nyuuzyou/wikis huggingface dataset:
Length of Lemniscate of Bernoulli

== Theorem ==
The total length of the lemniscate of Bernoulli given in polar coordinates as:
:$r^2 = a^2 \cos 2 \theta$
is given by:
where $F$ denotes the incomplete elliptic integral of the first kind.
== Proof ==
The arc length of a small length increment $\d s$ is given in polar co-ordinates by:
:$\paren {\d s}^2 = \paren {r \rd \theta}^2 + \paren {\d r}^2$
from which:
:$\dfrac {\d s} {\d \theta} = \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2}$
Half of one lobe of the lemniscate is achieved when $\theta$ goes from $0$ to $\pi / 4$.
Therefore the total length of the lemniscate of Bernoulli is given by:
:$\ds L = 4 \int_0^{\pi/4} \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2} \rd \theta$
First we show:
So:
Thus:
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems: Problem $7$
* : $1,31102 87771 46059 90523 \ldots$
Category:Lemniscate of Bernoulli
Category:Incomplete Elliptic Integral of the First Kind
Length of Logarithmic Spiral

== Theorem ==
Consider a logarithmic spiral $S$ given by the equation:
:$r = a e^{b \theta}$
Construct a tangent to $S$ at the point $Q = \tuple {a, 0}$.
Let the tangent cross the $y$-axis at $P$.
Then the length of $PQ$ equals the total length of $S$ from $P$ inwards to the origin.
== Proof ==
 
:
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.15$: Torricelli ()
Category:Logarithmic Spiral
Length of Logarithmic Spiral/Historical Note

== Theorem ==
The length of a logarithmic spiral was first found by in $1645$.
This was the first time anybody had found the length of a non-straight-line curve for anything other than a circle.
Before this had been done, few people could accept that this was possible to do.
For example, had stated in his in $1637$:
:Geometry should not include lines that are like strings, in that they are sometimes straight and sometimes curved, since the ratios between straight and curved lines are not known, and I believe cannot be discovered by human minds.
's response was:
: Who is so blind as not to see that, if there are two equal straight lines, one of which is then bent into a curve, that curve will be equal to the straight line?
== Sources ==
* : Chapter $\text {A}.15$: Torricelli ()
Category:Historical Notes
Length of Orbit of Subgroup Action on Left Coset Space

== Theorem ==
Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Let $K$ act on the left coset space $G / H^l$ by:
:$\forall \tuple {k, g H} \in K \times G / H^l: k * g H := \paren {k g} H$
The length of the orbit of $g H$ is $\index K {K \cap H^g}$.
== Proof ==
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 54 \gamma$
Category:Group Action on Coset Space
Length of Perimeter of Cardioid

== Theorem ==
Consider the cardioid $C$ embedded in a polar plane given by its polar equation:
:$r = 2 a \paren {1 + \cos \theta}$
where $a > 0$.
The length of the perimeter of $C$ is $16 a$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 11$: Special Plane Curves: Cardioid: $11.14$
* : $\S 9$: Special Plane Curves: Cardioid: $9.14.$
Category:Length of Perimeter of Cardioid
Category:Cardioids
Length of Perimeter of Cardioid/Proof 1

== Theorem ==
== Proof ==
Let $\LL$ denote the length of the perimeter of $C$.
The boundary of the $C$ is traced out where $-\pi \le \theta \le \pi$.
From Arc Length for Parametric Equations:
:$\ds \LL = \int_{\theta \mathop = -\pi}^{\theta \mathop = \pi} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
where, from Equation of Cardioid:
:$\begin {cases}
x & = 2 a \cos \theta \paren {1 + \cos \theta} \\
y & = 2 a \sin \theta \paren {1 + \cos \theta}
\end {cases}$
We have:
Thus:
Category:Length of Perimeter of Cardioid
Length of Perimeter of Cardioid/Proof 2

== Theorem ==
== Proof ==
Let $\LL$ denote the length of the perimeter of $C$.
The boundary of the $C$ is traced out where $-\pi \le \theta \le \pi$.
From Arc Length for Polar Curve:
:$\ds \LL = \int_{\theta \mathop = -\pi}^{\theta \mathop = \pi} \sqrt {r^2 + \paren {\frac {\d r} {\d \theta} }^2} \rd \theta$
where:
:$r = 2 a \paren {1 + \cos \theta}$
Note that we have: 
We therefore have: 
Category:Length of Perimeter of Cardioid
Length of Reciprocal of Product of Powers of 2 and 5

