nyuuzyou/wikis huggingface dataset:
Napier's Analogies/Tangent of Half Sum of Sides

== Napier's Analogies ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\tan \dfrac {a + b} 2 = \dfrac {\cos \frac {A - B} 2} {\cos \frac {A + B} 2} \tan \dfrac c 2$
== Proof ==
== Also see ==
* Delambre's Analogies
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $16$. Delambre's and Napier's analogies.
* 
* 
Category:Napier's Analogies
Napier's Cosine Rule for Quadrantal Triangles

== Theorem ==
:
Let $\triangle ABC$ be a quadrantal triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Let the side $c$ be a right angle.
Let the remaining parts of $\triangle ABC$ be arranged according to the exterior of this circle, where the symbol $\Box$ denotes a right angle.
:
Let one of the parts of this circle be called a middle part.
Let the two neighboring parts of the middle part be called adjacent parts.
Then the sine of the middle part equals the product of the cosine of the opposite parts.
== Proof ==
We are given that $c$ is a right angle.
Let the remaining parts of $\triangle ABC$ be arranged according to the exterior of the circle above, where the symbol $\Box$ denotes a right angle.
==== $\sin A$ ====
==== $\sin B$ ====
==== $\map \sin {\Box - a}$ ====
==== $\map \sin {C - \Box}$ ====
==== $\map \sin {\Box - b}$ ====
== Also see ==
* Napier's Tangent Rule for Quadrantal Triangles
* Napier's Cosine Rule for Right Spherical Triangles
* Napier's Tangent Rule for Right Spherical Triangles
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $10$. Right-angled and quadrantal triangles.
Category:Napier's Rules for Quadrantal Triangles
Napier's Cosine Rule for Right Spherical Triangles

== Theorem ==
:
Let $\triangle ABC$ be a right spherical triangle on the surface of a sphere whose center is $O$ such that the angle $\sphericalangle C$ is a right angle.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Let the remaining parts of $\triangle ABC$ be arranged according to the interior of this circle, where the symbol $\Box$ denotes a right angle.
:
Let one of the parts of this circle be called a middle part.
Let the two parts which do not neighbor the middle part be called opposite parts.
Then the sine of the middle part equals the product of the cosine of the opposite parts.
== Proof ==
We are given that $\sphericalangle C$ is a right angle.
Let the remaining parts of $\triangle ABC$ be arranged according to the interior of the circle above, where the symbol $\Box$ denotes a right angle.
==== $\sin a$ ====
==== $\sin b$ ====
==== $\map \sin {\Box - A}$ ====
==== $\map \sin {\Box - c}$ ====
==== $\map \sin {\Box - B}$ ====
== Also see ==
* Napier's Tangent Rule for Right Spherical Triangles
* Napier's Cosine Rule for Quadrantal Triangles
* Napier's Tangent Rule for Quadrantal Triangles
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.102$
* : Chapter $\text I$. Spherical Trigonometry: $10$. Right-angled and quadrantal triangles.
* 
* 
* : $\S 12$: Trigonometric Functions: $12.102.$
Category:Napier's Rules for Right Spherical Triangles
Napier's Rules for Quadrantal Triangles

== Theorem ==
Napier's Rules for Quadrantal Triangles are the special cases of the Spherical Law of Cosines for a quadrantal triangle.
Hence, let $\triangle ABC$ be a quadrantal triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Let side $c$ be a right angle.
Let the remaining parts of $\triangle ABC$ be arranged according to the exterior of this circle, where the symbol $\Box$ denotes a right angle.
:
Let one of the parts of this circle be called a middle part.
Let the two neighboring parts of the middle part be called adjacent parts.
Let the remaining two parts be called opposite parts.
Then:
:The sine of the middle part equals the product of the tangents of the adjacent parts.
:The sine of the middle part equals the product of the cosines of the opposite parts.
=== Tangents ===
=== Cosines ===
== Also see ==
* Napier's Rules of Circular Parts
* Napier's Rules for Right Spherical Triangles
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $10$. Right-angled and quadrantal triangles.
Category:Napier's Rules for Quadrantal Triangles
Category:Napier's Rules of Circular Parts
Category:Quadrantal Spherical Triangles
Napier's Rules for Right Spherical Triangles

