nyuuzyou/wikis huggingface dataset:
Minimal Element need not be Unique

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
It is possible for $S$ to have more than one minimal element.
== Proof ==
Proof by Counterexample:
Consider the set $S$ defined as:
:$S = \N \setminus \set {0, 1}$
That is, $S$ is the set of natural numbers with $0$ and $1$ removed.
Let $\preccurlyeq$ be the ordering on $S$ defined as:
:$\forall a, b \in S: a \preccurlyeq b \iff a \divides b$
where $a \divides b$ denotes that $a$ is a divisor of $b$.
From Divisor Relation on Positive Integers is Partial Ordering, $\struct {S, \preccurlyeq}$ is a partially ordered set.
Let $p \in S$ be a prime number.
Let $a \divides p$.
By definition of prime number, the only divisors of $p$ are $-p$, $-1$, $1$ and $p$.
Of these, only $p$ is an element of $S$.
Hence if $a \divides p$ it must be the case that $a = p$.
Hence $p$ is a minimal element of $S$ by definition.
Thus every prime number is a minimal element of $S$.
The result follows.
== Examples ==
== Also see ==
* Smallest Element is Unique
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
* 
* 
Category:Minimal Element need not be Unique
Category:Minimal Elements
Minimal Element need not be Unique/Examples/Arbitrary Example 1

== Example of Minimal Element need not be Unique ==
Consider the set $S = \set {a, b, c, d, e}$ with the partial ordering $\preccurlyeq$ defined as:
:${\preccurlyeq} := \set {\tuple {c, a}, \tuple {d, a}, \tuple {e, a}, \tuple {d, b}, \tuple {e, b}, \tuple {c, b}, \tuple {c, e} }$
This can be illustrated using the following Hasse diagram:
:
It can be seen by inspection that both $c$ and $d$ are minimal elements of the partially ordered set $\struct {S, \preccurlyeq}$.
== Sources ==
* 
* 
Category:Minimal Element need not be Unique
Minimal Element of Chain is Smallest Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $C$ be a chain in $S$.
Let $m$ be a minimal element of $C$.
Then $m$ is the smallest element of $C$.
== Proof ==
Let $x \in C$.
Since $m$ is minimal in $C$, $x \nprec m$.
Since $C$ is a chain, $x = m$ or $m \prec x$.
Thus for each $x \in C$, $m \preceq x$.
Therefore $m$ is the smallest element of $C$.
== Also see ==
* Maximal Element of Chain is Greatest Element
== Sources ==
* : $\S\text I.3$
Category:Minimal Elements
Category:Smallest Elements
Minimal Element of an Ordinal

== Theorem ==
The minimal element of any nonempty ordinal is the empty set.
That is, if $S$ is a nonempty ordinal, $\bigcap S = \O$
== Proof ==
Let $S$ be an ordinal.
Let the minimal element of $S$ be $s_0$.
This exists by dint of an ordinal being a woset.
From Ordering on Ordinal is Subset Relation, $S$ is well-ordered by $\subseteq$.
So, by definition of an ordinal:
:$s_0 = \set {s \in S: s \subset s_0}$
But as $s_0$ is minimal, there are no elements of $S$ which are a subset of it.
So:
:$\set {s \in S: s \subset s_0} = \O$
Hence the result.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Exercise $1.7.1$
Category:Ordinals
Minimal Negation Operator/Examples

== Examples of the Minimal Negation Operator ==
Let $\nu_k$ denote the minimal negation operator with $k$ arguments.
Expressed in disjunctive normal form, the first few instances of $\nu_k$ are as follows:
=== Example: $\nu_0$ ===
=== Example: $\nu_1$ ===
=== Example: $\nu_2$ ===
For $k > 2$ there is no immediate correspondence between $\nu_k$ and conventional logical operators.
Category:Examples of Minimal Negation Operator
Minimal Negation Operator/Examples/2

