _{j\in J}Y_{j}~\to ~Y_{k}$ denote the canonical projection onto $Y_{k},$ and
let $F_{k}~:=~\pi _{k}\circ F~:~X~\to ~Y_{k}$

so that $F~=~\left(F_{j}\right)_{j\in J},$ which is also the unique map satisfying: $\pi _{j}\circ F=F_{j}$ for all $j\in J.$ The map $\left(F_{j}\right)_{j\in J}~:~X~\to ~\prod _{j\in J}Y_{j}$ should not be confused with the Cartesian product $\prod _{j\in J}F_{j}$ of these maps, which is the map

\prod _{j\in J}F_{j}~:~\prod _{j\in J}X~\to ~\prod _{j\in J}Y_{j}

defined by sending

\left(x_{j}\right)_{j\in J}\in \prod _{j\in J}X

to

\left(F_{j}\left(x_{j}\right)\right)_{j\in J}.

Observation—If $F=\left(F_{j}\right)_{j\in J}~:~X~\to ~\prod _{j\in J}Y_{j}$ and $B~\subseteq ~\prod _{j\in J}Y_{j}$ then

F^{-1}(B)~\subseteq ~\bigcap _{j\in J}F_{j}^{-1}\left(\pi _{j}\left(B\right)\right).

If $B=\prod _{j\in J}\pi _{j}\left(B\right)$ then equality will hold:

F^{-1}(B)~=~\bigcap _{j\in J}F_{j}^{-1}\left(\pi _{j}\left(B\right)\right).

For equality to hold, it suffices for there to exist a family $\left(B_{j}\right)_{j\in J}$ of subsets $B_{j}\subseteq Y_{j}$ such that $B=\prod _{j\in J}B_{j},$ in which case:

F^{-1}\left(\prod _{j\in J}B_{j}\right)~=~\bigcap _{j\in J}F_{j}^{-1}\left(B_{j}\right)

and $\pi _{j}\left(B\right)=B_{j}$ for all $j\in J.$

Notes

^ Here "smallest" means relative to subset containment. So if $\Phi$ is any algebra of sets containing ${\mathcal {S}},$ then $\Phi _{\mathcal {S}}\subseteq \Phi .$
^ Since ${\mathcal {S}}\neq \varnothing ,$ there is some $S\in {\mathcal {S}}_{0}$ such that its complement also belongs to ${\mathcal {S}}_{0}.$ The intersection of these two sets implies that $\varnothing \in {\mathcal {S}}_{1}.$ The union of these two sets is equal to $X,$ which implies that $X\in \Phi _{\mathcal {S}}.$
^ The conclusion $X\setminus f^{-1}(S)=f^{-1}\left(Y\setminus S\right)$ can also be written as: $f^{-1}(T)^{\operatorname {C} }~=~f^{-1}\left(T^{\operatorname {C} }\right).$
^ Note that this condition $T=f^{-1}(f(T))$ depends entirely on $T$ and not on $S.$
^ $f\left(X\setminus S\right)~\supseteq ~Y\setminus f(S)$ can be rewritten as: $f\left(S^{\operatorname {C} }\right)~\supseteq ~f\left(S\right)^{\operatorname {C} }.$

Citations

^ ^a ^b ^c ^d Cite error: The named reference Encyclopedia of math was invoked but never defined (see the help page).
^ Kelley 1985, p. 85
^ See Munkres 2000, p. 21
^ See p.388 of Lee, John M. (2010). Introduction to Topological Manifolds, 2nd Ed.
^ ^a ^b See Halmos 1960, p. 39
^ ^a ^b See Munkres 2000, p. 19

