nyuuzyou/wikis huggingface dataset:
Comparison Test

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty b_n$ be a convergent series of positive real numbers.
Let $\sequence {a_n}$ be a sequence $\R$ or sequence in $\C$.
Let $\forall n \in \N_{>0}: \cmod {a_n} \le b_n$.
Then the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $\epsilon > 0$.
As $\ds \sum_{n \mathop = 1}^\infty b_n$ converges, its tail tends to zero.
So:
:$\ds \exists N: \forall n > N: \sum_{k \mathop = n + 1}^\infty b_k < \epsilon$
Let $\sequence {s_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$.
Then $\forall n > m > N$:
So $\sequence {s_n}$ is a Cauchy sequence.
The result follows from Real Number Line is Complete Metric Space or Complex Plane is Complete Metric Space.
== Also see ==
* Comparison Test for Divergence
== Sources ==
* : $\S 6.15$
* : $\S 1.2.2$: Summary of convergence tests: Theorem $1.3$
* : Part $\text {IV}$: Infinite Sequences and Infinite Series: Chapter $23$: Infinite Series
* 
* 
* 
* 
Category:Comparison Test
Category:Series
Category:Convergence Tests
Category:Named Theorems
Comparison Test/Corollary 1

== Corollary to Comparison Test ==
Let $\ds \sum_{n \mathop = 1}^\infty b_n$ be a convergent series of positive real numbers.
Let $\sequence {a_n}$ be a sequence in $\R$ or a sequence in $\C$.
Let $H \in \R$.
Let $\exists M: \forall n > M: \size {a_n} \le H b_n$.
Then the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges.
== Proof ==
Let $\epsilon > 0$.
Then $\dfrac \epsilon H > 0$.
As $\ds \sum_{n \mathop = 1}^\infty b_n$ converges, its tail tends to zero.
So:
:$\ds \exists N: \forall n > N: \sum_{k \mathop = n+1}^\infty b_k < \frac \epsilon H$
Let $\sequence {s_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$.
Then $\forall n > m > \max \set {M, N}$:
So $\sequence {s_n}$ is a Cauchy sequence and the result follows from:
:Real Number Line is Complete Metric Space
or:
:Complex Plane is Complete Metric Space.
== Sources ==
* : $\S 6.15$
Category:Comparison Test
Comparison Test/Corollary 2

== Corollary to Comparison Test ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two series of positive real numbers.
Let $\lim_{n \mathop \to \infty} \dfrac {a_n} {b_n} = k$ for some $k \in \R$.
Then either:
:both $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ are convergent
or:
:both $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ are divergent.
== Proof ==
== Sources ==
* 
* 
Category:Comparison Test
Comparison Test for Convergence of Power Series

== Theorem ==
Let $A = \ds \sum_{n \mathop \ge 0} a_n z^n$ and $B = \ds \sum_{n \mathop \ge 0} b_n z^n$ be power series in $\C$.
Let $R_A$ and $R_B$ be the radii of convergence of $A$ and $B$ respectively.
Let $\cmod {b_n} \le \cmod {a_n}$ for all $n \in \N$.
Then $R_A \le R_B$.
== Proof ==
 $R_A > R_B$.
Let $z_0 \in \C$ such that $R_B < \cmod {z_0} < R_A$.
Then $A$ is convergent at $z_0$ but $B$ is divergent at $z_0$.
But by the Comparison Test, if $A$ is convergent at $z_0$ then $B$ is also convergent at $z_0$.
From this contradiction it follows that there can be no such $z_0$.
That is:
:$R_A \le R_B$
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $4$
Category:Radius of Convergence
Category:Complex Power Series
Comparison Test for Divergence

