nyuuzyou/wikis huggingface dataset:
Two Intersecting Straight Lines are in One Plane

== Theorem ==
== Proof ==
:
Let the two straight lines $AB$ and $CD$ intersect at the point $E$.
Let $F$ and $G$ be arbitrary points on $EC$ and $EB$.
Let $CB$ and $FG$ be connected.
Let $FH$ and $GK$ be drawn across.
Suppose part of $\triangle ECB$ is in the plane of reference and the rest of it in another plane.
Then a part of one of the straight lines $EC$ and $EB$ will be in the plane of reference and the result in another plane.
But from , this cannot happen.
Therefore $\triangle ECB$ is all in one plane.
But whatever plane $\triangle ECB$ is in, each of the straight lines $EC$ and $EB$ are also in that plane.
And in whatever plane $EC$ and $EB$ are in, the straight lines $AB$ and $CD$ are also in that plane.
Therefore $AB$ and $CD$ are in one plane, and every triangle is in one 
plane.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Straight Lines
Category:Planes
Two Irrational Straight Lines arising from Medial Area from which Medial Area Subtracted

== Theorem ==
== Proof ==
:
Let $BC$ be a medial area.
Let the rational area $BD$ be subtracted from $BC$.
It needs to be demonstrated that the "side" of the remainder $EC$ is either:
: a second apotome of a medial straight line
or:
: a straight line which produces with a medial area a medial whole.
Let $FG$ be a rational straight line.
Let the rectangle $GH$ be applied to $FG$ equal to $BC$ producing $FH$ as breadth.
Let the area $GK$ equal to $BC$ be subtracted from $GH$.
Then the remainder $EC$ is equal to $LH$.
We have that:
:$BC$ and $BD$ are medial.
and:
:$BC$ is incommensurable with $BD$.
We also have:
:$BC = GH$
and:
:$BD = GK$
Therefore $GH$ and $GK$ are medial.
Therefore from :
:$FH$ and $FK$ are rational and incommensurable in length with $FG$.
We have that $BC$ is incommensurable with $BD$.
Therefore $GH$ is incommensurable with $GK$.
From:
:
and:
:
we have that:
:$HF$ is incommensurable in length with $FK$.
Therefore $FH$ and $FK$ are rational straight lines which are commensurable in square only.
Therefore $KH$ is an apotome and $KF$ is the annex to $KH$.
We have that:
: $HF^2 = FK^2 + \lambda^2$
where either:
:$\lambda$ is commensurable in length with $HF$
or:
:$\lambda$ is incommensurable in length with $HF$.
First suppose $\lambda$ is commensurable in length with $HF$.
We have that neither $FH$ and $FK$ is commensurable in length with the rational straight line $FG$.
Therefore $KH$ is a third apotome.
But from : 
:the "side" of $LH$ is a second apotome of a medial straight line.
Next suppose $\lambda$ is incommensurable in length with $HF$.
We have that neither $FH$ and $FK$ is commensurable in length with the rational straight line $FG$.
Therefore $KH$ is a sixth apotome.
But from : 
:the "side" of $LH$ is a straight line which produces with a medial area a medial whole.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Two Irrational Straight Lines arising from Medial Area from which Rational Area Subtracted

== Theorem ==
== Proof ==
:
Let $BC$ be a medial area.
Let the rational area $BD$ be subtracted from $BC$.
It needs to be demonstrated that the "side" of the remainder $EC$ is either:
:a first apotome of a medial straight line
or:
:a straight line which produces with a rational area a medial whole.
Let $FG$ be a rational straight line.
Let the rectangle $GH$ be applied to $FG$ equal to $BC$ producing $FH$ as breadth.
Let the area $GK$ equal to $BC$ be subtracted from $GH$.
Then the remainder $EC$ is equal to $LH$.
We have that $BC$ is medial and $BD$ is rational.
We also have:
:$BC = GH$
and:
:$BD = GK$
Therefore $GH$ is medial and $GK$ is rational.
But $GH$ and $GK$ are applied to a rational straight line $FG$.
Therefore from :
:$FK$ is rational and commensurable in length with $FG$
while from :
:$FH$ is rational and incommensurable in length with $FG$.
Therefore by : 
:$FH$ is incommensurable in length with $FK$.
Therefore $FH$ and $FK$ are rational straight lines which are commensurable in square only.
Therefore $KH$ is an apotome and $KF$ is the annex to $KH$.
We have that:
:$HF^2 = FK^2 + \lambda^2$
where either:
:$\lambda$ is commensurable in length with $HF$
or:
:$\lambda$ is incommensurable in length with $HF$.
First suppose $\lambda$ is commensurable in length with $HF$.
Then $FK$ is commensurable in length with the rational straight line $FG$.
Therefore $KH$ is a second apotome.
But from : 
:the "side" of $LH$ is a first apotome of a medial straight line.
Next suppose $\lambda$ is incommensurable in length with $HF$.
Then $FK$ is incommensurable in length with the rational straight line $FG$.
Therefore $KH$ is a fifth apotome.
But from : 
:the "side" of $LH$ is a straight line which produces with a rational area a medial whole.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution

