nyuuzyou/wikis huggingface dataset:
Lateral Surface Area of Circular Cylinder/Height

== Theorem ==
Let $\CC$ be a circular cylinder such that:
:the bases of $\CC$ are circles of radius $r$
:the height of $\CC$ is $h$
:the inclination of the generatrices of $\CC$ to the base of $\CC$ is $\theta$.
The area $\AA$ of the lateral surface of $\CC$ is given by the formula:
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Circular Cylinder of Radius $r$ and Slant Height $l$: $4.34$
* : $\S 7$: Geometric Formulas: Circular Cylinder of Radius $r$ and Slant Height $l$: $7.34.$
Category:Lateral Surface Area of Circular Cylinder
Lateral Surface Area of Circular Cylinder/Slant Height

== Theorem ==
Let $\CC$ be a circular cylinder such that:
:the bases of $\CC$ are circles of radius $r$
:the slant height of $\CC$ is $l$.
The area $\AA$ of the lateral surface of $\CC$ is given by the formula:
:$\AA = 2 \pi r l$
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Circular Cylinder of Radius $r$ and Slant Height $l$: $4.34$
* : $\S 7$: Geometric Formulas: Circular Cylinder of Radius $r$ and Slant Height $l$: $7.34.$
Category:Lateral Surface Area of Circular Cylinder
Lateral Surface Area of Cylinder

== Theorem ==
=== By Height and Cross-Sectional Perimeter ===
=== By Slant Height and Cross-Sectional Perimeter ===
=== By Height and Base Perimeter ===
=== By Slant Height and Base Perimeter ===
== Sources ==
* : $\S 4$: Geometric Formulas: Cylinder of Cross-sectional Area $A$ and Slant Height $l$
* : $\S 7$: Geometric Formulas: Cylinder of Cross-sectional Area $A$ and Slant Height $l$
Category:Lateral Surface Area of Cylinder
Category:Cylinders
Category:Area Formulas
Lateral Surface Area of Cylinder/Height and Base Perimeter

== Theorem ==
Let $\CC$ be a cylinder such that:
:the perimeter of the bases of $\CC$ is $p$
:the height of $\CC$ is $h$
The area $\AA$ of the lateral surface of $\CC$ is given by:
:$\AA = p h$
== Proof ==
== Sources ==
* : $\S 7$: Geometric Formulas: Cylinder of Cross-sectional Area $A$ and Slant Height $l$: $7.36.$
Category:Lateral Surface Area of Cylinder
Lateral Surface Area of Cylinder/Height and Cross-Sectional Perimeter

== Theorem ==
Let $\CC$ be a cylinder such that:
:the perimeter of a cross-section of $\CC$ at right angles to the generatrices is $p$
:the height of $\CC$ is $l$
:the inclination of the generatrices of $\CC$ to the base of $\CC$ is $\theta$.
The area $\AA$ of the lateral surface of $\CC$ is given by:
:$\AA = \dfrac {p h} {\sin \theta} = p h \cosec \theta$
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Cylinder of Cross-sectional Area $A$ and Slant Height $l$: $4.36$
Category:Lateral Surface Area of Cylinder
Lateral Surface Area of Cylinder/Slant Height and Base Perimeter

== Theorem ==
Let $\CC$ be a cylinder such that:
:the perimeter of the base of $\CC$ is $p$
:the slant height of $\CC$ is $l$
:the inclination of the generatrices of $\CC$ to the base of $\CC$ is $\theta$.
The area $\AA$ of the lateral surface of $\CC$ is given by:
:$\AA = p l \sin \theta$
== Proof ==
== Sources ==
* : $\S 7$: Geometric Formulas: Cylinder of Cross-sectional Area $A$ and Slant Height $l$: $7.36.$
Category:Lateral Surface Area of Cylinder
Lateral Surface Area of Cylinder/Slant Height and Cross-Sectional Perimeter