== Theorem ==
Let $n \in \Z$ be an integer.
Let $\dfrac 1 n$, when expressed as a decimal expansion, terminate after $m$ digits.
Then $n$ is of the form $2^p 5^q$, where $m$ is the greater of $p$ and $q$.
== Proof ==
Since $\dfrac 1 n$ terminates after $m$ digits:
:$\dfrac {10^m} n$ is an integer
:$\dfrac {10^{m - 1}} n$ is not an integer
From the first condition, we have $n = 2^p 5^q$ for some positive integers $p, q \le m$.
This gives $m \ge \max \set {p, q}$.
From the second condition, we cannot have both $p, q \le m - 1$.
Therefore at least one of $p, q$ is equal to $m$.
This gives $m \le \max \set {p, q}$.
These results give $m = \max \set {p, q}$.
== Sources ==
* : $5$
* : $5$
Category:Number Theory
Category:Reciprocals
Length of Side of Triangle as Cosines of Angles

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
:$c = a \cos B + b \cos A$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(38)$
Category:Length of Side of Triangle as Cosines of Angles
Category:Triangles
Category:Cosine Function
Length of Side of Triangle as Cosines of Angles/Proof

== Theorem ==
== Proof ==
:
Let a perpendicular be dropped from $C$ to $AB$ to meet $AB$ at $D$.
There are two cases:
=== Foot of Perpendicular is on $AB$ ===
See the diagram on the left.
=== Foot of Perpendicular is outside $AB$ ===
See the diagram on the right.
== Sources ==
* : $\text V$. Trigonometry: Solution of triangles
Category:Length of Side of Triangle as Cosines of Angles
Length of Subgroup Plus Length of Quotient Group

== Theorem ==
Let $G$ be a finite group.
Let $H$ be a normal subgroup of $G$.
Then:
:$\map l G = \map l H + \map l {G / H}$
where:
:$\map l G$ denotes the length of $G$
:$G / H$ denotes the quotient group of $G$ by $H$.
== Proof ==
Let $\HH$ be a composition series of $H$:
:$\set e = H_0 \subset H_1 \subset \dots \subset H_m = H$,
We can construct a composition series $\GG$ for $G$ which contains $H$:
:$\set e = H_0 \subset H_1 \subset \dots \subset H_m = H = G_0 \subset G_1 \subset \dots \subset G_n = G$
where, for $k = 0, \dots, n - 1$:
:$G_k \vartriangleleft G_{k + 1}$
and there exists no normal subgroup $G'$ of $G$ such that:
:$G_k \subset G' \subset G_{k + 1}$.
Now consider the sequence $\SS$:
:$H = G_0 \subset G_1 \subset \dots \subset G_n = G$
We can construct a normal series $\GG_1$ for $G / H$ from $\SS$:
:$H = G_0 / H \subset G_1 / H \subset \dots \subset G_n / H = G / H$
$\GG_1$ is indeed a normal series for $G / H$ because:
:$G_k \vartriangleleft G_{k + 1}$ by construction
:$H \vartriangleleft G_k$
:$\paren {G_{k + 1} / H} / \paren {G_k / H} \cong G_{k + 1} / G_k$ by Third Isomorphism Theorem.
 there exists a normal subgroup $G' / H$ of $G / H$ such that:
:$G_k / H \subset G' / H \subset G_{k + 1} / H$ 
for $k = 0, \dots, n - 1$.
Then the Third Isomorphism Theorem implies:
:$\paren {G' / H} / \paren {G_k / H} \cong G' / G_k$
Then $G_k$ is a proper normal subgroup of $G'$, contradicting the construction of $\GG$.
So no such $G' / H$ exists.
Hence, $\GG_1$ has no proper refinements and is a composition series.
Since the isomorphism $\paren {G_{k + 1} / H} / \paren {G_k / H} \cong G_{k + 1} / G_k$ holds for all $G_k, G_{k + 1}$:
:the length of $\SS$ is equal to $\map l {\GG_1} = \map l {G / H}$.
The result follows from the construction of $\GG$.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 73 \beta$
Category:Normal Subgroups
Category:Quotient Groups
Length of Tangent from Point to Circle center Origin

