nyuuzyou/wikis huggingface dataset:
Join Semilattice Ordered Subset Not Always Subsemilattice

== Theorem ==
Let $\struct {S, \circ}$ be a semilattice.
Let $\preceq$ be the ordering on $S$ defined by:
:$a \preceq b \iff \paren {a \circ b} = b$
Let $T$ be a subset of $S$.
Let the ordered subset $\struct {T, \preceq \restriction_T}$ be a join semilattice.
Let $\vee$ be the binary operation on $T$ defined by:
:for all $a, b \in T$: $a \vee b$ is the join of $a$ and $b$ $\preceq$.
Then:
:$\struct {T, \vee}$ may not be a subsemilattice of $\struct {S, \circ}$.
== Proof ==
Let $S = \set {p, q, r, s}$.
Let $\circ: S \times S \to S$ be defined by:
:$p \circ q = q \circ p = r$
:$p \circ r = r \circ p = r$
:$q \circ r = r \circ q = r$
:$p \circ s = s \circ p = s$
:$q \circ s = s \circ q = s$
:$r \circ s = s \circ r = s$
:$p \circ p = p$
:$q \circ q = q$
:$r \circ r = r$
:$s \circ s = s$
Then a manual check shows that $\struct {S, \circ}$ satisfies:
:
:
:
:
So $\struct {S, \circ}$ is a semilattice with ordering:
:${\preceq} = \set {\tuple {p, p}, \tuple {q, q}, \tuple {r, r}, \tuple {s, s}, \tuple {p, r}, \tuple {q,r}, \tuple {p, s}, \tuple {q,s}, \tuple {r, s} }$
Let $T = \set {p, q, s}$.
Then:
:${\preceq \restriction_T} = \set {\tuple {p, p}, \tuple {q, q}, \tuple {s, s}, \tuple {p, s}, \tuple {q, s} }$
Then a manual check shows that all joins exist in $\struct {T, \preceq \restriction_T}$:
:$p \vee q = q \vee p = s$
:$p \vee s = s \vee p = s$
:$q \vee s = s \vee q = s$
:$p \vee p = p$
:$q \vee q = q$
:$s \vee s = s$
So $\struct {T, \preceq \restriction_T}$ is a join semilattice $\preceq \restriction_T$.
Now:
:$p \vee q = s \ne r = p \circ q$
So $\vee$ is not the restriction of $\circ$ to $T$.
That is, $\struct {T, \vee}$ is not a subsemilattice of $\struct {S, \circ}$ by definition.
== Sources ==
* : Chapter $\text I$: Preliminaries, Definition $1.3$
Category:Subsemilattices
Join Semilattice has Smallest Element iff has Identity

== Theorem ==
Let $\struct{S, \vee, \preceq}$ be a join semilattice.
Let $s \in S$.
Then:
:$s$ is the smallest element of $S$
:$s$ is the identity in $\struct{S, \vee}$.
== Proof ==
By definition of smallest element:
:$s$ is the smallest element of $S$ 
:
:$\forall t \in S : s \preceq t$
We have:
By definition of identity:
:$s$ is the identity in $\struct{S, \vee}$
:$\forall t \in S :s \vee t = t = t \vee s$
The result follows.
Category:Join Semilattices
Join Semilattice is Dual to Meet Semilattice

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
The following are dual statements:
:$\struct{S, \vee, \preceq}$ is a join semilattice, where $\vee$ is the join for all $a,b \in S$
:$\struct{S, \wedge, \preceq}$ is a meet semilattice, where $\wedge$ is the meet for all $a,b \in S$
== Proof ==
By definition of join semilattice:
:$\struct{S, \vee, \preceq}$ is a join semilattice 
:
:$\forall a,b \in S : a \vee b \in S$, where $a \vee b$ is the join of $a$ and $b$
The dual of this statement is:
:$\forall a,b \in S : a \wedge b \in S$, where $a \wedge b$ is the meet of $a$ and $b$
by Dual Pairs (Order Theory).
By definition of meet semilattice:
:$\struct{S, \wedge, \preceq}$ is a meet semilattice 
== Also see ==
* Duality Principle (Order Theory)
Category:Join Semilattices
Category:Meet Semilattices
Category:Dual Pairs (Order Theory)
Join Semilattice is Ordered Structure