== Example of the Minimal Negation Operator ==
Let $\nu_k$ denote the minimal negation operator with $k$ arguments.
$\nu_2$ is the same operator as the exclusive or operator $\oplus$:
:$\map {\nu_2} {p, q} = p \oplus q$
Category:Examples of Minimal Negation Operator
Minimal Polynomial/Examples/Root (2 plus Root 3)

== Examples of Minimal Polynomials ==
Let $\theta = \sqrt {2 + \sqrt 3}$.
The minimal polynomial of $\theta$ in $\Q$ is $x^4 - 4 x + 1$.
Hence $\index {\map \Q \theta} \Q = 4$.
== Proof ==
From Algebraic Number: $\sqrt {2 + \sqrt 3}$, $\sqrt {2 + \sqrt 3}$ is algebraic over $\R$, and is a root of:
:$\map P x := x^4 - 4 x + 1$
The only possible divisor of $\map P x$ of degree $1$ are $x \pm 1$, which are seen not to actually divide $\map P x$.
Hence if $\map P x$ could be factorised, it would be as:
:$\paren {x^2 + a x + 1} \paren {x^2 + b x + 1}$
or:
:$\paren {x^2 - a x + 1} \paren {x^2 - b x + 1}$
and it is found by inspection that there are no $a, b \in \Z_{>0}$ which would fit.
So $\map P x$ has no divisors with integer coefficients.
It follows from Factors of Polynomial with Integer Coefficients have Integer Coefficients that $a$ and $b$ cannot be rational.
Hence $\map P x$ is irreducible over $\Q$.
Hence the result, by definition of minimal polynomial.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Example $78$
Category:Examples of Minimal Polynomials
Minimal Polynomial Exists

== Theorem ==
Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then there exists a minimal polynomial $f \in K \sqbrk x$ for $\alpha$ over $K$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Minimal Polynomial is Unique
Category:Minimal Polynomial Exists
Category:Minimal Polynomials
Minimal Polynomial Exists/Proof 1

== Theorem ==
== Proof ==
According to definition 1 of minimal polynomial, we ought to find $f \in K \sqbrk x$ such that:
:$f \in K \sqbrk x$ is a monic polynomial of smallest degree such that $\map f \alpha = 0$
Since $\alpha$ is algebraic over $K$, there is some $f \in K \sqbrk x$ such that $\map f \alpha = 0$.
Therefore we can define:
:$n = \min \set { \deg f : f \in K \sqbrk x, \map f \alpha = 0 }$
where $\deg f$ is the degree of $f$.
It follows that there is some $f \in K \sqbrk x$ such that $\deg f = n$.
Then $f$ is of smallest degree such that $\map f \alpha = 0$.
Let the first coefficient of $f$ be $a \in K$.
Since $K$ is a field, it follows that $g := f / a$ is monic.
Therefore $g$ is the sought monic polynomial of smallest degree such that $\map g \alpha = 0$.
Category:Minimal Polynomial Exists
Minimal Polynomial Exists/Proof 2

== Theorem ==
== Proof ==
According to definition 2 of minimal polynomial, we ought to find $f \in K \sqbrk x$ such that:
:$f \in K \sqbrk x$ is an irreducible, monic polynomial such that $\map f \alpha = 0$
Since $\alpha$ is algebraic over $K$, there is some $f \in K \sqbrk x$ such that $\map f \alpha = 0$.
By Polynomial Forms over Field form Unique Factorization Domain, $\map f x$ has a complete factorization:
:$\map f x = a \cdot \map {g_1} x \cdots \map {g_n} x$
where $a \in K, a \ne 0$, and the $g_i$ are irreducible and monic.
Since $\map f \alpha = 0$, it follows that for some $g_i$:
:$\map {g_i} \alpha = 0$
as required.
Category:Minimal Polynomial Exists
Minimal Polynomial is Irreducible