== Theorem ==
Let $\CC$ be a cylinder such that:
:the perimeter of a cross-section of $\CC$ at right angles to the generatrices is $p$
:the slant height of $\CC$ is $l$
The area $\AA$ of the lateral surface of $\CC$ is given by:
:$\AA = p l$
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Cylinder of Cross-sectional Area $A$ and Slant Height $l$: $4.36$
* 
* 
Category:Lateral Surface Area of Cylinder
Lateral Surface Area of Frustum of Right Circular Cone

== Theorem ==
Let $F$ be a frustum of a right circular cone.
The area $\AA$ of the lateral surface of $F$ is given as:
where:
:$r_1$ and $r_2$ are the radii of the bases of $F$
:$h$ is the altitude of $F$.
:$s$ is the slant height of $F$.
:
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Frustum of Right Circular Cone of Radii $a, b$ and Height $h$: $4.43$
* 
* 
* : $\S 7$: Geometric Formulas: Frustum of Right Circular Cone of Radii $a, b$ and Height $h$: $7.43.$
Category:Frustums
Category:Area Formulas
Lateral Surface Area of Right Circular Cone

== Theorem ==
Let $K$ be a right circular cone.
Let $r$ be the radius of the base of $K$.
Let $s$ be the slant height of $K$.
Then the area $\AA$ of the lateral surface of $K$ is given by:
:$\AA = \pi r s$
== Proof ==
== Also presented as ==
== Sources ==
* : $\S 4$: Geometric Formulas: Right Circular Cone of Radius $r$ and Height $h$: $4.38$
* 
* 
* : $\S 7$: Geometric Formulas: Right Circular Cone of Radius $r$ and Height $h$: $7.38.$
Category:Lateral Surface Area of Right Circular Cone
Category:Right Circular Cones
Category:Area Formulas
Lateral Surface Area of Right Circular Cone/Also presented as

== Lateral Surface Area of Right Circular Cone: Also presented as ==
The lateral surface area $\AA$ of a right circular cone can also be expressed in the form:
:$\AA = \pi r \sqrt {r^2 + h^2}$
where:
:$r$ is the radius of the base
:$h$ is the height.
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Right Circular Cone of Radius $r$ and Height $h$: $4.38$
* : $\S 7$: Geometric Formulas: Right Circular Cone of Radius $r$ and Height $h$: $7.38.$
Category:Lateral Surface Area of Right Circular Cone
Lateral Surface Area of Right Circular Cylinder

== Theorem ==
Let $\CC$ be a right circular cylinder:
:whose bases are circles of radius $r$
and
:whose height is $h$.
The area $\AA$ of the lateral surface of $\CC$ is given by the formula:
:$\AA = 2 \pi r h$
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Right Circular Cylinder of Radius $r$ and Height $h$: $4.32$
* : $\S 7$: Geometric Formulas: Right Circular Cylinder of Radius $r$ and Height $h$: $7.32.$
Category:Right Circular Cylinders
Category:Area Formulas
Latin Square/Examples/Order 3

== Example of Latin Square of order $3$ ==
This is an example of a Latin square of order $3$:
$\quad \begin {array} {|ccc|}
\hline
A & B & C \\
C & A & B \\
B & C & A \\
\hline
\end {array}$
== Sources ==
* 
* 
Category:Latin Squares
Latin Square/Examples/Order 4

== Example of Latin Square of order $4$ ==
This is an example of a Latin square of order $4$:
$\quad \begin {array} {|cccc|}
\hline
a & b & c & d \\
c & d & a & b \\
d & c & b & a \\
b & a & d & c \\
\hline
\end {array}$
Category:Examples of Latin Squares
Latin Square is not necessarily Cayley Table of Group

