nyuuzyou/wikis huggingface dataset:
Hasse Diagram/Examples

== Examples of Hasse Diagrams ==
=== Divisors of $12$ ===
=== Divisors of $24$ ===
=== Divisors of $30$ ===
=== Subsets of $\set {1, 2, 3}$ ===
=== Subsets of $\set {1, 2, 3, 4}$ ===
=== Subgroups of Symmetry Group of Rectangle ===
=== Subgroups of Symmetry Group of Square ===
=== Parallel Lines ===
== Informal Examples ==
=== British Monarchs ===
=== British Monarchs Line of Descent ===
=== Genealogy from Terah to Joseph ===
== Sources ==
* : $\S 1.2$
* : $\S 0.2$. Sets
* : Chapter $1$: Sets and mappings: $\S 1.5$: Ordered Sets
Category:Examples of Hasse Diagrams
Hasse Diagram/Examples/British Monarchs

== Example of Hasse Diagram ==
Recall the ordering on the British monarchy:
This Hasse diagram illustrates the restriction of $\MM$ to all $x$ of $K$ such that $x \mathrel \MM \text {Victoria}$ and $\text {Elizabeth II} \mathrel \MM x$.
The corresponding Hasse diagram illustrating the restriction of $\MM^{-1}$ to the same set is given below:
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Figure $12 \ (1)$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.1$
Category:Examples of Hasse Diagrams
Hasse Diagram/Examples/British Monarchs Descent

== Example of Hasse Diagram ==
Recall the partial ordering on the set of people:
This Hasse diagram illustrates the restriction of $\DD$ to the set of such that $x \mathrel \DD \text {Victoria}$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.2 \ \text{(b)}$
Category:Examples of Hasse Diagrams
Hasse Diagram/Examples/Divisors of 12

== Example of Hasse Diagram ==
: 
This Hasse diagram illustrates the "Divisor" ordering on the set $S = \set {1, 2, 3, 4, 6, 12}$, where $S$ is the set of all elements of $\N_{>0}$ which divide $12$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Figure $12 \ (5)$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.1$
Category:Examples of Hasse Diagrams
Hasse Diagram/Examples/Divisors of 24

== Example of Hasse Diagram ==
: 
This Hasse diagram illustrates the "Divisor" ordering on the set $S = \set {1, 2, 3, 4, 6, 8, 12, 24}$, where $S$ is the set of all elements of $\N_{>0}$ which divide $24$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.2 \ \text{(e)}$
Category:Examples of Hasse Diagrams
Hasse Diagram/Examples/Divisors of 30

== Example of Hasse Diagram ==
: 
This Hasse diagram illustrates the "Divisor" ordering on the set $S = \set {1, 2, 3, 5, 6, 10, 30}$.
That is, $S$ is the set of all elements of $\N_{>0}$ which divide $30$ except for $15$, which for the purposes of this example has been deliberately excluded.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.2 \ \text{(f)}$
Category:Examples of Hasse Diagrams
Hasse Diagram/Examples/Parallel Lines

== Example of Hasse Diagram ==
Recall this partial ordering on the set of lines:
This Hasse diagram illustrates the restriction of $\LL$ to the set of all infinite straight lines in the cartesian plane which are parallel to and one unit away from either the $x$-axis or the $y$-axis.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Figure $12 \ (3)$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.1$
Category:Examples of Hasse Diagrams
Hasse Diagram/Examples/Subgroups of Symmetry Group of Rectangle

== Example of Hasse Diagram ==
Consider the symmetry group of the rectangle:
This Hasse diagram illustrates the subgroup relation on $\map D 2$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Figure $12 \ (6)$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.1$
Category:Examples of Hasse Diagrams
Hasse Diagram/Examples/Subgroups of Symmetry Group of Square

== Example of Hasse Diagram ==
Consider the symmetry group of the square:
This Hasse diagram illustrates the subgroup relation on $\map D 4$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.2 \ \text{(d)}$
Category:Examples of Hasse Diagrams
Category:Symmetry Group of Square
Hasse Diagram/Examples/Subsets of 1, 2, 3

