nyuuzyou/wikis huggingface dataset:
Dual Matroid is Matroid

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Then the dual $M^*$ of $M$ is a matroid.
== Proof==
Let $\mathscr B$ be the set of bases of the matroid $M$.
From Matroid Bases Satisfy Formulation 4 Base Axiom:
:$\mathscr B$ satisfies formulation 4 base axiom.
By formulation 5 base axiom:
:$\mathscr B$ satisfies formulation 5 base axiom.
Let $\mathscr B^* = \set{S \setminus B : B \in \mathscr B}$.
From Subsets Satisfy Formulation 5 Matroid Base Axiom Iff Complements Satisfy Formulation 1:
:$\mathscr B^*$ satisfies formulation 1 base axiom.
From Matroid Bases Iff Satisfies Formulation 1 of Matroid Base Axiom:
:there exists a matroid $M^* = \struct{S, \mathscr I^*}$ whose bases is the set:
::$\mathscr B^* = \set{S \setminus B : B \in \mathscr B}$.
From Matroid is Uniquely Defined by Bases:
:$M^*$ is the unique matroid whose bases is the set:
::$\mathscr B^* = \set{S \setminus B : B \in \mathscr B}$.
The result follows.
== Sources ==
* Chapter $2.$ $\S 1.$ The Dual Matroid, Theorem $1$
* Chapter $2.$ Duality, $\S 2.1.$ The definition and basic properties, Theorem $2.1.1$
Category:Dual Matroids
Dual Operator is Bounded Linear Transformation

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $X^\ast$ and $Y^\ast$ be the normed duals of $X$ and $Y$ respectively.
Then the dual operator $T^\ast : Y^\ast \to X^\ast$ is a bounded linear transformation.
Further: 
:$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } = \norm T_{\map B {X, Y} }$
== Proof ==
Let $f, g \in Y^\ast$ and $\alpha, \beta \in \GF$.
Then we have, for each $x \in X$: 
So we have: 
:$\map {T^\ast} {\alpha f + \beta g} = \alpha T^\ast f + \beta T^\ast g$
In Dual Operator is Well-Defined, it was shown that for each $f \in Y^\ast$ and $x \in X$ we have: 
:$\cmod {\map {\paren {T^\ast f} } x} \le \norm f_{Y^\ast} \norm T_{\map B {X, Y} } \norm x_X$
So we have: 
:$\norm {T^\ast f}_{X^\ast} \le \norm T_{\map B {X, Y} } \norm f_{Y^\ast}$
for $f \in Y^\ast$. 
Hence: 
:$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } \le \norm T_{\map B {X, Y} }$
We finally need to show that $\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } = \norm T_{\map B {X, Y} }$.
Pick a sequence $\sequence {x_n}_{n \in \N}$ such that $\norm {x_n}_X = 1$ and: 
:$\ds \norm T_{\map B {X, Y} } - \frac 1 n \le \norm {T x_n}_Y \le \norm T_{\map B {X, Y} }$
so that: 
:$\norm {T x_n}_Y \to \norm T_{\map B {X, Y} }$
For each $n \in \N$, let $f_n \in Y^\ast$ be a support functional for $T x_n$.
Then we have: 
:$\map {\paren {T^\ast f_n} } {x_n} = \map {f_n} {T x_n} = \norm {T x_n}_Y$
so that: 
:$\ds \cmod {\map {\paren {T^\ast f_n} } {x_n} } \ge \paren {\norm T_{\map B {X, Y} } - \frac 1 n} \norm {x_n}_X$
We therefore have: 
:$\ds \paren {\norm T_{\map B {X, Y} } - \frac 1 n} \norm {f_n}_{Y^\ast} \le \norm {T^\ast f_n}_{X^\ast}$
so that: 
:$\ds \norm {T^\ast}_{\map B {Y^\ast, X^\ast} } \ge \norm T_{\map B {X, Y} } - \frac 1 n$
for each $n \in \N$.
Then: 
:$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } \ge \norm T_{\map B {X, Y} }$
With that, we obtain: 
:$\norm {T^\ast}_{\map B {Y^\ast, X^\ast} } = \norm T_{\map B {X, Y} }$
== Sources ==
* : Proposition $2.28$
Category:Dual Operators
Category:Bounded Linear Transformations
Dual Operator is Weak-* to Weak-* Continuous

