nyuuzyou/wikis huggingface dataset:
Transfinite Recursion Theorem/Formulation 1

== Theorem ==
Let $M$ be a class.
Let $g$ be a strictly progressing mapping on $M$.
Let $M$ be minimally superinductive under $g$.
For an arbitrary ordinal $\alpha$, let $M_\alpha$ be the $\alpha$th element of $M$ under the well-ordered class $\struct {M, \subseteq}$.
Then:
where:
:$\On$ denotes the class of all ordinals
:$K_{II}$ denotes the class of all limit ordinals.
== Proof 1 ==
== Proof 2 ==
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem/Formulation 1/Proof 1

== Theorem ==
== Proof ==
Let $\On$ denote the class of all ordinals.
By the Superinductive Class under Strictly Progressing Mapping is Proper Class, $M$ is not a set; it is a proper class.
From Minimally Superinductive Class is Well-Ordered under Subset Relation, $\struct {M, \subseteq}$ is indeed a well-ordered class.
By definition, $M$ is a $g$-tower.
Hence from $g$-Tower is Well-Ordered under Subset Relation:
:$(1): \quad \O$ is the smallest element of $M$.
:$(2): \quad$ The immediate successor of $x$ under the well-ordering $\subseteq$ is $\map g x$.
:$(3): \quad$ For a limit element $x$ of $M$:
::::$x = \bigcup x^\subset$
:::where $\bigcup x^\subset$ denotes the union of the lower section of $x$.
From the Counting Theorem, there exists a unique order isomorphism from $\On$ to $M$.
Let $M_\alpha$ denote the element of $M$ corresponding to $\alpha$ under this order isomorphism.
We refer to $M_\alpha$ as the "$\alpha$th element of $M$".
By the definition of order isomorphism, it is immediate that $M_0$ is the smallest element of $M$ under the well-ordering induced by $\subseteq$.
Hence:
:$M_0 = \O$
It is also immediate that $M_{\alpha^+}$ is the immediate successor of $M_\alpha$, that is:
:$M_{\alpha^+} = \map g {M_\alpha}$
Let $\lambda$ be a limit ordinal.
Then $M_\lambda$ is a limit element of $M$:.
Hence $M_\lambda$ is the union of the set of all $M_\alpha$ such that $\alpha < \lambda$.
That is:
:$\ds M_\lambda = \bigcup_{\alpha \mathop < \lambda} M_\alpha$
Hence the result.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Theorem $5.1$
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem/Formulation 1/Proof 2

== Theorem ==
== Proof ==
This is a special case of Transfinite Recursion Theorem: Formulation $4$.
Take $c$ to be $0$, that is $\O$.
Let us take the special case where $g$ is a strictly progressing mapping.
The result follows directly.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem/Formulation 2

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $S$ denote the class of all ordinal sequences.
Let $g$ be a mapping such that $S \subseteq \Dom g$.
Then there exists a unique mapping $F$ on $\On$ such that:
:$\forall \alpha \in \On: \map F \alpha = \map g {F \restriction \alpha}$
where $F \restriction \alpha$ denotes the restriction of $F$ to $\alpha$.
== Proof ==
First we establish the following
=== Lemma ===
By Characteristic of Extending Operation, there exists a mapping $F$ on $\On$ such that:
:$\forall \alpha \in \On: F \restriction \alpha^+ = \map E {F \restriction \alpha}$
where:
:$F \restriction \alpha$ denotes the restriction of $F$ to $\alpha$
:$\alpha^+$ denotes the successor ordinal of $\alpha$.
Then:
It remains to demonstrate uniqueness of $F$.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Theorem $5.5$
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem/Formulation 2/Lemma

