nyuuzyou/wikis huggingface dataset:
Empty Set is Closed/Normed Vector Space

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Then the empty set $\O$ is closed in $M$.
== Proof ==
From Normed Vector Space is Open in Itself, $X$ is open in $M$.
But:
:$\O = \relcomp X X$
where $\complement_X$ denotes the set complement relative to $X$.
The result follows by definition of closed set.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Empty Set is Closed
Category:Closed Sets (Normed Vector Spaces)
Empty Set is Closed/Topological Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then $\O$ is closed in $T$.
== Proof ==
From the definition of closed set, $U$ is open in $T = \struct {S, \tau}$ $S \setminus U$ is closed in $T$.
By definition of topological space, $S$ is open in $T$.
From Set Difference with Self is Empty Set:
:$S \setminus S = \O$
Hence $\O$ is closed in $T$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $4$
Category:Empty Set is Closed
Empty Set is Compact Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then the empty set $\O$ is a compact subspace of $T$.
== Proof ==
Recall the definition of compact subspace:
:$\struct {\O, \tau_\O}$ is compact in $T$ every open cover $\CC \subseteq \tau_\O$ for $\O$ has a finite subcover.
The only open cover for $\O$ that is contained in $\O$ is $\set \O$ itself.
This has only one finite subcover, and that is $\set \O$.
This is a finite subcover.
Hence the result, by definition of compact subspace.
Category:Compact Spaces
Category:Empty Set
Empty Set is Countable

== Theorem ==
The empty set $\O$ is countable.
== Proof ==
By Peano's Axioms, $\N_0 \sim \O$, where $\N_n$ denotes the initial segment of natural number $n$.
By definition, $\O$ is finite.
By definition, $\O$ is a countable set.
Category:Set Theory
Category:Empty Set
Category:Countable Sets
Empty Set is Element of Power Set

== Theorem ==
The empty set is an element of all power sets:
:$\forall S: \O \in \powerset S$
== Proof ==
== Also see ==
* Set is Element of its Power Set
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : $\S 1.8$. Sets of sets: Example $25$
* : Chapter $1$: Theory of Sets: $\S 2$: Sets and Subsets: Exercise $1 \ \text{(d)}$
Category:Empty Set
Category:Power Set
Empty Set is Element of Topology

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological space.
Then the empty set $\O$ is an open set of $T$.
== Proof ==
By Empty Set is Subset of All Sets:
:$\O \subseteq \tau$
By Union of Empty Set:
:$\O = \bigcup \O$
By :
:$\O \in \tau$
Category:Topology
Category:Empty Set
Empty Set is Initial Object

== Theorem ==
Let $\mathbf{Set}$ be the category of sets.
Then the empty set $\O$ is an initial object of $\mathbf{Set}$.
== Proof ==
Follows from Empty Mapping is Unique and the definition of initial object.
== Sources ==
* : $\S 2.2$: Example $2.11$: $1$
Category:Category of Sets
Category:Empty Set
Empty Set is Linearly Independent

== Theorem ==
The empty set is a linearly independent set.
== Proof ==
There are no sequences at all of $n$ terms of the empty set for any $n > 0$.
Hence the result holds vacuously.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases
Category:Empty Set
Category:Linear Independence
Empty Set is Nowhere Dense

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then the empty set $\O$ is nowhere dense in $T$.
== Proof ==
From Empty Set is Closed in Topological Space, $\O$ is closed in $T$.
From Closed Set Equals its Closure:
:$\O^- = \O$
where $\O^-$ is the the closure of $\O$.
From Empty Set is Element of Topology, $\O$ is open in $T$.
From the definition (trivially) we also have that:
:$\O^\circ = \O$
where $\O^\circ$ is the interior of $\O$.
So:
:$\struct {\O^-}^\circ = \O$
and so by definition $\O$ is nowhere dense in $T$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $5$
Category:Empty Set
Category:Nowhere Denseness
Empty Set is Open and Closed in Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then the empty set $\O$ is both open and closed in $M$.
== Proof ==
From Empty Set is Open in Metric Space, $\O$ is open in $M$.
From Empty Set is Closed in Metric Space, $\O$ is closed in $M$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets
Category:Open Sets (Metric Spaces)
Category:Closed Sets
Category:Empty Set
Category:Metric Spaces
Empty Set is Open in Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then the empty set $\O$ is an open set of $M$.
== Proof ==
By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set.
That is, there are no points in $S$ which have an open ball some of whose elements are not in $S$.
As there are no elements in $\O$, the result follows vacuously.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.4$
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.10$
Category:Empty Set
Category:Open Sets (Metric Spaces)
Category:Metric Spaces
Empty Set is Open in Neighborhood Space