== Theorem ==
While it is true that the Cayley table of a (finite) group is in the form of a Latin square it is not necessarily the case that a Latin square is the Cayley table of a group.
== Proof ==
Proof by Counterexample:
Let $\struct {S, \circ}$ be the algebraic structure defined by the following Cayley table:
:$\begin{array}{c|ccccc}
\circ & e & a & b & c & d \\
\hline
e & e & a & b & c & d \\
a & a & e & d & b & c \\
b & b & c & e & d & a \\
c & c & d & a & e & b \\
d & d & b & c & a & e \\
\end{array}$
By inspection it can be seen that the Cayley table for $\struct {S, \circ}$ is a Latin square.
However, we have that:
:$\paren {a \circ b} \circ c = d \circ c = a$
:$a \circ \paren {b \circ c} = a \circ d = c$
Thus $\circ$ is not an associative operation.
So, by definition, $\struct {S, \circ}$ is not a group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 5$: The Multiplication Table
Category:Cayley Tables
Latitudes on Mercator's Projection get Further Apart

== Theorem ==
Let $\MM$ be a Mercator's projection of Earth's surface such that the cylinder circumscribing Earth intersects it along the equator.
Then equally spaced lines of latitude on Earth map to parallel lines on $\MM$, but the further from the equator they are, the further apart they get..
== Proof ==
== Sources ==
* 
* 
Category:Mercator's Projection
Category:Latitude
Lattice (Ordered Set)/Examples/Divisor Relation

== Examples of Lattice in context of Ordered Sets ==
Let $\struct {\Z_{>0}, \divides}$ denote the order structure consisting of the (strictly) positive integers under the divisor relation $\divides$.
Then $\struct {\Z_{>0}, \divides}$ is a lattice.
== Proof ==
Let $a, b \in \Z_{>0}$.
Let $\gcd \set {a, b}$ denote the greatest common divisor of $a$ and $b$.
Let $\lcm \set {a, b}$ denote the lowest common multiple of $a$ and $b$.
By definition:
:the greatest common divisor of $a$ and $b$ is the infimum of $a$ and $b$ the divisor relation.
:the lowest common multiple of $a$ and $b$ is the supremum of $a$ and $b$ the divisor relation.
The result follows.
== Sources ==
* 
* 
Category:Examples of Orderings
Lattice Homomorphism is Both Meet and Join Semilattice Homomorphism

== Theorem ==
Let $L_1 = \struct{S_1, \vee_1, \wedge_1, \preceq_1}$ and $L_2 = \struct{S_2, \vee_2, \wedge_2, \preceq_2}$ be lattices.
Let $\phi: L_1 \to L_2$ be a lattice homomorphsm between $L_1$ and $L_2$.
Then:
:* $\phi: \struct{S_1, \vee_1, \preceq_1} \to \struct{S_2, \vee_2, \preceq_2}$
:* $\phi: \struct{S_1, \wedge_1, \preceq_1} \to \struct{S_2, \wedge_2, \preceq_2}$ 
are semilattice homomorphisms
== Proof ==
This follows immediately from:
:* Definition:Lattice (Order Theory)
:* Definition:Semilattice
:* Definition:Lattice Homomorphism
:* Definition:Semilattice Homomorphism
Category:Lattice Homomorphisms
Lattice Homomorphism is Order-Preserving

== Theorem ==
Let $L_1 = \struct{S_1, \vee_1, \wedge_1, \preceq_1}$ and $L_2 = \struct{S_2, \vee_2, \wedge_2, \preceq_2}$ be lattices.
Let $\phi: L_1 \to L_2$ be a lattice homomorphsm between $L_1$ and $L_2$.
Then:
:$\phi: \struct{S_1, \preceq_1} \to \struct{S_2, \preceq_2}$ is order-preserving
== Proof ==
From Lattice Homomorphism is Both Meet and Join Semilattice Homomorphism:
:$\phi: \struct{S_1, \vee_1, \preceq_1} \to \struct{S_2, \vee_2, \preceq_2}$ is a semilattice homomorphism
From Semilattice Homomorphism is Order-Preserving:
:$\phi: \struct{S_1, \preceq_1} \to \struct{S_2, \preceq_2}$ is order-preserving
Category:Lattice Homomorphisms
Lattice Ordering/Examples/Ancestry