== Example of Hasse Diagram ==
This Hasse diagram illustrates the "Subset" relation on the power set $\powerset S$ where $S = \set {1, 2, 3}$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Figure $12 \ (4)$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.1$
* : $\S 7$: Examples $7.3, \ 7.4$
Category:Examples of Hasse Diagrams
Hasse Diagram/Examples/Subsets of 1, 2, 3, 4

== Example of Hasse Diagram ==
This Hasse diagram illustrates the "Subset" relation on the power set $\powerset S$ where $S = \set {1, 2, 3, 4}$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.2 \ \text{(c)}$
Category:Examples of Hasse Diagrams
Hasse Diagram of Dual Ordering

== Theorem ==
Let $D$ be the Hasse diagram of an ordering $\RR$.
Then the Hasse diagram $D'$ of the dual ordering $\RR^{-1}$ of $\RR$ is obtained by turning $D$ upside down.
== Proof ==
Let $\tuple {x, y}$ be an ordered pair in $\RR$ which is represented by a line in $D$ joining $x$ and $y$ such that $x$ is below $y$.
Then $x \mathrel \RR y$ such that:
:$\not \exists z: x \mathrel \RR z \land \mathrel \RR y$
As $\RR^{-1}$ is the dual of $\RR$, we have that $y \mathrel {\RR^{-1} } x$ such that:
:$\not \exists z: y \mathrel {\RR^{-1} } z \land \mathrel {\RR^{-1} }$
and so there will be a line in $D'$ joining $x$ and $y$ such that $x$ is above $y$.
The result follows.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.2$
Category:Hasse Diagrams
Category:Dual Orderings
Hat-Check Distribution Gives Rise to Probability Mass Function

== Theorem ==
Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the hat-check distribution with parameter $n$ (where $n > 0$).
Then $X$ gives rise to a probability mass function.
== Proof ==
By definition:
:$\Img X = \set {0, 1, \ldots, n}$
:$\ds \map \Pr {X = k} = \dfrac 1 {\paren {n - k }!} \sum_{s \mathop = 0}^k \dfrac {\paren {-1}^s} {s!}$
Then:
So $X$ satisfies $\map \Pr \Omega = 1$, and hence the result.
Category:Hat-Check Distribution
Category:Hat-Check Problem
Hat-Check Problem

== Classic Problem ==
The traditional wording of the question is as follows.
A hat-check girl completely loses track of which of $n$ hats belong to which owners, and hands them back at random to their $n$ owners as the latter leave.
What is the probability $p_n$ that nobody receives their own hat back?
== Solution ==
Put into bald mathematical language, this boils down to:
For a set $S$ of $n$ elements, what is the number of derangements of $S$ divided by the number of permutations of $S$?
The answer is: approximately $\dfrac 1 e$, which can be demonstrated as follows.
Let $D_n$ be the number of derangements of a set of size $n$.
We have that:
:The Number of Permutations of a set of size $n$ is $n!$.
:The Closed Form for Number of Derangements on Finite Set of size $n$ is:
::$D_n = n! \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \dotsb + \paren {-1}^n \dfrac 1 {n!} }$
So:
Finally:
:$1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \dotsb$
converges to $\dfrac 1 e$ by Taylor Series Expansion for Exponential Function.
== Note ==
The accuracy of this estimate increases with increasing $n$.
Specifically, the error of the estimate is:
:$\ds \sum_{k \mathop = 0}^{\infty} \frac {\paren {-1}^k } {k!} - \sum_{k \mathop = 0}^n \frac {\paren {-1}^k } {k!} = \sum_{k \mathop = {n + 1} }^{\infty} \frac {\paren {-1}^k } {k!}$
Therefore, as $n \to \infty$, the error goes to $0$
:$\ds \lim_{n \mathop \to \infty} \paren {\sum_{k \mathop = 0}^{\infty} \frac {\paren {-1}^k } {k!} - \sum_{k \mathop = 0}^n \frac {\paren {-1}^k } {k!} } = 0$
If the venue in question is Hilbert's Hotel, then the answer is precise.
== Examples ==
== Also known as ==
This problem is also known as the envelope problem, couched in the language of an incompetent secretary placing letters at random into the envelopes without checking the address on the envelope matches the addressee of the letter.
== Sources ==
* : $0 \cdotp 367 \, 879 \, 441 \, 171 \, 442 \, 321 \, 595 \, 523 \, 770 \, 161 \, 460 \, 867 \, 445 \, 811 \, 131 \, 031$
* : $0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031$
Category:Hat-Check Problem
Category:Derangements
Category:Classic Problems
Category:Probability Theory
Category:Combinatorics
Hat-Check Problem/Examples/3