== Theorem ==
Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
on a closed interval $\closedint a b$.
Let $y_1$ and $y_2$ be linearly independent.
Then the general solution to $(1)$ is:
:$y = C_1 \map {y_1} x + C_2 \map {y_2} x$
where $C_1 \in \R$ and $C_2 \in \R$ are arbitrary constants.
== Proof ==
Let $\map y x$ be any particular solution to $(1)$ on $\closedint a b$.
It is to be shown that constants $C_1$ and $C_2$ can be found such that:
:$\map y x = C_1 \map {y_1} x + C_2 \map {y_2} x$
for all $x \in \closedint a b$.
By Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions:
:a particular solution to $(1)$ over $\closedint a b$ is completely determined by:
::its value
:and:
::the value of its derivative
at a single point.
From Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE:
:$C_1 \map {y_1} x + C_2 \map {y_2} x$
is a particular solution to $(1)$ over $\closedint a b$
We also have:
:$\map y x$
is a particular solution to $(1)$ over $\closedint a b$
Thus it is sufficient to prove that:
:$\exists x_0 \in \closedint a b: \exists C_1, C_2 \in \R$ such that:
::$ C_1 \map {y_1} {x_0} + C_2 \map {y_2} {x_0} = \map y {x_0}$
:and:
::$ C_1 \map { {y_1}'} {x_0} + C_2 \map { {y_2}'} {x_0} = \map y {x_0}$
For this system to be solvable for $C_1$ and $C_2$ it is necessary that:
:$\begin{vmatrix}
\map {y_1} x & \map {y_2} x \\
\map { {y_1}'} x & \map { {y_2}'} x \\
\end{vmatrix} = \map {y_1} x \map { {y_2}'} x - \map {y_2} x \map { {y_1}'} x \ne 0$
That is, that the Wronskian $\map W {y_1, y_2} \ne 0$ at $x_0$.
From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE:
:if $\map W {y_1, y_2} \ne 0$ at $x_0$, then $\map W {y_1, y_2} \ne 0$ for all $x \in \closedint a b$.
Hence it does not matter what point is taken for $x_0$; if the Wronskian is non-zero at one such point, it will be non-zero for all such points.
From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
:$W \left({y_1, y_2}\right) = 0$ for all $x \in \closedint a b$ $y_1$ and $y_2$ are linearly dependent.
But we have that $y_1$ and $y_2$ are linearly independent.
Hence:
:$\forall x \in \closedint a b: \map W {y_1, y_2} \ne 0$
and so:
:$\exists x_0 \in \closedint a b: \exists C_1, C_2 \in \R$ such that:
::$ C_1 \map {y_1} {x_0} + C_2 \map {y_2} {x_0} = \map y {x_0}$
:and:
::$ C_1 \map { {y_1}'} {x_0} + C_2 \map { {y_2}'} {x_0} = \map y {x_0}$
The result follows.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.1$ The reduced equation
* : $\S 3.15$: The General Solution of the Homogeneous Equation: Theorem $\text{A}$
Category:Homogeneous LSOODEs
Two Lines Meet at Unique Point

== Theorem ==
Let two straight line segments be constructed on a straight line segment from its endpoints so that they meet at a point.
Then there cannot be two other straight line segments equal to the former two respectively, constructed on the same straight line segment and on the same side of it, meeting at a different point.
== Proof ==
:
Let $AC$ and $CB$ be constructed on $AB$ meeting at $C$.
Let two other straight line segments $AD$ and $DB$ be constructed on $AB$, on the same side of it, meeting at $D$, such that $AC = AD$ and $CB = DB$.
 $C$ and $D$ are different points.
Let $CD$ be joined.
We have that:
:$AC = AD$
Hence from :
:$\angle ACD = \angle ADC$
From :
:$\angle ACD > \angle DCB$ because the whole is greater than the part.
Therefore:
:$\angle CDB > \angle DCB$
We have :
:$CB = DB$
Hence it follows that:
:$\angle CDB = \angle DCB$
But this contradicts:
:$\angle CDB > \angle DCB$
From this contradiction it follows that $C$ and $D$ cannot be different points.
Hence the result.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Lines
Two Lines Meeting which are Parallel to Two Other Lines Meeting contain Equal Angles