== Lemma for Transfinite Recursion Theorem: Formulation $2$ ==
Let $\On$ denote the class of all ordinals.
Let $S$ denote the class of all ordinal sequences.
Let $g$ be a mapping such that $S \subseteq \Dom g$.
There exists an extending operation $E$ such that for every mapping $F$ on $\On$:
:$\forall \alpha \in \On: \map {\paren {\map E {F \restriction \alpha} } } \alpha = \map g {F \restriction \alpha}$
== Proof ==
Let $\theta_\alpha$ be an arbitrary ordinal sequence of length $\alpha$.
Let $\map E {\theta_\alpha}$ be the extending operation defined as:
:$\map E {\theta_\alpha} := \theta_\alpha \cup \set {\tuple {\alpha, \map g {\theta_\alpha} } }$
Thus $\map E {\theta_\alpha}$ is an ordinal sequence of length $\alpha^+$.
Also, for $\beta \in \On$ such that $\beta < \alpha$:
:$\map {\paren {\map E {\theta_\alpha} } } \beta = \map {\theta_\alpha} \beta$
but:
:$\map {\paren {\map E {\theta_\alpha} } } \alpha = \map g {\theta_\alpha}$
Then for any mapping $F$ on $\On$:
:$F \restriction \alpha$ is an ordinal sequence of length $\alpha$.
Hence:
:$\map {\paren {\map E {F \restriction \alpha} } } \alpha = \map g {F \restriction \alpha}$
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Lemma $5.4$
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem/Formulation 3

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $h$ and $f$ be mappings defined for all sets.
Let $c$ be a set.
Then there exists a unique mapping $F$ on $\On$ such that:
where $K_{II}$ denotes the class of all limit ordinals.
== Proof ==
For an arbitrary ordinal sequence $\theta$, let us define $\map g \theta$ by the following conditions:
By the Transfinite Recursion Theorem: Formulation 2, there exists a unique mapping $F$ on $\On$ such that:
:$\forall \alpha \in \On: \map F \alpha = \map g {F \restriction \alpha}$
Hence we have:
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Theorem $5.6$
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem/Formulation 4

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $g$ be a mapping defined for all sets.
Let $c$ be a set.
Then there exists a unique $\On$-sequence $S_0, S_1, \dots, S_\alpha, \dots$ such that:
where $K_{II}$ denotes the class of all limit ordinals.
== Proof ==
This is a special case of the Transfinite Recursion Theorem: Formulation $3$.
Recall:
Set $\map f x = \bigcup x$, so that:
:$\map f {F \sqbrk \lambda}$
is then:
:$\ds \bigcup_{\alpha \mathop < \lambda} \map F \alpha$
Let $S_\alpha$ be the set $\map F \alpha$
and the given form of the Transfinite Recursion Theorem follows.
It remains to demonstrate uniqueness of $S_0, S_1, \dots, S_\alpha, \dots$.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Theorem $5.7$
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem/Formulation 5

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $h$ be a mapping.
Then there exists a unique mapping $F$ such that:
:$\forall \alpha \in \On: \map F \alpha = \map h {F \sqbrk \alpha}$
== Proof ==
This is a special case of the Transfinite Recursion Theorem: Formulation $2$.
Recall:
Consider $\map g x$.
If $x$ is a mapping, take $\map g x = \map h {\Img x}$.
If $x$ is not a mapping, take $\map g x = x$.
The result follows.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Theorem $5.8$
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem/Theorem 1