== Example of Ordering which is not Lattice Ordering ==
Recall the partial ordering on the set of people:
$D$ is not a lattice ordering.
== Proof ==
Let $a$ and $b$ be siblings.
Let $f$ and $m$ denote the father and mother of both $a$ and $b$.
Then both $f$ and $m$ are an upper bound of $\set {a, b}$.
But it is not necessarily the case that $f$ and $m$ share a common ancestor, unless you grant that either:
:every two people are somehow descended from the same proto-ancestor, maybe just the first single self-replicating chemical system that may be classified as life
or:
:both $f$ and $m$ are descended from Adam and Eve, and Eve of course is a descendant of Adam as she came from one of his ribs.
But be that as it may, consider the set:
:$S = \set {\text {President Buchanan}, \text {President Arthur} }$
Then as $\text {President Buchanan}$ never married and so had no descendants, $S$ has no lower bounds.
Hence $\set {\text {President Buchanan}, \text {President Arthur} }$ has no imfimum.
Hence, by definition, $D$ is not a lattice ordering. 
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Example $14.1$
Category:Examples of Lattice Orderings
Lattice Ordering/Examples/Parallel Lines

== Example of Ordering which is not Lattice Ordering ==
Recall the partial ordering on the set of straight lines:
$S$ is not a lattice ordering.
== Proof ==
Two straight lines admit a supremum and infimum when parallel.
Hence, by definition, $S$ is not a lattice ordering. 
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Example $14.2$
Category:Examples of Lattice Orderings
Lattice and Ring of Real-Valued Functions forms Ordered Ring

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\R$ denote the real number line.
Let $\struct {\R^S, +, \times}$ be the ring of real-valued functions from $S$ to $\R$.
Let $\struct {\R^S, \vee, \wedge, \le}$ be the lattice of real-valued functions from $S$ to $\R$.
Then:
:$\struct {\R^S, +, \times, \le}$ is an ordered ring
== Proof ==
From Structure Induced by Group Operation is Group:
:the zero of $\struct {\R^S, +, \times}$ is the constant mapping $0_{R^S} : S \to R$ defined by:
::$\forall s \in S : \map {0_{R^S}} s = 0_R$
It needs to be shown that the order $\le$ on the lattice of real-valued functions satisies the ring compatible ordering axioms:
=== $\paren {\text {OR} 1}$ ===
Let $f, g, h \in \R^S$.
Let $f \le g$.
Hence:
:$\forall s \in S: \map f s \le \map g s$
We have:
By definition of lattice of real-valued functions:
:$\paren {f + h} \le \paren {g + h}$
Since $f, g$ and $h$ were arbitrary:
:$\forall f, g, h \in \R^S : f \le g \implies \paren {f + h} \le \paren {g + h}$
=== $\paren {\text {OR} 2}$ ===
Let $f, g \in \R^S$.
Let $0_{\R^S} \le f, 0_{\R^S} \le g$.
Hence:
:$\forall s \in S: 0 \le \map f s, 0 \le \map g s$
We have:
By definition of lattice of real-valued functions:
:$0_{\R^S} \le f g$
Since $f$ and $g$ were arbitrary:
:$\forall f, g \in \R^S : 0_{\R^S} \le f, 0_{\R^S} \le g \implies 0_{\R^S} \le f g$
Thus the ring compatible ordering axioms are seen to hold for the order $\le$ on the lattice of real-valued functions.
== Sources ==
* : Chapter $1$: Functions on a Topological Space, $\S 1.2$
Category:Real-Valued Functions
Category:Ordered Rings
Lattice is Complete iff it Admits All Suprema