== Example of Hat-Check Problem ==
:$p_3 = \dfrac 1 3$
== Proof ==
When $n = 3$, there are only three hats to hand back and $6 = 3!$ ways to return the hats to the owners.
:$\begin{array}{c|ccc}
\ds \text {Correctly Returned} & \text {Adam} & \text {Bill} & \text {Clyde} \\
\hline
3 & A & B & C \\
\hline
1 & A & C & B \\
\hline
1 & C & B & A \\
\hline
1 & B & A & C \\
\hline
0 & B & C & A \\
\hline
0 & C & A & B \\
\hline
\end{array}$
From the table above, we see that there are only two permutations that irritate all parties involved.
Hence, $p_3 = \dfrac 2 6 = \dfrac 1 3$
Category:Examples of Hat-Check Problem
Hat-Check Problem/Examples/8

== Example of Hat-Check Problem ==
:$p_8 = \dfrac {2119} {5760}$
== Proof ==
When $n = 8$, there are eight hats to hand back.
Hence:
:$p_8$ is roughly $0.0000025$ (or less than $\dfrac 1 {9!}$) away from the estimate of $\dfrac 1 e$.
Category:Examples of Hat-Check Problem
Hausdorff's Maximal Principle

== Theorem ==
=== Formulation 1 ===
=== Formulation 2 ===
== Hausdorff's Maximal Principle and Axiom of Choice ==
=== Axiom of Choice implies Hausdorff's Maximal Principle ===
=== Hausdorff's Maximal Principle implies Axiom of Choice ===
== Also known as ==
== Also see ==
Category:Order Theory
Category:Hausdorff's Maximal Principle
Category:Equivalents of Axiom of Choice
Hausdorff's Maximal Principle/Also known as

== Hausdorff's Maximal Principle Also known as ==
Hausdorff's Maximal Principle is also known as the Hausdorff Maximal Principle.
Some sources call it the Hausdorff Maximality Principle or the Hausdorff Maximality Theorem.
Category:Hausdorff's Maximal Principle
Hausdorff's Maximal Principle/Formulation 1

== Theorem ==
Let $\struct {\PP, \preceq}$ be a non-empty partially ordered set.
Then there exists a maximal chain in $\PP$.
== Also known as ==
== Also see ==
== Sources ==
* : $\S 16$: Zorn's Lemma: Exercise $\text{(i)}$
* 
Category:Hausdorff's Maximal Principle
Hausdorff's Maximal Principle/Formulation 2

== Theorem ==
Let $A$ be a non-empty set of sets.
Let $S$ be the set of all chain of sets of $A$ (ordered under the subset relation).
Then every element of $S$ is a subset of a maximal element of $S$ under the subset relation.
== Also known as ==
== Also see ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Proposition $5.8$
Category:Hausdorff's Maximal Principle
Hausdorff's Maximal Principle implies Axiom of Choice

== Theorem ==
Let Hausdorff's Maximal Principle be accepted.
Then the Axiom of Choice holds.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Theorem $5.12$
Category:Hausdorff's Maximal Principle
Category:Axiom of Choice
Category:Hausdorff's Maximal Principle implies Axiom of Choice
Hausdorff's Maximal Principle implies Axiom of Choice/Lemma

