nyuuzyou/wikis huggingface dataset:
Reflexive Relation/Examples/Reflexive Relation on Cartesian Plane

== Examples of Use of Symmetric and Transitive Relation is not necessarily Reflexive ==
The subset of the Cartesian plane defined as:
:$\RR := \set {\tuple {x, y} \in \R^2: x \le y \le x + 1}$
determines a relation on $\R^2$ which is reflexive, but neither symmetric nor transitive.
== Proof ==
=== Reflexive Relation ===
We note that, by definition:
:$\forall x \in \R: \tuple {x, y} \in \RR$ such that $x = y$
and so:
:$\forall x \in \R: \tuple {x, x} \in \RR$
Hence $\RR$ is reflexive.
=== Non-Symmetric Relation ===
Proof by Counterexample:
But:
:$1 > 0$
and so:
:$\tuple {1, 0} \notin \RR$
thus demonstrating that $\RR$ is not symmetric.
=== Non-Transitive Relation ===
Proof by Counterexample:
but
:$\tuple {0, 2} \notin \RR$
thus demonstrating that $\RR$ is not transitive.
The relation $\RR$ is illustrated below:
:
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Relations
Category:Examples of Reflexive Relations
Reflexive Relation is Quasi-Reflexive

== Theorem ==
Let $\RR$ be a reflexive relation on a set $S$.
Then $\RR$ is a quasi-reflexive relation on $S$.
== Proof ==
By definition of reflexive relation:
:$\forall x \in S: \tuple {x, x} \in \RR$
Hence by definition of domain:
:$x \in \Dom \RR$
and hence by definition of field and Set is Subset of Union:
:$x \in \Field \RR$
That is:
:$\forall x \in \Field \RR: \tuple {x, x} \in \RR$
Hence the result by definition of quasi-reflexive relation.
Category:Quasi-Reflexive Relations
Category:Reflexive Relations
Reflexive Relation is Serial

== Theorem ==
Every reflexive relation is also a serial relation.
== Proof ==
Let $\RR \subseteq S \times S$ be a relation in $S$.
We have that $\RR$ is serial :
:$\forall x \in S: \exists y \in S: \tuple {x, y} \in \RR$
That is, every element relates to at least one element.
We have that $\RR$ is reflexive :
:$\forall x \in S: \tuple {x, x} \in \RR$
Hence if $\RR$ is reflexive, every $x$ is related to itself, thereby fulfilling the criterion for $\RR$ to be serial.
Category:Reflexive Relations
Category:Serial Relations
Reflexive Relation on Set of Cardinality 2 is Transitive

== Theorem ==
Let $S$ be a set whose cardinality is equal to $2$:
:$\card S = 2$
Let $\odot \subseteq S \times S$ be a reflexive relation on $S$.
Then $\odot$ is also transitive.
== Proof ==
, let $S = \set {a, b}$.
Let $\odot$ be reflexive.
By definition of reflexive relation:
:$\Delta_S \subseteq \odot$
where $\Delta_S$ is the diagonal relation:
:$\Delta_S = \set {\tuple {x, x}: x \in S}$
That is:
:$\set {\tuple {a, a}, \tuple {b, b} } \subseteq \odot$
Suppose $\set {\tuple {a, a}, \tuple {b, b} } = \odot$.
Then by Diagonal Relation is Equivalence, $\odot$ is transitive.
Suppose there exists $\tuple {x, y} \in \odot$ such that $\tuple x \ne y$ and $\tuple {y, x} \notin \odot$.
Then:
:$x \odot x, x \odot y \implies x \odot y$
and:
:$x \odot y, y \odot y \implies x \odot y$
Now suppose there exists $\tuple {x, y} \in \odot$ such that $\tuple x \ne y$ and $\tuple {y, x} \in \odot$.
Then we have:
:$x \odot y, y \odot x \implies x \odot x$
and:
:$y \odot x, x \odot y \implies y \odot y$
which hold because $\odot$ is reflexive
Hence in all cases, a reflexive relation on $S$ is also transitive.
Category:Reflexive Relations
Category:Transitive Relations
Reflexive Relation on Singleton is Well-Ordering