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty b_n$ be a divergent series of positive real numbers.
Let $\sequence {a_n}$ be a sequence in $\R$.
Let:
:$\forall n \in \N_{>0}: b_n \le a_n$
Then the series $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
== Proof ==
This is the contrapositive of the Comparison Test. 
Hence the result, from the Rule of Transposition.
== Also see ==
* Push Theorem
== Sources ==
* : $\S 1.2.2$: Summary of convergence tests: Theorem $1.3$
* 
* : $\S 9.4$
* 
Category:Comparison Test
Comparison Test for Improper Integral

== Theorem ==
Let $I = \openint a b$ be an open real interval.
Let $\phi$ be a real function which is continuous on $I$ and also non-negative on $I$.
Let $f$ be a real function which is continuous on $I$.
Let $f$ satisfy:
:$\forall x \in I: \size {\map f x} \le \map \phi x$
If the improper integral of $\phi$ over $I$ exists, then so does that of $f$.
== Proof ==
, we consider the case $I = \openint 0 \to$ such that $\ds l = \int_0^{\mathop \to +\infty} \map \phi x \rd x$ exists.
Let:
:$\ds a_n = \int_{n - 1}^n \map f x \rd x$
:$\ds b_n = \int_{n - 1}^n \map \phi x \rd x$
for $n = 1, 2, \ldots$
Then the series:
:$\ds \sum_{n \mathop = 1}^\infty b_n$
is a convergent series of positive terms whose sum is $l$.
Also:
$\size {a_n} \le b_n$
for $n = 1, 2, \ldots$
Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ a convergent series by the Comparison Test.
That is:
:$\ds m = \lim_{N \mathop \to \infty} \int_0^N \map f x \rd x$
exists.
Given any $X > 0$, $N$ can be taken to be the smallest natural number which satisfies $N > X$.
Then:
It follows that:
:$\ds \lim_{X \mathop \to \infty} \int_1^X \map f x \rd x = m$
as required.
The other cases are treated similarly.
== Sources ==
* : $\S 13.29$
* : Appendix: $\S 18.8$: Integration
Category:Comparison Test
Category:Convergence Tests
Category:Improper Integrals
Comparison of Sides of Five Platonic Figures/Lemma

== Lemma to Comparison of Sides of Five Platonic Figures ==
== Proof ==
:
Let $ABCDE$ be a regular pentagon.
Let the circle $ABCDE$ be circumscribed around the pentagon $ABCDE$.
Let $F$ be the center of the circle $ABCDE$.
Let $FA, FB, FC, FD, FE$ be drawn.
The straight lines $FA, FB, FC, FD, FE$ bisect the vertices $A, B, C, D, E$ respectively.
The angles at $F$ form a total of $4$ right angles, and are equal.
Therefore one of them, for example $\angle AFB$ equals one right angle less a fifth.
Therefore $\angle FAB + \angle ABF$ consists of one right angle plus a fifth.
But:
:$\angle FAB = \angle FBC$
and so $\angle ABC$ equals one right angle plus a fifth.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Pentagons
Compass and Straightedge Construction for Regular Heptagon does not exist/Proof 1

== Theorem ==
== Proof ==
By definition, the regular heptagon has $7$ sides.
$7$ is a prime number which is not a fermat prime.
The result follows Construction of Regular Prime $p$-Gon Exists iff $p$ is Fermat Prime.
== Sources ==
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
Category:Compass and Straightedge Construction for Regular Heptagon does not exist
Compass and Straightedge Construction for Regular Heptagon does not exist/Proof 2