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\struct {S, \vee, \preceq}$ is an ordered structure.
That is, $\preceq$ is compatible with $\vee$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Meet Semilattice is Ordered Structure
Category:Join Semilattices
Category:Ordered Structures
Category:Join Semilattice is Ordered Structure
Join Semilattice is Ordered Structure/Proof 1

== Theorem ==
== Proof ==
For $\struct {S, \vee, \preceq}$ to be an ordered structure is equivalent to, for all $a, b, c \in S$:
:$a \preceq b \implies a \vee c \preceq b \vee c$
:$a \preceq b \implies c \vee a \preceq c \vee b$
Since Join is Commutative, it suffices to prove the first of these implications.
By definition of join:
:$a \vee c = \sup \set {a, c}$
where $\sup$ denotes supremum.
By Join Succeeds Operands:
:$b \preceq b \vee c$
:$c \preceq b \vee c$
Now also $a \preceq b$, and by transitivity of $\preceq$ we find that:
:$a \preceq b \vee c$
Thus $b \vee c$ is an upper bound for $\set {a, c}$.
Hence:
:$a \vee c \preceq b \vee c$
by definition of supremum.
Category:Join Semilattice is Ordered Structure
Join Semilattice is Ordered Structure/Proof 2

== Theorem ==
== Proof ==
Let $a, b, c \in S$.
Let $a \preceq b$.
By the definition of join semilattice:
:$a \vee b = b$
Thus:
:$\paren {a \vee b} \vee c = b \vee c$
Since $\vee$ is associative, commutative, and idempotent:
:$\paren {a \vee c} \vee \paren {b \vee c} = b \vee c$
Therefore, $a \vee c \preceq b \vee c$.
From Join is Commutative, we conclude that:
:$c \vee a \preceq c \vee b$
Category:Join Semilattice is Ordered Structure
Join Semilattice is Semilattice

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\struct {S, \vee}$ is a semilattice.
== Proof ==
Recall the definition of join semilattice:
By definition of join semilattice, $\vee$ is closed.
The other three defining properties for a semilattice follow respectively from:
:Join is Commutative
:Join is Associative
:Join is Idempotent
Hence $\struct {S, \vee}$ is a semilattice.
== Also see ==
* Meet Semilattice is Semilattice
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.22 \ \text {(a)}$
* : $\text {II} \ \S 1$ Example $(7)$
* : Chapter $\text I$: Preliminaries, Definition $1.3$
Category:Join Semilattices
Category:Semilattices
Join Succeeds Operands

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $a, b \in S$ admit a join $a \vee b \in S$.
Then:
:$a \preceq a \vee b$
:$b \preceq a \vee b$
That is, $a \vee b$ succeeds its operands $a$ and $b$.
== Proof ==
By definition of join:
:$a \vee b = \sup \set {a, b}$
where $\sup$ denotes supremum.
Since a supremum is an upper bound:
:$a \preceq \sup \set {a, b}$
:$b \preceq \sup \set {a, b}$
as desired.
== Also see ==
* Meet Precedes Operands
Category:Join Operation
Join and Meet in Inclusion Ordered Set of Topology

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $L = \left({\tau, \preceq}\right)$ be an inclusion ordered set of $\tau$.
Let $X, Y \in \tau$.
Then $X \vee Y = X \cup Y$ and $X \wedge Y = X \cap Y$
== Proof ==
By definition of topological space:
:$X \cup Y, X \cap Y \in \tau$
Thus by Join in Inclusion Ordered Set ans Meet in Inclusion Ordered Set:
:the result holds.
== Sources ==
* 
Category:Join and Meet
Category:Topology
Join in Inclusion Ordered Set

== Theorem ==
Let $P = \left({X, \subseteq}\right)$ be an inclusion ordered set.
Let $A, B \in X$ such that
:$A \cup B \in X$
Then $A \vee B = A \cup B$
== Proof ==
By Set is Subset of Union:
:$A \subseteq A \cup B$ and $B \subseteq A \cup B$
By definition:
:$A \cup B$ is upper bound for $\left\{ {A, B}\right\}$
We will prove that
:$\forall C \in X: C$ is upper bound for $\left\{ {A, B}\right\} \implies A \cup B \subseteq C$
Let $C \in X$ such that
:$C$ is upper bound for $\left\{ {A, B}\right\}$.
By definition of upper bound:
:$A \subseteq C$ and $B \subseteq C$
Thus by Union is Smallest Superset:
:$A \cup B \subseteq C$
By definition of supremum:
:$\sup \left\{ {A, B}\right\} = A \cup B$
Thus by definition of join:
:$A \vee B = A \cup B$
== Sources ==
* 
Category:Join and Meet
Join is Associative

