nyuuzyou/wikis huggingface dataset:
Complement of Countable Set in Topological Space where Open Neighborhood is Uncountable is Everywhere Dense

== Theorem ==
Let $\struct {X, \tau}$ be a topological space such that:
:every open set $U \in \tau$ is uncountable.
Let $A \subseteq X$ be countable.
Then $X \setminus A$ is everywhere dense in $\struct {X, \tau}$. 
=== Corollary ===
== Proof ==
Let $a \in X$. 
Let $U$ be an open neighborhood of $a$ in $\struct {X, \tau}$. 
Then $U$ is uncountable by hypothesis.
From Uncountable Set less Countable Set is Uncountable, $U \cap \paren {X \setminus A}$ is uncountable.
In particular, $U \cap \paren {X \setminus A} \ne \O$.
Hence we have $a \in \paren {X \setminus A}^-$, where $\paren {X \setminus A}^-$ denotes the closure of $X \setminus A$.
So $X \setminus \paren {X \setminus A}^-$.
Since we have $\paren {X \setminus A}^-$ automatically, we obtain $\paren {X \setminus A}^- = X$.
So $X \setminus A$ is everywhere dense in $\struct {X, \tau}$. 
Category:Topological Spaces
Category:Complement of Countable Set in Topological Space where Open Neighborhood is Uncountable is Everywhere Dense
Complement of Countable Set in Topological Space where Open Neighborhood is Uncountable is Everywhere Dense/Corollary

== Corollary ==
Consider the real numbers $\R$ as a normed vector space with the Euclidean norm.
Let $A \subseteq \R$ be countable.
Then $\R \setminus A$ is everywhere dense in $\R$. 
== Proof ==
Let $U \subseteq \R$ be an open set.
Then there exists $a < b$ such that $\openint a b \subseteq U$. 
Taking $\epsilon > 0$ such that $a - \epsilon < b + \epsilon$, we have: 
:$\closedint {a + \epsilon} {b - \epsilon} \subseteq \openint a b \subseteq U$ 
From Closed Interval in Reals is Uncountable, $\closedint {a + \epsilon} {b - \epsilon}$ is uncountable.
Hence $U$ is uncountable.
Hence every open subset of $\R$ is uncountable.
Hence Complement of Countable Set in Topological Space where Open Neighborhood is Uncountable is Everywhere Dense is admissable, and $\R \setminus A$ is everywhere dense in $\R$.
Category:Complement of Countable Set in Topological Space where Open Neighborhood is Uncountable is Everywhere Dense
Complement of Direct Image Mapping of Injection equals Direct Image of Complement

== Theorem ==
Let $f: S \to T$ be an injection.
Let $f^\to: \powerset S \to \powerset T$ denote the direct image mapping of $f$.
Then:
:$\forall A \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } A = \map {\paren {f^\to \circ \complement_S} } A$
where $\circ$ denotes composition of mappings.
== Proof ==
As $f$ is an injection, it is a fortiori a one-to-many relation.
From Image of Set Difference under Relation: Corollary 2:
:$\forall A \in \powerset S: \map {\paren {\complement_{\Img \RR} \circ \RR^\to} } A = \map {\paren {\RR^\to \circ \complement_S} } A$
where $\RR \subseteq S \times T$ is a one-to-many relation on $S \times T$.
Substituting $f$ for $\RR$ gives the result:
:$\forall A \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } A = \map {\paren {f^\to \circ \complement_S} } A$
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $3$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(c)}$
Category:Direct Image Mappings
Category:Injections
Category:Relative Complement
Complement of Element is Irreducible implies Element is Meet Irreducible

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $P = \struct {\tau, \preceq}$ be an ordered set
where $\mathord \preceq = \mathord \subseteq \cap \paren {\tau \times \tau}$
Let $A \in \tau$.
Then $\relcomp S A$ is irreducible implies
$A$ is meet irreducible in $P$
where $\relcomp S A$ denotes the relative complement of $A$ relative to $S$.
== Proof ==
Assume that
:$\relcomp S A$ is irreducible.
Let $x, y \in \tau$ such that
:$A = x \wedge y$
By definition of topological space:
:$x \cap y \in \tau$
By Meet in Inclusion Ordered Set:
:$x \wedge y = x \cap y$
By De Morgan's Laws: Complement of Intersection:
:$\relcomp S A = \relcomp S x \cup \relcomp S y$
By definition:
:$\relcomp S x$ and $\relcomp S y$ are closed.
By definition of irreducible:
:$\relcomp S A = \relcomp S x$ or $\relcomp S A = \relcomp S y$
Thus by Relative Complement of Relative Complement:
:$A = x$ or $A = y$
== Sources ==
* 
* 
Category:Topology
Category:Meet Irreducible Elements
Complement of Empty Set is Universal Set