== Theorem ==
== Proof ==
Construction of a regular heptagon is the equivalent of constructing the point $\tuple {\cos \dfrac {2 \pi} 7, \sin \dfrac {2 \pi} 7}$ from the points $\tuple {0, 0}$ and $\tuple {1, 0}$
Let $\epsilon = \map \exp {\dfrac {2 \pi} 7}$.
Then $\epsilon$ is a root of $x^7 - 1$.
We have:
But we have:
Thus $2 c$ is a root of the polynomial $x^3 + x^2 - 2 x - 1$
But from Irreducible Polynomial: $x^3 + x^2 - 2 x - 1$ in Rationals:
:$x^3 + x^2 - 2 x - 1$ is irreducible over $\Q$.
Thus by Algebraic Element of Degree 3 is not Element of Field Extension of Degree Power of 2, $\cos \dfrac {2 \pi} 7$ is not an element of any extension of $\Q$ of degree $2^m$.
The result follows from Point in Plane is Constructible iff Coordinates in Extension of Degree Power of 2.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 40$. Construction with Ruler and Compasses: Example $79$
Category:Trisecting the Angle
Category:Compass and Straightedge Construction for Regular Heptagon does not exist
Compatibility of Atlases is Equivalence Relation

== Theorem ==
Let $M$ be a topological space.
Let $d$ and $k$ be natural numbers.
Let $\AA$ denote the set of all $d$-dimensional atlases of class $\CC^k$ on $M$.
Define a relation $\sim$ on $\AA$ by putting, for any two $\CC^k$-atlases $\FF$ and $\GG$:
:$\FF \sim \GG$ $\FF$ and $\GG$ are $C^k$-compatible.
Then $\sim$ is an equivalence relation on $\AA$.
== Proof ==
It is to be shown that $\sim$ is reflexive, symmetric and transitive.
=== Reflexive ===
Let $\FF \in \AA$ be a $C^k$-atlas.
That $\FF$ is compatible with itself is precisely condition $(2)$ of the definition of $C^k$-atlas.
=== Symmetric ===
Let $\FF$ and $\GG$ be $C^k$-atlases and suppose that $\FF \sim \GG$.
Let $\struct {U, \phi} \in \FF$ and let $\struct {V, \psi} \in \GG$ be charts.
Then :
:$\phi \circ \psi^{-1}: \psi \sqbrk {U \cap V} \to \phi \sqbrk {U \cap V}$
is a $C^k$ mapping.
Because $\phi$ and $\psi$ are homeomorphisms, we have that:
:$\psi \circ \phi^{-1}: \phi \sqbrk {U \cap V} \to \psi \sqbrk {U \cap V}$
is also a homeomorphism, and in particular continuous.
By the Inverse Function Theorem, $\psi \circ \phi^{-1}$ is also a $C^k$ mapping.
Since the charts were arbitrary, we conclude that $\GG \sim \FF$.
=== Transitive ===
Let $\FF$, $\GG$ and $\HH$ be $C^k$-atlases, and suppose that $\FF \sim \GG$ and $\GG \sim \HH$.
Category:Compatible Atlases
Category:Examples of Equivalence Relations
Complement Operation between Union and Intersection Power Structures is Isomorphism

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Let $\struct {\powerset S, \cap, \subseteq}$ be the ordered semigroup formed from the set intersection operation and subset relation.
Let $\struct {\powerset S, \cap, \supseteq}$ be the ordered semigroup formed from the set intersection operation and superset relation.
Let $\struct {\powerset S, \cup, \subseteq}$ be the ordered semigroup formed from the set union operation and subset relation.
Let $\struct {\powerset S, \cup, \supseteq}$ be the ordered semigroup formed from the set union operation and superset relation.
Let $\complement: \powerset S \to \powerset S$ be the complement operation on $\powerset S$:
:$\forall X \in \powerset S: \map \complement X = S \setminus X$
where $S \setminus X$ denotes the set difference of $X$ with $S$.
Then $\complement$ is an ordered semigroup isomorphism from:
:$\struct {\powerset S, \cap, \subseteq}$ to $\struct {\powerset S, \cup, \supseteq}$
and:
:$\struct {\powerset S, \cap, \supseteq}$ to $\struct {\powerset S, \cup, \subseteq}$
== Proof ==
From:
:Power Set with Intersection and Subset Relation is Ordered Semigroup
:Power Set with Intersection and Superset Relation is Ordered Semigroup
:Power Set with Union and Subset Relation is Ordered Semigroup
:Power Set with Union and Superset Relation is Ordered Semigroup
each of the given ordered structures is indeed an ordered semigroup.
From Relative Complement Mapping on Powerset is Bijection, $\complement$ is a bijection.
From Relative Complement of Relative Complement it follows that $\complement$ is an involution.
From De Morgan's Laws for Relative Complements:
:$\map \complement {T_1 \cap T_2} = \map \complement {T_1} \cup \map \complement {T_2}$
and:
:$\map \complement {T_1 \cup T_2} = \map \complement {T_1} \cap \map \complement {T_2}$
hence demonstrating that $\complement$ exhibits the morphism property from both $\struct {\powerset S, \cap}$ to $\struct {\powerset S, \cup}$ and back again.
Then from Relative Complement inverts Subsets we have that $\complement$ is an order isomorphism between $\struct {\powerset S, \subseteq}$ and $\struct {\powerset S, \supseteq}$ in both directions.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.8$
Category:Ordered Semigroup Isomorphisms
Category:Power Structures
Complement Relative to Subset is Subset of Complement Relative to Superset