== Theorem ==
Let $S$ be a non-empty set.
Let $\Sigma$ be the set of all choice functions of all non-empty subsets of $S$.
$\Sigma$ is closed under chain unions.
== Proof ==
Let $f$ and $g$ be choice functions.
Then:
:$f \subseteq g$
:
:$\Dom f \subseteq \Dom g$
:$\forall x \in \Dom f: \map f x = \map g x$
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Theorem $5.12$
Category:Hausdorff's Maximal Principle implies Axiom of Choice
Hausdorff's Maximal Principle implies Axiom of Choice/Proof

== Theorem ==
== Proof ==
Let Hausdorff's Maximal Principle be accepted.
Let $S$ be a non-empty set.
Let $\Sigma$ be the set of all choice functions of all non-empty subsets of $S$.
From Countable Set has Choice Function every finite subset of $S$ has a choice function.
Hence $\Sigma$ is a non-empty set.
=== Lemma ===
By Hausdorff's Maximal Principle it follows that $\Sigma$ has a maximal element $f$.
Hence $f$ is seen to be a choice function for $S$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Exercise $5.1$
Category:Hausdorff's Maximal Principle implies Axiom of Choice
Hausdorff's Maximal Principle implies Kuratowski's Lemma

== Theorem ==
Let Hausdorff's Maximal Principle be accepted as true.
Then Kuratowski's Lemma holds.
== Proof 1 ==
== Proof 2 ==
Category:Hausdorff's Maximal Principle
Category:Kuratowski's Lemma
Category:Hausdorff's Maximal Principle implies Kuratowski's Lemma
Hausdorff's Maximal Principle implies Kuratowski's Lemma/Proof 1

== Theorem ==
== Proof ==
Recall Hausdorff's Maximal Principle:
Recall Kuratowski's Lemma:
So, let us assume Hausdorff's Maximal Principle.
Let $S$ be a non-empty set of sets which is closed under chain unions.
$\set b$ is trivially a chain.
Hence by Hausdorff's Maximal Principle $\set b$ is a subset of a maximal chain of elements of $S$.
Hence $b$ is an element of a maximal chain $C$ of elements of $S$.
We have that $S$ is closed under chain unions.
Hence $\bigcup C \in S$.
Then $b \subseteq \bigcup C$.
If $\bigcup C$ were a proper subset of some element $x \in S$, $C$ would be a proper subset of the chain $C \cup \set x$.
But $C$ is maximal so that cannot happen.
Hence $\bigcup C$ is maximal .
Thus it is seen that Kuratowski's Lemma holds.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Proposition $5.10$
Category:Hausdorff's Maximal Principle implies Kuratowski's Lemma
Hausdorff's Maximal Principle implies Kuratowski's Lemma/Proof 2

== Theorem ==
== Proof ==
We have:
* Hausdorff's Maximal Principle implies Axiom of Choice
* Axiom of Choice implies Kuratowski's Lemma
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Hausdorff's Maximal Principle implies Kuratowski's Lemma
Hausdorff's Maximal Principle implies Zermelo's Well-Ordering Theorem

== Theorem ==
Let Hausdorff's Maximal Principle hold.
Then Zermelo's Well-Ordering Theorem holds.
== Proof ==
Let $X$ be a non-empty set.
Let $X$ contain at least two elements; otherwise, $X$ can be trivially well-ordered.
We use Hausdorff's Maximal Principle to construct a well-ordering on $X$.
We note that there exists at least the following strict well-orderings on subsets of $X$: 
For any singleton $\set x$, $x$ is vacuously well-ordered with respect to every other element in the subset, of which there are none.
Any doubleton $\set {y, z}$ can be strictly well-ordered by defining $x < y$ and $y \not < x$.
Let $\AA$ be the set of all ordered pairs $\struct {A, <}$ such that $A$ is a subset of $X$ and $<$ is a strict well-ordering of $A$.
Define $\prec$ as:
:$\struct {A, <} \prec \struct {A', <'}$
 