== Theorem ==
:
Let $\triangle ABC$ be a right spherical triangle on the surface of a sphere whose center is $O$ such that the angle $\sphericalangle C$ is a right angle.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Let the remaining parts of $\triangle ABC$ be arranged according to the interior of this circle, where the symbol $\Box$ denotes a right angle.
:
Let one of the parts of this circle be called a middle part.
Let the two neighboring parts of the middle part be called adjacent parts.
Let the remaining two parts be called opposite parts.
Then:
:The sine of the middle part equals the product of the tangents of the adjacent parts.
:The sine of the middle part equals the product of the cosines of the opposite parts.
=== Tangents ===
=== Cosines ===
== Also see ==
* Napier's Rules of Circular Parts
* Napier's Rules for Quadrantal Triangles
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $10$. Right-angled and quadrantal triangles.
* 
* 
* 
Category:Napier's Rules for Right Spherical Triangles
Category:Napier's Rules of Circular Parts
Napier's Rules of Circular Parts

== Theorem ==
=== Napier's Rules for Right Spherical Triangles ===
=== Napier's Rules for Quadrantal Triangles ===
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $10$. Right-angled and quadrantal triangles.
* 
* 
Category:Napier's Rules of Circular Parts
Category:Spherical Trigonometry
Napier's Tangent Rule for Quadrantal Triangles

== Theorem ==
:
Let $\triangle ABC$ be a quadrantal triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Let the side $c$ be a right angle.
Let the remaining parts of $\triangle ABC$ be arranged according to the exterior of this circle, where the symbol $\Box$ denotes a right angle.
:
Let one of the parts of this circle be called a middle part.
Let the two neighboring parts of the middle part be called adjacent parts.
Then the sine of the middle part equals the product of the tangents of the adjacent parts.
== Proof ==
We are given that $c$ is a right angle.
Let the remaining parts of $\triangle ABC$ be arranged according to the exterior of the circle above, where the symbol $\Box$ denotes a right angle.
==== $\sin A$ ====
==== $\sin B$ ====
==== $\map \sin {\Box - a}$ ====
==== $\map \sin {C - \Box}$ ====
==== $\map \sin {\Box - b}$ ====
== Also see ==
* Napier's Cosine Rule for Quadrantal Triangles
* Napier's Tangent Rule for Right Spherical Triangles
* Napier's Cosine Rule for Right Spherical Triangles
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $10$. Right-angled and quadrantal triangles.
Category:Napier's Rules for Quadrantal Triangles
Napier's Tangent Rule for Right Spherical Triangles

== Theorem ==
:
Let $\triangle ABC$ be a right spherical triangle on the surface of a sphere whose center is $O$ such that the angle $\sphericalangle C$ is a right angle.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Let the remaining parts of $\triangle ABC$ be arranged according to the interior of this circle, where the symbol $\Box$ denotes a right angle.
:
Let one of the parts of this circle be called a middle part.
Let the two neighboring parts of the middle part be called adjacent parts.
Then the sine of the middle part equals the product of the tangents of the adjacent parts.
== Proof ==
We are given that $\sphericalangle C$ is a right angle.
Let the remaining parts of $\triangle ABC$ be arranged according to the interior of the circle above, where the symbol $\Box$ denotes a right angle.
==== $\sin a$ ====
==== $\sin b$ ====
==== $\map \sin {\Box - A}$ ====
==== $\map \sin {\Box - c}$ ====
==== $\map \sin {\Box - B}$ ====
== Also see ==
* Napier's Cosine Rule for Right Spherical Triangles
* Napier's Cosine Rule for Quadrantal Triangles
* Napier's Tangent Rule for Quadrantal Triangles
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.101$
* : Chapter $\text I$. Spherical Trigonometry: $10$. Right-angled and quadrantal triangles.
* 
* 
* : $\S 12$: Trigonometric Functions: $12.101.$
Category:Napier's Rules for Right Spherical Triangles
Napoleon's Theorem