== Theorem ==
The complement of the empty set is the universal set:
:$\map \complement \O = \mathbb U$
== Proof ==
Substitute $\mathbb U$ for $S$ in $\relcomp S \O = S$ from Relative Complement of Empty Set.
== Also see ==
* Complement of Universal Set is Empty Set
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B i}$
* : $\S 6.6$: Subsets
* : $\S 2$
* : $\text {(viii)}$
Category:Set Complement
Category:Empty Set
Category:Universal Set
Complement of F-Sigma Set is G-Delta Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be an $F_\sigma$ set of $T$.
Then its complement $S \setminus X$ is a $G_\delta$ set of $S$.
== Proof ==
Let $X$ be an $F_\sigma$ set of $T$.
Then $X = \ds \bigcup \VV$ where $\VV$ is a countable set of closed sets in $T$.
Then from De Morgan's Laws: Difference with Union we have:
:$\ds S \setminus X = S \setminus \bigcup \VV = \bigcap_{V \mathop \in \VV} \paren {S \setminus V}$
By definition of closed set, each of the $S \setminus V$ are open sets.
So $\ds \bigcap_{V \mathop \in \VV} \paren {S \setminus V}$ is a countable intersection of open sets in $T$.
Hence $S \setminus X$ is, by definition, a $G_\delta$ set of $T$.
== Also see ==
* Complement of $G_\delta$ Set is $F_\sigma$ Set
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
* : Problems: Section $1: \ 3$
Category:F-Sigma Sets
Category:G-Delta Sets
Complement of G-Delta Set is F-Sigma Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be an $G_\delta$ set of $T$.
Then its complement $S \setminus X$ is an $F_\sigma$ set of $T$.
== Proof ==
Let $X$ be a $G_\delta$ set of $T$.
Let $X = \bigcap \UU$ where $\UU$ is a countable set of open sets in $T$.
Then from De Morgan's Laws: Difference with Intersection we have:
:$\ds S \setminus X = S \setminus \bigcap \UU = \bigcup_{U \mathop \in \UU} \paren {S \setminus U}$
By definition of closed set, each of the $S \setminus U$ are closed sets.
So $\ds \bigcup_{U \mathop \in \UU} \paren {S \setminus U}$ is a countable union of closed sets in $T$.
Hence $S \setminus X$ is, by definition, an $F_\sigma$ set of $T$.
== Also see ==
* Complement of $F_\sigma$ Set is $G_\delta$ Set
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
* : Problems: Section $1: \ 3$
Category:F-Sigma Sets
Category:G-Delta Sets
Complement of Horizontal Section of Set is Horizontal Section of Complement

== Theorem ==
Let $X$ and $Y$ be sets.
Let $E \subseteq X \times Y$. 
Let $y \in Y$. 
Then: 
:$\paren {\paren {X \times Y} \setminus E}^y = X \setminus E^y$
where:
:$\paren {\paren {X \times Y} \setminus E}^y$ is the $y$-horizontal section of the set difference $\paren {X \times Y} \setminus E$
:$E^y$ is the $y$-horizontal section of $E$. 
 
== Proof ==
Note that from the definition of set difference, we have that: 
:$x \in X \setminus E^y$
: 
:$x \in X$ and $x \not \in E^y$.
That is, from the definition of the $y$-horizontal section: 
:$x \in X$ and $\tuple {x, y} \not \in E$.
This is equivalent to: 
:$\tuple {x, y} \in \paren {X \times Y} \setminus E$
From the definition of the $y$-horizontal section, this is then equivalent to: 
:$x \in \paren {\paren {X \times Y} \setminus E}^y$
So we have:
:$x \in X \setminus E^y$ $x \in \paren {\paren {X \times Y} \setminus E}^y$.
So:
:$\paren {\paren {X \times Y} \setminus E}^y = X \setminus E^y$
Category:Horizontal Section of Sets
Complement of Inaccessible by Directed Suprema Subset is Closed under Directed Suprema