:$\struct {A, <}$ equals an initial segment of $\struct {A', <'}$.
Let $\BB$ be a set of ordered pairs in $\AA$ such that $\BB$ is ordered by $\prec$.
Let $B'$ be the union of the sets $B$ for all $\struct {B, <} \in \BB$.
Let $<'$ be the union of the relations $<$ for all $\struct {B, <}$.
By Equality to Initial Segment Imposes Well-Ordering, $\struct {B', <'}$ is strictly well-ordered set.
By Hausdorff's Maximal Principle, there exists a maximal chain defined by $<'$ imposed on a maximal subset of $X$.
We claim that this maximal chain is imposed on the entirety of $X$.
 that $\struct {B', <'}$ is a maximal chain but $B' \ne X$.
Then consider $B' \cup \set x$, with $x \in X \setminus B'$, with the extension of $<'$ such that:
:$\forall a \in B': a <' x$. 
Then $B' = S_x$, the initial segment in $B' \cup \set x$ determined by $x$.
But by the definition of $\prec$, we then have:
:$\struct {B' \cup \set x, \text {extension of } <'} \prec \struct {B', <'}$
But then by the definition of $B'$, we would have $B' \cup \set x \subseteq B'$, contrary to the assumption that $x \in X \setminus B'$.
From this contradiction infer that $B' = X$.
That is, all of $X$ can be well-ordered.
== Also see ==
* Zermelo's Well-Ordering Theorem implies Hausdorff's Maximal Principle
== Sources ==
* $\S 1.11$ Supplementary Exercise $6$
Category:Zermelo's Well-Ordering Theorem
Category:Hausdorff's Maximal Principle
Hausdorff's Maximal Principle implies Zorn's Lemma

== Theorem ==
Hausdorff's Maximal Principle implies Zorn's Lemma.
== Proof ==
Let $\struct {\PP, \preceq}$ be a partially ordered set.
Hausdorff's Maximal Principle states that there is a maximal chain in $\struct {\PP, \preceq}$.
If this maximal chain has an upper bound in $\PP$, then such an upper bound is a maximal element of $\PP$ itself, by the maximality of the chain.
== Sources ==
* : (P.2)
Category:Hausdorff's Maximal Principle
Category:Zorn's Lemma
Hausdorff's Maximal Principle is equivalent to Axiom of Choice

== Theorem ==
Every ordered set has a maximal chain the axiom of choice holds.
== Proof ==
* Axiom of Choice implies Hausdorff's Maximal Principle
* Hausdorff's Maximal Principle implies Zermelo's Well-Ordering Theorem
* Zermelo's Well-Ordering Theorem is Equivalent to Axiom of Choice
== Sources ==
* : $\S 16$: Zorn's Lemma: Exercise $\text{(i)}$
Category:Axiom of Choice
Category:Hausdorff's Maximal Principle
Hausdorff-Besicovitch Dimension is Well-Defined

== Theorem ==
Let $F \subseteq \R^n$ be a subset of the $n$-dimensional Euclidean space.
Let $\map {\HH^s} F$ be the $s$-dimensional Hausdorff measure on $\R^n$ of $F$ for each $s \in \R_{\ge 0}$.
Then:
In particular, the Hausdorff-Besicovitch dimension is well-defined.
== Proof ==
Category:Hausdorff-Besicovitch Dimension
Hausdorff Condition is Preserved under Homeomorphism

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_2$ (Hausdorff) space, then so is $T_B$.
== Proof ==
By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_2$ (Hausdorff) Space is Preserved under Closed Bijection.
== Sources ==
* : $4$: The Hausdorff condition: $4.2$: Separation axioms: Proposition $4.2.4 \ \text{(d)}$
* : $4$: The Hausdorff condition: Exercise $4.3: 2$
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Homeomorphisms (Topological Spaces)
Category:Hausdorff Spaces
Hausdorff Paradox

== Theorem ==
There exists a disjoint decomposition of the sphere $\mathbb S^2$ into four sets $A, B, C, Q$ such that $A, B, C, B \cup C$ are all congruent and $Q$ is countable.
== Proof 1 ==
== Proof 2 ==
Whether you view this result as a veridical paradox or an antinomy depends your acceptance or otherwise of the .
== Sources ==
* 
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Theorem $1.1$
Category:Hausdorff Paradox
Category:Topology
Category:Veridical Paradoxes
Category:Antinomies
Hausdorff Paradox/Lemma 1