== Theorem ==
Let $\triangle ABC$ be an arbitrary triangle.
Let $\triangle ABF$, $\triangle BCD$ and $\triangle ACE$ be equilateral triangles constructed on $AB$, $BC$ and $AC$ respectively on the exterior of $\triangle ABC$.
Let $O_1$, $O_2$ and $O_3$ be the incenters of $\triangle ABF$, $\triangle BCD$ and $\triangle ACE$.
Then $\triangle O_1 O_2 O_3$ is an equilateral triangle.
:
=== Variant 1 ===
=== Variant 2 ===
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
== Sources ==
* 
* 
* 
Category:Napoleon's Theorem
Category:Triangles
Napoleon's Theorem/Historical Note

== Historical Note on Napoleon's Theorem ==
The attribution of Napoleon's Theorem to is in fact dubious.
An early appearance is by in $1825$.
An earlier appearance is in one of the papers for candidates for the Gold Medal in the General Examination of the University of Dublin in October $1820$.
It is known that did have some understanding of mathematics, so the suggestion that he may have discovered this theorem is not impossible.
It is rediscovered over and over again by enthusiastic amateurs.
== Sources ==
* 
* 
Category:Historical Notes
Napoleon's Theorem/Lemma 1

== Lemma for Napoleon's Theorem ==
Let $T = \triangle ABC$ be an equilateral triangle in the Cartesian plane $\CC$.
Let the sides of $\triangle ABC$ be the vectors $\mathbf u$, $\mathbf v$, and $\mathbf w$.
Let the interior of $T$ lie to the left of the vector path.
Let $\mathbf v$ be rotated by an angle of $60 \degrees$ anticlockwise.
Let the rotated vector be $\mathbf v'$.
Then:
:$\mathbf u + \mathbf v' = \mathbf 0$
where $\mathbf 0$ denotes the zero vector.
:
== Proof ==
Let: $v = \norm {\mathbf v}$, $v' = \norm {\mathbf v'}$, and $u = \norm {\mathbf u}$.
By Vector Magnitude is Invariant Under Rotation:
:$v = v'$
By definition of equilateral triangle:
:$v = u$
By definition of equilateral triangle the angle between $\mathbf u$ and $\mathbf v$ is $120 \degrees$.
By definition of rotation the angle between $\mathbf u$ and $\mathbf v'$ is $180 \degrees$.
That is, the direction of $\mathbf v'$ is opposite to that of $\mathbf u$.
It follows that:
Category:Napoleon's Theorem
Napoleon's Theorem/Lemma 2

== Lemma ==
Let $T = \triangle ABC$ be an equilateral triangle in the plane $\CC$.
Let $\mathbf{v}$ be a vector in $\CC$ with magnitude $\dfrac 1 3 \norm {AB}$ and direction $\vec {BA}$.
Let a unit rotation be anticlockwise by $60^{\circ}$, and denote vector $\mathbf{v}$ after this rotation as $\mathbf{v}'$.
Then the vector path from $B$ to the incenter $O$ of $T$ is $\mathbf{v} + -\mathbf{v}' '$ and the vector path from the incenter $O$ to $A$ is $\mathbf{v}' + \mathbf{v}$.
== Proof ==
The angle between $\mathbf{v}$ and $\mathbf{v}'$ is $60^{\circ}$.
If $\mathbf{v}$ is rotated again:
:the angle between $\mathbf{v}$ and $\mathbf{v}' '$ is $120^{\circ}$.
By the definition of vector, for an arbitrary vector $\mathbf{w}$:
:$\mathbf{w} + (-\mathbf{w}) = \mathbf{0}$, where $\mathbf{0}$ is the zero vector.
This accounts for all of the vectors in the figure.
By these definitions:
:$-\mathbf{v} = \mathbf{v}$
We define vector paths from the vertices of an equilateral triangle to its incenter.
By Orthocenter, Centroid and Circumcenter Coincide iff Triangle is Equilateral: 
:the incenter of $T$ is also the centroid.
By Medians of Triangle Meet at Centroid:
:the centroid divides the altitude in two parts: the part nearest the base is $\dfrac 1 3$ of the whole.
By construction:
:it is evident that the black dotted point in the figure is the centroid of $T$.
To see this, note that:
:the point lies on the intersection of medians of the triangle
also:
:the distance from the base to the point is $1/3$ the total height.
Also by construction:
:the correctness of the paths is evident.
:$\mathbf{v} + (- \mathbf{v}' ')$ is a vector path from $B$ to $O$.
:$\mathbf{v}' + \mathbf{v}$ is a vector path from $O$ to $A$.
Category:Napoleon's Theorem
Napoleon's Theorem/Proof 1