== Theorem ==
Let $S = \set s$ be a singleton.
Let $\RR$ be a reflexive relation on $S$.
Then $\RR$ is a well-ordering on $S$.
== Proof ==
Let $S = \set s$.
By definition of reflexive relation:
:$s \mathrel \RR s$
It trivially holds that:
:$\forall a, b \in S: a \mathrel \RR b \land b \mathrel \RR a \implies a = b$
and so $\RR$ is antisymmetric.
It also trivially holds that:
:$\forall a, b, c \in S: a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c$
Thus $\RR$ is an ordering on $S$.
We also have trivially that:
:$\forall a, b \in S: a \mathrel \RR b \lor b \mathrel \RR a$
and so $\RR$ is a total ordering on $S$.
Finally from Finite Totally Ordered Set is Well-Ordered:
:$\struct {S, \RR}$ is a well-ordered set.
Hence the result.
Category:Singletons
Category:Reflexive Relations
Category:Well-Orderings
Reflexive Riesz Lemma

== Theorem ==
Let $X$ be a reflexive normed vector space.
Let $Y$ be a proper closed linear subspace of $X$. 
Then $\exists x_\alpha \in X$ such that: 
:$\norm {x_\alpha} = 1$
with:
:$\ds \map d {x_\alpha, Y} = 1$
where $d$ denotes distance to a set.
== Proof ==
By Existence of Distance Functional, there exists a bounded linear functional $f$ on $X$ such that:
:$\map f y = 0$ for each $y \in Y$
with norm $1$:
:$\norm f = 1$
By the easy direction (sufficiency) of James's Theorem, $\exists x_\alpha \in X$ such that:
:$\norm {x_\alpha} = 1$
and $f$ attains its norm at $x_\alpha$:
:$\size {\map f {x_\alpha} } = 1$
Since $\map f y = 0$ for each $y \in Y$, we have:
:$\forall y \in Y: 1 = \size {\map f {x_\alpha} } = \size {\map f {x_\alpha - y} } \le \norm {x_\alpha - y}$
Taking the infimum over all $y \in Y$, we obtain:
:$\ds 1 \le \inf_{y \mathop \in Y} \norm {x_\alpha - y}$
But $\norm {x_\alpha} = 1$, we obtain:
:$\ds 1 = \inf_{y \mathop \in Y} \norm {x_\alpha - y}$
By definition of distance to a set:
:$\ds \map d {x_\alpha, Y} = 1$
Hence the result.
== Also see ==
* Riesz's Lemma
Category:Riesz's Lemma
Category:Functional Analysis
Reflexive Transitive Closure/Examples/Arbitrary Example 1

== Example of Reflexive Transitive Closure ==
Let $S = \set {1, 2, 3}$ be a set.
Let $\RR$ be the relation on $S$ defined as:
:$\RR = \set {\tuple {1, 2}, \tuple {2, 2}, \tuple {2, 3} }$
The reflexive transitive closure $\RR^*$ of $\RR$ is given by:
:$\RR^* = \set {\tuple {1, 1}, \tuple {1, 2}, \tuple {2, 2}, \tuple {2, 3}, \tuple {1, 3}, \tuple {3, 3} }$
== Sources ==
* : Chapter $1$: Preliminaries: $1.5$ Relations: Closures of Relations
Category:Examples of Reflexive Transitive Closures
Reflexive Transitive Closure/Examples/Arbitrary Example 2

