nyuuzyou/wikis huggingface dataset:
Regular Paracompact Space is not necessarily Metrizable/Proof 1

== Theorem ==
== Proof ==
Let $T$ be the Sorgenfrey line.
From Sorgenfrey Line satisfies all Separation Axioms, $T$ is a regular space.
From Sorgenfrey Line is Paracompact, $T$ is a paracompact space.
From Sorgenfrey Line is not Metrizable, $T$ is not a metrizable space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Metrizability
Category:Regular Paracompact Space is not necessarily Metrizable
Regular Paracompact Space is not necessarily Metrizable/Proof 2

== Theorem ==
== Proof ==
Let $T$ be the radial interval topology.
From Radial Interval Topology is Completely Normal, $T$ is a completely normal space.
Hence from Sequence of Implications of Separation Axioms, $T$ is a regular space.
From Radial Interval Topology is Paracompact, $T$ is a paracompact space.
From Radial Interval Topology is not Metrizable, $T$ is not a metrizable space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Metrizability
Category:Regular Paracompact Space is not necessarily Metrizable
Regular Polygon can be Circumscribed around Circle

== Theorem ==
Let $P$ be a regular polygon.
Then it is possible to circumscribe $P$ around a circle $C$, whose center is the same as the center of $P$.
== Proof ==
Let $P$ be a regular polygon.
 it is not possible to circumscribe $P$ around a circle $C$, whose center is the same as the center of $P$.
Let $AB$, $BC$ and $CD$ be sides of $P$ such that $AB$ and $BC$ are on the tangent to a circle $K$ such that $CD$ is not tangent to that circumference.
There can only be one such circle tangent to both $AB$ and $BC$.
This has to be possible, or all sides of $P$ would be tangent to $K$.
That would make $P$ cyclic which contradicts our supposition about $P$.
Let the center of $K$ be $O$.
As $CD$ is not tangent to $K$, then $OC \ne OD$.
Hence $\angle ABC \ne \angle BCD$ and so $P$ is not equiangular.
Hence $P$ is not a regular polygon.
This contradicts our assertion about $P$.
Hence by Proof by Contradiction it follows that $P$ is cyclic.
Category:Regular Polygons
Regular Polygon is Cyclic

== Theorem ==
Let $P$ be a regular polygon.
Then $P$ is a cyclic polygon.
== Proof ==
Let $P$ be a regular polygon.
 $P$ is not cyclic.
Let $AB$, $BC$ and $CD$ be sides of $P$ such that $A$, $B$ and $C$ are on the circumference of a circle $K$ such that $D$ is not on that circumference.
This has to be possible, or all vertices of $P$ would lie on $K$.
That would make $P$ cyclic which contradicts our supposition about $P$.
Let the center of $K$ be $O$.
As $D$ is not on the circumference of that circle, then $OC \ne OD$.
Hence $\angle ABC \ne \angle BCD$ and so $P$ is not equiangular.
Hence $P$ is not a regular polygon.
This contradicts our assertion about $P$.
Hence by Proof by Contradiction it follows that $P$ is cyclic.
== Sources ==
* 
* 
Category:Regular Polygons
Regular Polygon is composed of Isosceles Triangles

== Theorem ==
Let $P$ be a regular $n$-gon.
Let $O$ be the center of $P$.
Then there exists a triangulation of $P$ into $n$ congruent isosceles triangles.
The three vertices of each triangle are $O$ and two adjacent vertices of $P$.
== Proof ==
From Regular Polygon is Cyclic, it follows that $O$ is the center of the circumcircle of $P$.
Let $V_1 , V_2$ be adjacent vertices of $P$.
By definition of circumcircle, $V_1 O$ and $V_2 O$ are radii of the circumcircle.
As $V_1 O$ and $V_2 O$ have the same length, it follows that $\triangle V_1 V_2 O$ is an isosceles triangles.
As there are $n$ pairs of adjacent vertices in $P$, corresponding to the $n$ sides of $P$, it follows that there are $n$ triangles of the type $\triangle V_1 V_2 O$.
All of these triangles consist of one side of $P$, as well as two sides of the type $V_1 O, V_2 O$ which they share with one of the adjacent triangles.
Hence, the triangles form a triangulation of $P$.
By definition of regular polygon, all sides of $P$ have the same length.
From Triangle Side-Side-Side Congruence, it follows that all the triangles are congruent.
Category:Regular Polygons
Category:Isosceles Triangles
Regular Prism is Uniform Polyhedron/Mistake