== Theorem ==
Let $A, B, C$ be sets such that $A \subseteq B \subseteq C$.
Then:
:$\relcomp B A \subseteq \relcomp C A$
== Proof ==
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $2 \ \text{(a)}$
Category:Subsets
Category:Relative Complement
Complement Union with Superset is Universe

== Theorem ==
:$S \subseteq T \iff \map \complement S \cup T = \mathbb U$
where:
:$S \subseteq T$ denotes that $S$ is a subset of $T$
:$S \cup T$ denotes the union of $S$ and $T$
:$\complement$ denotes set complement
:$\mathbb U$ denotes the universal set.
=== Corollary ===
== Proof ==
== Also see ==
* Intersection with Complement is Empty iff Subset
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $1$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B vi}$
Category:Subsets
Category:Set Union
Category:Set Complement
Category:Empty Set
Complement Union with Superset is Universe/Corollary

== Corollary to Complement Union with Superset is Universe ==
:$S \cup T = \mathbb U \iff \map \complement S \subseteq T$
where:
:$S \subseteq T$ denotes that $S$ is a subset of $T$
:$S \cup T$ denotes the union of $S$ and $T$
:$\complement$ denotes set complement
:$\mathbb U$ denotes the universal set.
== Proof ==
Let $X = \map \complement S$.
Then:
== Also see ==
* Empty Intersection iff Subset of Complement
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $11 \ \text{(a)}$
* : Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(d)}$
Category:Subsets
Category:Set Union
Category:Set Complement
Category:Empty Set
Complement in Boolean Algebra is Unique

== Theorem ==
Let $\left({S, \vee, \wedge}\right)$ be a Boolean algebra.
Then for all $a \in S$, there is a unique $b \in S$ such that:
:$a \wedge b = \bot, a \vee b = \top$
i.e., a valid choice for $\neg a$ as in axiom $(BA \ 4)$ for Boolean algebras.
== Proof ==
Suppose $b, c \in S$ both satisfy the identities.
Then:
That is, $b = c$.
The result follows.
Category:Boolean Algebras
Complement in Distributive Lattice is Unique

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a bounded distributive lattice.
Then every $a \in S$ admits at most one complement.
=== Corollary ===
== Proof ==
Let $a \in S$, and suppose that $b, c \in S$ are complements for $a$.
Then:
Interchanging $c$ and $b$ in the above gives that $c = c \wedge b$ as well.
Hence $b = c$, as desired.
Category:Distributive Lattices
Category:Bounded Lattices
Complement in Distributive Lattice is Unique/Corollary

== Corollary to Complement in Distributive Lattice is Unique ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a Boolean lattice.
Then every $a \in S$ has a unique complement $\neg a$.
== Proof ==
By definition, a Boolean lattice is a complemented distributive lattice.
A complemented lattice is bounded by definition.
By Complement in Distributive Lattice is Unique, it follows that every $a \in S$ has at most one complement.
Since $\struct {S, \vee, \wedge, \preceq}$ is complemented, the result follows.
Category:Distributive Lattices
Category:Boolean Lattices
Complement of Bottom/Boolean Algebra