== Theorem ==
Let $\struct {S, \NN}$ be a neighborhood space.
Then the empty set $\O$ is an open set of $\struct {S, \NN}$.
== Proof ==
Suppose $\O$ were not an open set of $\struct {S, \NN}$.
Then $\exists x \in \O$ such that $\O$ is not a neighborhood of $\O$.
By definition of empty set, such an $x$ does not exist.
Hence the result.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Lemma $3.6$
Category:Neighborhood Spaces
Category:Empty Set
Category:Open Sets
Empty Set is Open in Normed Vector Space

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the empty set $\O$ is an open set of $M$.
== Proof ==
By definition, an open set $S \subseteq X$ is one where every point inside it is an element of an open ball contained entirely within that set.
That is, there are no points in $S$ which have an open ball some of whose elements are not in $S$.
As there are no elements in $\O$, the result follows vacuously.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Empty Set
Category:Open Sets (Normed Vector Spaces)
Empty Set is Submagma of Magma

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Then:
:$\struct {\O, \circ}$ is a submagma of $\struct {S, \circ}$
where $\O$ is the empty set.
== Proof ==
By definition, a magma is an algebraic structure $\struct {S, \circ}$ where $\circ$ is closed.
That is:
:$\forall x, y \in S: x \circ y \in S$
By definition, $\struct {T, \circ}$ is a submagma of $S$ if:
:$\forall x, y \in T: x \circ y \in T$
But:
:$\not \exists x, y \in \O: x \circ y \notin \O$
it follows vacuously that:
:$\forall x, y \in \O: x \circ y \in \O$
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.3$
Category:Magmas
Category:Submagmas
Category:Empty Set
Empty Set is Subset of All Sets

== Theorem ==
The empty set $\O$ is a subset of every set (including itself).
That is:
:$\forall S: \O \subseteq S$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection
* : A Note on Set-Theoretic Concepts
* : $\S 1.1$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $1$: Set Theory: $1.2$: Sets and subsets
* : $\S 0.2$. Sets
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B ii}$
* : $1$ Set Theory: $1$. Sets and Functions: $1.1$: Basic definitions
* : Appendix: Elementary set and number theory
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 3$
* : $\S 1.3$: Subsets: Remark $3.3$
* : Chapter $1$: Sets: Subsets
* : $\S 2$: Sets and functions: Sets
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
* : Chapter $1$: Theory of Sets: $\S 2$: Sets and Subsets
* : $\S 1.2$: Sets and Equivalence Relations
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.5 \ \text{(iv)}$
* : $2$. Graphs: Sets
* 
* 
* 
Category:Subsets
Category:Empty Set
Category:Empty Set is Subset of All Sets
Empty Set is Subset of All Sets/Proof 1

== Theorem ==
== Proof ==
By the definition of subset, $\O \subseteq S$ means:
:$\forall x: \paren {x \in \O \implies x \in S}$
By the definition of the empty set:
:$\forall x: \neg \paren {x \in \O}$
Thus $\O \subseteq S$ is vacuously true.
== Sources ==
* : $\S 3$: Unordered Pairs
* : $\S 1.3$. Intersection
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
* : $\S 6.4$: Subsets
* : Appendix $\text A$: Sets and Functions: $\text{A}.1$: Sets
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
* : Appendix $\text{A}.2$: Theorem $\text{A}.5$
Category:Empty Set is Subset of All Sets
Empty Set is Subset of All Sets/Proof 2

== Theorem ==
== Proof ==
$S \subseteq T$ means:
:every element of $S$ is also in $T$
or, equivalently:
:every element that is not in $T$ is not in $S$ either.
Thus:
which means there is no element in $S$ which is not also in $T$.
There are no elements of $\O$, from the definition of the empty set.
Therefore $\O$ has no elements that are not also in any other set.
Thus, from the above, all elements of $\O$ are all (vacuously) in every other set.
== Sources ==
* : $\S 3$: Unordered Pairs
* : Appendix $\text{A}.1$: Sets and Subsets
Category:Empty Set is Subset of All Sets
Empty Set is Subset of Power Set