== Theorem ==
== Proof ==
For simplicity of notation, we relabel the incenters $O_1, O_2, O_3$ as $P, Q, R$.
By Line from Vertex of Triangle to Incenter is Angle Bisector:
:$CQ$ bisects $\angle ACE$
:$CP$ bisects $\angle BCD$
Given:
:$\angle ACE = \angle BCD = 60^{\circ}$
Since $\triangle CBD$ and $\triangle ACE$ are both equilateral, the four smaller angles are all equal to $30^{\circ}$:
:$\angle ACQ = \angle QCE = \angle BCP = \angle PCD = 30^{\circ}$
By addition:
:$\angle BCE = \angle BCA + 60^{\circ}$
But also by addition:
:$\angle PCQ = \angle BCA + 60^{\circ}$
So by Common Notion 1:
:$\angle BCE = \angle PCQ$
Let $\angle BCE = \angle PCQ = \gamma$
By Law of Cosines:
:$(1) \ PQ^2 = CP^2 + CQ^2 - 2 \ CP \cdot CQ \cdot \cos {\gamma}$
By the same argument:
:$\angle BAE = \angle RAQ = \angle BAC + 60^{\circ}$
Let $\angle BAE = \angle RAQ = \alpha$.
We also can obtain $CP$ and $CQ$ in terms of the sides of the original triangle.
Draw the altitude from $Q$ to $AC$.
Since $\angle ACQ = 30^{\circ}$:
:$CA/2 : CQ = \map \cos {30^{\circ}} = \dfrac {\sqrt{3}} 2$
Simplifying:
:$CA = \sqrt{3} CQ$
:$CA^2 = 3 CQ^2$
In the same way:
:$BC/2 : CP = \map \cos {30^{\circ}} = \dfrac {\sqrt{3}} 2$
Simplifying:
:$BC = \sqrt{3} CP$
:$BC^2 = 3 CP^2$
Multiplying:
:$3 CP \cdot CQ = BC \cdot CA$
Multiply (1) by $3$:
:$3PQ^2 = 3CP^2 + 3CQ^2 - 2 \cdot 3 \cdot CP \cdot CQ \cdot \cos {\gamma}$
and substitute:
:$(2) \ 3PQ^2 = BC^2 + CA^2 - 2 BC \cdot CA \cdot \cos {\gamma}$
Recall that $\alpha = \angle BAE$.
We can make all the same arguments for $QR$ so by symmetry:
:$(3) \ 3 QR^2 = AB^2 + AC^2 - 2 AB \cdot AC \cos \alpha$
Finally consider $BE$ as part of $\triangle ABE$. 
By Law of Cosines:
:$BE^2 = AB^2 + AE^2 - 2 AB \cdot AE \cdot \cos \alpha$
But $AE = AC$ so:
:$(4) \ BE^2 = AB^2 + AC^2 - 2 AB \cdot AC \cdot \cos \alpha$
Consider $BE$ as part of $\triangle BCE$. 
By Law of Cosines:
:$BE^2 = BC^2 + CE^2 - 2 BC \cdot CE \cdot \cos \gamma$
Since $AC = CE$:
:$(5) \ BE^2 = BC^2 + AC^2 - 2 BC \cdot AC \cdot \cos \gamma$
By Common Notion 1, since the LHS of (4) and (5) are equal, the RHS are also equal.
The RHS of (4) is the same as the RHS of (3).
The RHS of (5) is the same as the RHS of (2).
So then too by Common Notion 1 the LHS of (2) and (3) are equal:
:$3 \cdot PQ^2 = 3 \cdot QR^2$
Simplifying:
:$PQ = QR$
We can make the same argument for any pair of points from $P, Q, R$.
Therefore:
:$PQ = QR = PR$
$\triangle PQR$ satisfies the definition of an equilateral triangle.
Category:Triangles
Category:Napoleon's Theorem
Napoleon's Theorem/Proof 2