== Theorem ==
Let $L = \struct {S, \preceq}$ be an up-complete ordered set.
Let $X$ be an inaccessible by directed suprema subset of $S$.
Then $\relcomp S X$ is closed under directed suprema.
== Proof ==
Let $D$ be a directed subset of $S$ such that
:$D \subseteq \relcomp S X$
By Empty Intersection iff Subset of Relative Complement:
:$D \cap X = \O$
By definition of inaccessible by directed suprema:
:$\sup D \notin X$
Thus by definition of relative complement:
:$\sup D \in \relcomp S X$
== Also see ==
* Complement of Closed under Directed Suprema Subset is Inaccessible by Directed Suprema
== Sources ==
* 
Category:Order Theory
Complement of Interior equals Closure of Complement

This page also covers Interior of Complement equals Complement of Closure.
== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ and $H^\circ$ denote the interior of $H$.
Let $\map \complement H$ be the complement of $H$ in $T$:
:$\map \complement H = T \setminus H$
Then:
:$\map \complement {H^\circ} = \paren {\map \complement H}^-$
and similarly:
:$\paren {\map \complement H}^\circ = \map \complement {H^-}$
These can alternatively be written:
:$T \setminus H^\circ = \paren {T \setminus H}^-$
:$\paren {T \setminus H}^\circ = T \setminus H^-$
which, it can be argued, is easier to follow.
== Proof ==
Let $\tau$ be the topology on $T$.
Let $\mathbb K = \set {K \in \tau: K \subseteq H}$.
Then:
By the definition of closed set, $K$ is open in $T$ $T \setminus K$ is closed in $T$.
Also, from Set Complement inverts Subsets we have that $T \setminus K \supseteq T \setminus H$.
Now consider the set $\mathbb K'$ defined as:
:$\mathbb K' := \set {K' \subseteq T: \paren {T \setminus H} \subseteq K', K' \text { closed in } T}$
From the above we see that:
:$K \in \mathbb K \iff T \setminus K \in \mathbb K'$
Thus:
Then we note that:
and so:
== Also see ==
* Interior equals Complement of Closure of Complement
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Closures
Category:Set Interiors
Complement of Interval Defined by Absolute Value

== Theorem ==
Let $\xi, \delta \in \R$ be real numbers.
Let $\delta > 0$.
Then:
=== Complement of Open Interval Defined by Absolute Value ===
=== Complement of Closed Interval Defined by Absolute Value ===
== Also see ==
* Interval Defined by Absolute Value
Category:Real Intervals
Category:Absolute Value Function
Complement of Irreducible Topological Subset is Prime Element

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $X$ be an irreducible subset of $S$ such that:
:$\relcomp S X \in \tau$
Let $L = \struct {\tau, \preceq}$ be an inclusion ordered set of topology $\tau$.
Then $\relcomp S X$ is prime element in $L$.
== Proof ==
Let $Y, Z \in \tau$ such that
:$Y \wedge Z \preceq \relcomp S X$
By definition of topological space:
:$Y \cap Z \in \tau$
By Meet in Inclusion Ordered Set:
:$Y \cap Z = Y \wedge Z$
By definition of inclusion ordered set:
:$Y \cap Z \subseteq \relcomp S X$
By Relative Complement inverts Subsets and Relative Complement of Relative Complement:
:$X \subseteq \relcomp S {Y \cap Z}$
By De Morgan's Laws: Relative Complement of Intersection:
:$X \subseteq \relcomp S Y \cup \relcomp S Z$
By Intersection with Subset is Subset:
:$X = \paren {\relcomp S Y \cup \relcomp S Z} \cap X$
By Intersection Distributes over Union:
:$X = \paren {\relcomp S Y \cap X} \cup \paren {\relcomp S Z \cap X}$
By definition of closed set and Relative Complement of Relative Complement:
:$X$, $\relcomp S Y$, and $\relcomp S Z$ are closed sets.
By Intersection of Closed Sets is Closed in Topological Space:
:$\relcomp S Y \cap X$, $\relcomp S Z \cap X$ are closed sets.
By definition of irreducible:
:$\relcomp S Y \cap X = X$ or $\relcomp S Z \cap X = X$
By Intersection with Subset is Subset:
:$X \subseteq \relcomp S Y$ or $X \subseteq \relcomp S Z$
By Relative Complement inverts Subsets and Relative Complement of Relative Complement:
:$Y \subseteq \relcomp S X$ or $Z \subseteq \relcomp S X$
Thus by definition of inclusion ordered set:
:$Y \preceq \relcomp S X$ or $Z \preceq \relcomp S X$
== Sources ==
* 
* 
Category:Prime Elements
Category:Irreducible Spaces
Complement of Limit Inferior is Limit Superior of Complements