== Example of Reflexive Transitive Closure ==
Let $S = \set {1, 2, 3, 4, 5}$ be a set.
Let $\RR$ be the relation on $S$ defined as:
:$\RR = \set {\tuple {1, 2}, \tuple {2, 3}, \tuple {3, 4}, \tuple {5, 4} }$
The reflexive transitive closure $\RR^*$ of $\RR$ is given by:
:$\RR^* = \set {\tuple {1, 2}, \tuple {1, 3}, \tuple {1, 4}, \tuple {2, 3}, \tuple {2, 4}, \tuple {3, 4}, \tuple {5, 4}, \tuple {1, 1}, \tuple {2, 2}, \tuple {3, 3}, \tuple {4, 4}, \tuple {5, 5} }$
== Proof ==
From Transitive Closure of Relation: Arbitrary Example $2$, the transitive closure $\RR^+$ of $\RR$ is given by:
:$\RR^+ = \set {\tuple {1, 2}, \tuple {1, 3}, \tuple {1, 4}, \tuple {2, 3}, \tuple {2, 4}, \tuple {3, 4}, \tuple {5, 4} }$
By definition of reflexive transitive closure:
:$\RR^* = \paren {\RR^+}^=$
where $\paren {\RR^+}^=$ denotes the reflexive closure of $\RR^+$.
The diagonal relation on $S$ is given by:
:$\Delta_S = \set {\tuple {1, 1}, \tuple {2, 2}, \tuple {3, 3}, \tuple {4, 4}, \tuple {5, 5} }$
The result then follows by definition of reflexive closure:
:$\paren {\RR^+}^= := \RR^+ \cup \Delta_S$
== Sources ==
* : Chapter $1$: Preliminaries: Exercises: $1.7$
Category:Examples of Reflexive Transitive Closures
Reflexive and Symmetric Relation is not necessarily Transitive

== Theorem ==
Let $S$ be a set.
Let $\alpha \subseteq S \times S$ be a relation on $S$.
Let $\alpha$ be both reflexive and symmetric.
Then it is not necessarily the case that $\alpha$ is also transitive.
== Proof 1 ==
== Proof 2 ==
Category:Reflexive Relations
Category:Symmetric Relations
Category:Transitive Relations
Category:Reflexive and Symmetric Relation is not necessarily Transitive
Reflexive and Symmetric Relation is not necessarily Transitive/Proof 1

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b, c}$.
Let:
: $\alpha = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {b, a}, \tuple {b, c}, \tuple {c, b} }$
By inspection it is seen that $\alpha$ is both reflexive and symmetric.
However, we have:
:$a \mathrel \alpha b$ and $b \mathrel \alpha c$
but it is not the case that $a \mathrel \alpha c$.
Hence $\alpha$ is both reflexive and symmetric but not transitive.
Category:Reflexive and Symmetric Relation is not necessarily Transitive
Reflexive and Symmetric Relation is not necessarily Transitive/Proof 2

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $S = \Z$ be the set of integers.
Let $n \in \Z$ such that $n > 1$.
Let $\alpha$ be the relation on $S$ defined as:
:$\forall x, y \in S: x \mathrel \alpha y \iff \size {x - y} < n$
where $\size x$ denotes the absolute value of $x$.
It is seen that:
:$\forall x \in \Z: \size {x - x} = 0 < n$
and so:
:$\forall x \in \Z: x \mathrel \alpha x$
Thus $\alpha$ is reflexive
Then it is seen that:
:$\forall x, y \in \Z: \size {x - y} = \size {y - x}$
and so if $\size {x - y} < n$ then so is $\size {y - x} < n$.
That is, if $x \mathrel \alpha y$ then $y \mathrel \alpha x$.
Thus $\alpha$ is symmetric.
Now let $x = -\paren{n - 1}, y = 0, z = n - 1$.
We have:
and so both $x \mathrel \alpha y$ and $y \mathrel \alpha z$.
But:
Thus:
:$\neg x \mathrel \alpha z$
and so $\alpha$ is not transitive.
Hence $\alpha$ is both reflexive and symmetric but not transitive.
== Sources ==
* : Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $3$
Category:Reflexive and Symmetric Relation is not necessarily Transitive
Reflexive and Transitive Relation is Idempotent