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\vee$ is associative.
== Proof ==
Let $a, b, c \in S$ be arbitrary.
Then:
Hence the result.
== Also see ==
* Meet is Associative
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.23 \ \text {(a)}$
Category:Join Operation
Category:Examples of Associative Operations
Join is Commutative

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\vee$ is commutative.
== Proof ==
Let $a, b \in S$ be arbitrary.
Then:
Hence the result.
== Also see ==
* Meet is Commutative
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.23 \ \text {(a)}$
Category:Join Operation
Category:Examples of Commutative Operations
Join is Dual to Meet

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $a, b, c \in S$.
The following are dual statements:
:$c = a \vee b$, the join of $a$ and $b$
:$c = a \wedge b$ the meet of $a$ and $b$
== Proof ==
By definition of join, $c = a \vee b$ :
:$c = \sup \set {a, b}$
where $\sup$ denotes supremum.
The dual of this statement is:
:$c = \inf \set {a, b}$
where $\inf$ denotes infimum, by Dual Pairs (Order Theory).
By definition of meet, this means $c = a \wedge b$.
== Also see ==
* Duality Principle (Order Theory)
Category:Join and Meet
Category:Dual Pairs (Order Theory)
Join is Idempotent

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\vee$ is idempotent.
== Proof ==
Let $a \in S$ be arbitrary.
Then:
Hence the result.
== Also see ==
* Meet is Idempotent
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.23 \ \text {(a)}$
Category:Join Operation
Category:Examples of Idempotence‎
Join is Way Below if Operands are Way Below

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Let $x, y, z \in S$ such that
:$x \ll z$ and $y \ll z$
where $\ll$ denotes the way below relation.
Then
:$x \vee y \ll z$
== Proof ==
Let $D$ be a directed subset of $S$ such that
:$D$ admits a supremum
and
:$z \preceq \sup D$
By definition of way below relation:
:$\exists d_1 \in D: x \preceq d_1$
and
:$\exists d_2 \in D: y \preceq d_2$
By definition of directed subset:
:$\exists d \in D: d_1 \preceq d$ and $d_2 \preceq d$
By definition of transitivity:
:$x \preceq d$ and $y \preceq d$
Thus by definition of supremum:
:$x \vee y \preceq d$
Thus by definition way below relation:
:$x \vee y \ll z$
== Sources ==
* 
* 
Category:Way Below Relation
Join of Compact Spaces is Compact

== Theorem ==
Let $A, B$ be compact topological spaces.
Then:
:$A \ast B$ is compact
where $A \ast B$ denotes the join of $A$ and $B$.
== Proof ==
By Closed Real Interval is Compact:
:$\closedint 0 1$
is compact.
Thus, by Topological Product of Compact Spaces:
:$A \times B \times \closedint 0 1$
is compact.
Therefore, by Quotient Space of Compact Space is Compact:
:$A \ast B = \paren {A \times B \times \closedint 0 1} / \RR$
is compact.
Category:Compact Spaces
Category:Quotient Spaces
Join of Finite Sub-Sigma-Algebras Generates Join of Finite Partitions

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\BB, \CC \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\map \xi {\BB \vee \CC} = \map \xi \BB \vee \map \xi \CC$
where:
:$\map\xi\cdot$ denotes the generated finite partition
:$\vee$ on the denotes the join of finite sub-$\sigma$-algebras
:$\vee$ on the denotes the join of finite partitions
== Proof ==
== Sources ==
* $4.1$: Partitions and Subalgebras
Category:Probability Theory
Category:Ergodic Theory
Join of Sets of Integer Multiples is Set of Integer Multiples of GCD