== Theorem ==
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of sets.
Then:
:$\ds \map \complement {\liminf_{n \mathop \to \infty} \ E_n} = \limsup_{n \mathop \to \infty} \ \map \complement {E_n}$
where $\liminf$ and $\limsup$ denote the limit inferior and limit superior, respectively.
== Proof ==
== Sources ==
* : $\S 1.1$: Rings and Algebras: Problem $1.1.1$
Category:Limits Superior of Set Sequences
Category:Limits Inferior of Set Sequences
Complement of Limit Superior is Limit Inferior of Complements

== Theorem ==
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of sets.
Then:
:$\ds \map \complement {\limsup_{n \mathop \to \infty} \ E_n} = \liminf_{n \mathop \to \infty} \ \map \complement {E_n}$
where $\limsup$ and $\liminf$ denote the limit superior and limit inferior, respectively.
== Proof ==
== Sources ==
* : $\S 1.1$: Rings and Algebras: Problem $1.1.1$
Category:Limits Superior of Set Sequences
Category:Limits Inferior of Set Sequences
Complement of Lower Closure is Prime Element in Inclusion Ordered Set of Scott Sigma

== Theorem ==
Let $L = \struct {S, \preceq, \tau}$ be a complete Scott topological lattice.
Let $D = \struct {\map \sigma L, \precsim}$ be an inclusion ordered set of the Scott sigma of $L$.
Let $x \in S$.
Then:
:$\relcomp S {x^\preceq}$ is a prime element in $D$
and:
:$\relcomp S {x^\preceq} \ne S$
== Proof ==
By Scott Topology equals to Scott Sigma:
:$\tau = \map \sigma L$
By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice:
:$x^\preceq = \set x^-$
where $\set x^-$ denotes the topological closure of $\set x$.
By Topological Closure of Singleton is Irreducible:
:$x^\preceq$ is topologically irreducible
Thus by Complement of Irreducible Topological Subset is Prime Element:
:$\relcomp S {x^\preceq}$ is a prime element in $D$.
By definitions of reflexivity and lower closure of element:
:$x \in x^\preceq$
By definitions of relative complement and difference:
:$x \notin \relcomp S {x^\preceq}$
Hence by definition set equality:
:$\relcomp S {x^\preceq} \ne S$
== Sources ==
* 
* 
Category:Prime Elements
Category:Topological Order Theory
Complement of Lower Closure of Element is Open in Scott Topological Ordered Set

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a relational structure with Scott topology
where $\struct {S, \preceq}$ is an up-complete ordered set.
Let $x \in S$.
Then $\relcomp S {x^\preceq}$ is topologically open,
where
:$x^\preceq$ denotes the lower closure of $x$,
:$\relcomp S {x^\preceq}$ denotes the relative complement of $x^\preceq$.
== Proof ==
By Lower Closure of Element is Topologically Closed in Scott Topological Ordered Set:
:$x^\preceq$ is closed.
By definition of closed set:
:$\relcomp S {x^\preceq} \in \tau$
Thus by definition:
:$\relcomp S {x^\preceq}$ is a open set.
== Sources ==
* 
* 
Category:Topological Order Theory
Category:Open Sets
Complement of Lower Section is Upper Section

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $L$ be a lower section.
Then $S \setminus L$ is an upper section.
== Proof ==
Let $u \in S \setminus L$.
Let $s \in S$ such that $u \preceq s$.
 $s \notin S \setminus L$.
Then $s \in L$.
By definition of lower section, $u \in L$, a contradiction.
Hence $s \in S \setminus L$.
Since this holds for all such $u$ and $s$, $S \setminus L$ is an upper section.
== Also see ==
* Complement of Upper Section is Lower Section
== Sources ==
* 
Category:Upper Sections
Category:Lower Sections
Complement of Normal Subgroup is Isomorphic to Quotient Group