== Theorem ==
Let $G$ be a (class) mapping from $\On^{\On}$ to $\On$.
Let $K$ be a class of mappings $f$ that satisfy: 
:the domain of $f$ is some ordinal $y$
:$\forall x \in y: \map f x = \map G {f {\restriction_x} }$
where $f {\restriction_x}$ denotes the restriction of $f$ to $x$.
Let $F = \bigcup K$ be the union of $K$.
Then:
:$(1): \quad F$ is a mapping with domain $\On$
:$(2): \quad \forall x \in \On: \map F x = \map G {F {\restriction_x} }$
:$(3): \quad F$ is unique. That is, if another mapping $A$ has the above two properties, then $A = F$.
=== Corollary ===
== Proof ==
First a lemma:
=== Lemma ===
=== $K$ is a chain ===
First, it is necessary to show that $K$ is a chain of mappings under the subset relation.
Take any $f, g \in K$.
Set $B$ equal to the domain of $f$ and set $C$ equal to the domain of $g$.
From Relation between Two Ordinals: 
:$B \subseteq C \lor C \subseteq B$
From Lemma 1:
:$\forall x \in B \cap C: \map f x = \map g x$
So depending on whether $B \subseteq C$ or $C \subseteq B$:
:$f \subseteq g \lor g \subseteq f$
From Subset Relation is Ordering, it follows that $\subseteq$ is a total ordering on $K$.
=== Domain of $F$ is an ordinal ===
$K$ is a chain of mappings.
Since the union of a chain of mappings is a mapping, $F$ is a mapping.
Moreover, $\forall x \in B: \map f x = \map F x$, because $F$ is an extension of the mapping $f$.
The domain of $F$ is a subset of the class of all ordinals because it is the union of a collection of mappings whose domains are themselves subsets of ordinals.
From this it follows that $\Dom F$ is an ordinal, since it is a transitive subset of $\On$.
=== Domain of $F$ is $\On$ ===
Assume the domain of $F$ is a set.
Since $F$ is a mapping, the range of $F$ is a set and $F$ is a set.
Then $\Dom F = \gamma$ for some ordinal $\gamma$.
Define $C$ as follows: 
:$C = F \cup \set {\tuple {\gamma, \map G {F {\restriction_\gamma} } } }$
Then $C$ is a mapping with domain $\gamma^+$ and satisfies:
:$\forall x < \gamma^+: \map C x = \map G {F {\restriction_\gamma} }$
Therefore $C$ is a member of $K$ (note that $F {\restriction_\gamma} = C {\restriction_\gamma}$).
: $\ds C \in K \implies C \subseteq \bigcup K$
So $C \subseteq F$.
But if $F$ is a set, $F \subsetneq C$, a contradiction.
Therefore, $\Dom F$ cannot be a set and must therefore be the set of all ordinals $\On$.
=== Value of $F$ is $\map G {F {\restriction_\alpha} }$ ===
:$\forall x \in \On: \exists f \in K: \map f x = \map F x$
Since $f \in K$: 
:$\map f x = \map G {f {\restriction_x} }$
But $\forall y < x: \map f y = \map F y$, so by Equality of Restrictions: 
:$f {\restriction_x} = F {\restriction_x}$
Therefore:
:$\map F \alpha = \map G {F {\restriction_\alpha} }$
=== $F$ is Unique ===
Finally, assume there is another mapping $H$ that satisfies the first two properties.
Then $H$ is equal to $F$ by transfinite induction:
So for all ordinals:
:$\map H y = \map F y$
Thus:
:$H = F$
== Also see ==
* Well-Founded Recursion
* Transfinite Induction
* Burali-Forti Paradox
== Sources ==
* : $\S 7.40$
* 
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem/Theorem 2

== Theorem ==
Let $\Dom x$ denote the domain of $x$.
Let $\Img x$ denote the image of the mapping $x$.
Let $G$ be a class of ordered pairs $\tuple {x, y}$ satisfying at least one of the following conditions:
:$(1): \quad x = \O$ and $y = a$
:$(2): \quad \exists \beta: \Dom x = \beta^+$ and $y = \map H {\map x {\bigcup \Dom x} }$
:$(3): \quad \Dom x$ is a limit ordinal and $y = \bigcup \Rng x$.
Let $\map F \alpha = \map G {F \restriction \alpha}$ for all ordinals $\alpha$.
Then:
:$F$ is a mapping and the domain of $F$ is the class of all ordinals, $\On$.
:$\map F \O = a$
:$\map F {\beta^+} = \map H {\map F \beta}$
:For limit ordinals $\beta$, $\ds \map F \beta = \bigcup_{\gamma \mathop \in \beta} \map F \gamma$
:$F$ is unique.
::That is, if there is another function $A$ satisfying the above three properties, then $A = F$.
== Proof ==
We can proceed in the fourth part by Transfinite Induction.
=== Basis for the Induction ===
This proves the basis for the induction.
=== Induction Step ===
This proves the induction step.
=== Limit Case ===
This proves the limit case.
== Sources ==
* : $\S 7.42$
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem/Uniqueness