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation on a set $S$.
Let $\RR$ be both reflexive and transitive.
Then $\RR$ is idempotent, in the sense that:
:$\RR \circ \RR = \RR$
where $\circ$ denotes composition of relations.
== Proof ==
Let $\RR$ be both reflexive and transitive.
By definition of transitive relation:
:$\RR \circ \RR \subseteq \RR$
Let $\tuple {x, y} \in \RR$.
By definition of reflexive relation:
:$\tuple {y, y} \in \RR$
By definition of composition of relations:
:$\tuple {x, y} \in \RR \circ \RR$
Hence:
:$\RR \subseteq \RR \circ \RR$
Thus by definition of set equality:
:$\RR \circ \RR = \RR$
== Sources ==
* : Chapter $0$: Relations
Category:Reflexive Relations
Category:Transitive Relations
Category:Idempotence
Reflexive and Transitive Relation is not necessarily Symmetric

== Theorem ==
Let $S$ be a set.
Let $\alpha \subseteq S \times S$ be a relation on $S$.
Let $\alpha$ be both reflexive and transitive.
Then it is not necessarily the case that $\alpha$ is also symmetric.
== Proof 1 ==
== Proof 2 ==
Category:Reflexive Relations
Category:Symmetric Relations
Category:Transitive Relations
Category:Reflexive and Transitive Relation is not necessarily Symmetric
Reflexive and Transitive Relation is not necessarily Symmetric/Proof 1

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b, c}$.
Let:
:$\alpha = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {b, c}, \tuple {a, c} }$
By inspection it is seen that $\alpha$ is both reflexive and transitive.
However, we have:
:$a \mathrel \alpha b$
but it is not the case that $b \mathrel \alpha a$.
Hence $\alpha$ is both reflexive and transitive but not symmetric.
Category:Reflexive and Transitive Relation is not necessarily Symmetric
Reflexive and Transitive Relation is not necessarily Symmetric/Proof 2

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $S = \Z$ be the set of integers.
Let $\alpha$ be the relation on $S$ defined as:
:$\forall x, y \in S: x \mathrel \alpha y \iff x \le y$
It is seen that:
:$\forall x \in \Z: x \le x$
and so:
:$\forall x \in \Z: x \mathrel \alpha x$
Thus $\alpha$ is reflexive.
Then it is seen that:
:$\forall x, y, z \in \Z: x \le y, y \le z \implies x \le z$
Thus $\alpha$ is transitive.
Now let $x = 1$ and $y = 2$.
Then:
: $x \le y$ but it is not the case that $y \le x$
and so $\alpha$ is not symmetric.
Hence $\alpha$ is both reflexive and transitive but not symmetric.
== Sources ==
* : Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $3$
Category:Reflexive and Transitive Relation is not necessarily Symmetric
Regiomontanus' Angle Maximization Problem

== Theorem ==
Let $AB$ be a line segment.
Let $AB$ be produced to $P$.
Let $PQ$ be constructed perpendicular to $AB$.
Then the angle $AQB$ is greatest when $PQ$ is tangent to a circle passing through $A$, $B$ and $Q$.
== Proof ==
:
There exists a unique circle $C$ tangent to $PQ$ passing through $A$ and $B$.
From Angles in Same Segment of Circle are Equal, the angle subtended by $AB$ from any point on $C$ is the same as angle $\angle AQB$.
All other points on $PQ$ that are not $Q$ itself are outside $C$.
Hence the angle subtended by $AB$ to such points is smaller.
Hence the result.
== Historical Note ==
== Sources ==
* 
* : The Best View of a Statue: $93$
Category:Circles
Category:Classic Problems
Regiomontanus' Angle Maximization Problem/Historical Note