== Theorem ==
== Proof ==
=== ===
=== ===
==== Vectors ====
Side $a = BC$ lies opposite vertex $A$ of $\triangle ABC$.
Let the vector $\mathbf{a}$ have magnitude $\dfrac 1 3 a$ in the direction of $\vec {BC}$.
$O_1$ is the incenter of the $\triangle BCD$ with side $a$ from $\triangle ABC$.
Side $b = CA$ lies opposite vertex $B$ of $\triangle ABC$.
Let the vector $\mathbf{b}$ have magnitude $\dfrac 1 3 b$ in the direction of $\vec {CA}$.
$O_2$ is the incenter of the $\triangle CAE$ with side $b$ from $\triangle ABC$.
Side $c = AB$ lies opposite vertex $C$ of $\triangle ABC$.
Let the vector $\mathbf{c}$ have magnitude $\dfrac 1 3 c$ in the direction of $\vec {AB}$.
$O_3$ is the incenter of the $\triangle ABF$ with side $c$ from of $\triangle ABC$.
==== Construction of Vector Paths ====
Let $\mathbf{p}$ be the vector path from $O_1$ to $O_2$ through vertex $C$.
By :
:The first part of $\mathbf{p}$ from $O_1$ to $C$ is $\mathbf{a}' + \mathbf{a}$.
Also by :
:The second part of $\mathbf{p}$ from $C$ to $O_2$ is $\mathbf{b} + - \mathbf{b}' '$.
By addition:
:$\mathbf{p} = \mathbf{a}' + \mathbf{a} + \mathbf{b} + - \mathbf{b}' '$
Let $\mathbf{q}$ be the vector path from $O_2$ to $O_3$ through vertex $A$.
By :
:The first part of $\mathbf{q}$ from $O_2$ to $A$ is $\mathbf{b}' + \mathbf{b}$.
By :
:The second part of $\mathbf{q}$ from $A$ to $O_3$ is $\mathbf{c} + - \mathbf{c}' '$.
By addition:
:$\mathbf{q} = \mathbf{b}' + \mathbf{b} + \mathbf{c} + - \mathbf{c}' '$
==== Test for Equilateral Triangle ====
By the test from , we construct $\mathbf{p} + \mathbf{q}'$: 
:If the result is $\mathbf{p} + \mathbf{q}' = \mathbf{0}$, the two vectors $\mathbf{p}$ and $\mathbf{q}$ are sides of an equilateral triangle.
:$\mathbf{q}' = \mathbf{b}' ' + \mathbf{b}' + \mathbf{c}' - \mathbf{c}' ' '$
By definition of plane rotation and that $\theta = \frac 1 6$ of a complete rotation:
:$- \mathbf{c}' ' ' = \mathbf{c}$
Substituting:
::$\mathbf{q}' = \mathbf{b}' ' + \mathbf{b}' + \mathbf{c}' + \mathbf{c}$
Then:
The result follows.
Category:Napoleon's Theorem
Napoleon's Theorem/Variant 1