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ and $\struct {Y^\ast, \norm {\, \cdot \,}_{Y^\ast} }$ be the normed duals of $X$ and $Y$ respectively.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$.
Then $T^\ast$ is $\struct {w^\ast, w^\ast}$-continuous.
== Proof ==
From Continuity in Initial Topology, it is enough to show that for each $\Phi \in \struct {X^\ast, w^\ast}^\ast$, we have: 
:$\Phi \circ T^\ast : \struct {Y^\ast, w^\ast} \to \GF$ is Continuous.
From Characterization of Continuity of Linear Functional in Weak-* Topology‎, we have $\Phi = x^\wedge$ for some $x \in X$, where $x^\wedge$ is the evaluation linear transformation evaluated at $x$.
For $f \in Y^\ast$, we have: 
So we have: 
:$x^\wedge \circ T^\ast = \paren {T x}^\wedge$
with $T x \in Y$. 
From Characterization of Continuity of Linear Functional in Weak-* Topology ‎, $\paren {T x}^\ast : \struct {Y^\ast, w^\ast} \to \GF$ is continuous.
So $x^\wedge \circ T^\ast : \struct {Y^\ast, w^\ast} \to \GF$ is continuous for each $x \in X$.
So from Continuity in Initial Topology, $T^\ast$ is $\struct {w^\ast, w^\ast}$-continuous.
Category:Dual Operators
Category:Weak-* Topologies
Dual Operator is Well-Defined

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $X^\ast$ and $Y^\ast$ be the normed duals of $X$ and $Y$ respectively.
Then the dual operator $T : Y^\ast \to X^\ast$ is well-defined.
== Proof ==
We first show that $T^\ast$ is well-defined as a mapping $Y^\ast \to X^\ast$. 
That is, we want to show that $T^\ast f \in X^\ast$ for each $f \in Y^\ast$.
For $x, y \in X$ and $\lambda, \mu \in \GF$, we have: 
So $T^\ast f$ is a linear functional.
We show that $T^\ast f$ is bounded.
For each $x \in X$, we have:
So $T^\ast f$ is bounded.
== Sources ==
* : Definition $2.27$
Category:Dual Operators
Dual Operator of Composition

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$, $Y$ and $Z$ be normed vector spaces over $\GF$.
Let $X^\ast$, $Y^\ast$ and $Z^\ast$ be the normed dual spaces of $X$, $Y$ and $Z$ respectively. 
Let $T : X \to Y$ and $S : Y \to Z$ be bounded linear transformations.
Let $T^\ast : Y^\ast \to X^\ast$ and $S^\ast : Z^\ast \to Y^\ast$ be the dual operators of $T$ and $S$ respectively. 
Then: 
:$\paren {S T}^\ast = T^\ast S^\ast$
where $\paren {S T}^\ast$ is the dual operator of $S T : X \to Z$. 
== Proof ==
Let $f \in Z^\ast$.
Then, we have: 
== Sources ==
* : Proposition $2.28$
Category:Dual Operators
Dual Ordering is Ordering

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\succeq$ denote the dual ordering of $\preceq$.
Then $\succeq$ is an ordering on $S$.
== Proof ==
By definition of ordering, $\preceq$ is reflexive, transitive and antisymmetric.
By definition, $\succeq$ is the inverse relation to $\preceq$.
By Inverse of Reflexive Relation is Reflexive, $\succeq$ is reflexive.
By Inverse of Antisymmetric Relation is Antisymmetric, $\succeq$ is antisymmetric.
By Inverse of Transitive Relation is Transitive, $\succeq$ is transitive.
Thus by definition $\succeq$ is an ordering on $S$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.2$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.8$
* : $\S \text{I}.2$: Theorem $2$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $27 \ \text {(a)}$
* : $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.14 \ \text{(b)}$
Category:Dual Orderings
Dual Pairs (Order Theory)