== Theorem ==
Let $G$ be a mapping.
Let $f$ be a mapping with a domain $y$ where $y$ is an ordinal.
Let $f$ satisfy the condition that:
:$\forall x \in y: \map f x = \map G {f \restriction x}$
where $f \restriction x$ denotes the restriction of $f$ to $x$.
Let $g$ be a mapping with a domain $z$ where $z$ is an ordinal.
Let $g$ satisfy the condition that:
:$\forall x \in z: \map g x = \map G {g \restriction x}$
Let $y \subseteq z$.
Then:
:$\forall x \in y: \map f x = \map g x$
== Proof ==
Proof by transfinite induction:
Suppose that:
:$\forall x \in \alpha: \map f x = \map g x$
for some arbitrary ordinal $\alpha < y$.
Then $\alpha < z$.
Hence:
So applying induction:
:$\forall \alpha < y: \map f \alpha = \map g \alpha$
== Sources ==
* 
Category:Transfinite Recursion Theorem
Transfinite Recursion Theorem without Axiom of Replacement implies Counting Theorem

== Theorem ==
Let the Transfinite Recursion Theorem (Formulation $5$) be accepted as axiomatic.
Let the Axiom of Replacement not be accepted.
Then the Counting Theorem holds.
That is, it is possible to prove the Counting Theorem by using the Transfinite Recursion Theorem (Formulation $5$) as an axiom in place of the Axiom of Replacement.
== Proof ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Exercise $5.1$
Category:Transfinite Recursion Theorem
Category:Counting Theorem
Category:Axiom of Replacement
Transformation Law for Connection Coefficients

== Theorem ==
Let $M$ be a smooth manifold with or without boundary.
Let $TM$ be the tangent bundle of $M$.
Let $\nabla$ be a connection in $TM$.
Let $U \subseteq M$ be an open subset.
Let $\tuple {E_i}$ and $\tuple {\tilde E_j}$ be smooth local frames for $TM$ on $U$.
Suppose $\tuple {E_i}$ and $\tuple {\tilde E_j}$ are related by:
:$\tilde E_i = A^j_i E_j$
where Einstein summation convention has been imposed.
where $\paren {A^j_i}$ is a matrix of functions.
Let $\Gamma^k_{ij}$ and $\tilde \Gamma^k_{ij}$ denote connection coefficients $\tuple {E_i}$ and $\tuple {\tilde E_i}$.
Then:
:$\ds \tilde \Gamma^k_{ij} = \paren{A^{-1}}^k_p A^q_i A^r_j \Gamma^p_{qr} + \paren{A^{-1}}^k_p A^q_i \map {E_q} {A^p_j}$
== Proof ==
== Sources ==
* : $\S 4$: Connections. Connections in the Tangent Bundle
Category:Smooth Manifolds
Category:Connections
Transformation Matrix/Examples/Cartesian Plane

== Example of Transformation Matrix ==
Let $\CC_\alpha$ and $\CC_\beta$ be cartesian coordinate systems over a plane.
Let $\begin {pmatrix} x \\ y \end {pmatrix}$ be the column vector representing an arbitrary point $P$ under $\CC_\alpha$.
Let $\begin {pmatrix} x' \\ y' \end {pmatrix}$ be the column vector representing $P$ in $\CC_\beta$.
Then:
:$\begin {pmatrix} x' \\ y' \end {pmatrix} = \mathbf T \begin {pmatrix} x \\ y \end {pmatrix}$
where $\mathbf T$ denotes the transformation matrix of the coordinate transformation from $\CC_\alpha$ to $\CC_\beta$.
== Sources ==
* 
* 
Category:Examples of Transformation Matrices
Transformation Matrix/Examples/Extension in Direction of X-Axis

== Example of Transformation Matrix ==
Let $\CC$ be a cartesian plane.
Let $P = \begin {pmatrix} x \\ y \end {pmatrix}$ be the column vector representing an arbitrary point $P$ in $\CC$.
Let $P' = \begin {pmatrix} x' \\ y' \end {pmatrix}$ be the column vector representing $P$ after an extension in the positive direction of the $x$-axis.
Let $\mathbf T$ denote the transformation matrix to convert $P$ to $P'$.
Then:
:$\mathbf T = \begin {pmatrix} k & 0 \\ 0 & 1 \end {pmatrix}$
== Proof ==
== Sources ==
* 
* 
Category:Examples of Transformation Matrices
Transformation Matrix/Examples/Plane Reflection in X-Axis