== Theorem ==
Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then the minimal polynomial in $\alpha$ over $K$ is unique and irreducible.
== Proof ==
Let $\map f x$ be a minimal polynomial of $\alpha$ over $K$ of degree $n$.
From Minimal Polynomial is Unique we have that $\map f x$ is unique.
Since $K$ is a field, we may assume that the coefficient of $x^n$ is $1$.
 that $f$ is not irreducible.
Then there exist non-constant polynomials $g, h \in K \sqbrk x$ such that $f = g h$.
By definition of $f$ as the minimal polynomial in $\alpha$:
:$0 = \map f \alpha = \map g \alpha \, \map h \alpha$
From Field is Integral Domain, $L$ is an integral domain.
Therefore, as $\map g \alpha \, \map h \alpha \in L$, either $\map g \alpha = 0$ or $\map h \alpha = 0$.
This contradicts the minimality of the degree of $f$.
Hence the result, by Proof by Contradiction.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $71 \ \text{(i)}$
Category:Minimal Polynomials
Minimal Polynomial is Unique

== Theorem ==
Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then the minimal polynomial of $\alpha$ over $K$ is unique.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Minimal Polynomial Exists
Category:Minimal Polynomial is Unique
Category:Minimal Polynomials
Minimal Polynomial is Unique/Proof 1

== Theorem ==
== Proof ==
Let $f$ be a minimal polynomial of $\alpha$ over $K$ according to definition 1.
By Minimal Polynomial is Irreducible, we have that $f$ is irreducible over $K$. 
Let $g$ be another polynomial in $K \sqbrk x$ such that $\map g \alpha = 0$.
By the definition of minimal polynomial, $\map \deg f \le \map \deg g$, where $\deg$ denotes degree.
By Division Theorem for Polynomial Forms over Field, there exist polynomials $q, r \in K \sqbrk x$ such that:
:$g = q f + r$ and $\map \deg r < \map \deg f$.
Suppose $\map \deg r > 0$.
Then evaluating both sides of the equation above at $\alpha$, we obtain $\map r \alpha = 0$.
This contradicts the minimality of the degree of $f$.
Thus, $r$ is constant and equal to $0$.
We have now shown that $f$ divides all polynomials in $K \sqbrk x$ which vanish at $\alpha$.
By the monic restriction, it then follows that $f$ is unique.
Category:Minimal Polynomial is Unique
Minimal Polynomial is Unique/Proof 2

== Theorem ==
== Proof ==
Let $f, g \in K \sqbrk x$ be minimal polynomials for $\alpha$ according to definition 2.
That is, let $f$ and $g$ be irreducible monic polynomials in $K \sqbrk x$ with $\map f \alpha = \map g \alpha = 0$.
Suppose $f$ and $g$ are distinct.
Then $f$ and $g$ are coprime.
Thus there exist polynomials $a, b \in K \sqbrk x$ with $a f + b g = 1$.
Taking the evaluation homomorphism in $\alpha$, we obtain the contradiction that $0=1$.
Category:Minimal Polynomial is Unique
Minimal Polynomial of Element with Finite Orbit under Group of Automorphisms over Fixed Field in terms of Orbit