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Then:
:$\neg \bot = \top$
== Proof ==
Since $\bot$ is the identity for $\vee$, the first condition for $\neg \bot$:
:$\bot \vee \neg \bot = \top$
implies that $\neg \bot = \top$ is the only possibility.
Since $\top$ is the identity for $\wedge$, it follows that:
:$\bot \wedge \top = \bot$
and we conclude that:
:$\neg \bot = \top$
as desired.
== Also see ==
* Complement of Top (Boolean Algebras)
== Sources ==
* : $\S 2$: Exercise $2$
Category:Boolean Algebras
Complement of Bottom/Bounded Lattice

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a bounded lattice.
Then the bottom $\bot$ has a unique complement, namely $\top$, top.
== Proof ==
We know that $\bot$ is the identity for $\vee$.
Therefore, from the condition that:
:$\bot \vee a = \top$
for a complement $a$ of $\bot$, it follows that $a = \top$ is the only possibility.
Since also:
:$\bot \wedge \top = \bot$
as $\top$ is the identity for $\wedge$, the result follows.
== Also see ==
* Complement of Top
Category:Bounded Lattices
Complement of Bounded Set in Complex Plane has at most One Unbounded Component

== Theorem ==
Let $S \subseteq \C$ be bounded.
Then $\C \setminus S$ has at most one unbounded component.
== Proof ==
If $\C \setminus S$ has no unbounded components, we are done.
Suppose that $\C \setminus S$ has at least one unbounded component.
We must show that it has at most one. 
Since $S$ is bounded, there exists $R > 0$ such that:
:$\cmod z \le R$ for all $z \in S$.
Let $C$ be an unbounded component of $\C \setminus S$.
Since $C$ is unbounded, there exists $z_0 \in C$ with $\cmod {z_0} > R$.
Hence:
:$\ds z_0 \in C \cap \set {z \in \C : \cmod z > R}$
From Complement of Closed Disk in Complex Plane is Path-Connected, we have that:
:$\set {z \in \C : \cmod z > R}$ is path-connected.
From Path-Connected Space is Connected:
:$\set {z \in \C : \cmod z > R}$ is connected.
Hence from Union of Connected Sets with Common Point is Connected, we have that:
:$C \cup \set {z \in \C : \cmod z > R}$ is connected.
From definition 3 of a component, we have that:
:$C$ is a maximal connected set.
Since:
:$C \subseteq C \cup \set {z \in \C : \cmod z > R}$
we therefore have:
:$C = C \cup \set {z \in \C : \cmod z > R}$
So we have:
:$\set {z \in \C : \cmod z > R} \subseteq C$
Let $C'$ be a component of $\C \setminus S$ that is distinct from $C$. 
Then $C'$ is disjoint from $C$.
In particular, $C'$ is disjoint from $\set {z \in \C : \cmod z > R}$.
So, we have:
:$C' \subseteq \C \setminus C \subseteq \set {z \in \C : \cmod z \le R}$
That is:
:$C'$ is bounded.
Hence:
:$C$ is the only unbounded component of $\C \setminus S$.
This was the demand.
Category:Components
Complement of Clopen Set is Clopen

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a clopen set of $T$.
Let $\relcomp S H$ denote the complement of $H$ relative to $S$.
Then $\relcomp S H$ is also a clopen set of $T$.
== Proof ==
By definition of clopen, $H$ is open in $T$.
By definition of closed set, $\relcomp S H$ is closed in $T$.
By definition of clopen, $H$ is closed in $T$. 
By definition of closed set, $\relcomp S H$ is open in $T$.
Thus $\relcomp S H$ is both open in $T$ and closed in $T$.
Hence the result, by definition of clopen set.
Category:Relative Complement
Category:Clopen Sets
Complement of Closed Interval Defined by Absolute Value