== Source Work ==
:
:polyhedron
== Mistake ==
:Right prisms and antiprisms that have regular polygons as bases are also uniform.
== Correction ==
So, while the bases are regular polygons , the lateral faces are not necessarily regular.
Hence it is important to state that such a prism must be a uniform prism:
Similarly, an antiprism has lateral faces which, while triangular, are not necessarily equilateral.
In order to rescue this result, it is important to state that such an antiprism must in fact be a uniform antiprism:
== Also see ==
* Uniform Prism is Uniform Polyhedron
* Uniform Antiprism is Uniform Polyhedron
== Sources ==
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Regular Representation of Invertible Element is Permutation

== Theorem ==
Let $\struct {S, \circ}$ be a monoid.
Let $a \in S$ be invertible.
Then the left regular representation $\lambda_a$ and the right regular representation $\rho_a$ are permutations of $S$.
== Proof ==
Suppose $a \in \struct {S, \circ}$ is invertible.
A permutations is a bijection from a set to itself.
As $\lambda_a: S \to S$ and $\rho_a: S \to S$ are defined from $S$ to $S$, all we need to do is show that they are bijections.
To do that we can show that they are both injective and surjective.
=== Injectivity ===
From Invertible Element of Monoid is Cancellable, as $a$ is invertible, it is also cancellable.
From Cancellable iff Regular Representations Injective, it follows that both $\lambda_a$ and $\rho_a$ are injective.
=== Surjectivity ===
Let $y \in S$.
Then:
Thus both $\lambda_a$ and $\rho_a$ are surjective.
So $\lambda_a$ and $\rho_a$ are injective and surjective, and therefore bijections, and thus permutations of $S$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Theorem $7.1$
Category:Monoids
Category:Regular Representations
Regular Representation on Subgroup is Bijection to Coset

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
=== Left Coset ===
=== Right Coset ===
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem
Category:Cosets
Category:Regular Representations
Category:Regular Representation on Subgroup is Bijection to Coset
Regular Representation on Subgroup is Bijection to Coset/Left

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $y H$ denote the left coset of $H$ by $y$.
The mapping $\lambda_x: H \to x H$, where $\lambda_x$ is the left regular representation of $H$ with respect to $x$, is a bijection from $H$ to $x H$.
== Proof ==
Let $h \in H$.
Then, by definition of left regular representation:
:$\map {\lambda_x} h = x h \in x H$
Thus:
:$\forall h \in H: \map {\lambda_x} h \in x H$
So $\lambda_x: H \to x H$ is a mapping.
A permutation is a bijection.
As Regular Representations in Group are Permutations, it follows that $\lambda_x$ is a bijection.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem
Category:Regular Representation on Subgroup is Bijection to Coset
Regular Representation on Subgroup is Bijection to Coset/Right

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $H y$ denote the right coset of $H$ by $y$.
The mapping $\rho_x: H \to H x$, where $\rho_x$ is the right regular representation of $H$ with respect to $x$, is a bijection from $H$ to $H x$.
== Proof ==
Let $h \in H$.
Then:
:$\map {\rho_x} h = h x \in H x$
Thus:
:$\forall h \in H: \map {\rho_x} h \in H x$
demonstrating that $\rho_x: H \to H x$ is a mapping.
A permutation is a bijection by definition.
As Regular Representations in Group are Permutations, it follows that $\rho_x$ is a bijection.
Category:Regular Representation on Subgroup is Bijection to Coset
Regular Representation wrt Cancellable Element on Finite Semigroup is Bijection