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Then $\struct {S, \preceq}$ is a complete lattice 
:$\forall X \subseteq S: X$ admits a supremum.
== Proof ==
=== Sufficient Condition ===
Let $\struct {S, \preceq}$ be a complete lattice.
Thus by definition of complete lattice:
:$\forall X \subseteq S: X$ admits a supremum.
=== Necessary Condition ===
Let $\struct {S, \preceq}$ be such that
:$\forall X \subseteq S: X$ admits a supremum.
We will prove that
:$\forall X \subseteq S: X$ admits both a supremum and an infimum.
Let $X \subseteq S$.
Thus by assumption:
:$X$ admits a supremum.
Define:
:$Y := \leftset {s \in S: s}$ is a lower bound for $\rightset X$
By assumption:
:$Y$ admits a supremum
We will prove that
:$\sup Y$ is a lower bound for $X$
Let $x \in X$.
By definition of lower bound:
:$x$ is an upper bound for $Y$
By definition of supremum:
:$\sup Y \preceq x$
Thus by definition:
:$\sup Y$ is a lower bound for $X$
We will prove that
:$\forall s \in S: s$ is a lower bound for $X \implies s \preceq \sup Y$
Let $s \in S$ such that:
:$s$ is a lower bound for $X$.
By definition of $Y$:
:$s \in Y$
By definition of supremum:
:$\sup Y$ is an upper bound for $Y$.
Thus by definition of upper bound:
:$s \preceq \sup Y$
Thus by definition
:$X$ admits an infimum
By definition
:$\struct {S, \preceq}$ is a lattice.
Thus the result follows by definition of complete lattice.
== Sources ==
* 
* 
Category:Complete Lattices
Lattice of Bounded Continuous Functions is Sublattice of Continuous Real-Valued Functions

== Theorem ==
Let $\struct {S, \tau }$ be a topological space.
Let $\R$ denote the real number line.
Let $\struct{\map C {S, R}, \vee, \wedge, \le}$ be the lattice of continuous real-valued functions from $S$.
Let $\struct{\map {C^*} {S, R}, \vee, \wedge, \le}$ be the lattice of bounded continuous real-valued functions from $S$.
Then:
:$\struct{\map {C^*} {S, R}, \vee, \wedge, \le}$ is a sublattice of $\struct{\map C {S, R}, \vee, \wedge, \le}$
== Proof ==
To show that $\struct{\map {C^*} {S, R}, \vee, \wedge, \le}$ is a sublattice of $\struct {\map C {S, R}, \vee, \wedge, \le}$ it is sufficient to show that $\map {C^*} {S, R}$ is closed under $\vee$ and $\wedge$.
From Maximum Rule for Bounded Continuous Real-Valued Functions:
:$\map {C^*} {S, R}$ is closed under $\vee$
From Minimum Rule for Bounded Continuous Real-Valued Functions:
:$\map {C^*} {S, R}$ is closed under $\wedge$
Category:Lattices of Bounded Continuous Real-Valued Functions
Lattice of Continuous Functions is Sublattice of All Real-Valued Functions

== Theorem ==
Let $\struct {S, \tau }$ be a topological space.
Let $\R$ denote the real number line.
Let $\struct {\R^S, \vee, \wedge, \le}$ be the lattice of real-valued functions from $S$.
Let $\struct{\map C {S, R}, \vee, \wedge, \le}$ be the lattice of continuous real-valued functions from $S$.
Then:
:$\struct{\map C {S, R}, \vee, \wedge, \le}$ is a sublattice of $\struct {\R^S, \vee, \wedge, \le}$
== Proof ==
To show that $\struct{\map C {S, R}, \vee, \wedge, \le}$ is a sublattice of $\struct {\R^S, \vee, \wedge, \le}$ it is sufficient to show that $\map C {S, R}$ is closed under $\vee$ and $\wedge$.
From Maximum Rule for Continuous Real-Valued Functions:
:$\map C {S, R}$ is closed under $\vee$
From Minimum Rule for Continuous Real-Valued Functions:
:$\map C {S, R}$ is closed under $\wedge$
Category:Lattices of Continuous Real-Valued Functions
Lattice of Power Set is Algebraic