== Theorem ==
Let $E$ be a field.
Let $G \le \Aut E$ be a subgroup of its automorphism group.
Let $F = \map {\operatorname {Fix_E} } G$ be its fixed field.
Let $\alpha \in E$ have a finite orbit under $G$.
Then $\alpha$ is algebraic over $F$ and the product of polynomials
:$\ds \map p x = \prod_{\beta \mathop \in \Lambda} \paren {x - \beta}$
is the minimal polynomial of $\alpha$ over $F$.
== Proof ==
By Product over Finite Set with Zero Factor, we have $\map p \alpha = 0$.
By definition, $p \in E \sqbrk x$.
=== $p$ has coefficients in $F$ ===
We show that $p \in F \sqbrk x$.
Let $\sigma \in G$, and denote the induced automorphism of $E[x]$ still by $\sigma$.
We show that $\map \sigma p = p$.
We have:
Because this is true for all $\sigma \in G$, indeed the coefficients of $p$ are in $\Fix G = F$.
Thus $p \in F \sqbrk x$.
=== $p$ is the minimal polynomial ===
By Product of Monic Polynomials is Monic, $p$ is monic.
In particular, $p$ is nonzero.
Thus $\alpha$ is algebraic over $F$.
Let $f$ be its minimal polynomial over $F$.
We show that $f = p$.
By definition and because $\map p \alpha = 0$, $f$ divides $p$.
By Group of Automorphisms is Contained in Automorphism Group over Fixed Field, $G \le \Aut {E/F}$.
By Automorphism Group of Field Extension Permutes Roots of Minimal Polynomial, each $\beta \in \Lambda$ is a root of $f$.
By Polynomial Factor Theorem, each $x - \beta$ divides $f$.
By Product of Pairwise Coprime Divisors of Polynomial over Field is Divisor, $p$ divides $f$.
By Monic Polynomials are Equal iff Divide Each Other, $p = f$.
Category:Field Extensions
Minimal Polynomial of Square Matrix/Examples/Multiple of Identity

== Example of Minimal Polynomial of Square Matrix ==
Let $\mathbf A$ be a square matrix of order $n$ such that:
:$\mathbf A = k \mathbf I_n$
where $\mathbf I_n$ denotes the identity matrix of order $n$
Then the minimal polynomial of $\mathbf A$ is:
:$\mathbf x - k \mathbf I_n$
== Proof ==
We have that:
$\mathbf A = k \mathbf I_n$
from which:
:$\mathbf x - k \mathbf I_n = \mathbf 0$
Hence the result by definition of minimal polynomial.
== Sources ==
* 
Category:Examples of Minimal Polynomials of Square Matrices
Minimal Polynomial of Square Matrix/Examples/Order 2

== Example of Minimal Polynomial of Square Matrix ==
Let $\mathbf A$ be a square matrix of order $2$.
Let $\mathbf A$ be neither:
:a scalar multiple of the identity matrix of order $2$
nor:
:the zero matrix of order $2$.
Then the minimal polynomial of $\mathbf A$ is:
:$\mathbf x^2 - a \mathbf x + b \mathbf I$
where:
:$a = \map \tr {\mathbf A}$ denotes the trace of $A$ 
:$b = \map \det {\mathbf A}$ denotes the determinant of $A$
:$\mathbf I$ denotes the $2 \times 2$ identity matrix.
== Proof ==
First we demonstrate that $\mathbf A^2 - a \mathbf A + b \mathbf I = \mathbf 0$.
Let $\mathbf A = \begin {pmatrix} a & b \\ c & d \end {pmatrix}$.
Then:
It follows after algebra that $\mathbf A^2 - a \mathbf A + b \mathbf I = \mathbf 0$.
== Sources ==
* 
Category:Examples of Minimal Polynomials of Square Matrices
Minimal Polynomial of Square Matrix is Divisor of Characteristic Polynomial

== Theorem ==
Let $\mathbf A$ be a square matrix of order $n$.
The minimal polynomial of $\mathbf A$ is a divisor of the characteristic polynomial of $\mathbf A$.
== Proof ==
== Sources ==
* 
Category:Minimal Polynomials of Square Matrices
Category:Characteristic Polynomial of Matrix
Minimal Prime/Examples

== Examples of Minimal Primes ==
$881$ is a minimal prime in base $10$ because there is no prime among the proper subsequences of the digits: $8$, $1$, $88$, $81$.
The subsequence does not have to consist of consecutive digits.
Hence $149$ is not a minimal prime in base $10$, because $19$ is prime.
However, by the definition of subsequence, the digits do have to be in the same order.
So, for example, $991$ is still a minimal prime in base $10$, even though a subset of the digits can form the shorter prime $19$ by changing the order.
Category:Minimal Primes
Minimal Prime/Sequence