== Theorem ==
Let $\triangle ABC$ be an arbitrary triangle.
Let $\triangle ABF$, $\triangle BCD$ and $\triangle ACE$ be equilateral triangles constructed on $AB$, $BC$ and $AC$ respectively toward the interior of $\triangle ABC$.
Let $O_1$, $O_2$ and $O_3$ be the incenters of $\triangle ABF$, $\triangle ACE$ and $\triangle BCD$ respectively.
Then $\triangle O_1 O_2 O_3$ is an equilateral triangle.
:
== Proof ==
=== ===
=== ===
The notation differs for the variant Theorem: $O_1$ is now $O_3$ and $O_3$ is now $O_1$. 
We will use vectors in the same order as in Napoleon's Theorem/Proof 2, starting with $\mathbf{a}$.
As a result, the vector path will be clockwise, some small adjustments will be necessary..
==== Vectors ====
Side $a = BC$ lies opposite vertex $A$ of $\triangle ABC$.
Let the vector $\mathbf{a}$ have magnitude $\dfrac 1 3 a$ in the direction of $\vec {BC}$.
$O_3$ is the incenter of the $\triangle BCD$ with side $a$ from $\triangle ABC$.
Side $b = CA$ lies opposite vertex $B$ of $\triangle ABC$.
Let the vector $\mathbf{b}$ have magnitude $\dfrac 1 3 b$ in the direction of $\vec {CA}$.
$O_2$ is the incenter of the $\triangle CAE$ with side $b$ from $\triangle ABC$.
Side $c = AB$ lies opposite vertex $C$ of $\triangle ABC$.
Let the vector $\mathbf{c}$ have magnitude $\dfrac 1 3 c$ in the direction of $\vec {AB}$ from $\triangle ABC$.
$O_1$ is the incenter of the $\triangle ABF$ with side $c$
==== Construction of Vector Paths ====
Let $\mathbf{p}$ be the vector path from $O_3$ to $O_2$ through vertex $C$.
Note that the equilateral triangles are reflected. This makes the assignment of path elements different.
By :
:The first part of $\mathbf{p}$ from $O_3$ to $C$ is $- \mathbf{a}' ' + \mathbf{a}$.
Also by :
:The second part of $\mathbf{p}$ from $C$ to $O_2$ is $\mathbf{b} + \mathbf{b}'$.
By addition:
:$\mathbf{p} = - \mathbf{a}' ' + \mathbf{a} + \mathbf{b} + \mathbf{b}'$
:
Let $\mathbf{q}$ be the vector path from $O_2$ to $O_1$ through vertex $A$.
By :
:The first part of $\mathbf{q}$ from $O_2$ to $A$ is $- \mathbf{b}' ' + \mathbf{b}$.
By :
:The second part of $\mathbf{q}$ from $A$ to $O_3$ is $\mathbf{c} + \mathbf{c}'$.
By addition:
:$\mathbf{q} = - \mathbf{b}' ' + \mathbf{b} + \mathbf{c} + \mathbf{c}'$
==== Test for Equilateral Triangle ====
The test from must be modified because the vector path from $O_3$ to $O_2$ to $O_1$ goes clockwise.
Therefore, we construct $\mathbf{p}' + \mathbf{q}$: 
:If the result is $\mathbf{p}' + \mathbf{q} = \mathbf{0}$, the two vectors $\mathbf{p}$ and $\mathbf{q}$ are sides of an equilateral triangle.
:$\mathbf{p}' = - \mathbf{a}' ' ' + \mathbf{a}' + \mathbf{b}' + \mathbf{b}' '$
By definition of plane rotation and that $\theta = \frac 1 6$ of a complete rotation:
:$- \mathbf{a}' ' ' = \mathbf{a}$
Substituting:
:$\mathbf{p}' = \mathbf{a} + \mathbf{a}' + \mathbf{b}' + \mathbf{b}' '$
Then:
The result follows.
== Historical Note ==
== Sources ==
* 
* 
Category:Napoleon's Theorem
Category:Triangles
Natural Basis of Product Topology