References

Artin, Michael (1991). Algebra. Prentice Hall. ISBN 81-203-0871-9.
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Courant, Richard, Herbert Robbins, Ian Stewart, What is mathematics?: An Elementary Approach to Ideas and Methods, Oxford University Press US, 1996. ISBN 978-0-19-510519-3. "SUPPLEMENT TO CHAPTER II THE ALGEBRA OF SETS".
Dixmier, Jacques (1984). General Topology. Undergraduate Texts in Mathematics. Translated by Berberian, S. K. New York: Springer-Verlag. ISBN 978-0-387-90972-1. OCLC 10277303.
Dolecki, Szymon; Mynard, Frédéric (2016). Convergence Foundations Of Topology. New Jersey: World Scientific Publishing Company. ISBN 978-981-4571-52-4. OCLC 945169917.
Dugundji, James (1966). Topology. Boston: Allyn and Bacon. ISBN 978-0-697-06889-7. OCLC 395340485.
Halmos, Paul R. (1960). Naive set theory. The University Series in Undergraduate Mathematics. van Nostrand Company. Zbl 0087.04403.
Joshi, K. D. (1983). Introduction to General Topology. New York: John Wiley and Sons Ltd. ISBN 978-0-85226-444-7. OCLC 9218750.
Kelley, John L. (1985). General Topology. Graduate Texts in Mathematics. Vol. 27 (2 ed.). Birkhäuser. ISBN 978-0-387-90125-1.
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Munkres, James R. (2000). Topology (2nd ed.). Upper Saddle River, NJ: Prentice Hall, Inc. ISBN 978-0-13-181629-9. OCLC 42683260.
Schechter, Eric (1996). Handbook of Analysis and Its Foundations. San Diego, CA: Academic Press. ISBN 978-0-12-622760-4. OCLC 175294365.
Schubert, Horst (1968). Topology. London: Macdonald & Co. ISBN 978-0-356-02077-8. OCLC 463753.
Stoll, Robert R.; Set Theory and Logic, Mineola, N.Y.: Dover Publications (1979) ISBN 0-486-63829-4. "The Algebra of Sets", pp 16—23.
Willard, Stephen (2004) [1970]. General Topology. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-43479-7. OCLC 115240.
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External links

Operations on Sets at ProvenMath

[2] Here "smallest" means relative to subset containment. So if $\Phi$ is any algebra of sets containing ${\mathcal {S}},$ then $\Phi _{\mathcal {S}}\subseteq \Phi .$

[3] Since ${\mathcal {S}}\neq \varnothing ,$ there is some $S\in {\mathcal {S}}_{0}$ such that its complement also belongs to ${\mathcal {S}}_{0}.$ The intersection of these two sets implies that $\varnothing \in {\mathcal {S}}_{1}.$ The union of these two sets is equal to $X,$ which implies that $X\in \Phi _{\mathcal {S}}.$

[4] The conclusion $X\setminus f^{-1}(S)=f^{-1}\left(Y\setminus S\right)$ can also be written as: $f^{-1}(T)^{\operatorname {C} }~=~f^{-1}\left(T^{\operatorname {C} }\right).$

[7] Note that this condition $T=f^{-1}(f(T))$ depends entirely on $T$ and not on $S.$

[8] $f\left(X\setminus S\right)~\supseteq ~Y\setminus f(S)$ can be rewritten as: $f\left(S^{\operatorname {C} }\right)~\supseteq ~f\left(S\right)^{\operatorname {C} }.$

[Encyclopedia_of_math-1] Cite error: The named reference Encyclopedia of math was invoked but never defined (see the help page).

[kelley-1985-5] Kelley 1985, p. 85

[munkres-2000-p21-6] See Munkres 2000, p. 21

[lee-2010-p388-9] See p.388 of Lee, John M. (2010). Introduction to Topological Manifolds, 2nd Ed.

[halmos-1960-p39-10] See Halmos 1960, p. 39

[munkres-2000-p19-11] See Munkres 2000, p. 19

[1]

[note 1]

[note 2]

[note 3]

[2]

[3]

[note 4]

[note 5]

[4]

[5]

[6]