== Sequence ==
The complete sequence of minimal primes base $10$ is:
:$2$, $3$, $5$, $7$, $11$, $19$, $41$, $61$, $89$, $409$, $449$, $499$, $881$, $991$, $6469$, $6949$, $9001$, $9049$, $9649$, $9949$, $60 \, 649$, $666 \, 649$, $946 \, 669$, $60 \, 000 \, 049$, $66 \, 000 \, 049$, $66 \, 600 \, 049$
Category:Minimal Primes
Minimal Ring Generated by System of Sets

== Theorem ==
Let $\SS$ be a non-empty system of sets.
Then there is a unique ring of sets $\map \RR \SS$ which:
:$(1): \quad$ contains $\SS$
:$(2): \quad$ is contained by every ring of sets which also contains $\SS$.
This ring of sets $\map \RR \SS$ is called the minimal ring generated by $\SS$.
== Proof ==
=== Uniqueness ===
Suppose there were two such rings of sets $\map \RR \SS$ and $\map \RR \SS'$.
Then by definition their intersection $\map \RR \SS \cap \map \RR \SS'$ would also contain $\SS$.
By Intersection of Rings of Sets, $\map \RR \SS \cap \map \RR \SS'$ is also a ring of sets.
From Intersection is Subset, $\map \RR \SS \cap \map \RR \SS' \subseteq \map \RR \SS$ and $\map \RR \SS \cap \map \RR \SS' \subseteq \map \RR \SS'$.
Hence either $\map \RR \SS$ or $\map \RR \SS'$ can not be minimal.
So if $\map \RR \SS$ exists, it has to be unique.
=== Existence ===
Consider the union:
:$\ds X = \bigcap_{A \mathop \in \SS} A$
of all sets in $\SS$.
Now consider the power set $\powerset X$ of all subsets of $X$.
From Power Set is Algebra of Sets and by definition of algebra of sets, $\powerset X$ is a ring of sets containing $\SS$.
Now let $\Sigma$ be the set of all rings of sets contained in $\powerset X$ which also contain $\SS$.
Then consider the intersection:
:$\ds \map \RR \SS = \bigcap_{\TT \mathop \in \Sigma} \TT$
of all these rings of sets.
It is clear that $\map \RR \SS$ has the properties specified.
In particular, as all $\TT \in \Sigma$ contain $\SS$, then so does $\map \RR \SS$.
Finally, note that if $\RR'$ is a ring of sets containing $\SS$, then:
:$\RR^* = \RR' \cap \powerset X$
is a ring of sets in $\Sigma$.
Hence $\map \RR \SS \subseteq \RR^* \subseteq \RR'$ as we needed to show.
Category:Set Systems
Minimal Smooth Surface Spanned by Contour

== Theorem ==
Let $\map z {x, y}: \R^2 \to \R$ be a real-valued function.
Let $\Gamma$ be a closed contour in $3$-dimensional Euclidean space.
Then the smooth surface of least area spanned by the contour $\Gamma$ has to satisfy the following Euler's equation:
:$r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} = 0$
where:
with subscript denoting respective partial derivatives.
In other words, its mean curvature has to vanish.
== Proof ==
The surface area for a smooth surface embedded in $3$-dimensional Euclidean space is given by:
:$\ds A \sqbrk z = \iint_\Gamma \sqrt {1 + z_x^2 + z_y^2} \rd x \rd y$
It follows that:
By Euler's equation:
:$\dfrac {r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} } {\paren {1 + p^2 + q^2}^{\frac 3 2} } = 0$
Due to the smoothness of the surface, $1 + p^2 + q^2$ is bounded.
Hence, the following equation is sufficient:
:$r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} = 0$
Introduce the following change of variables:
Then Euler's equation can be rewritten as:
:$\dfrac {E g - 2 F f + G e} {2 \paren {E G - F^2} } = 0$
By definition, mean curvature is:
:$M = \dfrac {E g - 2 F f + G e} {2 \paren {E G - F^2} }$
Hence:
:$M = 0$
== Sources ==
* : $\S 1.5$: The Case of Several Variables
Category:Calculus of Variations
Minimal Smooth Surface of Revolution