== Theorem ==
== Proof ==
:
Let $AB$ and $BC$ be straight lines which meet one another.
Let $DE$ and $EF$ be straight lines which meet one another such that $AB$ is parallel to $DE$ and $BC$ is parallel to $EF$.
It is to be demonstrated that $\angle ABC = \angle DEF$.
Let $BA, BC, ED, EF$ be cut off equal to one another.
Let $AC, CF, BE, AD, DF$ be joined.
We have that:
:$BA = ED$
and
:$BA \parallel ED$
Therefore from :
:$AD = BE$
and
:$AD \parallel BE$
For the same reason:
:$CF = BE$
and
:$CF \parallel BE$
So each of $AD$ and $CF$ is equal and parallel to $BE$.
But from :
:$AD = CF$
and
:$AD \parallel CF$
We have that $AC$ and $DF$ join $AD$ and $CF$.
Therefore from :
:$AC = DF$
and
:$AC \parallel DF$
So we have that $AB$ and $BC$ are equal and parallel to $DE$ and $EF$.
We also have that $AC = DF$.
So from :
:$\angle ABC = \angle DEF$
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Solid Geometry
Two Lines Perpendicular to Same Plane are Parallel

== Theorem ==
== Proof ==
:
Let $AB$ and $CD$ be two straight lines at right angles to the plane of reference.
It is to be demonstrated that $AB$ is parallel to $CD$.
Let $AB$ and $CD$ meet the plane of reference at $B$ and $D$ respectively.
Let the straight line $BD$ be joined.
Let $DE$ be drawn in the plane of reference at right angles to $BD$ such that $DE = AB$.
Let $BE$, $AE$ and $AD$ be joined.
We have that $AB$ is at right angles to the plane of reference.
So by :
:$AB$ is at right angles to every straight line which meets it and is in the plane of reference.
But each of $BD$ and $BE$ is in the plane of reference and meets $AB$.
Therefore $\angle ABD$ and $\angle ABE$ are both right angles.
For the same reason $\angle CDB$ and $\angle CDE$ are both right angles.
We have that $AB = DE$, and that $BD$ is common.
Thus the two sides $AB$ and $BD$ of $\triangle ABD$ equal the two sides $BD$ and $BE$ of $\triangle EDB$.
The triangles $\triangle ABD$ and $\triangle EDB$ both include right angles.
Therefore by :
:$AD = BE$
We have that:
:$AB = DE$
and:
:$AD = BE$
Thus the two sides $AB$ and $BE$ of $\triangle ABE$ equal the two sides $ED$ and $DA$ of $\triangle EDA$.
We also have that $AE$ is common.
So from :
:$\angle ABE = \angle EDA$
But $\angle ABE$ is a right angle.
Therefore $ED$ is at right angles to $DA$.
But $ED$ is also at right angles to the straight lines $BD$ and $DC$.
Therefore $ED$ is set up at right angles to the three straight lines $BD$, $DA$ and $DC$ at their meeting points.
Therefore from :
:$BD$, $DA$ and $DC$ are in the same plane.
But from :
:the triangle $DAB$ is in one plane.
Therefore $AB$ is in the same plane as $DB$ and $DA$.
Therefore the straight lines $AB$, $BD$ and $DC$ are in one plane.
Also, each of $\angle ABD$ and $\angle BDC$ is a right angle.
Therefore from Supplementary Interior Angles implies Parallel Lines:
:$AB$ is parallel to $CD$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Perpendiculars
Category:Parallel Lines
Category:Planes
Two Ninths as Pandigital Fraction

== Theorem ==
There are $2$ ways $\dfrac 2 9$ can be expressed as a pandigital fraction:
:$\dfrac 2 9 = \dfrac {3924} {17658}$
:$\dfrac 2 9 = \dfrac {7596} {34182}$
== Proof ==
Can be verified by brute force.
Category:Pandigital Fractions
Two Non-Intersecting Circles have Four Common Tangents