== Theorem ==
Let $m, n \in \Z$.
Let $m \Z$ denote the set of integer multiples of $m$
Let $r \in \Z$ such that:
:$m \Z \subseteq r \Z$
and:
:$n \Z \subseteq r \Z$
Then:
:$\gcd \set {m, n} \Z \subseteq r \Z$
where $\gcd$ denotes greatest common divisor.
== Proof ==
From Set of Integer Multiples is Integral Ideal, each of $m \Z$, $n \Z$, $r \Z$ and $\gcd \set {m, n} \Z$ are integral ideals.
Let $c \in \gcd \set {m, n} \Z$.
By definition of integral ideal:
:$\gcd \set {m, n} \divides c$
By Set of Integer Combinations equals Set of Multiples of GCD:
:$\exists x, y \in \Z: c = x m + y n$
But as $m \Z \subseteq r \Z$ and $n \Z \subseteq r \Z$:
:$m \in r \Z$ and $n \in \r Z$
Thus by definition of integral ideal:
:$x m + y n \in r \Z$
So:
:$c \in \gcd \set {m, n} \Z \implies c \in r \Z$
and the result follows by definition of subset.
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $11$
Category:Sets of Integer Multiples
Category:Greatest Common Divisor
Join of Subgroups is Group Generated by Union

== Theorem ==
Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Let $S$ be the set of words of $H \cup K$.
Then $S$ is a subgroup of $K$ such that:
:$S = \gen {H \cup K} = H \vee K$
where $H \vee K$ denotes the join of $H$ and $K$.
== Proof ==
By definition, the set of words in $H \cup K$ is:
:$S = \map W {H \cup K} := \set {s_1 \circ s_2 \circ \cdots \circ s_n: n \in \N_{>0}: s_i \in H \cup K 1 \le i \le n}$
Let $h \in H$.
Then setting $n = 1$ in the above definition and letting $s_1 = h$ it follows that $H \subseteq S$.
Similarly it is seen that $K \subseteq S$
So from Union is Smallest Superset it follows that $H \cup K \subseteq S$.
From Set of Words Generates Group it follows that $S$ is the subgroup of $G$ generated by $H \cup K$.
== Sources ==
* : $\S 1.9$: Theorem $19$
Category:Group Theory
Join with Complement is Top

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 2.
Then:
:$\exists \top \in S: \forall a \in S: a \vee \neg a = \top$
where $\vee$ denotes the join operation in $S$.
This element $\top$ is unique for any given $S$, and is named top.
== Proof ==
Let $\exists r, s \in S: r \vee \neg r = a, \ s \vee \neg s = b$
Then:
Thus, whatever $r$ and $s$ may be:
:$r \vee \neg r = s \vee \neg s$
This unique element can be assigned the symbol $\top$ and named top as required.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.8$: Problems: $1 \ \text{B}$
Category:Boolean Algebras
Joining Arcs makes Another Arc

== Theorem ==
Let $T$ be a topological space.
Let $\mathbb I \subseteq \R$ be the closed unit interval $\closedint 0 1$.
Let $a, b, c$ be three distinct points of $T$.
Let $f, g: \mathbb I \to T$ be arcs in $T$ from $a$ to $b$ and from $b$ to $c$ respectively.
Let $h: \mathbb I \to T$ be the mapping given by:
:$\map h x = \begin{cases}
\map f {2 x} & : x \in \closedint 0 {\dfrac 1 2} \\
\map g {2 x - 1} & : x \in \closedint {\dfrac 1 2} 1
\end{cases}$
Then either:
:$h$ is an arc in $T$
or
:There exists some restriction of $h$ which, possibly after reparametrisation, is an arc in $T$.
== Proof ==
From Arc in Topological Space is Path, $f$ and $g$ are also paths in $T$.
So by Joining Paths makes Another Path it follows that $h$ is a path in $T$.
Now if $\Img f \cap \Img g = b$ it can be seen that:
:$\forall x \in \Img h : x = \begin{cases}
\map f y & \text {for some $y \in \closedint 0 {\dfrac 1 2}$}, \text{ or} \\
\map g z & \text {for some $z \in \closedint {\dfrac 1 2} 1$}
\end{cases}$
and it follows that $h$ is an injection, and therefore an arc.
On the other hand, suppose:
:$\exists y \in \closedint 0 {\dfrac 1 2}: \exists z \in \closedint {\dfrac 1 2} 1: \map f y = \map g z$
such that $\map f y \ne b$.
Category:Topology
Joining Paths makes Another Path