== Theorem ==
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds X := \prod_{i \mathop \in I} X_i$
Then the natural basis on $X$ is the set $\BB$ of cartesian products of the form $\ds \prod_{i \mathop \in I} U_i$ where:
:for all $i \in I : U_i \in \tau_i$
:for all but finitely many indices $i : U_i = X_i$
=== Finite Product ===
== Proof ==
Let $\NN$ denote the natural basis on $X$.
By definition of the natural basis, $\NN$ is the basis generated by:
:$\SS = \set {\pr_i^{-1} \sqbrk U: i \in I, \, U \in \tau_i}$
where for each $i \in I$, $\pr_i: X \to X_i$ denotes the $i$th projection on $X$:
:$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$
=== Lemma 1 ===
=== Lemma 2 ===
From Synthetic Basis formed from Synthetic Sub-Basis:
:$\NN = \set {\bigcap \FF: \FF \subseteq \SS, \, \FF \text{ is finite} }$
From Lemma 1 and Lemma 2 it follows that:
:$\NN \subseteq \BB$.
=== Lemma 3 ===
From Lemma 3 and Synthetic Basis formed from Synthetic Sub-Basis:
:$\BB \subseteq \NN$
The result follows from the definition of set equality.
Category:Product Topology
Category:Natural Basis of Product Topology
Natural Basis of Product Topology/Finite Product

== Theorem ==
Let $n \in \N$.
For all $k \in \set {1, \ldots, n}$, let $\struct {X_k, \tau_k}$ be topological spaces.
Let $\ds X = \prod_{k \mathop = 1}^n X_k$ be the (finite) cartesian product of $X_1, \ldots, X_n$.
Then the natural basis on $X$ is:
:$\BB = \set {\ds \prod_{k \mathop = 1}^n U_k : \forall k : U_k \in \tau_k}$
== Proof ==
From Natural Basis of Product Topology, the natural basis on $X$ is the set $\BB$ of cartesian products of the form:
:$\ds \prod_{i \mathop \in I} U_i$
where:
:for all $k = 1, \dots, n: U_k \in \tau_i$
:for all but finitely many indices $k : U_k = X_k$
Let:
:$\BB' = \set {\ds \prod_{k \mathop = 1}^n U_k : \forall k : U_k \in \tau_k}$
By definition of $\BB$:
:$\BB \subseteq \BB'$
Let $B \in \BB'$, that is: 
:$B = \ds \prod_{k \mathop = 1}^n U_k$
where $\forall k: U_k \in \tau_k$.
Let $J = \set {k: U_k \neq X_k}$.
Since $J \subseteq \set{ 1, \dots, n}$ is finite then:
:for all but finitely many indices $k : U_k = X_k$
Thus:
:$B \in \BB$
Since $B$ was arbitrary, then:
:$\BB' \subseteq \BB$
It follows that:
:$\BB = \BB'$
from the definition of set equality.
Category:Natural Basis of Product Topology
Natural Basis of Product Topology/Lemma 1

== Lemma for Natural Basis of Product Topology ==
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds X := \prod_{i \mathop \in I} X_i$
For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$th projection on $X$:
:$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$
Let:
:$\SS = \set {\pr_i^{-1} \sqbrk U: i \in I, \, U \in \tau_i}$
Let $\BB$ be the set of cartesian products of the form:
:$\ds \prod_{i \mathop \in I} U_i$
where:
:for all $i \in I : U_i \in \tau_i$
:for all but finitely many indices $i : U_i = X_i$
Then:
 
:$\SS \subseteq \BB$
== Proof ==
Let $i \in I, \, U \in \tau_i$.
By Inverse Image Mapping Induced by Projection:
where:
:$V_i = U$
:$\forall j \ne i : V_j = X_j$
By definition of $\BB$:
:$\pr_i^{-1} \sqbrk U \in \BB$
Since $i$ and $U$ were arbitrary:
:$\SS \subseteq \BB$ 
Category:Natural Basis of Product Topology