== Theorem ==
Let $\CC$ be a circle of radius $r$ whose center is at the origin $O$ of a Cartesian plane.
Let $P = \tuple {x, y}$ be a point in the plane of $\CC$ in the exterior of $\CC$.
Let $PT$ be a tangent to $\CC$ from $P$ such that $T$ is the point of tangency.
Then the length of $PT$ is given by:
:$PT^2 = x^2 + y^2 - r^2$
== Proof ==
Let $\NN$ be the normal to $\CC$ at the point $T$.
From Normal to Circle passes through Center, $\NN$ passes through $O$.
By definition of the normal to $\CC$, $\NN$ is perpendicular to $PT$.
Hence $OT$, $PT$ and $OP$ form a right triangle whose hypotenuse is $OP$.
As $OT$ is a line segment coinciding with a radius of $\CC$:
:$OT = r$
The length of $OP$ comes from the Distance Formula:
:$OP = \sqrt {\paren {x - 0}^2 + \paren {y - 0}^2} = \sqrt {x^2 + y^2}$
Hence we have:
== Sources ==
* : Chapter $\text {III}$. The Circle: $11$. Geometrical meaning of the expression $x^2 + y^2 - r^2$
Category:Circles
Category:Tangents
Length of Tangent to Astroid between Axes equals Radius of Stator

== Theorem ==
Let $C_1$ be a circle of radius $b$ roll without slipping around the inside of a circle $C_2$ of radius $a = 4 b$.
Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.
Let $P$ be a point on the circumference of $C_1$.
Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.
Let $H$ be the astroid formed by the locus of $P$.
The segment of the tangent to $H$ between the $x$-axis and the $y$-axis is constant and equals the radius of the stator $C_2$.
== Proof ==
:
From Equation of Astroid, $H$ can be expressed as:
:$\begin {cases} x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end {cases}$
Thus the slope of the tangent to $H$ at $\tuple {x, y}$ is:
Thus the equation of the tangent to $H$ is given by:
:$y - a \sin^3 \theta = -\tan \theta \paren {x - a \cos^3 \theta}$
The $x$-intercept is found by setting $y = 0$ and solving for $x$:
Similarly, the $y$-intercept is found by setting $x = 0$ and solving for $y$, which gives:
:$y = a \sin \theta$
The length of the part of the tangent to $H$ between the $x$-axis and the $y$-axis is given by:
which is constant.
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid
* 
Category:Astroids
Length of Tangent to Excircle

== Theorem ==
The excircle on one side of a triangle is tangent to that side and divides it into the same two lengths as the tangent to the incircle, but in opposite order.
== Proof ==
<onlyinclude>
:
Let the sides opposite vertex $A, B, C$ be $a,b,c$.
Let the semiperimeter of $\triangle ABC$ be $s$.
Construct the incircle of $\triangle ABC$ with center $M$.
Let $\triangle ABC$ be tangent to the incircle at $H$.
Construct the excircle of $\triangle ABC$ on side $c$ with center $N$.
Let $\triangle ABC$ be tangent to the excircle at $T$.
By Tangent Points of Incircle in Terms of Semiperimeter:
:$AF = AH = s - a$
:$CF = CG = s - c$
:$BG = BH = s - b$
By addition:
:$AB = AH + BH = \paren {s - a} + \paren {s - b}$
:$AB = AT + BT = \paren {s - a} + \paren {s - b}$
By Tangents to Circle from Point are of Equal Length:
:$AD = AT$
:$BE = BT$
By addition:
:$AT + BT = c$
Substituting:
:$AD + BE = c$
Rearranging:
:$BE = c - AD$
By Tangents to Circle from Point are of Equal Length, the tangents drawn to the excircle from $C$ are equal:
:$AD + b = BE + a$
By addition:
:$AD + b = c - AD + a$
:$2AD = a + c - b$
:$AD = \dfrac {a + c - b} 2$
By the definition of $s$ above:
:$\paren {s - b} = \dfrac {a + c + b} 2 - b = \dfrac {a + c - b} 2$
Combining results:
:$AD = s - b$
We find one pair of tangents to the incircle by substitution:
:$AT = AD = s - b$
Since $c = \paren {s - a} + \paren {s - b} = AD + BE$:
:$BE = s - a$
We find a second pair of tangents to the incircle by substitution:
:$BT = BE = s - a$
The third pair of tangents $CD$ and $CE$ are
:$CD = CE = 3 s - \paren {a + b + c} = s$
Category:Incircles of Triangles
Category:Excircles of Triangles
Length of Union of Chain of Ordinal Sequences