== Definition ==
Let $G$ be a group with identity $e$.
Let $N$ be a normal subgroup of $G$.
Let $K$ be a complement of $N$.
Then the quotient group $G / N$ is isomorphic to $K$.
== Proof ==
Let $\phi: K \to G / N$ be the mapping defined as:
:$\forall k \in K: \map \phi k = k N$
We have:
Thus $\phi$ is shown to be a homomorphism.
By definition of $K$ and $N$ we have that:
:$\forall g \in G: \exists k \in K, n \in N: g = k n$
That is, every element of $g$ is the product of an element of $K$ with an element of $N$.
Let $x N, y N \in G / N$.
Then:
demonstrating that $\phi$ is an injection.
Then we have:
demonstrating that $\phi$ is a surjection.
Thus $\phi$ is an injective and surjective homomorphism.
Hence, by definition, $\phi$ is an isomorphism.
Hence the result.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $22$
Category:Normal Subgroups
Category:Subgroup Complements
Complement of Open Interval Defined by Absolute Value

== Theorem ==
Let $\xi, \delta \in \R$ be real numbers.
Let $\delta > 0$.
Then:
:$\set {x \in \R: \size {\xi - x} \ge \delta} = \R \setminus \openint {\xi - \delta} {\xi + \delta}$
where:
:$\openint {\xi - \delta} {\xi + \delta}$ is the open real interval between $\xi - \delta$ and $\xi + \delta$
:$\setminus$ denotes the set difference operator.
== Proof ==
== Also presented as ==
:$\set {x \in \R: \size {x - \xi} > \delta} = \R \setminus \openint {\xi - \delta} {\xi + \delta}$
which is immediate from:
:$\size {x - \xi} = \size {\xi - x}$
== Also see ==
* Complement of Closed Interval Defined by Absolute Value
* Open Interval Defined by Absolute Value
* Closed Interval Defined by Absolute Value
Category:Real Intervals
Category:Absolute Value Function
Complement of Preimage equals Preimage of Complement

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $T_1$ be a subset of $T$.
Then:
:$\relcomp S {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$
where:
:$\complement_S$ (in this context) denotes relative complement
:$f^{-1} \sqbrk {T_1}$ denotes preimage.
That is:
:$S \setminus f^{-1} \sqbrk {T_1} = f^{-1} \sqbrk {T \setminus T_1}$
This can be expressed in the language and notation of inverse image mappings as:
:$\forall T_1 \in \powerset T: \relcomp S {\map {f^\gets} {T_1} } = \map {f^\gets} {\relcomp T {T_1} }$
== Proof ==
From One-to-Many Image of Set Difference: Corollary 2 we have:
:$\relcomp {\Img \RR} {\RR \sqbrk {S_1} } = \RR \sqbrk {\relcomp S {S_1} }$
where:
:$S_1 \subseteq S$
:$\RR \subseteq T \times S$ is a one-to-many relation on $T \times S$.
Hence as $f^{-1}: T \to S$ is a one-to-many relation:
:$\relcomp {\Preimg f} {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$
But from Preimage of Mapping equals Domain, we have that:
:$\Preimg f = S$
Hence:
:$\relcomp S {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$
== Sources ==
* : $\S 10$: Inverses and Composites
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.13 \ \text{(d)}$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1 \ \text{(jw)}$
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $4 \ \text{(c})$
* : Notation and Terminology
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $6 \ \text {(c)}$
Category:Mapping Theory
Category:Relative Complement
Complement of Prime Ideal of Ring is Multiplicatively Closed

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $P \subset R$ be a prime ideal of $R$.
Then its complement $R \setminus P$ is multiplicatively closed.
== Proof ==
Since $P$ is a proper ideal by , we have:
:$1_R \in R \setminus P$
where $1_R$ is the unity of $R$.
 $R \setminus P$ is not multiplicatively closed.
That is:
:$\exists a, b \in R \setminus P: a b \notin R \setminus P$
This means:
:$a b \in P$
This contradicts the assertion that $P$ is a prime ideal of $R$.
== Proof 2 ==
This follows immediately from .
== Also see ==
* Definition:Localization of Ring at Prime Ideal
Category:Prime Ideals of Rings
Category:Localization of Rings
Complement of Primitive Recursive Set

== Theorem ==
Let $S \subseteq \N$ be primitive recursive.
Then its relative complement $\N \setminus S$ of $S$ in $\N$ is primitive recursive.
== Proof ==
By definition, we have that the characteristic function $\map {\chi_{\N \mathop \setminus S} } n = 1$ $\map {\chi_S} n = 0$.
So:
:$\map {\chi_{\N \mathop \setminus S} } n = \map {\chi_{\set 0} } {\map {\chi_S} n}$
Thus $\chi_{\N \mathop \setminus S}$ is obtained by substitution from $\chi_{\set 0}$ and $\chi_S$.
The result follows from Set Containing Only Zero is Primitive Recursive.
Category:Primitive Recursive Functions
Category:Relative Complement
Complement of Reflexive Relation