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $a, b \in S$, and let $T \subseteq S$.
Then the following phrases about, and concepts pertaining to $\left({S, \preceq}\right)$ are dual to one another:
:
$b \preceq a$ 		$a \preceq b$
$a$ succeeds $b$ 		$a$ precedes $b$
$a$ strictly succeeds $b$ 		$a$ strictly precedes $b$
$a$ is an upper bound for $T$ 		$a$ is a lower bound for $T$
$a$ is a supremum for $T$ 		$a$ is an infimum for $T$
$a$ is a maximal element of $T$ 		$a$ is a minimal element of $T$
$a$ is the greatest element 		$a$ is the smallest element
the weak lower closure $a^\preceq$ of $a$ 		the weak upper closure $a^\succeq$ of $a$
the strict lower closure $a^\prec$ of $a$ 		the strict upper closure $a^\succ$ of $a$
the strict lower closure $T^\prec$ of $T$ 		the strict upper closure $T^\succ$ of $T$
the join $a \vee b$ of $a$ and $b$ 		the meet $a \wedge b$ of $a$ and $b$
$T$ is a lower section in $S$ 		$T$ is an upper section in $S$
$\struct{S, \vee, \preceq}$ is a join semilattice 		$\struct{S, \wedge, \preceq}$ is a meet semilattice
$\struct{S, \preceq}$ is a complete join semilattice 		$\struct{S, \preceq}$ is a complete meet semilattice
$a$ is a join irreducible element 		$a$ is a meet irreducible element
$a$ is a join prime element 		$a$ is a meet prime element
$T$ is a filter of $\struct {S, \preceq}$ 		$T$ is an ideal of $\struct {S, \preceq}$
$T$ is a completely prime ideal of $\struct{S, \preceq}$		$T$ is a completely prime filter of $\struct{S, \preceq}$
== Proof ==
Let $\succeq$ be the dual ordering of $\preceq$.
By definition of dual statement:
:$b \preceq a$
is dual to:
:$b \succeq a$
and by definition of dual ordering, this is equivalent to:
:$a \preceq b$
By virtue of Dual of Dual Statement (Order Theory), the converse follows.
The other claims are proved on the following pages, in order:
:Succeed is Dual to Precede
:Strictly Succeed is Dual to Strictly Precede
:Upper Bound is Dual to Lower Bound
:Supremum is Dual to Infimum
:Maximal Element is Dual to Minimal Element
:Greatest Element is Dual to Smallest Element
:Weak Lower Closure is Dual to Weak Upper Closure
:Strict Lower Closure is Dual to Strict Upper Closure
:Join is Dual to Meet
:Lower Section is Dual to Upper Section
:Join Semilattice is Dual to Meet Semilattice
:Complete Join Semilattice is Dual to Complete Meet Semilattice
:Join Irreducible Element is Dual of Meet Irreducible Element
:Join Prime Element is Dual of Meet Prime Element
:Ideal is Dual of Filter (Order Theory)
:Completely Prime Ideal is Dual of Completely Prime Filter
== Also see ==
* Duality Principle
Category:Order Theory
Category:Dual Pairs (Order Theory)
Dual Representation Preserves Natural Pairings Between Vectors and Their Duals