== Theorem ==
Let $T$ be a topological space.
Let $I \subseteq \R$ be the closed real interval $\closedint 0 1$.
Let $f, g: I \to T$ be paths in $T$ from $a$ to $b$ and from $b$ to $c$ respectively.
Let $h: I \to T$ be the mapping given by:
$\map h x = \begin{cases}
\map f {2x} & : x \in \closedint 0 {\dfrac 1 2} \\
\map g {2x - 1} & : x \in \closedint {\dfrac 1 2} 1
\end{cases}$
Then $h$ is a path in $T$.
== Proof ==
First we see that $h$ is well-defined, because on $\closedint 0 {\dfrac 1 2} \cap \closedint {\dfrac 1 2} 1 = \set {\dfrac 1 2}$ we have $\map f 1 = b = \map g 0$.
Now $\mathbin h {\restriction_{\closedint 0 {\frac 1 2} } } \mathop = f \circ k$ where $k: \closedint 0 {\dfrac 1 2} \to \closedint 0 1$ is given by $\map k x = 2 x$.
So by Composite of Continuous Mappings is Continuous, $\mathbin h {\restriction_{\closedint 0 {\frac 1 2} } }$ is continuous.
Similarly, $\mathbin h {\restriction_{\closedint {\frac 1 2} 1} }$ is continuous.
By Pasting Lemma for Finite Union of Closed Sets, it follows that $h$ is continuous.
Finally, $\map h 0 = \map f 0 = a$ and $\map h 1 = \map g 1 = c$.
== Sources ==
* : $6.3$: Path-connectedness: Lemma $6.3.3$
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Topology
Category:Path-Connected Sets
Joins of Homeomorphic Spaces are Homeomorphic

== Theorem ==
Let $X, Y, X', Y'$ be topological spaces.
Suppose that:
:$X \sim X'$
:$Y \sim Y'$
where $\sim$ denotes homeomorphic spaces.
Then:
:$(X \ast Y) \sim (X' \ast Y')$
where $\ast$ denotes the join.
== Proof ==
By definition of homeomorphic, let:
:$\phi_X : X \to X'$
:$\phi_Y : Y \to Y'$
be homeomorphisms.
Additionally, let:
:$I_{\closedint 0 1} : \closedint 0 1 \to \closedint 0 1$
be the identity mapping.
By Identity Mapping is Homeomorphism, we have that $I_{\closedint 0 1}$ is a homeomorphism.
Let $\phi : X \times Y \times \closedint 0 1 \to X' \times Y' \times \closedint 0 1$ be defined as:
:$\map \phi {x, y, t} = \tuple {\map {\phi_X} x, \map {\phi_Y} y, \map {I_{\closedint 0 1} } t}$
By Products of Homeomorphic Spaces are Homeomorphic:
:$\phi$ is a homeomorphism.
Consider the equivalence relations:
:$\RR \subseteq \paren {X \times Y \times \closedint 0 1}^2$
:$\RR' \subseteq \paren {X' \times Y' \times \closedint 0 1}^2$
each induced by the mapping:
:$\map R {x, y, t} = \begin{cases} x & : t = 0 \\ \tuple {x, y, t} & : t \in \openint 0 1 \\ y & : t = 1 \end{cases}$
which is taken over the corresponding domain and codomain as in the definition of join.
By Quotients of Homeomorphic Spaces are Homeomorphic, the result will follow if we can show:
:$\forall a, b \in X \times Y \times \closedint 0 1: \map \RR {a, b} \iff \map {\RR'} {\map \phi a, \map \phi b}$
By definition of the induced relation, it suffices to show that:
:$\map R {x, y, t} = \map R {x', y', t'} \iff \map R {\map {\phi_X} x, \map {\phi_Y} y, t} = \map R {\map {\phi_X} {x'}, \map {\phi_Y} {y'}, t'}$
We have:
Therefore, by the remarks above:
:$\paren {X \ast Y} \sim \paren {X' \ast Y'}$
Category:Homeomorphisms (Topological Spaces)
Jordan's Inequality

== Theorem ==
:
:$\dfrac {2 x} \pi \le \sin x \le x$
for all $x$ in the interval $\closedint 0 {\dfrac \pi 2}$
== Proof ==
The inequality is true by Sine Inequality.
The inequality is true for $x = 0$ and $x = \dfrac \pi 2$, where we have equality.
Now consider $x \in \openint 0 {\dfrac \pi 2}$.
From Shape of Sine Function, $\sin x$ is concave on the interval $\closedint 0 \pi$.
Letting $x_1 = 0$, $x_2 = x$ and $x_3 = \dfrac \pi 2$ in the definition of Concave Real Function we obtain:
:$\dfrac {\sin x} x \ge \dfrac {1 - \sin x} {\dfrac \pi 2 - x}$
Rearranging gives the desired inequality.
== Sources ==
* 
Category:Sine Function
Jordan's Lemma