== Theorem ==
The empty set is a subset of all power sets:
:$\forall S: \O \subseteq \powerset S$
== Proof ==
Follows directly from Empty Set is Subset of All Sets.
== Also see ==
* Empty Set is Element of Power Set
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 2$: Sets and Subsets: Exercise $1 \ \text{(e)}$
Category:Empty Set
Category:Power Set
Category:Subsets
Empty Set is Unique

== Theorem ==
The empty set is unique.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Exercise $1.1: \ 1$
* : $1$ Enumerability
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions
* 
* 
Category:Empty Set
Category:Empty Set is Unique
Empty Set is Unique/Proof 1

== Theorem ==
== Proof ==
Let $\O$ and $\O'$ both be empty sets.
From Empty Set is Subset of All Sets, $\O \subseteq \O'$, because $\O$ is empty.
Likewise, we have $\O' \subseteq \O$, since $\O'$ is empty.
Together, by the definition of set equality, this implies that $\O = \O'$.
Thus there is only one empty set.
== Sources ==
* : $\S 1.3$: Subsets
* : $\S 6.5$: Subsets
* : $2$. Graphs: Sets
Category:Empty Set is Unique
Empty Set is Unique/Proof 2

== Theorem ==
== Proof ==
Let $A$ and $B$ both be empty sets.
Thus:
:$\forall x: \neg \paren {x \in A}$
and:
:$\forall x: \neg \paren {x \in B}$
Hence:
:$x \notin A \iff x \notin B$
and so:
:$x \in A \iff x \in B$
vacuously.
From the Axiom of Extension:
:$\forall x: \paren {x \in A \iff x \in B} \iff A = B$
Hence the result.
== Sources ==
* : $\S 3$: Unordered Pairs
* 
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.4$
* : $1$. Notation, Conventions: $4$
Category:Empty Set is Unique
Empty Set is Well-Ordered

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Let $\O$ denote the empty set.
Let $\RR_\O$ denote the restriction of $\RR$ to $\O$.
Then $\struct {\O, \RR_\O}$ is a well-ordered set.
== Proof 1 ==
== Proof 2 ==
Category:Empty Set
Category:Well-Orderings
Category:Empty Set is Well-Ordered
End-Extension of Ordered Set/Examples/Closed Real Interval

== Example of End-Extension of Ordered Set ==
The closed real interval $\closedint 0 1$ is:
:an end-extension of the half-open interval $\hointr 0 1$
:but not an end-extension of the open interval $\openint 0 1$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Examples of End-Extensions
Endomorphism from Integers to Multiples

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be a mapping.
Then $\phi$ is a group endomorphism :
:$\exists k \in \Z: \forall n \in \Z: \map \phi n = k n$
== Proof ==
=== Necessary Condition ===
Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be an endomorphism.
Let $k = \map \phi 1$.
We have that $n = \overbrace {1 + \cdots + 1}^n$ for any positive integer $n$.
Thus:
Also:
Thus:
:$\forall n \in \Z: \map \phi n = k n$
=== Sufficient Condition ===
Let $k \in \Z$ such that:
: $\forall n \in \Z: \map \phi n = k n$
Then:
Thus $\phi: \struct {\Z, +} \to \struct {\Z, +}$ is a group homomorphism from $\Z$ to $\Z$.
Hence by definition $\phi$ is a group endomorphism.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{L}$
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 61$
Category:Integers
Category:Group Endomorphisms
Endomorphisms on Entropic Structure whose Pointwise Product is Identity Automorphism