== Example of Transformation Matrix ==
Let $\CC$ be a cartesian plane.
Let $P = \begin {pmatrix} x \\ y \end {pmatrix}$ be the column vector representing an arbitrary point $P$ in $\CC$.
Let $P' = \begin {pmatrix} x' \\ y' \end {pmatrix}$ be the column vector representing $P$ after a reflection in the $x$-axis.
Let $\mathbf T$ denote the transformation matrix to convert $P$ to $P'$.
Then:
:$\mathbf T = \begin {pmatrix} 1 & 0 \\ 0 & -1 \end {pmatrix}$
== Proof ==
== Sources ==
* 
* 
Category:Examples of Transformation Matrices
Transformation of P-Norm

== Theorem ==
Let $p, q \ge 1$ be real numbers.
Let ${\ell^p}_\R$ denote the $p$-sequence space on $\R$.
Let $\norm {\mathbf x}_p$ denote the $p$-norm of $\mathbf x$.
Let $\mathbf x = \sequence {x_n} \in {\ell^{p q} }_\R$.
Suppose further that $\mathbf x^p = \sequence { {x_n}^p} \in {\ell^q}_\R$.
Then:
:$\norm {\mathbf x^p}_q = \paren {\norm {\mathbf x}_{p q} }^p$
== Proof ==
Category:Functional Analysis
Category:Norm Theory
Category:P-Sequence Metrics
Category:P-Norms
Transformation of Unit Matrix into Inverse

== Theorem ==
Let $\mathbf A$ be a square matrix of order $n$ of the matrix space $\map {\MM_\R} n$.
Let $\mathbf I$ be the unit matrix of order $n$.
Suppose there exists a sequence of elementary row operations that reduces $\mathbf A$ to $\mathbf I$.
Then $\mathbf A$ is nonsingular.
Futhermore, the same sequence, when performed on $\mathbf I$, results in the inverse of $\mathbf A$.
== Proof ==
For ease of presentation, let $\breve {\mathbf X}$ be the inverse of $\mathbf X$.
We have that $\mathbf A$ can be transformed into $\mathbf I$ by a sequence of elementary row operations.
By repeated application of Elementary Row Operations as Matrix Multiplications, we can write this assertion as:
From Elementary Row Matrix is Nonsingular:
:$\mathbf E_1, \dotsc, \mathbf E_t \in \GL {n, \R}$
We can multiply on the left both sides of this equation by:
By repeated application of Elementary Row Operations as Matrix Multiplications, each $\mathbf E_n$ on the corresponds to an elementary row operation.
Hence the result.
Category:Unit Matrices
Category:Inverse Matrices
Category:Elementary Row Operations
Transitive Chaining

== Theorem ==
Let $\RR$ be a transitive relation on a set $S$.
Let $n \in \N$ be a natural number.
Let $n \ge 2$.
Let $\sequence {x_k}_{k \mathop \in \set {1, 2, \dots, n} }$ be a sequence of $n$ terms.
For each $k \in \set {1, 2, \dots, n - 1}$, let $x_k \mathrel \RR x_{k + 1}$.
That is, let $x_1 \mathrel \RR x_2$, $x_2 \mathrel \RR x_3, \dotsc, x_n - 1 \mathrel \RR x_n$.
Then:
:$x_1 \mathrel \RR x_n$
== Proof ==
The proof proceeds by induction on $n$, the number of terms in the sequence.
We first define a propositional function, $P$, as follows:
For each $n \in \N$ such that $n \ge 2$, let $\map P n$ be the proposition that if both of the following hold:
:$\sequence {x_k}_{k \mathop \in \set {1, 2, \dots, n} }$ is a sequence of $n$ terms
:$\forall k \in \set {1, 2, \dots, n - 1}: x_k \mathrel \RR x_{k + 1}$
then:
:$x_1 \mathrel \RR x_n$
=== Basis for the Induction ===
The case $n = 2$ is verified as follows:
In this case, the sequence has only two elements, $x_1$ and $x_2$.
Thus by the premise, $x_1 \mathrel \RR x_2$, so $\map P 2$ holds.
This is the basis for the induction.
=== Induction Hypothesis ===
Fix $n \in \N$ with $n \ge 2$.
Suppose that $\map P n$ holds.
This is our induction hypothesis.
=== Induction Step ===
This is our induction step:
Let $\sequence {x_k}_{k \mathop \in \set {1, 2, \dots, n, n + 1} }$ be a sequence of $n + 1$ terms.
For each $k \in \set {1, 2, \dots, n - 1, n}$, let $x_k \mathrel \RR x_{k + 1}$.
In particular:
:$x_n \mathrel \RR x_{n + 1}$
By the induction hypothesis:
:$x_1 \mathrel \RR x_n$.
Thus since $\RR$ is transitive:
:$x_1 \mathrel \RR x_{n + 1}$
We conclude that $\map P {n + 1}$ holds.
The result follows by the Principle of Mathematical Induction.
Category:Transitive Relations
Transitive Class/Examples