== Theorem == 
Let $r > 0$ be a real number.
Let:
:$C_r = \set {r e^{i \theta}: 0 \le \theta \le \pi}$
Let $g : C_r \to \C$ be a continuous function.
Define $f : C_r \to \C$ by: 
:$\map f z = e^{i a z} \map g z$
for each $z \in C_r$, for some real number $a > 0$. 
Then:
:$\ds \size {\int_{C_r} \map f z \rd z} \le \frac \pi a \paren {\max_{0 \mathop \le \theta \mathop \le \pi} \size {\map g {r e^{i \theta} } } }$
== Proof == 
We have: 
From Definite Integral of Constant Multiple of Real Function, we can write:
:$\ds r \int_0^\pi e^{-a r \sin \theta} \max_{0 \le \theta \le \pi} \size {\map g {r e^{i \theta} } } \rd \theta = r \paren {\max_{0 \le \theta \le \pi} \size {\map g {r e^{i \theta} } } } \int_0^\pi e^{-a r \sin \theta} \rd \theta$
We now focus on: 
:$\ds \int_0^\pi e^{-a r \sin \theta} \rd \theta$
We have: 
Note that we also have, from Jordan's Inequality: 
:$\sin \theta \ge \dfrac {2 \theta} \pi$
We therefore have: 
So:
== Also see ==
* Estimation Lemma for Contour Integrals 
Category:Complex Contour Integrals
Category:Complex Analysis
Jordan-Hölder Theorem

== Theorem ==
Let $G$ be a finite group.
Let $\HH_1$ and $\HH_2$ be two composition series for $G$.
Then:
:$\HH_1$ and $\HH_2$ have the same length
:Corresponding factors of $\HH_1$ and $\HH_2$ are isomorphic.
== Proof ==
By the Schreier-Zassenhaus Theorem, two normal series have refinements of equal length whose factors are isomorphic.
But from the definition of composition series, $\HH_1$ and $\HH_2$ have no proper refinements.
Hence any such refinements must be identical to $\HH_1$ and $\HH_2$ themselves.
== Historical Note ==
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 73$. The Jordan-Hölder THeorem
* : $1$: Introduction to Finite Group Theory: $1.9$
* 
Category:Composition Series
Jordan-Schönflies Theorem

== Theorem ==
Let $\gamma : \closedint 0 1 \to \R^2$ be a Jordan curve.
Let $\Img \gamma$ denote the image of $\gamma$, $\Int \gamma$ denote the interior of $\gamma$, and $\Ext \gamma$ denote the exterior of $\gamma$.
Let $\mathbb S^1$ denote the unit circle whose center is at the origin $\mathbf 0$ of the Euclidean space $\R^2$..
Let $\map {B_1} { \mathbf 0 }$ denote the open ball in $\R^2$ with radius $1$ and center $\mathbf 0$, and let $\map {B_1^-} { \mathbf 0 }$ denote the closed ball in $\R^2$ with radius $1$ and center $\mathbf 0$.
Then there exists a homeomorphism $\phi : \R^2 \to \R^2$ such that:
:The restriction of $\phi$ to $\Img \gamma \times \mathbb S^1$ is a homeomorphism between $\Img \gamma$ and $\mathbb S^1$.
:The restriction of $\phi$ to $\Int \gamma \times \map {B_1} { \mathbf 0 }$ is a homeomorphism between $\Int \gamma$ and $\map {B_1} { \mathbf 0 }$.
:The restriction of $\phi$ to $\Ext \gamma \times \R^2 \setminus \map {B_1^-} { \mathbf 0 }$ is a homeomorphism between $\Ext \gamma$ and $\R^2 \setminus \map {B_1^-} { \mathbf 0 }$.
== Proof ==
== Sources ==
* 
Category:Jordan Curves
Jordan Content is Monotone

== Theorem ==
Let $A, B \subseteq \R^n$ be bounded subspaces of Euclidean $n$-space.
Suppose that $A \subseteq B$.
Further suppose that the Jordan content of both $A$ and $B$ exists.
Then:
:$\map m A \le \map m B$
where $m$ denotes the Jordan content.
== Proof ==
By definition of the Jordan content:
:$\map m A = \map {m^*} A$
:$\map m B = \map {m^*} B$
where $m^*$ denotes the outer Jordan content.
The result follows from Outer Jordan Content is Monotone.
Category:Jordan Content