== Theorem ==
Let $\left({S, \circ}\right)$ be a finite semigroup.
Let $a \in S$ be cancellable.
Then:
:the left regular representation $\lambda_a$
and:
:the right regular representation $\rho_a$
of $\left({S, \circ}\right)$ with respect to $a$ are both bijections.
=== Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection ===
=== Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection ===
== Proof ==
By Cancellable iff Regular Representations Injective, $\lambda_a$ and $\rho_a$ are injections.
We have that $S$ is finite.
From Injection from Finite Set to Itself is Surjection, both $\lambda_a$ and $\rho_a$ are surjections.
Thus $\lambda_a$ and $\rho_a$ are injective and surjective, and therefore bijections.
Category:Semigroups
Category:Regular Representations
Category:Cancellability
Regular Representations in Group are Permutations

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $a \in G$ be any element of $G$.
Then the left regular representation $\lambda_a$ and the right regular representation $\rho_a$ are permutations of $G$.
== Proof ==
This follows directly from the fact that all elements of a group are by definition invertible.
Therefore the result Regular Representation of Invertible Element is Permutation applies.
== Sources ==
* : $\S 1.6$: Example $18$
* : Chapter $2$: The Definition of Group Structure: $\S 28 \beta$
* : $\S 35.8$: Elementary consequences of the group axioms
Category:Group Theory
Category:Regular Representations
Regular Representations in Semigroup are Permutations then Structure is Group

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
For $a \in S$, let $\lambda_a: S \to S$ and $\rho_a: S \to S$ denote the left regular representation and right regular representation with respect to $a$ respectively:
For all $a$ in $S$, let $\lambda_a$ be a permutation on $S$.
Let there exist $b$ in $S$ such that $\rho_b$ is a permutation on $S$.
Then $\struct {S, \circ}$ is a group.
== Proof ==
We have that $\lambda_a$ be a permutation on $S$ for all $a \in S$.
In particular this applies to $b$.
So:
:$\lambda_b$ is a permutation on $S$
:$\rho_b$ is a permutation on $S$
and so from Regular Representations wrt Element are Permutations then Element is Invertible:
:$\struct {S, \circ}$ has an identity element
:$b$ is invertible in $\struct {S, \circ}$.
It remains to be shown that $a$ is invertible in $S$ for all $a \in S$.
Let the identity element of $\struct {S, \circ}$ be $e$.
We have that $\lambda_a$ be a permutation on $S$ for all $a \in S$.
Hence:
Thus for all $a$ in $S$, $a$ has a left inverse $a'$.
Now as $e$ is an identity element of $S$, it is by definition a left identity.
Hence from Left Inverse for All is Right Inverse, $a'$ is also a right inverse for $a$.
That is, every $a \in S$ has an element which is both a left inverse and a right inverse.
That is, every $a \in S$ has an inverse.
So we have that $\struct {S, \circ}$ is a semigroup such that:
:$\struct {S, \circ}$ has an identity element
:every element of $S$ has an inverse.
Hence, by definition, $\struct {S, \circ}$ is a group.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.14$
Category:Regular Representations
Category:Semigroups
Category:Group Theory
Regular Representations wrt Element are Permutations then Element is Invertible