== Theorem ==
Let $\struct {S, \odot}$ be an entropic structure.
Let $\alpha$ and $\beta$ be endomorphisms on $S$ such that:
:$\alpha \odot \beta$ is the identity automorphism on $S$
where $\alpha \odot \beta$ denotes the pointwise product of $\alpha$ with $\beta$:
:$\forall x \in S: \map {\paren {\alpha \odot \beta} } x = \map \alpha x \odot \map \beta x$
Let $\otimes$ be the operation on $S$ defined as:
:$\forall x, y \in S: x \otimes y := \map \alpha x \odot \map \beta y$
Then $\struct {S, \otimes}$ is an entropic idempotent structure, and hence self-distributive.
== Proof ==
Let $a, b, c, d \in S$ be arbitrary.
Hence $\struct {S, \otimes}$ is an idempotent structure.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.27 \ \text {(c)}$
Category:Examples of Idempotence
Category:Examples of Entropic Operations
Category:Examples of Self-Distributive Operations
Endorelation/Examples/Arbitrary Relation

== Examples of Endorelation ==
Let $V_0 = \set {a, b, c}$.
A possible endorelation on $V_0$ is:
:$R = \set {\tuple {a, a}, \tuple {b, b}, \tuple {b, c}, \tuple {c, b} }$
== Sources ==
* : Appendix $\text{A}.2$: Cartesian Products and Relations
Category:Examples of Endorelations
Endorelation/Examples/Divisor

== Example of Endorelation ==
Let $\Z_{>0}$ denote the set of (strictly) positive integers.
For $a, b \in \Z$, let $a \divides b$ denote the statement:
:$a$ is a divisor of $b$.
Then $\divides$ is an endorelation on $\Z$.
Thus, $\tuple {1, 1}$, $\tuple {2, 6}$, $\tuple {3, 12}$ are all elements of the endorelation $\tuple {\Z^2, \divides}$.
== Sources ==
* 
* 
Category:Examples of Endorelations
Endorelation/Examples/Properties of Arbitrary Relation 1

== Examples of Endorelation ==
Let $V = \set {u, v, w, x}$.
Let $E$ be the relation on $V$ defined as:
:$E = \set {\tuple {u, v}, \tuple {v, u}, \tuple {v, w}, \tuple {w, v} }$
Then $E$ is:
:antireflexive
:symmetric
:non-transitive.
== Proof ==
For all $a \in V$, we have that:
:$\tuple {a, a} \notin E$
and so $E$ is antireflexive.
For all $\tuple {a, b} \in E$, we have that:
:$\tuple {b, a} \in E$
and so $E$ is symmetric.
We have:
:$\tuple {u, v} \in E$ and $\tuple {v, w} \in E$, but not $ \tuple {u, w} \in E$
and so $E$ is not non-transitive.
== Sources ==
* : Appendix $\text{A}.2$: Cartesian Products and Relations: Example $\text{A}.1$
Category:Examples of Endorelations
Endorelation/Examples/Properties of Arbitrary Relation 2

== Examples of Endorelation ==
Let $V = \set {a, b, c, d}$.
Let $R$ be the relation on $V$ defined as:
:$E = \set {\tuple {a, a}, \tuple {a, b}, \tuple {a, c}, \tuple {a, d}, \tuple {b, b}, \tuple {b, c}, \tuple {b, d}, \tuple {c, c}, \tuple {c, d}, \tuple {d, d} }$
Then $E$ is:
:reflexive
:antisymmetric
:transitive.
== Proof ==
For all $x \in V$, we have that:
:$\tuple {x, x} \in R$
and so $R$ is reflexive.
For all $\tuple {x, y} \in R$, we have that:
:$\tuple {y, x} \in R \iff x = y$
and so $R$ is antisymmetric.
We have:
:$\tuple {x, y} \in R$ and $\tuple {y, z} \in R$ implies that $ \tuple {x, z} \in R$
and so $R$ is not transitive.
== Sources ==
* : Appendix $\text{A}.2$: Cartesian Products and Relations: Problem Set $\text{A}.2$: $8$
Category:Examples of Endorelations
Endorelation/Examples/Properties of Arbitrary Relation 3