==Definition==
Where $\rho$ is a representation of $G$ on a finite-dimensional vector space $V$ -- that is, $\rho$ is a homomorphism from $G$ to $GL(V)$ -- and the dual space $V^*$ is the set of homomorphisms of $V$, let the dual representation $\rho:V^* \rightarrow V^*$, a permutation of $V^*$, be defined by the formula $\rho^*(g)=\rho(g^{-1})^T$.
Recognizing the natural pairing between elements of $V$ and $V^*$, we will refer to a pairing function, $n(v^*)=v$, between these elements, which we will also refer to by the notation $\langle v^*, v\rangle$.
==Theorem==
We want to show that for all $g \in G$, $v \in V$, and $v^* \in V^*$ such that $\langle v^*, v\rangle$, $ \langle \rho^*(g)(v^*),\rho(g)(v) \rangle$; in other words, that $n(\rho^*(g)(v^*)) = \rho(g)(v)$. To that end, we will use properties and definitions to rewrite $n(\rho^*(g)(v^*))$.
Applying transitivity, we have $n(\rho^*(g)(v^*)) = \rho(g)v$; that is, $\langle \rho^*(g)(v^*) = \rho(g)v \rangle$ as intended.
Dual of Dual Matroid Equals Matroid

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $M^*$ denote the dual of $M$.
Then:
:$\paren{M^*}^* = M$
That is, the dual $\paren{M^*}^*$ of $M^*$ is $M$.
== Proof==
Let $\mathscr B$ denote the set of bases of $M$.
Let $\paren{\mathscr B^*}^*$ denote the set of bases of $\paren{M^*}^*$.
By definition of dual matroid:
:
We have:
Let $\paren{\mathscr I^*}^*$ denote the independent subsets of $\paren{M^*}^*$.
We have:
By definition of matroid:
:$\paren{M^*}^* = M$
== Sources ==
* Chapter $2.$ $\S 1.$ The Dual Matroid, Theorem $1$
* Chapter $2.$ Duality, $\S 2.1.$ The definition and basic properties, Theorem $2.1.3$
Category:Dual Matroids
Dual of Dual Ordering

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\struct {S, \succeq}$ be its dual.
Then the dual of $\struct {S, \succeq}$ is again $\struct {S, \preceq}$.
== Proof ==
Denote with $\preceq'$ the dual of $\succeq$.
By definition of dual ordering, we thus have for all $a, b \in S$:
:$a \preceq b$ $b \succeq a$
:$b \succeq a$ $a \preceq' b$
Hence $a \preceq b$ $a \preceq' b$.
The result follows from Equality of Relations.
== Sources ==
* 
Category:Dual Orderings
Dual of Dual Statement (Order Theory)

== Theorem ==
Let $\Sigma$ be a statement about ordered sets.
Let $\Sigma^*$ be its dual statement.
Then $\Sigma$ is also the dual statement of $\Sigma^*$.
== Proof ==
By definition, the dual statement $\Sigma^*$ is formed by replacing the ordering $\preceq$ with its dual $\succeq$.
By Dual of Dual Ordering, applying this operation twice results in the original sentence $\Sigma$ again.
Category:Order Theory
Dual of Lattice Ordering is Lattice Ordering