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $\lambda_a: S \to S$ and $\rho_a: S \to S$ be the left regular representation and right regular representation with respect to $a$ respectively:
Let both $\lambda_a$ and $\rho_a$ be permutations on $S$.
Then there exists an identity element for $\circ$ and $a$ is invertible.
== Proof ==
We have that $\rho_a$ is a permutation on $S$.
Hence:
Then we have:
which demonstrates that $g$ is a right identity for $\circ$.
In the same way, we have that $\lambda_a$ is also a permutation on $S$.
Hence:
Then we have:
which demonstrates that $g$ is a left identity for $\circ$.
So, by definition, $g$ is an identity element for $\circ$.
Again, we have that $\rho_a$ is a permutation on $S$, and so:
and that $\lambda_a$ is also a permutation on $S$, and so:
So $h$ is both a left inverse and a right inverse for $a$.
Hence by definition $h$ is an inverse for $a$.
Hence $a$ is invertible by definition.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.13$
* 
Category:Regular Representations
Category:Semigroups
Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is regular and second-countable.
Then $T$ is homeomorphic to a subspace of the Hilbert cube.
== Proof ==
From Second-Countable Space is Lindelöf: 
:$T$ is a Lindelöf space
From Regular Lindelöf Space is Normal Space:
:$T$ is a normal space
By definition of second-countable:
:there exists a countable basis $\BB$ for $\tau$
Let:
:$\AA = \set{\tuple{U,V} : U, V \in \BB : U^- \subseteq V}$
where $U^-$ denotes the closure of $U$ in $T$.
=== Lemma 1 ===
From Urysohn's Lemma:
:for all $\tuple{U, V} \in \AA$ there exists a Urysohn function $f_{U,V} : S \to \closedint 0 1$ for $U^-$ and $S \setminus V$
By definition of Urysohn function:
:$\forall \tuple{U,V}\in \AA : f_{U,V}$ is continuous
=== Lemma 2 ===
Let $\closedint 0 1^\AA$ be the product space of countable copies of $\closedint 0 1$ indexed by $\AA$, that is:
:$I^\AA = \ds \prod_{\tuple{U, V} \in \AA} \closedint 0 1$
Let $f: S \to \closedint 0 1^\AA$ be the evaluation mapping induced by the family $\family{f_{U,V}}_{\tuple{U, V} \in \AA}$.
That is, $f$ is the mapping defined by:
:$\forall s \in S: \map f s = \family{\map {f_{U,V}} s}_{\tuple{U,V}\in \AA}$
From Topological Evaluation Mapping is Continuous:
:$f$ is continuous
From Evaluation Mapping on T1 Space is Embedding if Mappings Separate Points from Closed Sets:
:$f$ is an embedding
By definition of embedding:
:$T$ is homeomorphic to a subspace of $I^\AA$
From Hilbert Cube is Homeomorphic to Countable Infinite Product of Real Number Unit Intervals:
:$I^\AA$ is homeomorphic to the Hilbert cube $\struct{I^\omega, d_2}$ 
where $d_2$ is a metric
From Composite of Homeomorphisms is Homeomorphism:
:$T$ is homeomorphic to a subspace of the Hilbert cube $\struct{I^\omega, d_2}$ 
== Sources ==
* : Chapter $4$: Embedding and Metrization
* : Chapter $7$: Metrizable Spaces: $\S23$: Metrization: Definition $23.1$
Category:Regular Spaces
Category:Second-Countable Spaces
Category:Hilbert Cube
Category:Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube
Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube/Lemma 1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ b a countable basis for $\tau$
Let:
:$\AA = \set{\tuple{U,V} : U, V \in \BB : U^- \subseteq V}$
where $U^-$ denotes the closure of $U$ in $T$.
Then: 
:$\AA$ is countable
== Proof ==
We have:
:$\AA \subseteq \BB \times \BB$
where $\BB \times \BB$ is the Cartesian product of $\BB$ with itself.
From Cartesian Product of Countable Sets is Countable:
:$\BB \times \BB$ is countable
From Subset of Countable Set is Countable:
:$\AA$ is countable
Category:Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube
Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube/Lemma 2