== Historical Note on Regiomontanus' Angle Maximization Problem ==
 first posed this problem in $1471$ to .
He posed it in the context of choosing the optimum place to stand to view a statue positioned above eye level.
Too close and it will appear heavily foreshortened, too far away and it will just appear small.
It is notable as the first extremal problem since Heron's Principle of Reflection.
The problem has been reinvented several times, and has a contemporary application: it gives the best place to take a conversion kick in the game of .
== Sources ==
* 
* : The Best View of a Statue: $93$
Category:Historical Notes
Region Less One Point is Region

== Theorem ==
Let $M = \struct {A, d}$ be a dense-in-itself metric space.
Let $R \subseteq M$ be a region of $M$.
Let $x \in R$.
Then $R \setminus \set x$ is also a region of $M$.
== Proof ==
From the definition, a region is a non-empty, open, path-connected subset of $M$.
First, note that as $R$ is open it can not be a singleton from Finite Subspace of Dense-in-itself Metric Space is not Open.
Therefore $R \setminus \set x$ is not empty.
Next, we see that from Open Set Less One Point is Open that $R$ is open.
Now, let $\alpha, \beta \in R$.
As $R$ is path-connected, we can join $\alpha$ and $\beta$ with a path $\Gamma$.
If $x \notin \Gamma$, then $\Gamma$ is also a path in $R \setminus \set x$, and the proof is complete.
If $x \in \Gamma$, then we consider the open $\epsilon$-ball $\map {B_\epsilon} x$ of $x$ for some $\epsilon$ such that $\map {B_\epsilon} x \subseteq R$.
Category:Metric Spaces
Regression Coefficients of Normally Distributed Random Variable

== Theorem ==
Let $X$ be a random variable.
Let $x$ be a given value of $X$.
Let $Y$ be a random variable with a normal distribution.
Let the variance $\sigma^2$ of $Y$ (usually unknown) be independent of $x$.
Let $S$ be a sample of $n$ independent pairs of observations $\tuple {x_i, y_i}$ for $i = 1, 2, \ldots, n$.
The maximum likelihood estimators $\beta_0$ and $\beta_1$ are given by:
where:
== Proof ==
The maximum likelihood estimators are obtained by the method of least squares.
That is, the aim is to minimize $\ds \sum_i \paren {y_i - \beta_0 - \beta_1 x_i}^2$.
== Sources ==
* 
* 
Category:Regression Coefficients
Category:Normal Distribution
Regular Antiprism is Uniform Polyhedron

== Theorem ==
Let $P$ be a regular antiprism.
Then $P$ is a uniform polyhedron.
== Proof ==
That all the faces of $P$ are regular polygons follows from the definition of regular antiprism.
From Regular Antiprism is Isogonal, $P$ is isogonal.
Hence the result.
== Sources ==
* 
* 
Category:Regular Antiprisms
Category:Uniform Polyhedra
Regular Curve in Riemannian Manifold has Unit-Speed Forward Reparametrization

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold with or without boundary.
Let $I \subseteq \R$ be a real interval.
Let $\gamma : I \to M$ be a regular curve.
Then $\gamma$ has a unit-speed forward reparametrization.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Definitions
Category:Riemannian Geometry
Regular Dodecahedron is Dual of Regular Icosahedron

== Theorem ==
The regular dodecahedron is the dual of the regular icosahedron.
== Proof ==
== Sources ==
* $12$
* : $20$
* : $30$
* : $12$
* : $20$
* : $30$
Category:Regular Dodecahedra
Category:Regular Icosahedra
Regular Dodecahedron is Dual of Regular Icosahedron/Mistake

== Source Work ==
:
:The Dictionary
::$30$
:
:The Dictionary
::$30$
== Mistake ==
:The dodecahedron and its dual, the icosahedron, each have $30$ edges. 
== Correction ==
It is important not to neglect the crucial point that both figures must be regular.
Otherwise, for example, consider the dodecahedron which is the bipyramid consisting of two hexagonal pyramids placed base to base.
Its dual is the hexagonal prism, which has $8$ faces, and so is not an icosahedron at all. 
== Sources ==
* : $30$
* : $30$
Category:Mistakes/Curious and Interesting Numbers
Regular Graph/Examples/Incomplete 0-Regular