== Theorem ==
Let $\map y x$ be a real mapping in 2-dimensional real Euclidean space.
Let $y$ pass through the points $\tuple {x_0, y_0}$ and $\tuple {x_1, y_1}$.
Consider a surface of revolution constructed by rotating $y$ around the $x$-axis.
Suppose this surface is smooth for any $x$ between $x_0$ and $x_1$.
Then its surface area is minimized by the following curve, known as a catenoid:
:$y = C \map \cosh {\dfrac {x + C_1} C}$
Furthermore, its area is:
:$A = \paren {x_1 - x_0} C \pi + \dfrac {\pi C^2} 2 \paren {\map \sinh {\dfrac {2 \paren {x_1 + C_1} } C} - \map \sinh {\dfrac {2 \paren {x_0 + C_1} } C} }$
== Proof ==
The area functional of the surface of revolution is:
:$\ds A \sqbrk y = 2 \pi \int_{x_0}^{x_1} y \sqrt {1 + y'^2} \rd x$
The integrand does not depend on $x$.
By Euler's Equation:
:$F - y' F_{y'} = C$
that is:
:$y \sqrt {1 + y'^2} - \dfrac {y y'^2} {\sqrt {1 + y'^2} } = C$
which is equivalent to:
:$y = C \sqrt {1 + y'^2}$
Solving for $y'$ yields:
 
:$y' = \sqrt {\dfrac {y^2 - C^2} {C^2} }$
Consider this as a differential equation for $\map x y$:
:$\dfrac {\d x} {\d y} = \dfrac C {\sqrt {y^2 - C^2} }$
Integrate this once:
:$x + C_1 = C \ln {\dfrac {y + \sqrt {y^2 - C^2} } C}$
Solving for $y$ results in:
:$y = C \cosh {\dfrac {x + C_1} C}$
To find the area, substitute this function into the formula for the area:
== Sources ==
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
Category:Calculus of Variations
Minimal Uncountable Well-Ordered Set Unique up to Isomorphism

== Theorem ==
Let $\Omega, \Omega'$ be minimal uncountable well-ordered sets.
Then $\Omega$ is order isomorphic to $\Omega'$.
That is, the minimal uncountable well-ordered set is unique up to order isomorphism.
== Proof ==
From Wosets are Isomorphic to Each Other or Initial Segments, precisely one of the following holds:
:$\Omega$ is order isomorphic to $\Omega'$
or:
:$\Omega$ is order isomorphic to an initial segment in $\Omega'$
or:
:$\Omega'$ is order isomorphic to an initial segment in $\Omega$.
By the definition of minimal uncountable well-ordered set, $\Omega$ and $\Omega'$ are uncountable.
The initial segments of $\Omega$ and $\Omega'$ are countable.
An uncountable set cannot be order isomorphic to a countable set.
Thus $\Omega$ must be order isomorphic to $\Omega'$.
The result follows from the definition of uniqueness.
== Sources ==
* $\S \text P.17$
Category:Well-Orderings
Category:Uncountable Sets
Minimal WRT Restriction