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation.
Then $\RR$ is reflexive its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is antireflexive.
Likewise, $\RR$ is antireflexive its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is reflexive.
== Proof ==
Let $\RR \subseteq S \times T$ be reflexive.
Then:
:$\forall x \in S: \tuple {x, x} \in \RR$
By the definition of complement of $\RR$:
:$\tuple {x, y} \in \RR \implies \tuple {x, y} \notin \relcomp {S \times S} \RR$
The same applies to $\tuple {x, x}$, and thus:
:$\forall x \in S: \tuple {x, x} \notin \relcomp {S \times S} \RR$
Thus $\relcomp {S \times S} \RR$ is antireflexive.
Similarly, by definition:
:$\forall x \in S: \tuple {x, x} \notin \RR \implies \neg \tuple {x, x} \notin \relcomp {S \times S} \RR$
By Double Negation it follows that:
:$\tuple {x, x} \in \relcomp {S \times S} \RR$
The converses of these results follow from the fact that:
:$\relcomp {S \times S} {\relcomp {S \times S} \RR} = \RR$
by Relative Complement of Relative Complement.
Category:Reflexive Relations
Category:Relative Complement
Complement of Relation Compatible with Group is Compatible

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\RR$ be a relation on $G$.
Let $\RR$ be compatible with $\circ$.
Let $\QQ = \complement_{G \times G} \RR$, so that:
:$\forall a, b \in G: a \mathrel \QQ b \iff \neg \paren {a \mathrel \RR b}$
Then $\QQ$ is a relation compatible with $\circ$.
== Proof ==
Let $x, y, z \in G$.
Suppose that $\neg \paren {\paren {x \circ z} \mathrel \QQ \paren {y \circ z} }$.
Then by the definition of $\QQ$:
:$\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
Because $\RR$ is compatible with $\circ$:
:$\paren {x \circ z} \circ z^{-1} \mathrel \RR \paren {y \circ z} \circ z^{-1}$
By and the :
:$x \mathrel \RR y$
so by the definition of $\QQ$:
:$\neg \paren {x \mathrel \QQ y}$
By the Rule of Transposition:
:$\forall x, y, z \in G: x \mathrel \QQ y \implies \paren {x \circ z} \mathrel \QQ \paren {y \circ z}$
A similar argument shows that:
:$\forall x, y, z \in G: x \mathrel \QQ y \implies \paren {z \circ x} \mathrel \QQ \paren {z \circ y}$
Thus, by definition, $\QQ$ is a relation compatible with $\circ$.
Category:Compatible Relations
Category:Group Theory
Complement of Relative Complement is Union with Complement

== Theorem ==
Let $A, B, C$ be sets such that $A \subseteq B \subseteq C$.
Then:
:$\relcomp C {\relcomp B A} = A \cup \relcomp C B$
== Proof ==
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $2 \ \text{(b)}$
Category:Subsets
Category:Relative Complement
Complement of Set of Rational Pairs in Real Euclidean Plane is Arc-Connected

== Theorem ==
Let $\struct {\R^2, d}$ be the real number plane with the usual (Euclidean) topology.
Let $S \subseteq \R^2$ be the subset of $\R^2$ defined as:
:$\forall x, y \in \R^2: \tuple {x, y} \in S \iff x, y \in \Q$
Hence let $A := \R^2 \setminus S$:
:$\tuple {x, y} \in A$ either $x$ or $y$ or both is irrational.
Then $A$ is arc-connected.
== Proof ==
Let $\tuple {a, b} \in A$.
Consider any point $\tuple {x_1, y_1} \in A$ whose coordinates are both irrational.
By definition, either $a$ or $b$ is irrational.
, suppose $a$ is irrational.
Then the union of the straight lines $x = a, y = y_1$ is an arc-connected subset of $A$ connecting $\tuple {x_1, y_1}$ to $\tuple {a, b}$.
Hence any point in $A$ can be connected to $\tuple {x_1, y_1}$ by an arc.
Hence the result, by definition of arc-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $33$. Special Subsets of the Plane: $2$
Category:Euclidean Space
Category:Arc-Connected Spaces