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a lattice.
Let $\preccurlyeq$ be the lattice ordering on $\struct {S, \preccurlyeq}$.
Then its dual ordering $\succcurlyeq$ is also a lattice ordering.
== Proof ==
Let $\struct {S, \preccurlyeq}$ be a lattice.
It is to be shown that:
:for all $x, y \in S$, the ordered set $\struct {\set {x, y}, \succcurlyeq}$ admits both a supremum and an infimum. 
Let $x, y \in S$.
Then $\struct {\set {x, y}, \preccurlyeq}$ admits both a supremum and an infimum.
Let $c = \map \sup {\set {x, y}, \preccurlyeq}$.
Then by definition of supremum:
:$\forall s \in \set {x, y}: s \preccurlyeq c$
:$\forall d \in S: c \preccurlyeq d$
where $d$ is an upper bound of $\struct {\set {x, y}, \preccurlyeq} \subseteq S$.
Hence by definition of dual ordering:
:$\forall s \in \set {x, y}: c \succcurlyeq s$
:$\forall d \in S: d \succcurlyeq c$
where $d$ is an upper bound of $\struct {\set {x, y}, \preccurlyeq} \subseteq S$.
By Upper Bound is Lower Bound for Inverse Ordering, $d$ is a lower bound of $\struct {\set {x, y}, \succcurlyeq} \subseteq S$.
So by definition of infimum:
:$c = \map \inf {\set {x, y}, \succcurlyeq}$
That is, $\struct {\set {x, y}, \succcurlyeq}$ admits an infimum.
Let $c = \map \inf {\set {x, y}, \preccurlyeq}$.
Then by definition of infimum:
:$\forall s \in \set {x, y}: c \preccurlyeq s$
:$\forall d \in S: d \preccurlyeq c$
where $d$ is a lower bound of $\struct {\set {x, y}, \preccurlyeq} \subseteq S$.
Hence by definition of dual ordering:
:$\forall s \in \set{x, y}: s \succcurlyeq c$
:$\forall d \in S: c \succcurlyeq d$
where $d$ is a lower bound of $\struct {\set {x, y}, \preccurlyeq} \subseteq S$.
By Lower Bound is Upper Bound for Inverse Ordering, $d$ is an upper bound of $\struct {\set {x, y}, \succcurlyeq} \subseteq S$.
So by definition of supremum:
:$c = \map \sup {\set {x, y}, \succcurlyeq}$
That is, $\struct {\set {x, y}, \succcurlyeq}$ admits a supremum.
Hence $\struct {\set {x, y}, \succcurlyeq}$ admits both a supremum and an infimum.
That is, $\succcurlyeq$ is a lattice ordering.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.4$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.10$
Category:Lattice Theory
Category:Dual Orderings
Dual of Order Type is Well-Defined Mapping

== Theorem ==
The dual operation on order types is a well-defined mapping.
== Proof ==
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.
Let $\struct {S_1, \succcurlyeq_1}$ and $\struct {S_2, \succcurlyeq_2}$ denote the dual ordered sets of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$.
Let $\struct {S_1, \preccurlyeq_1} \cong \struct {S_2, \preccurlyeq_2}$, where $\cong$ denotes order isomorphism.
Let $\varrho = \map \ot {S_1, \preccurlyeq_1}$ denote the order type of $\struct {S_1, \preccurlyeq_1}$.
By definition of order type:
:$\varrho = \map \ot {S_1, \preccurlyeq_1} = \map \ot {S_2, \preccurlyeq_2}$
as $\struct {S_1, \preccurlyeq_1} \cong \struct {S_2, \preccurlyeq_2}$
Let $\varrho^*$ denote the denote the dual of $\varrho$.
Thus by definition:
:$\varrho^* = \map \ot {S_1, \succcurlyeq_1}$
From Order Types of Duals of Isomorphic Sets are Equal:
:$\varrho^* = \map \ot {S_2, \succcurlyeq_2}$
The result follows by definition of well-defined mapping.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Dual Orderings
Dual of Ordered Semigroup is Ordered Semigroup

== Theorem ==
Let $\struct {S, \circ, \preccurlyeq}$ be an ordered semigroup.
Then its dual $\struct {S, \circ, \succcurlyeq}$ is also an ordered semigroup.
== Proof ==
From Dual Ordering is Ordering, we have that $\struct {S, \succcurlyeq}$ is an ordered set.
We also note from the definition that $\struct {S, \circ}$ is a semigroup.
It remains to be demonstrated that $\succcurlyeq$ is compatible with $\circ$.
Recall that $\struct {S, \circ, \preccurlyeq}$ is an ordered semigroup.
Hence $\preccurlyeq$ is compatible with $\circ$.
Let $x, y \in S$ be arbitrary such that $x \succcurlyeq y$.
We have:
and similarly:
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.3$
Category:Ordered Semigroups
Category:Dual Orderings
Dual of Preordered Set is Preordered Set