== Theorem ==
Let $A$ be a set or class.
Let $\RR$ be a relation on $A$.
Let $B$ be a subset or subclass of $A$.
Let $\RR'$ be the restriction of $\RR$ to $B$.
Let $m \in B$.
Then:
:$m$ is a strictly minimal element under $\RR$ in $B$
:
:$m$ is a strictly minimal element under $\RR'$ in $B$.
== Proof ==
=== Sufficient Condition ===
Let $m$ be a strictly minimal element under $\RR$ in $B$.
Let $x$ be any element of $B$.
 that $x \mathrel {\RR'} m$.
Then since $\RR' \subseteq \RR$:
:$x \mathrel \RR m$
contradicting the fact that $m$ is a strictly minimal element under $\RR$ in $B$.
Thus:
:$\lnot \paren {x \mathrel {\RR'} m}$
As this holds for all $x \in B$, $m$ is a strictly minimal element under $\RR'$ in $B$.
=== Necessary Condition ===
Let $m$ be a strictly minimal element under $\RR'$ in $B$.
Let $x \in B$.
 that $x \mathrel \RR m$.
Then $x, m \in B$.
Therefore:
:$\tuple {x, m} \in B \times B$
Thus:
:$\tuple {x, m} \in \RR \cap \paren {B \times B} = \RR'$
so $x \mathrel {\RR'} m$
This contradicts the fact that $m$ is a strictly minimal element under $\RR'$ in $B$.
Thus:
:$\lnot \paren {x \mathrel \RR m}$
As this holds for all $x \in B$, it follows that $m$ is a strictly minimal element under $\RR$ in $B$.
Category:Restrictions
Minimally Closed Class under Progressing Mapping

== Theorem ==
Statement of Conditions:
Then the following results hold:
=== Minimally Closed Class under Progressing Mapping induces Nest ===
=== Bounded Subset of Minimally Closed Class under Progressing Mapping has Greatest Element ===
=== Fixed Point of Progressing Mapping on Minimally Closed Class is Greatest Element ===
=== Minimally Closed Class under Progressing Mapping is Well-Ordered ===
=== Smallest Element of Minimally Closed Class under Progressing Mapping ===
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.17$
Category:Minimally Closed Classes
Category:Progressing Mappings
Category:Minimally Closed Classes under Progressing Mapping
Minimally Closed Class under Progressing Mapping induces Nest

== Theorem ==
For all $x, y \in N$:
:either $\map g x \subseteq y$ or $y \subseteq x$
and $N$ forms a nest:
:$\forall x, y \in N: x \subseteq y$ or $y \subseteq x$
== Proof ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.17 \ (1)$
Category:Minimally Closed Classes under Progressing Mapping
Minimally Closed Class under Progressing Mapping induces Nest/Proof

== Theorem ==
== Proof ==
Let $\RR$ be the relation on $N$ defined as:
:$\forall x, y \in N: \map \RR {x, y} \iff \map g x \subseteq y \lor y \subseteq x$
We are given that $g$ is a progressing mapping.
From the Progressing Function Lemma, we have that:
In what follows, $(1)$ is not needed.
From the Double Induction Principle for Minimally Closed Class, if $\RR$ is a relation on $N$ which satisfies:
then $\map \RR {x, y}$ holds for all $x, y \in N$.
As $\Dom g = N$ it is seen that $(\text D 2)$ is fulfilled by $(2)$ directly.
It remains to be demonstrated that $(\text D 1)$ holds.
Indeed, from Smallest Element of Minimally Closed Class under Progressing Mapping, we have that:
:$\forall x \in N: b \subseteq x$
By the Rule of Addition:
:$\forall x \in N: \map g x \subseteq b \lor b \subseteq x$
That is:
:$\forall x \in N: \map \RR {x, b}$
Thus by the Double Induction Principle for Minimally Closed Class:
:$\forall x, y \in N: \map \RR {x, y}$
That is:
:$\forall x, y \in N: \map g x \subseteq y \lor y \subseteq x$
As $g$ is a progressing mapping, we have:
:$x \subseteq \map g x$
Hence from:
:$\map g x \subseteq y \lor y \subseteq x$
we have:
:$x \subseteq y \lor y \subseteq x$
Hence the result by definition of a nest.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Exercise $4.1$
Category:Minimally Closed Classes under Progressing Mapping