== Theorem ==
Let $C$ be a chain of ordinal sequences.
Let $\lambda$ be a limit ordinal such that all elements of $C$ are of length (strictly) less than $\lambda$.
Suppose that for every ordinal $\alpha < \lambda$ there exists $\theta \in C$ such that:
:$\size \theta = \alpha$
where $\size \theta$ denotes the length of $\theta$.
Then:
:$\size {\bigcup C} = \lambda$
where $\bigcup C$ denotes the union of $C$.
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Lemma $5.2 \ (1)$
Category:Ordinal Sequences
Category:Chains (Order Theory)
Length of Velocity of Smooth Curve is Constant if Curve is Geodesic

== Theorem ==
Let $\struct {M, g}$ be a Riemannian or pseudo-Riemannian manifold with or without boundary.
Let $\nabla$ be a metric connection on $M$.
Let $I \subseteq \R$ be an open real interval.
Let $\gamma : I \to M$ be a smooth curve.
Let $\gamma'$ be the velocity of $\gamma$.
Let $\size {\, \cdot \,}$ be the Riemannian or pseudo-Riemannian inner product norm.
Then $\size {\map {\gamma'} t}$ is constant if $\gamma$ is geodesic.
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Connections on Abstract Riemannian Manifolds
Category:Geodesics
Category:Covariant Derivatives
Length of Velocity of Smooth Curve is Constant iff Covariant Derivative of Velocity along Curve is Orthogonal to Velocity

== Theorem ==
Let $\struct {M, g}$ be a Riemannian or pseudo-Riemannian manifold with or without boundary.
Let $\nabla$ be a metric connection on $M$.
Let $I \subseteq \R$ be an open real interval.
Let $\gamma : I \to M$ be a smooth curve.
Let $\gamma'$ be the velocity of $\gamma$.
Suppose $D_t$ is the covariant derivative along $\gamma$.
Let $\size {\, \cdot \,}$ be the Riemannian or pseudo-Riemannian inner product norm.
Then $\size {\map {\gamma'} t}$ is constant $D_t \map {\gamma'} t$ is orthogonal to $\map {\gamma'} t$ for all $t \in I$.
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Connections on Abstract Riemannian Manifolds
Category:Connections
Category:Covariant Derivatives
Letter L and Letter T are not Homeomorphic

== Theorem ==
Let $\R^2$ denote the real number plane under the Euclidean topology.
Let $\mathsf L \subseteq \R^2$ denote the letter $L$:
:$\mathsf L := \closedint 0 1 \times \set 0 \cup \set 0 \times \closedint 0 1$
Let $\mathsf T \subseteq \R^2$ denote the letter $T$:
:$\mathsf T := \closedint {-1} 1 \times \set 0 \cup \set 0 \times \closedint 0 1$
Then $\mathsf L$ and $\mathsf T$ are not homeomorphic.
== Proof ==
 $f: \mathsf T \to \mathsf L$ is a homeomorphism.
Let $g$ be the restriction of $f$ to $\mathsf T \setminus \set \bszero$, where $\bszero := \tuple {0, 0}$ denotes the origin of $\R^2$.
Then from Restriction of Homeomorphism is Homeomorphism, $g$ is also a homeomorphism.
But $\bszero$ is the junction point of $\mathsf T$, which means that $\mathsf T \setminus \set \bszero$ consists of $3$ disjoint half-open intervals.
However, no matter where $\map f x$ is located in $\mathsf L$, the set $\mathsf L \setminus \set {\map f x}$ consists either of $1$ or $2$ half-open intervals.
This contradicts the supposition that such a homeomorphism $f$ exists.
Hence, by Proof by Contradiction, $\mathsf L$ and $\mathsf T$ are not homeomorphic.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Examples $3.6.2 \ \text{(d)}$
Category:Letter L
Category:Letter T
Letter L and Letter T are not Homeomorphic/Mistake

== Source Work ==
:
:$3$: Continuity generalized: topological spaces:
::$3.6$: Homeomorphisms:
:::Examples $3.6.2 \ \text{(d)}$
== Mistake ==
:The letter $\mathsf L$ and the letter $\mathsf T$ (defined as in $\text{(c)}$) are not homeomorphic.
== Correction ==
The letter $\mathsf T$ is actually not defined in $\text{(c)}$.
In fact, nowhere in can a definition for the letter $\mathsf T$ actually be found.
The definition as provided on has been created from the viewpoint of an intuitive understanding of the material and a plausible interpretation.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Examples $3.6.2 \ \text{(d)}$
Category:Mistakes/Introduction to Metric and Topological Spaces
Letters of Names of Numbers in Alphabetical Order/French/Historical Note