== Theorem ==
Let $P = \struct {S, \preceq}$ be a preordered set.
Then dual of $P$, $P^{-1} = \struct {S, \succeq}$ is also a preordered set.
== Proof ==
By Inverse of Reflexive Relation is Reflexive:
:$\succeq$ is reflexive.
By Inverse of Transitive Relation is Transitive:
:$\succeq$ is transitive.
Hence $\succeq$ is a preordering.
== Sources ==
* 
* 
Category:Dual Orderings
Dual of Total Ordering is Total Ordering

== Theorem ==
Let $\preccurlyeq$ be a total ordering.
Then its dual ordering $\succcurlyeq$ is also a total ordering.
== Proof ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
From Dual Ordering is Ordering we have that $\succcurlyeq$ is an ordering.
Let $x, y \in S$.
Then $x \preccurlyeq y$ or $y \preccurlyeq x$ since $\preccurlyeq$ is total.
By definition of dual ordering, also $y \succcurlyeq x$ or $x \succcurlyeq y$.
Hence the result, by definition of total ordering.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.4$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.10$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $27 \ \text {(b)}$
Category:Total Orderings
Category:Dual Orderings
Dual of Well-Ordering is not necessarily Well-Ordering

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a well-ordered set.
Then its dual $\struct {S, \preccurlyeq}$ is not necessarily also a well-ordered set.
== Proof ==
Consider the ordered structure $\struct {\N, \le}$.
From the Well-Ordering Principle, $\struct {\N, \le}$ is a well-ordered set..
Consider the dual $\struct {\N, \ge}$ of $\struct {\N, \le}$.
Let this be expressed as:
:$\struct {\N, \preccurlyeq} := \struct {\N, \ge}$
so as to enhance the clarification of the nature of the ordering in question.
From Dual of Total Ordering is Total Ordering, $\struct {\N, \preccurlyeq}$ is a totally ordered set.
 $\struct {\N, \preccurlyeq}$ is a well-ordered set.
Then there exists a minimal element of $\struct {\N, \preccurlyeq}$:
:$\exists m \in \N: \forall x \in \N: x \preccurlyeq m \implies x = m$
But consider $y = m + 1$.
We have:
:$y \ge m$ but $y \ne m$
That is:
:$y \preccurlyeq m$ but $y \ne m$
and so $m$ cannot be a minimal element of $\struct {\N, \preccurlyeq}$ after all.
It follows that $\struct {\N, \preccurlyeq}$ is not a well-ordered set.
The result follows by Proof by Contradiction
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.10$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $27 \ \text {(c)}$
Category:Well-Orderings
Category:Dual Orderings
Duality Law for Stirling Numbers

== Theorem ==
For all integers $n, m \in \Z$:
:$\ds {n \brace m} = {-m \brack -n}$
where:
:$\ds {n \brace m}$ denotes a Stirling number of the second kind
:$\ds {n \brack m}$ denotes an unsigned Stirling number of the first kind.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(58)$
Category:Stirling Numbers
Duality Principle

* Set Theory:
** Duality Principle for Sets
* Order Theory:
** Duality Principle (Order Theory)
* Abstract Algebra:
** Duality Principle (Boolean Algebras)
** Duality Principle (Huntington Algebras)
* Category Theory:
** Duality Principle (Category Theory)
* Projective Geometry:
** Duality Principle (Projective Geometry)
== Also known as ==
== Also see ==
* Definition:Dual
Duality Principle (Boolean Algebras)

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Then any theorem in $\struct {S, \vee, \wedge, \neg}$ remains valid if both $\vee$ and $\wedge$ are interchanged, and also $\bot$ and $\top$ are interchanged throughout the whole theorem.
== Proof ==
Let us take the axioms of a Boolean algebra $\struct {S, \wedge, \vee, \neg}$:
It can be seen by inspection that exchanging $\wedge$ and $\vee$, and $\bot$ and $\top$ throughout, does not change the axioms.
Thus, what you get is a Boolean algebra again.
Hence the result.
== Sources ==
* : $\S 1.5$: Theorem $1.14$
* 
* 
* 
* 
Category:Boolean Algebras
Category:Named Theorems
Duality Principle (Category Theory)/Conceptual Duality