== Theorem ==
Let $X$ be a set.
Let $L = \struct {\powerset X, \cup, \cap, \preceq}$ be the lattice of power set of $X$
where $\mathord \preceq = \mathord \subseteq \cap \paren {\powerset X \times \powerset X}$
Then $L$ is algebraic.
== Proof ==
We will prove that
:$\forall x \in \powerset X: x^{\mathrm{compact} }$ is directed.
Let $x \in \powerset X$.
By Empty Set is Bottom of Lattice of Power Set:
:$\O = \bot$
where $\bot$ denotes the bottom of $L$.
By Bottom is Way Below Any Element:
:$\bot \ll \bot$
where $\ll$ is the way below relation.
By definition:
:$\bot$ is compact.
By definition of the smallest element:
:$\bot \preceq x$
By definition of compact closure:
:$\bot \in x^{\mathrm{compact} }$
By definition:
:$x^{\mathrm{compact} }$ is non-empty.
Thus by Non-Empty Compact Closure is Directed:
:$x^{\mathrm{compact} }$ is directed.
By Power Set is Complete Lattice:
:$L$ is complete lattice.
Thus by definition of complete lattice:
:$L$ is up-complete.
It remains to prove that:
:$L$ satisfies the axiom of $K$-approximation.
Let $x \in \powerset X$.
We will prove that:
:$\forall a \in \powerset X: a$ is upper bound for $x^{\mathrm{compact} } \implies x \preceq a$
Let $a \in \powerset X$ such that:
:$a$ is upper bound for $x^{\mathrm{compact} }$
We will prove that:
:$x \subseteq a$
Let $t \in x$.
By definition of power set:
:$x \subseteq X$
By definitions of subset and singleton:
:$\set t \subseteq X$ and $\set t \subseteq x$
By definition of power set:
:$\set t \in \powerset X$
By Singleton is Finite:
:$\set t$ is finite.
By definition of $\operatorname {Fin}$:
:$\set t \in \map {\operatorname {Fin} } x$
where $\map {\operatorname {Fin} } x$ denotes the set of all finite subset of $x$.
By Compact Closure is Set of Finite Subsets in Lattice of Power Set:
:$\set t \in x^{\mathrm{compact} }$
By definition of upper bound:
:$\set t \preceq a$
By definition of $\preceq$:
:$\set t \subseteq a$
Thus by definitions of subset and singleton:
:$t \in a$
Thus by definition of $\preceq$:
:$x \preceq a$
By definition of compact closure:
:$\forall y \in x^{\mathrm{compact} }: y \preceq x$
By definition of upper bound:
:$x$ is upper bound for $x^{\mathrm{compact} }$
Thus by definition of supremum:
:$x = \map \sup {x^{\mathrm{compact} } }$
== Sources ==
* 
* 
Category:Continuous Lattices
Category:Power Set
Lattice of Power Set is Arithmetic

== Theorem ==
Let $X$ be a set.
Let $P = \struct {\powerset X, \cup, \cap, \subseteq}$ be a lattice of power set.
Then $P$ is an arithmetic ordered set.
== Proof ==
Define $C = \struct {\map K P, \preceq}$ as an ordered subset of $P$
where $\map K P$ denotes the compact subset of $P$.
Thus by Lattice of Power Set is Algebraic:
:$P$ is algebraic.
It remains to prove that:
:$\map K P$ is meet closed.
Let $x, y \in \map K P$.
By definition of compact subset:
:$x$ is compact.
By Element is Finite iff Element is Compact in Lattice of Power Set
:$x$ is finite.
By Intersection is Subset:
:$x \cap y \subseteq x$
By Subset of Finite Set is Finite:
:$x \cap y$ is finite.
By Element is Finite iff Element is Compact in Lattice of Power Set
:$x \cap y$ is compact.
Thus by definition of compact subset:
:$x \cap y \in \map K P$
== Sources ==
* 
* 
Category:Continuous Lattices
Lattice of Real-Valued Functions forms Distributive Lattice