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is regular and second-countable.
Let $\BB$ be a countable basis for $\tau$.
Let $\AA = \set{\tuple{U,V} : U, V \in \BB : U^- \subseteq V}$ where $U^-$ denotes the closure of $U$ in $T$.
For all $\tuple{U, V} \in \AA$, let:
:$f_{U,V} : S \to \closedint 0 1$ be a Urysohn function for $U^-$ and $S \setminus V$
Then:
:the family of continuous mappings $\family{f_{U,V}}_{\tuple{U,V} \in \AA}$ separates points from closed sets
== Proof ==
Let $F$ be a closed subset of $T$.
Let $x \in S \setminus F$.
By definition of closed subset:
:$S \setminus F$ is open in $S$
Let $V = S \setminus F$.
By definition of regular space:
:$\exists U \in \tau: x \subseteq U, U^- \subseteq V$
Hence $\tuple{U, V} \in \AA$.
Consider the Urysohn function $f_{U, V}$ for $U^-$ and $S \setminus V$:
By definition of Urysohn function:
:$f_{U, V} \sqbrk {U^-} = \set 0, f_{U, V} \sqbrk {S \setminus V} = \set 1$ 
Hence:
:$\map {f_{U, V}} x = 0, f_{U, V} \sqbrk F = \set 1$
Since $\set 1^- = \set 1$, then:
:$\map {f_{U, V}} x = 0 \notin \set 1 = f_{U, V} \sqbrk F^-$
By definition, the family of continuous mappings $\family{f_{U,V}}_{\tuple{U,V} \in \AA}$ separates points from closed sets
Category:Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube
Regular Space is Completely Hausdorff Space

== Theorem ==
Let $\struct {S, \tau}$ be a regular space.
Then $\struct {S, \tau}$ is also a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
== Proof ==
Let $T = \struct {S, \tau}$ be a regular space.
Let $x, y \in S$ with $x \ne y$.
From the definition of a completely Hausdorff Space:
:$T$ is completely Hausdorff if $x$ and $y$ are separated by closed neighborhoods.
From the definition:
:$\struct {S, \tau}$ is a $T_3$ space
:$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space.
From Regular Space is $T_2$ Space, there exist disjoint open sets containing $x$ and $y$ respectively:
:$\exists U_x, U_y \in \tau: x \in U_x, y \in U_y, U_x \cap U_y = \O$
From definition of $T_3$ space, each open set contains a closed neighborhood around each of its points, specifically for $x$ and $y$:
:$\exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U_x$
:$\exists N_y: \relcomp S {N_y} \in \tau: \exists V_y \in \tau: y \in V_y \subseteq N_y \subseteq U_y$
From Subsets of Disjoint Sets are Disjoint:
:$N_x \cap N_y = \O$
Hence $x$ and $y$ are separated by closed neighborhoods.
The result follows.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Hausdorff Spaces
Category:Regular Spaces
Category:Completely Hausdorff Spaces
Regular Space is Preserved under Homeomorphism

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a regular space, then so is $T_B$.
== Proof ==
We have that $\struct {S, \tau}$ is a regular space :
:$\struct {S, \tau}$ is a $T_3$ space
:$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space.
From $T_3$ Space is Preserved under Homeomorphism:
:If $T_A$ is a $T_3$ space, then so is $T_B$.
From $T_0$ (Kolmogorov) Space is Preserved under Homeomorphism:
:If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Homeomorphisms (Topological Spaces)
Category:Regular Spaces
Regular Space is Semiregular Space

== Theorem ==
Let $\struct {S, \tau}$ be a regular space.
Then $\struct {S, \tau}$ is also a semiregular space.
== Proof ==
Let $T = \struct {S, \tau}$ be a regular space.
From the definition:
:$\struct {S, \tau}$ is a $T_3$ space
:$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space.
We also have that a $T_3$ Space is Semiregular.
Hence the result, by definition of semiregular space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties
Category:Semiregular Spaces
Category:Regular Spaces
Regular Space is T2 Space