== Examples of Regular Graphs ==
The $0$-regular graphs which are not complete are the edgeless graphs $N_n$ of order $n$ for $n > 1$.
For example, $N_2$:
:
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $10 \ \text {(a)}$
Category:Examples of Regular Graphs
Regular Graph/Examples/Incomplete 2-Regular

== Examples of Regular Graphs ==
The $2$-regular graphs which are not complete are the cycle graphs $C_n$ of order $n$ for $n > 3$.
For example, $C_4$:
:
Note that $C_3$ is both $2$-regular and complete.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $10 \ \text {(c)}$
Category:Examples of Regular Graphs
Regular Graph is Tree iff Complete Graph of Order 2

== Theorem ==
Let $G$ be a non-edgeless regular graph.
Then $G$ is a tree $G$ is $K_2$, the complete graph of order $2$.
== Proof ==
=== Necessary Condition ===
Let $G$ be a non-edgeless regular graph which is also a tree.
From Finite Tree has Leaf Nodes it follows that $G$ has at least two vertices of degree $1$.
Therefore, for $G$ to be regular it need to be $1$-regular.
Suppose $G$ has $3$ or more vertices.
Let $u, v, w$ be such vertices of $G$.
Suppose each of $\set {u, v}$, $\set {v, w}$ and $\set {u, w}$ were adjacent.
Then there would be a cycle in $G$: $\tuple {u, v, w, u}$ and so $G$ by definition would not be a tree.
So at least one pair $\set {u, v}$, $\set {v, w}$ and $\set {u, w}$ is not adjacent.
 suppose this pair is $\set {u, v}$.
By definition, a tree is connected.
Therefore there exists a path $P$ between $u$ and $v$.
As $u$ and $v$ are not adjacent, there exists another vertex $y$ of $G$ through which $P$ passes.
In order for this to be the case, the degree of $y$ needs to be greater than $1$.
It follows that $G$ can not be $1$-regular.
So to be regular, $G$ can contain only $2$ vertices.
By Finite Connected Simple Graph is Tree iff Size is One Less than Order it follows that $G$ contains $2$ vertices connected together by a single edge.
By definition, this is the complete graph of order $2$.
=== Sufficient Condition ===
$K_2$, the complete graph of order $2$, consists of $2$ vertices connected together by a single edge.
From Complete Graph is Regular it follows that $K_2$ is a non-edgeless regular graph.
From Circuit of Simple Graph has Three Edges or More it follows that $K_2$ contains no circuits.
It follows by definition that $K_2$ is a tree.
== Sources ==
* : $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees: Problem $3$
Category:Tree Theory
Category:Regular Graphs
Category:Complete Graphs
Regular Heptagon is Smallest with no Compass and Straightedge Construction

== Theorem ==
The regular heptagon is the smallest regular polygon (smallest in the sense of having fewest sides) that cannot be constructed using a compass and straightedge construction.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $7$
* : $7$
Category:Regular Heptagon is Smallest with no Compass and Straightedge Construction
Category:Compass and Straightedge Constructions
Category:Regular Polygons
Category:7
Regular Heptagon is Smallest with no Compass and Straightedge Construction/Proof 1

== Theorem ==
== Proof ==
The fact that it is impossible to construct a regular heptagon using a compass and straightedge construction is demonstrated in Compass and Straightedge Construction for Regular Heptagon does not exist.
From Construction of Equilateral Triangle, an equilateral triangle can be constructed.
From Inscribing Square in Circle, for example, a square can be constructed.
From Inscribing Regular Pentagon in Circle, a regular pentagon can be constructed.
From Inscribing Regular Hexagon in Circle, a regular hexagon can be constructed.
Thus all regular polygons with $3$ to $6$ sides can be so constructed, but not one with $7$ sides.
Category:Regular Heptagon is Smallest with no Compass and Straightedge Construction
Regular Heptagon is Smallest with no Compass and Straightedge Construction/Proof 2