== Theorem ==
Let $C_1$ and $C_2$ be circles embedded in the plane such that:
:$C_1$ and $C_2$ do not intersect
:one is not inside the other.
Then there are $4$ common tangents to $C_1$ and $C_2$:
:$2$ of the common tangents have the circles on the same side of the tangent: the external tangents
:$2$ of the common tangents have the circles on opposite sides of the tangent: the internal tangents.
== Proof ==
:
== Also see ==
* Definition:External Tangent
* Definition:Internal Tangent
== Sources ==
* 
* 
Category:Common Tangents
Category:Circles
Two Non-Negative Integers have Zeckendorf Representations of which one is Shifted Representation of the Other

== Theorem ==
Let $m, n \in \Z_{\ge 0}$ be non-negative integers.
Then there exists a unique set of integers:
:$\left\{ {k_1, k_2, \ldots, k_r}\right\}$
where:
: $k_1 \gg k_2 \gg \cdots \gg k_r$
where $a \gg b$ denotes that $a - b > 1$
such that:
:$m = F_{k_1} + F_{k_2} + \cdots + F_{k_r}$
and:
:$n = F_{k_1 + 1} + F_{k_2 + 1} + \cdots + F_{k_r + 1}$
Note that:
: each of the $k$'s may be negative
and:
:$r$ may equal $0$.
== Proof ==
== Sources ==
* 
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $42$
Category:Zeckendorf Representation
Two Paths between Vertices in Cycle Graph

== Theorem ==
Let $G$ be a simple graph.
Let $u, v$ be vertices in $G$ such that $u \ne v$.
Then:
:for any two vertices $u, v$ in $G$ such that $u \ne v$ there exists exactly two paths between $u$ and $v$
:
:$G$ is a cycle graph.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
Let $G = \struct {V, E} = C_n$ be the cycle graph of order $n$.
Let $V = \set {v_1, v_2, \ldots, v_n}$
Let $u, v \in C_n$.
Then both $u, v$ are on the cycle $C = \tuple {u, u_1, u_2, \ldots, u_p, v, v_1, v_2, \ldots, v_q, u}$ where $p + q + 2 = n$.
By definition of cycle graph this is the only cycle in $G$.
Let $e$ be the edge $u u_1$ of $G$ (or, if $u$ and $v$ are adjacent, let $e = u v$).
From Size of Cycle Graph equals Order, there are exactly $n$ edges in $G$.
Then the edge deletion $G \setminus e$ has $n - 1$ edges.
By Finite Connected Simple Graph is Tree iff Size is One Less than Order it follows that $G \setminus e$ is a tree.
Thus by Path in Tree is Unique there is exactly $1$ path from $u$ to $v$ in $G - e$, that is:
:$P_1 = \tuple {u, v_q, v_{q - 1}, \ldots, v_2, v_1, v}$
By replacing $e$, there exist a second path from $u$ to $v$ in $G$:
:$P_2 = \tuple {u, u_1, u_2, \ldots, u_{p - 1}, u_p, v}$
or if $u$ and $v$ are adjacent:
:$P_2 = \tuple {u, v}$
== Sources ==
* : $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees: Problem $5$
Category:Cycle Graphs
Two People with Same Number of Hairs on Head

== Problem ==
Why must there certainly be at least $2$ people in the world with exactly the same number of hairs on their head?
== Solution ==
It can be assumed that the most hairs a person has on their head may be of the order of a couple of hundreds of thousands.
The number of people in the world is (as of $2020$) approaching $7$ billion.
The result follows by the Pigeonhole Principle.
So, select individuals arbitrarily, and count the hairs on their head.
Sooner or later, you will pick someone who has the same number of hairs that you have already selected.
This will happen by the time you have selected a few hundred thousand people, at the very latest.
That is, way before you have counted everybody in the world.
== Historical Note ==
== Sources ==
* 
* : Henry van Etten: $124$
Category:Combinatorics
Category:Classic Problems
Two Planes have Line in Common

== Theorem ==
Two distinct planes have exactly one (straight) line in common.
== Proof ==
Take two distinct lines in plane $1$.
From Propositions of Incidence: Plane and Line, they each meet plane $2$ in one point each, say at $A$ and $B$.
Thus $A$ and $B$ both lie in both planes.
Thus the line defined by $A$ and $B$ lies in both planes.
== Sources ==
* : Chapter $1$: Introduction: The Propositions of Incidence: $1.2$: The projective method: The propositions of incidence
Category:Projective Geometry
Two Planes have Line in Common/Hilbert's Axioms