== Historical Note on Letters of Names of Numbers in Alphabetical Order: French ==
The integers the letters of whose names appear in alphabetical order when written in French were noticed in $1998$ by on reading in by ($1997$) that $40$ was the only such number in English.
It is more than probable that others had noticed it before this, but no evidence of this has yet been unearthed.
It crops up online on occasion, for example https://www.mathsisfun.com/puzzles/alphabetical-order-solution.html
The result even appears in contemporary literature: The Curse of the Ice Serpent by Jon Mayhew (2015) features a puzzle whose solution depends on it.
Category:Historical Notes
Levi-Civita Connection in Coordinates

== Theorem ==
Let $\struct {M, g}$ be a Riemannian or pseudo-Riemannian manifold with or without boundary.
Let $\nabla$ is the Levi-Civita connection of $\struct {M, g}$.
Let $g$ be a Riemannian or pseudo-Riemannian metric, which locally reads:
:$g = g_{ij} \rd x^i \otimes \rd x^j$
where $\paren {g_{ij}}$ is a matrix of smooth functions.
Let $g^{ij}$ be the inverse of $\paren {g_{ij}}$. 
Suppose $\Gamma^k_{ij}$ are the connection coefficients of $\nabla$.
Then in any smooth coordinate chart for $M$ we have:
:$\Gamma^k_{ij} = \dfrac 1 2 g^{kl} \paren {\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij} }$
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Symmetric Connections
Category:Levi-Civita Connection
Levi-Civita Connection in Local Frame

== Theorem ==
Let $\struct {M, g}$ be a Riemannian or pseudo-Riemannian manifold with or without boundary.
Let $U \subseteq M$ be an open subset of $M$.
Let $\sqbrk {\cdot, \cdot}$ be the Lie bracket.
Let $\tuple {E_i}$ be a smooth local frame.
Let $c^k_{ij} : U \to \R$ be smooth functions such that:
:$\sqbrk {E_i, E_j} = c^k_{ij} E_k$
Let $\nabla$ is the Levi-Civita connection of $\struct {M, g}$.
Let $g$ be a Riemannian or pseudo-Riemannian metric, which locally reads:
:$g = g_{ij} \rd x^i \otimes \rd x^j$
where $\paren {g_{ij}}$ is a matrix of smooth functions.
Let $g^{ij}$ be the inverse of $\paren {g_{ij}}$. 
Suppose $\Gamma^k_{ij}$ are the connection coefficients of $\nabla$.
Then in the frame $\tuple {E_i}$ we have:
:$\Gamma^k_{ij} = \dfrac 1 2 g^{kl} \paren {E_i g_{jl} + E_j g_{il} - E_l g_{ij} - g_{jm} c^m_{il} - g_{lm} c^m_{ji} + g_{im} c^m_{lj} }$
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Symmetric Connections
Category:Levi-Civita Connection
Levi-Civita Connection in Local Orthonormal Frame

== Theorem ==
Let $\struct {M, g}$ be a Riemannian with or without boundary.
Let $U \subseteq M$ be an open subset of $M$.
Let $\sqbrk {\cdot, \cdot}$ be the Lie bracket.
Let $\tuple {E_i}$ be a smooth local orthonormal frame.
Let $c^k_{ij} : U \to \R$ be smooth functions such that:
:$\sqbrk {E_i, E_j} = c^k_{ij} E_k$
Let $\nabla$ is the Levi-Civita connection of $\struct {M, g}$.
Let $g$ be a Riemannian or pseudo-Riemannian metric, which locally reads:
:$g = g_{ij} \rd x^i \otimes \rd x^j$
where $\paren {g_{ij}}$ is a matrix of smooth functions.
Let $g^{ij}$ be the inverse of $\paren {g_{ij}}$. 
Suppose $\Gamma^k_{ij}$ are the connection coefficients of $\nabla$.
Then in the frame $\tuple {E_i}$ we have:
:$\Gamma^k_{ij} = \dfrac 1 2 \paren {c^k_{ij} - c^j_{ik} - c^i_{jk} }$
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Symmetric Connections
Category:Levi-Civita Connection
Levi-Civita Connection in Terms of Vector Fields