== Examples of Endorelation ==
Let $V = \set {a, b, c, d}$.
Let $R$ be the relation on $V$ defined as:
:$r = \set {\tuple {a, a}, \tuple {a, b}, \tuple {b, b}, \tuple {b, c}, \tuple {c, b}, \tuple {c, c} }$
Then $E$ is:
:non-reflexive
:non-symmetric
:non-transitive.
== Proof ==
We have that:
:$\tuple {d, d} \notin R$
and so $R$ is not reflexive.
We also have that:
:$\tuple {a, b} \in R$
but:
:$\tuple {b, a} \notin R$
and so $R$ is not symmetric.
We have that:
:$\tuple {b, c} \in R$
but:
:$\tuple {c, b} \notin R$
and so $R$ is not asymmetric.
It follows that $R$ is non-symmetric.
We have:
:$\tuple {a, b} \in R$ and $\tuple {b, c} \in R$
but we also have that:
:$\tuple {a, c} \notin R$
and so $R$ is not non-transitive.
== Sources ==
* : Appendix $\text{A}.2$: Cartesian Products and Relations: Problem Set $\text{A}.2$: $9$
Category:Examples of Endorelations
Entire Function Bounded by Polynomial is Polynomial

== Theorem ==
Let $f : \C \to \C$ be an entire function such that: 
:$\cmod {\map f z} \le M {\cmod z}^k$
for all $z \in \C$, for some $k \in \N$ and real $M > 0$. 
Then $f$ is a polynomial of degree at most $k$. 
== Proof ==
Let $r > 0$ be a real number.
Let $D = \map {B_r} 0$ be the open ball with centre $0$ of radius $r$.
By Holomorphic Function is Analytic, we have that: 
:$\ds \map f z = \sum_{n \mathop = 0}^\infty a_n z^n$
for all $z \in \C$, where: 
:$\ds a_n = \frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f t} {t^{n + 1} } \rd t$
Note that for $t \in \partial D$, we have: 
:$\cmod {\map f t} \le M {\cmod t}^k = M r^k$
so for all $t \in \partial D$:
:$\cmod {\dfrac {\map f t} {t^{n + 1} } } \le \dfrac {M r^k} {r^{n + 1} }$
From the Estimation Lemma for Contour Integrals::
Then for $n > k$, we have: 
:$\dfrac M {r^{n - k} } \to 0$
as $r \to \infty$.
Since $r$ was arbitrary, we have: 
:$a_n = 0$
for all $n > k$. 
That is: 
:$\ds \map f z = \sum_{n \mathop = 0}^k a_n z^n$
for all $z \in \C$.
That is, $f$ is a polynomial of degree at most $k$.
Category:Complex Analysis
Entire Function with Bounded Real Part is Constant

== Theorem ==
Let $f : \C \to \C$ be an entire function. 
Let the real part of $f$ be bounded.
That is, there exists a positive real number $M$ such that: 
:$\cmod {\map \Re {\map f z} } < M$
for all $z \in \C$, where $\map \Re {\map f z}$ denotes the real part of $\map f z$.
Then $f$ is constant.
=== Corollary ===
== Proof ==
Let $g : \C \to \C$ be a complex function with:
:$\ds \map g z = e^{\map f z}$
By Derivative of Complex Composite Function, $g$ is entire with derivative:
:$\ds \map {g'} z = \map {f'} z e^{\map f z}$
We have:
So $g$ is a bounded entire function.
By Liouville's Theorem, $g$ is therefore a constant function.
We therefore have, by Derivative of Constant: Complex:
:$\map {g'} z = 0$
for all $z \in \C$.
That is:
:$\map {f'} z e^{\map f z} = 0$
Since the exponential function is non-zero, we must have: 
:$\map {f'} z = 0$
for all $z \in \C$.
From Zero Derivative implies Constant Complex Function, we then have that $f$ is constant on $\C$. 
Category:Complex Analysis
Entire Function with Bounded Real Part is Constant/Corollary

== Corollary ==
Let $f : \C \to \C$ be an entire function. 
Let the imaginary part of $f$ be bounded.
That is, there exists a positive real number $M$ such that: 
:$\cmod {\map \Im {\map f z} } < M$
for all $z \in \C$, where $\map \Im {\map f z}$ denotes the imaginary part of $\map f z$.
Then $f$ is constant.
== Proof ==
Since $\cmod i = 1$, $i f$ is bounded.
From Combination Theorem for Complex Derivatives: Multiple Rule, $i f$ is entire. 
From Real Part of Imaginary Unit times Element of *-Algebra, we have:
:$\cmod {\map \Re {i \map f z} } = \cmod {\map \Im {\map f z} } < M$
From Entire Function with Bounded Real Part is Constant, there exists $c \in \C$ such that $i \map f z = c$.
Hence $\map f z = -i c$.
So $f$ is constant.
Category:Entire Function with Bounded Real Part is Constant
Entropic Idempotent Structure is Self-Distributive