== Theorem ==
Let $\struct {S, \tau}$ be a regular space.
Then $\struct {S, \tau}$ is also a $T_2$ (Hausdorff) space.
== Proof ==
Let $T = \struct {S, \tau}$ be a regular space.
From the definition of regular space:
:$\struct {S, \tau}$ is a $T_3$ space
:$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space.
Let $x, y \in S$.
As $T$ is $T_0$, it follows that either:
:$\exists V \in \tau: x \in V, y \notin V$
or:
:$\exists V \in \tau: y \in V, x \notin V$
that is, there exists $V$, an open set, containing one but not the other.
, suppose that $\exists V \in \tau: y \in V, x \notin V$.
Then by definition of relative complement:
:$x \in \relcomp S V$
Let $F := \relcomp S V$.
As $V$ is open, by definition of closed set we have that $F = \relcomp S V$ is closed.
That is:
:$\relcomp S V \in \tau$
As $y \in V$ it follows that $y \notin F$, that is:
:$y \in \relcomp S F$
Now $\struct {S, \tau}$ is a $T_3$ space, and so:
:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$
So we have that:
:$x \in F \subseteq U \implies x \in U$
:$y \notin F, y \in V$
such that $U \cap V = \O$.
So:
:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
which is precisely the definition of a $T_2$ (Hausdorff) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Regular and Normal Spaces
Category:Regular Spaces
Category:Hausdorff Spaces
Regular Space with Sigma-Locally Finite Basis is Normal Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a regular space.
Let $\BB$ be a $\sigma$-locally finite basis.
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 $T_3$ Space with Sigma-Locally Finite Basis is $T_4$ Space:
:$T$ is a $T_4$ space
By definition, $T$ is a normal space.
Category:Regular Spaces
Category:Normal Spaces
Regular Space with Sigma-Locally Finite Basis is Perfectly Normal Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a regular topological space.
Let $\BB$ be a $\sigma$-locally finite basis.
Then:
:$T$ is a perfectly normal space
== Proof ==
By definition of regular topological space:
:$T$ is a $T_3$ space
:$T$ is a $T_0$ (Kolmogorov) space
From T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space:
:$T$ is a perfectly $T_4$ 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
By definition, $T$ is a perfectly normal space.
Category:Regular Spaces
Category:Perfectly Normal Spaces
Related Angles/Examples/20 Degrees

== Examples of Related Angles ==
The angles $20 \degrees$, $160 \degrees$, $200 \degrees$ and $340 \degrees$ are related angles.
== Proof ==
We note that:
It follows from:
:Sine of Supplementary Angle that $\sin 160 \degrees = \sin 20 \degrees$
:Cosine of Supplementary Angle that $\cos 160 \degrees = -\cos 20 \degrees$
:Tangent of Supplementary Angle that $\tan 160 \degrees = -\tan 20 \degrees$
:Sine of Angle plus Straight Angle that $\sin 200 \degrees = -\sin 20 \degrees$
:Cosine of Angle plus Straight Angle that $\cos 200 \degrees = -\cos 20 \degrees$
:Tangent of Angle plus Straight Angle that $\tan 200 \degrees = \tan 20 \degrees$
:Sine of Explementary Angle that $\sin 340 \degrees = -\sin 20 \degrees$
:Cosine of Explementary Angle that $\cos 340 \degrees = \cos 20 \degrees$
:Tangent of Explementary Angle that $\tan 340 \degrees = -\tan 20 \degrees$
== Sources ==
* 
* 
Category:Examples of Related Angles
Relation/Examples/Between

== Example of Relation ==
Let $A$, $B$ and $C$ be sets of points on a plane.
Let $\RR \subseteq A \times B \times C$ denote the subset of $A \times B \times C$ defined as:
:$\forall a \in A, b \in B, c \in C: \tuple {a, b, c \in \R}$ $a$ lies between $b$ and $c$
Then $\RR$ is an example of a ternary relation on $A \times B \times C$.
== Sources ==
* 
* 
Category:Examples of Ternary Relations
Category:Examples of Relations