== Hausdorff Paradox: Lemma ==
Let $G$ be the free product of the groups $G_1 = \set {e_1, \phi}$ and $G_2 = \set {e_2, \psi, \psi^2}$.
Let $U := \mathbb D^3 \subset \R^3$ be a unit ball in real Euclidean space of $3$ dimensions.
Let $\phi$ and $\psi$ be represented by the axes of rotation $a_\phi$ and $a_\psi$ passing through the center of $U$ such that:
:$\phi$ is a rotation by $180 \degrees$, that is $\pi$ radians about $a_\phi$
:$\psi$ is a rotation by $120 \degrees$, that is $\dfrac {2 \pi} 3$ radians about $a_\psi$
Hence consider $G$ as the group of all rotations generated by $\phi$ and $\psi$.
The identity of $G$ is then the identity mapping $\mathbf I_3$.
Then $a_\phi$ and $a_\psi$ can be determined in such a way that distinct elements of $G$ represent distinct rotations generated by $\phi$ and $\psi$.
== Proof Outline ==
== Proof ==
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Lemma $1.3$
Category:Hausdorff Paradox
Hausdorff Paradox/Lemma 1/Proof Outline

== Proof Outline for Hausdorff Paradox: Lemma ==
== Proof Outline ==
We determine the angle $\theta$ between $a_\phi$ and $a_\psi$ such that no element of $G$ other than its identity $\mathbf I_3$ represents the identity rotation.
Let us consider a typical element $\alpha$ of $G$:
:$\alpha = \phi \circ \psi^{\pm 1} \circ \cdots \circ \phi \circ \psi^{\pm 1}$
Using the properties of orthogonal transformations and elementary trigonometry, we seek to prove that the equation:
:$\alpha = \mathbf I_3$
has only finitely many solutions.
Consequently, except for a countable set of values, we may select any angle $\theta$ that satisfies the requirements.
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Lemma $1.3$
Category:Hausdorff Paradox
Hausdorff Paradox/Lemma 2

== Hausdorff Paradox: Lemma ==
Let $G$ be the free product of the groups $G_1 = \set {e_1, \phi}$ and $G_2 = \set {e_2, \psi, \psi^2}$.
Let $U := \mathbb D^3 \subset \R^3$ be a unit ball in real Euclidean space of $3$ dimensions.
Let $\phi$ and $\psi$ be represented by the axes of rotation $a_\phi$ and $a_\psi$ passing through the center of $U$ such that:
:$\phi$ is a rotation by $180 \degrees$, that is $\pi$ radians about $a_\phi$
:$\psi$ is a rotation by $120 \degrees$, that is $\dfrac {2 \pi} 3$ radians about $a_\psi$
Hence consider $G$ as the group of all rotations generated by $\phi$ and $\psi$.
The identity of $G$ is then the identity mapping $\mathbf I_3$.
The group $G$ can be partitioned into $3$ disjoint sets:
:$G = A \cup B \cup C$
such that:
== Proof ==
We construct $A$, $B$ and $C$ by recursion on the lengths of the elements of $G$.
Let:
Then we construct the Cayley table for $G$ as follows:
:$\begin {array} {c|ccc}
\circ & \alpha \in A & \alpha \in B & \alpha \in C \\
\hline
\text {$\alpha$ ends with $\psi^{\pm 1}$}: & \alpha \phi \in B & \alpha \phi \in A & \alpha \phi \in A \\
\text {$\alpha$ ends with $\phi$}: & \alpha \psi \in B & \alpha \psi \in C & \alpha \psi \in A \\
\text {$\alpha$ ends with $\phi$}: & \alpha \psi^{-1} \in C & \alpha \psi^{-1} \in A & \alpha \psi^{-1} \in B \\
\end {array}$
This guarantees that the conditions $(1)$, $(2)$ and $(3)$ are satisfied.
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere: Lemma $1.4$
Category:Hausdorff Paradox