== Theorem ==
Two planes $\alpha, \beta$ have no point in common or a straight line $a$ in common.
== Proof ==
By Law of Excluded Middle, $\alpha$ and $\beta$ have no point in common, or some point $A$ in common.
In the former case, the theorem holds trivially.
In the latter, by Axiom $\text I, 6$, there is a second point $B$ lying on both $\alpha$ and $\beta$.
By Axiom $I, 1$, there is a straight line $a$ containing both $A$ and $B$.
Then, by Axiom $I, 5$, every point of $a$ lies on both $\alpha$ and $\beta$.
== Sources ==
* : $\text I \S 2$. Group $1$: Axioms of connection
Category:Hilbert's Axioms
Two Ring is Boolean Ring

== Theorem ==
Let $2$ be the two ring.
Then $2$ is a Boolean ring.
== Proof ==
From Ring of Integers Modulo m is Ring, $2$ is a ring with unity.
Furthermore, the identities:
:$0 \cdot 0 = 0$
:$1 \cdot 1 = 1$
show that $2$ is also an idempotent ring.
Hence the result, by definition of Boolean ring.
== Sources ==
* : $\S 1$
Category:Boolean Rings
Two Sevenths as Pandigital Fraction

== Theorem ==
There are $6$ ways $\dfrac 2 7$ can be expressed as a pandigital fraction:
:$\dfrac 2 7 = \dfrac {3654} {12789}$
:$\dfrac 2 7 = \dfrac {3674} {12859}$
:$\dfrac 2 7 = \dfrac {5342} {18697}$
:$\dfrac 2 7 = \dfrac {7418} {25963}$
:$\dfrac 2 7 = \dfrac {9786} {34251}$
:$\dfrac 2 7 = \dfrac {9862} {34517}$
== Proof ==
Can be verified by brute force.
Category:Pandigital Fractions
Two Stage Radioactive Decay

== Theorem ==
Let $R_A$ be a radioactive isotope which decays into another radioactive isotope $R_B$ with the rate constant $k_1$.
Let $R_B$ decay into a third element $R_C$ with the rate constant $k_2$.
Let the initial quantity of $R_A$ be $x_0$.
Let the amounts of $R_A$ and $R_B$ present at time $t$ be $x$ and $y$ respectively.
Then:
:$y = \begin{cases}
\dfrac {k_1 x_0} {k_2 - k_1} \left({e^{-k_1 t} - e^{-k_2 t} }\right) & : k_1 \ne k_2 \\
& \\
k_1 x_0 t e^{-k_1 t} & : k_1 = k_2
\end{cases}$
== Proof ==
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $26$
Category:Decay Equation
Two Straight Lines make Equal Opposite Angles

== Theorem ==
=== Porism ===
== Proof ==
Let $AB$ and $CD$ be two straight lines that cut each other at the point $E$.
Since the straight line $AE$ stands on the straight line $CD$, the angles $\angle AED$ and $\angle AEC$ make two right angles.
Since the straight line $DE$ stands on the straight line $AB$, the angles $\angle AED$ and $\angle BED$ make two right angles.
But $\angle AED$ and $\angle AEC$ also make two right angles.
So by Common Notion 1 and the fact that all right angles are congruent, $\angle AED + \angle AEC = \angle AED + \angle BED$.
Let $\angle AED$ be subtracted from each.
Then by Common Notion 3 it follows that $\angle AEC = \angle BED$.
Similarly it can be shown that $\angle BEC = \angle AED$.
== Also known as ==
This result is also called the vertical angle theorem.
The arises from the fact that the angles proven equal are known as vertical angles.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.3$
* : Chapter $\text {A}.1$: Thales (ca. B.C.)
Category:Angles
Two divides Power Plus One iff Odd

== Theorem ==
Let $q, n \in \Z_{>0}$.
Then:
:$2 \divides \paren {q^n + 1}$
 $q$ is odd.
In the above, $\divides$ denotes divisibility.
== Proof ==
By Parity of Integer equals Parity of Positive Power, $q^n$ is even $q$ is even.
Thus it follows that $q^n + 1$ is even $q$ is odd.
The result follows by definition of even integer.
== Sources ==
* : $\S 2.4$: The rational numbers and some finite fields: Further Exercises $8$
Category:Number Theory
Two is Boolean Algebra