== Theorem ==
Let $\struct {S, \odot}$ be an algebraic structure such that $\odot$ is both idempotent and entropic.
Then $\struct {S, \odot}$ is a self-distributive structure.
== Proof ==
and:
Hence the result by definition of self-distributive structure.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.27 \ \text {(a)}$
Category:Idempotence
Category:Entropic Structures
Category:Self-Distributive Operations
Entropic Structure with Identity is Commutative Monoid

== Theorem ==
Let $\struct {S, \odot}$ be a magma.
Let $\struct {S, \odot}$ be an entropic structure:
:$\forall a, b, c, d \in S: \paren {a \odot b} \odot \paren {c \odot d} = \paren {a \odot c} \odot \paren {b \odot d}$
Let $\struct {S, \odot}$ have an identity element $e$.
Then $\struct {S, \odot}$ is a commutative monoid.
== Proof ==
We have that $\struct {S, \odot}$ is a magma.
Thus $\struct {S, \odot}$ is closed, and is fulfilled.
Then:
Thus $\odot$ is an associative operation and is fulfilled.
We are given that $e$ is an identity element for $\struct {S, \odot}$.
Hence is fulfilled .
Thus we have that $\struct {S, \odot}$ is a monoid.
Then:
Thus $\struct {S, \odot}$ is a commutative monoid.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.12 \ \text{(e)}$
Category:Entropic Structures
Category:Commutative Monoids
Entropy of Join of Open Covers is Less than Sum of their Entropies

== Theorem ==
Let $X$ be a compact topological space.
Let $\alpha, \beta$ be open covers for $X$.
Let $\alpha \vee \beta$ be the join of $\alpha$ and $\beta$.
Let $\map H \cdot$ denote the entropy.
Then:
:$\map H {\alpha \vee \beta} \le \map H \alpha + \map H \beta$
== Proof==
== Sources ==
* Chapter $7$: Topological Entropy
Category:Definitions/Ergodic Theory
Entropy of Measure-Preserving Transformation with respect to Finite Sub-Sigma Algebra is Well-Defined

== Theorem ==
Let $\struct {X, \BB, \mu}$ be a probability space.
Let $T: X \to X$ be a $\mu$-preserving transformation.
Let $\AA \subseteq \BB$ be a finite sub-$\sigma$-algebra.
Then the entropy of $T$ with respect to $\AA$:
:$\ds \map h {T, \AA} := \lim_{n \mathop \to \infty} \frac 1 n \map H {\bigvee_{k \mathop = 0}^{n - 1} T^{-k} \AA}$
is well-defined.
== Proof ==
Let:
:$\ds a_n := \map H {\bigvee_{k \mathop = 0}^{n - 1} T^{-k} \AA}$
We need to show that the limit:
:$\ds \lim_{n \mathop \to \infty} \frac {a_n} n$
exists.
In view of Fekete's Subadditive Lemma, it suffices to show the subadditivity of $\sequence {a_n}$.
To this end, let $m, n \ge 1$.
Then:
Category:Definitions/Ergodic Theory
Envelope (Plane Geometry)/Examples/Circles with Centers on Circle

== Example of Envelope in context of Plane Geometry ==
Let $\CC$ be a circle with radius $r$ with center at $O$.
Let $\SS$ be the set of circles with radius $a$ whose centers all lie on the circumference of $\CC$.
Then the envelope of $\SS$ consists of:
:a circle with radius $r + a$ with center at $O$
:a circle with radius $\size {r - a}$ with center at $O$.
:
== Proof ==
== Sources ==
* 
* 
Category:Examples of Envelopes (Plane Geometry)
Envelope (Solid Geometry)/Examples/Spheres with Centers on Sphere

== Example of Envelope in context of Solid Geometry ==
Let $\SS$ be a sphere with radius $r$ with center at $O$.
Let $\FF$ be the set of spheres with radius $a$ whose centers all lie on the surface of $\SS$.
Then the envelope of $\FF$ consists of:
:a sphere with radius $r + a$ with center at $O$
:a sphere with radius $\size {r - a}$ with center at $O$.
== Proof ==
== Sources ==
* 
* 
Category:Examples of Envelopes (Solid Geometry)