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

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $F \subseteq S$ be a closed set in $T$.
Then:
:$S \setminus F \in \tau$
== Proof ==
By definition of closed set:
:$\exists U \in \tau : F = S \setminus U$
From Relative Complement inverts Subsets of Relative Complement:
:$U = S \setminus F$
The result follows.
Category:Closed Sets
Category:Open Sets
Complement of Closed under Directed Suprema Subset is Inaccessible by Directed Suprema

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

== Theorem ==
The complement of the complement of a set is the set itself:
:$\map \complement {\map \complement S} = S$
== Proof ==
Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $\relcomp S {\relcomp S T} = T$ from Relative Complement of Relative Complement.
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \beta$
* : Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement
* : $\S 6.6$: Subsets
* : $\S 2$
* : $\text {(ix)}$
Category:Set Complement
Complement of Complement (Boolean Algebras)

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Then for all $a \in S$:
:$\map \neg {\neg a} = a$
== Proof ==
Follows directly from Complement in Boolean Algebra is Unique.
== Sources ==
* : $\S 2$: Exercise $2$
Category:Boolean Algebras
Complement of Complement in Uniquely Complemented Lattice

== Theorem ==
Let $\struct {S, \wedge, \vee, \preceq}$ be a uniquely complemented lattice.
For each $x \in S$, let $\neg x$ be the complement of $x$.
Then for each $x \in S$:
:$\neg \neg x = x$
== Proof ==
By the definition of a complement of $x$:
:$\neg x \vee x = \top$
:$\neg x \wedge x = \bot$
Since $\vee$ and $\wedge$ are commutative:
:$x \vee \neg x = \top$
:$x \wedge \neg x = \bot$
Thus by the definition of complement, $x$ is a complement of $\neg x$.
By the definition of a uniquely complemented lattice, $x = \neg \neg x$.
Category:Uniquely Complemented Lattices
Complement of Complete Bipartite Graph

== Theorem ==
Let $K_{p, q}$ be a complete bipartite graph.
The complement of $K_{p, q}$ consists of a disconnected graph with two components:
:The complete graph $K_p$
:The complete graph $K_q$.
== Proof ==
By definition, the complete bipartite graph $K_{p, q}$ consists of two sets of vertices: $A$ of cardinality $p$, and $B$ of cardinality $q$, such that:
:Every vertex in $A$ is adjacent to every vertex in $B$
:No vertex in $A$ is adjacent to any other vertex in $A$
:No vertex in $B$ is adjacent to any other vertex in $B$.
The complement of $K_{p, q}$ therefore must be a graph $G$ such that:
:No vertex in $A$ is adjacent to any vertex in $B$
:Every vertex in $A$ is adjacent to every other vertex in $A$
:Every vertex in $B$ is adjacent to every other vertex in $B$.
The second and third of these conditions describes the complete graphs $K_p$ and $K_q$.
From the first of these conditions, it follows that $G$ comes in two disconnected graph components.
Hence the result.
Category:Bipartite Graphs
Complement of Complete Graph is Edgeless Graph

== Theorem ==
Let $K_n$ denote the complete graph of order $n$.
Then the complement of $K_n$ is the $n$-edgeless graph $N_n$.
== Proof ==
By definition of complete graph, each vertex of $K_n$ is adjacent to every other vertex of $K_n$.
Let $\overline {K_n}$ denote the complement of $K_n$.
Let $v$ be a vertex of $\overline {K_n}$.
By definition of complement, $v$ is adjacent to all the vertices of $\overline {K_n}$ that it is not in $K_n$.
But that means $v$ is adjacent to no vertices of $\overline {K_n}$.
That means $v$ is a $0$-regular graph.
The result follows from Graph is 0-Regular iff Edgeless.
== Sources ==
* 
Category:Complete Graphs
Category:Edgeless Graphs