== Theorem ==
Let $\Sigma$ be a statement about metacategories, be it in natural language or otherwise.
Suppose that $\Sigma$ holds for all metacategories.
Then so does its dual statement $\Sigma^*$.
== Proof ==
From Dual Category of Dual Category, any metacategory $\mathbf C$ may be regarded as a dual category.
Thus it is sufficient to verify $\Sigma^*$ holds in all dual categories.
In any dual category $\mathbf C^{\mathrm {op}}$, we see that:
:$\operatorname{dom} f^{\mathrm {op}} = \paren {\operatorname{cod} f}^{\mathrm {op}}$
:$\operatorname{cod} f^{\mathrm {op}} = \paren {\operatorname{dom} f}^{\mathrm {op}}$
:$f^\mathrm {op} \circ g^\mathrm {op} = \paren {g \circ f}^{\mathrm {op}}$
Thus, the interpretation of $\Sigma^*$ in $\mathbf C^{\mathrm {op}}$ comes down to the interpretation of $\Sigma$ in $\mathbf C$.
That is, except for amending every symbol $\star$ of $\mathbf C$ with a superscript "$\mathrm {op}$", yielding $\star^\mathrm {op}$.
This operation does not change the validity of $\Sigma$ in any possible way, and $\Sigma$ was assumed to hold in $\mathbf C$.
Hence we conclude that $\Sigma^*$ must hold in $\mathbf C^{\mathrm {op}}$ as well.
The result follows.
== Sources ==
* : $\S 3.1$: Proposition $3.2$
Category:Duality Principle (Category Theory)
Duality Principle (Category Theory)/Formal Duality

== Theorem ==
=== Morphisms-Only Category Theory ===
Let $\Sigma$ be a statement in the language of category theory.
Suppose $\Sigma$ is provable from the axioms for morphisms-only category theory $\mathrm{MOCT}$:
:$\mathrm{MOCT} \vdash \Sigma$
Then the dual statement $\Sigma^*$ is also provable from these axioms, i.e.:
:$\mathrm{MOCT} \vdash \Sigma^*$
=== Object Category Theory ===
Let $\mathrm{CT}$ be the collection of seven axioms on Characterization of Metacategory via Equations.
Suppose a statement $\Sigma$ about metacategories follows from the axioms $\mathrm{CT}$.
Then so does its dual statement $\Sigma^*$.
== Proof ==
=== Proof for Morphisms-Only Category Theory ===
From a formal perspective, if one would have derived (for some collection of statements $\Delta$):
:$\Delta \vdash \Sigma$
without using the axioms, then because $R_\circ$, $\operatorname{dom}$ and $\operatorname{cod}$ would still be undefined, it follows that necessarily also:
:$\Delta^* \vdash \Sigma^*$
In this correspondence, taking $\Delta$ to be $\mathrm{CT}$, it follows that:
:$\mathrm{CT} \vdash \Sigma$
implies:
:$\mathrm{CT}^* \vdash \Sigma^*$
As the Category Axioms are Self-Dual, in that $\mathrm{CT}^* = \mathrm{CT}$, we obtain the result.
=== Proof for Object Category Theory ===
== Sources ==
* : $\S 3.1$: Proposition $3.1$
Category:Duality Principle (Category Theory)
Duality Principle (Order Theory)

== Theorem ==
Let $\Sigma$ be a statement about ordered sets (whether in natural or a formal language).
Let $\Sigma^*$ be the dual statement of $\Sigma$.
=== Local Duality ===
=== Global Duality ===
== Also see ==
* Definition:Dual Statement (Order Theory)
* Dual Pairs, which allows for easier determination of dual statements.
* Duality Principle (Category Theory), a more general duality principle.
Category:Order Theory
Category:Named Theorems