== Theorem ==
Let $\mathbf 2$ denote two.
Then $\mathbf 2$ is a Boolean algebra.
== Proof ==
It is useful to first state the Cayley tables for the three logical operations $\lor$, $\land$ and $\neg$:
:$\begin{array}{c|cc}
 \lor & \bot & \top \\ \hline
 \bot & \bot & \top \\
 \top & \top & \top
\end{array} \qquad \begin{array}{c|cc}
 \land & \bot & \top \\ \hline
 \bot & \bot & \bot \\
 \top & \bot & \top
\end{array} \qquad \begin{array}{c|cc}
 & \bot & \top \\ \hline
 \neg & \top & \bot
\end{array}$
Let us now verify the axioms for a Boolean algebra in turn.
=== $(BA \ 0)$: Closure ===
It is immediate from the Cayley tables that $S$ is closed under $\lor$, $\land$ and $\neg$.
=== $(BA \ 1)$: Commutativity ===
Follows from the Rule of Commutation.
=== $(BA \ 2)$: Distributivity ===
Follows from the Rule of Distribution
=== $(BA \ 3)$: Identities ===
Follows from Conjunction with Tautology and Disjunction with Contradiction.
=== $(BA \ 4)$: Complements ===
Follows from Contradiction is Negation of Tautology and Tautology is Negation of Contradiction.
Having verified all axioms, we conclude $\mathbf 2$ is a Boolean algebra.
== Sources ==
* : $\S 2$: Exercise $3$
Category:Boolean Algebras
Two is Boolean Lattice

== Theorem ==
Let $\struct{\mathbf 2, \vee, \wedge, \neg, \preceq}$ be (Boolean lattice) two.
Then:
:$\struct{\mathbf 2, \vee, \wedge, \neg, \preceq}$ is a Boolean lattice.
== Proof ==
By definition of Boolean lattice two:
:$\struct{\mathbf 2, \vee, \wedge, \neg}$ is Boolean algebra two
From Two is Boolean Algebra:
:$\struct{\mathbf 2, \vee, \wedge, \neg}$ is a Boolean algebra
From Boolean Algebra is Equivalent to Boolean Lattice:
:$\struct{\mathbf 2, \vee, \wedge, \neg, \preccurlyeq}$ is a Boolean lattice
where:
:$\forall a, b \in \mathbf 2 : a \preccurlyeq b$ $a = a \wedge b$
By definition of Boolean algebra two:
:$\wedge$ is given by the Cayley table:
:$\begin{array}{c|cc}
 \land & \bot & \top \\ \hline
 \bot & \bot & \bot \\
 \top & \bot & \top
\end{array}$
By manual checking it is seen that:
:$\preccurlyeq = \set{\tuple{\bot, \bot}, \tuple{\bot, \top}, \tuple{\top, \top}}$
By definition of Boolean lattice two:
:$\preceq = \set{\tuple{\bot, \bot}, \tuple{\bot, \top}, \tuple{\top, \top}}$
Hence:
:$\preceq = \preccurlyeq$.
The result follows.
Category:Boolean Lattices
Two is a Locale

== Theorem ==
Let $\struct{\mathbf 2, \vee, \wedge, \neg, \preceq}$ denote (Boolean lattice) two.
Then:
:$\struct{\mathbf 2, \vee, \wedge, \preceq}$ is a locale.
== Proof ==
Follows immediately from:
* Two is Boolean Lattice
* Finite Boolean Lattice is a Locale
Category:Boolean Lattices
Category:Locales
Tychonoff's Theorem

== Theorem ==
=== General Theorem ===
=== Tychonoff's Theorem for Hausdorff Spaces ===
== Also known as ==
== Also see ==
* Tychonoff's Theorem Without Choice, a version that holds under more restrictive conditions but does not require the Axiom of Choice.
== Historical Note ==
== Sources ==
* : $5$: Compact spaces: $5.6$: Compactness and Constructions: Remark $5.6.3$
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
* 
Category:Tychonoff's Theorem
Category:Compact Spaces
Category:Product Spaces
Tychonoff's Theorem/General Case

== Theorem ==
Let $I$ be an indexing set.
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty topological spaces.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Then $X$ is compact each $X_i$ is compact.
== Proof 1 ==
 
== Proof 2 ==
== Also see ==
* Tychonoff's Theorem for Hausdorff Spaces, a weaker result requiring only BPI instead of the full AoC.
* Tychonoff's Theorem Without Choice, a weaker result that holds in pure ZF theory.
== Sources ==
* 
Category:Tychonoff's Theorem