== Theorem ==
== Proof ==
A theorem states that:
"The n-gon is constructible by compass and straightedge construction if and only if $n = 2^kn_0$, with $k \in \mathbb{Z}_{\geq0}$ and $n_0$ the product of any number of distinct Fermat primes."
Note that 3 and 5 are both Fermat primes, so their respective n-gon are both constructible; and $4$ is just a power of $2$, so the 4-gon also constructible. 
Lastly, since 6 is the product of 2 and 3, the 6-gon is constructible as well. Hence, we remain with 7, which does not satisfy the theorem's criteria; and thus the 7-gon is the smallest regular polygon that cannot be constructed using a compass and straightedge construction.
Category:Regular Heptagon is Smallest with no Compass and Straightedge Construction
Regular Hexagon is composed of Equilateral Triangles

== Theorem ==
Let $P$ be a regular hexagon.
Let $O$ be the center of $P$.
Then there exists a triangulation of $P$ into six congruent equilateral triangles.
The three vertices of each triangle are $O$ and two adjacent vertices of $P$.
== Proof ==
Let $V_2$ be a vertex of $P$.
Let $V_1, V_3$ be the two adjacent vertices of $V_1$.
From Regular Polygon is composed of Isosceles Triangles, it follows that there exists a triangulation of $P$ into six congruent isosceles triangles of the type $V_1 V_2 O$.
As $\angle V_1 V_2 O$ and $\angle V_3 V_2 O$ are corresponding angles of the congruent isosceles triangles $\triangle V_1 V_2 O$ and $\triangle V_3 V_2 O$, it follows that $\angle V_1 V_2 O = \angle V_3 V_2 O$.
Let $V_2 O$ denote one side of the triangles $\triangle V_1 V_2 O$ and $\triangle V_1 V_3 O$.
As the angles $\angle V_1 V_2 O$ and $\angle V_3 V_2 O$ are constructed by intersecting the angle $\angle V_1 V_2 V_3$ with $V_2 O$, it follows that:
As $\triangle V_1 V_2 O$ is an isosceles triangle, it follows that:
:$\angle V_2 V_1 O = 60 \degrees$
From Sum of Angles of Triangle equals Two Right Angles used on $\triangle V_1 V_2 O$, it follows that:
:$\angle V_1 O V_2 = 180 \degrees - \angle V_1 V_2 O - \angle V_2 V_1 O = 60 \degrees$
From Equiangular Triangle is Equilateral, it follows that $\triangle V_1 V_2 O$ is an equilateral triangle.
As the triangles in the triangulation of $P$ are congruent, it follows that they are all equilateral.
== Illustration ==
:
Here, the angle $\alpha$ corresponds to $\angle V_1 O V_2$.
The side $s$ corresponds to $V_1 V_2$.
Category:Hexagons
Category:Equilateral Triangles
Regular Lindelöf Space is Normal Space

== Theorem ==
Let $T = \struct{S, \tau}$ be a regular Lindelöf topological space.
Then:
:$T$ is a normal space.
== Proof ==
By definition of regular space:
:$T$ is a $T_3$ space
:$T$ is a $T_0$ (Kolmogorov) space
From Regular Space is $T_2$ Space:
:$T$ is a $T_2$ space
From $T_2$ Space is $T_1$ Space:
:$T$ is a $T_1$ space
From $T_3$ Lindelöf Space is $T_4$ Space:
:$T$ is a $T_4$ space
By definition, $T$ is a normal space.
Category:Regular Spaces
Category:Lindelöf Spaces
Category:Normal Spaces
Regular Paracompact Space is not necessarily Metrizable

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is regular and paracompact.
Then it is not necessarily the case that $T$ is metrizable.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Metrizability
Category:Metrizable Topologies
Category:Paracompact Spaces
Category:Regular Spaces
Category:Regular Paracompact Space is not necessarily Metrizable