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\R$ denote the real number line.
Let $\struct {\R^S, \vee, \wedge, \le}$ be the lattice of real-valued functions from $S$ where:
:$\forall f, g \in \R^S : f \vee g : S \to \R$ is defined by:
::$\forall s \in S : \map {\paren{f \vee g}} s = \max \set{\map f s, \map g s}$
:$\forall f, g \in \R^S : f \wedge g : S \to \R$ is defined by:
::$\forall s \in S : \map {\paren{f \wedge g}} s = \min \set{\map f s, \map g s}$
:$\forall f, g \in \R^S : f \le g$ :
::$\forall x \in S : \map f x \le \map g x$
Then:
:$\struct {\R^S, \vee, \wedge, \le}$ is a distributive lattice.
== Proof ==
From Usual Ordering on Real Numbers is Total Ordering:
:$\struct {\R, \le}$ is totally ordered
where $\le$ denotes the usual ordering on $\R$.
From Totally Ordered Set is Lattice
:$\struct{\R, \le}$ is a lattice.
By definition of join:
:$\forall x, y \in \R : x \vee y = \sup \set {x, y}$
where $x \vee y$ denotes the join of $x$ and $y$ in $\struct {\R, \le}$.
From Max Operation Yields Supremum of Parameters:
:$\forall x, y \in \R : \map \max {x, y} = \sup \set {x, y} = x \vee y$
By definition of meet:
:$\forall x, y \in \R : x \wedge y = \inf \set {x, y}$
where $x \wedge y$ denotes the meet of $x$ and $y$ in $\struct {\R, \le}$.
From Min Operation Yields Infimum of Parameters:
:$\forall x, y \in \R : \map \min {x, y} = \inf \set {x, y} = x \wedge y$
From Structure Induced by Lattice Operations is Lattice:
:$\struct {\R^S, \vee, \wedge, \le}$ is a lattice.
From Max and Min Operations are Distributive over Each Other:
:$\struct {\R, \vee, \wedge, \le}$ is a distributive lattice.
From Structure Induced by Distributive Operation is Distributive:
:$\struct {\R^S, \vee, \wedge, \le}$ is a distributive lattice.
Category:Distributive Lattices
Category:Real-Valued Functions
Lattice of Subgroups forming Totally Ordered Set is Indecomposable

== Theorem ==
Let $\struct {G, \circ}$ be a Group.
Let $\mathbb G$ be the set of subgroups of $G$.
Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by the subset ordering on $\mathbb G$.
Let $\struct {\mathbb G, \subseteq}$ be totally ordered.
Then $\struct {G, \circ}$ is an indecomposable group.
== Proof ==
First we note that from Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ is indeed a complete lattice.
As postulated, let $\struct {\mathbb G, \subseteq}$ be totally ordered.
 there exists a decomposition of $\struct {G, \circ}$.
Then:
:$\exists H, K \in \mathbb G: H \cap K = \set e$
where $\struct {H, \circ}$ and $\struct {K, \circ}$ are non-trivial normal subgroups of $\struct {G, \circ}$.
But then neither $H \subseteq K$ nor $K \subseteq H$.
That is, $\struct {\mathbb G, \subseteq}$ is not totally ordered.
Hence the result by Proof by Contradiction.
== Also see ==
* Indecomposable Lattice of Subgroups does not necessarily form Totally Ordered Set: the converse does not hold.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.16 \ \text {(a)}$
Category:Decomposable Groups
Laurent Expansion of Function about Pole

== Theorem ==
Let $f$ be a complex function with a pole at $z_0 \in \C$ of order $k$.
The Laurent expansion of $f$ about $z_0$ can be expressed as:
:$\map f z = \ds \sum_{n \mathop = -k}^\infty a_n \paren {z - z_0}^n$
== Proof ==
== Sources ==
* 
* 
Category:Laurent Series
Category:Poles (Complex Analysis)
Laurent Expansion of Isolated Essential Singularity

== Theorem ==
Let $f$ be a complex function with an essential singularity at $z_0 \in \C$.
Let $z_0$ also be an isolated singularity.
Then there exists a Laurent expansion for $f$ with a principal part with infinitely many terms.
== Proof ==
== Examples ==
== Sources ==
* 
* 
Category:Laurent Expansion of Isolated Essential Singularity
Category:Essential Singularities
Category:Isolated Singularities
Category:Laurent Series