== Theorem ==
Let $\struct {M, g}$ be a Riemannian or pseudo-Riemannian manifold with or without boundary.
Let $\map {\mathfrak X} M$ be the space of smooth vector fields on $M$.
Let $\sqbrk {\cdot, \cdot}$ be the Lie bracket.
Let $\innerprod \cdot \cdot$ be the Riemannian or pseudo-Riemannian scalar product.
Suppose $\nabla$ is the Levi-Civita connection of $\struct {M, g}$.
Then:
:$\forall X, Y, Z \in \map {\mathfrak X} M : \innerprod {\nabla_X Y} Z = \dfrac 1 2 \paren {X \innerprod Y Z + Y \innerprod Z X - Z \innerprod X Y - \innerprod Y {\sqbrk {X, Z} } - \innerprod Z {\sqbrk {Y, X} } + \innerprod X {\sqbrk {Z, Y} } }$
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Symmetric Connections
Category:Levi-Civita Connection
Lexicographic Order/Examples/Unit Square with Open Side

== Example of Lexicographic Order ==
Consider the lexicographic product of the real intervals $\hointr 0 1$ and $\closedint 0 1$ under the usual ordering:
:$\struct {T, \preccurlyeq_l} := \struct {\hointr 0 1, \le} \otimes^l \struct {\closedint 0 1, \le}$
$\struct {T, \preccurlyeq_l}$ has one minimal element:
:$\tuple {0, 0}$
which is also the smallest element: of $\struct {T, \preccurlyeq_l}$.
$\struct {T, \preccurlyeq_l}$ has no greatest element and no maximal elements.
== Proof ==
Recall the definition of the lexicographic order:
::$\tuple {a, b} \preccurlyeq_l \tuple {c, d} \iff \tuple {a \prec_1 b} \lor \paren {a = b \land c \preccurlyeq_2 d}$
Let $\tuple {a, b} \in T$ be a minimal element of $T$.
:$\forall \tuple {x, y} \in T: \tuple {x, y} \preccurlyeq_l \tuple {a, b} \implies \tuple {x, y} = \tuple {a, b}$
Suppose $a \ne 0$.
Then:
:$\tuple {0, b} \preccurlyeq_l \tuple {a, b}$ but $\tuple {0, b} \ne \tuple {a, b}$
Similarly, suppose $b \ne 0$.
Then:
:$\tuple {a, 0} \preccurlyeq_l \tuple {a, b}$ but $\tuple {a, 0} \ne \tuple {a, b}$
Hence if $\tuple {a, b}$ is a minimal element of $T$ it follows that:
:$\tuple {a, b} = \tuple {0, 0}$
Next we note that:
:$\forall \tuple {c, d} \in T: 0 \le c, 0 \le d$
and so:
:$\forall \tuple {c, d} \in T: \tuple {0, 0} \preccurlyeq_l \tuple {c, d}$
Hence by definition $\tuple {0, 0}$ is a smallest element of $T$.
By Smallest Element is Unique, $\tuple {0, 0}$ is then the smallest element of $T$.
 $T$ has a maximal element $\tuple {u, v}$.
Then we have:
:$\tuple {u, v} \preccurlyeq_l \tuple {x, y} \implies \tuple {u, v} = \tuple {x, y}$
and in particular:
:$\tuple {u, v} \preccurlyeq_l \tuple {x, v} \implies \tuple {u, v} = \tuple {x, v}$
That is:
:$u \le x \implies u = x$
But we have that:
:$u \in \hointr 0 1$
and so:
:$u < 1$
So take:
:$x = \dfrac u 1 + \dfrac 1 1 = \dfrac {u + 1} 2$
Hence from Mediant is Between:
:$u < x$
and so it is not the case that $u \le x \implies u = x$.
Hence it is not the case that:
:$\tuple {u, v} \preccurlyeq_l \tuple {x, y} \implies \tuple {u, v} = \tuple {x, y}$
and so $\tuple {u, v}$ is not a maximal element of $T$.
So by Proof by Contradiction it follows that there is no maximal element of $T$.
 $\tuple {u, v}$ is a greatest element of $T$.
From Greatest Element is Maximal, $\tuple {u, v}$ would have to be a maximal element of $T$.
But from above $T$ has no maximal element.
Hence by Proof by Contradiction it follows that there is no greatest element of $T$.
Hence the result.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $31$
Category:Examples of Order Products