== Examples of Transitive Classes ==
=== Empty Class ===
=== Singleton of Empty Class ===
=== Class $\set {\O, \set \O}$ ===
=== Class $\set {\O, \set \O, \set {\O, \set \O} }$ ===
=== Class $\set {\O, \set {\O, \set \O} }$ ===
=== Singleton of Singleton of Empty Class is not Transitive ===
Category:Examples of Transitive Classes
Transitive Class/Examples/Class with Empty Class and its Singleton

== Example of Transitive Class ==
Let $\O$ denote the empty class.
Then the class:
:$\set {\O, \set \O}$
is transitive.
== Proof ==
Consider the elements of $\set {\O, \set \O}$ in turn.
There are no elements of $\O$.
There is exactly one elements of $\set \O$, and that is $\O$.
We see that $\O$ is itself an element of $\set {\O, \set \O}$.
Thus all elements of elements of $\set {\O, \set \O}$ are themselves elements of $\set {\O, \set \O}$.
Hence by definition $\set {\O, \set \O}$ is transitive.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity
Category:Examples of Transitive Classes
Transitive Class/Examples/Ordinal 3

== Example of Transitive Class ==
Let $\O$ denote the empty class.
Consider the ordinal $3$, defined as:
:$\mathcal 3 := \set {\O, \set \O, \set {\O, \set \O} }$
$\mathcal 3$ is transitive.
== Proof ==
Consider the $3$ elements of $\mathcal 3$ in turn.
There are no elements of $\O$.
There is exactly one elements of $\set \O$, and that is $\O$.
We see that $\O$ is itself an element of $\set {\O, \set \O, \set {\O, \set \O} }$.
The elements of $\set {\O, \set \O}$, are $\O$ and $\set \O$.
We see that both are element of $\mathcal 3$.
Thus all elements of elements of $\mathcal 3$ are themselves elements of $\mathcal 3$.
Hence by definition $\mathcal 3$ is transitive.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity
Category:Examples of Transitive Classes
Transitive Class/Examples/Ordinal 3 without 1 is not Transitive

== Example of Non-Transitive Class ==
Let $\O$ denote the empty class.
Consider the class $S$, defined as:
:$S := \set {\O, \set {\O, \set \O} }$
$S$ is not transitive.
== Proof ==
$S$ has $2$ elements: $\O$ and $\set {\O, \set \O}$.
Note that one of the elements of $\set {\O, \set \O}$ is $\set \O$.
But $\set \O$ is not itself an element of $S$.
Thus not all elements of elements of $S$ are themselves elements of $S$.
Hence by definition $S$ is not transitive.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity
Category:Examples of Transitive Classes
Transitive Class/Examples/Singleton of Singleton of Empty Class is not Transitive

== Example of Non-Transitive Class ==
Let $\O$ denote the empty set.
Consider the class $S$, defined as:
:$S := \set {\set \O}$
$S$ is not transitive.
== Proof ==
$S$ has $1$ element: $\set \O$.
$\set \O$ in turn has one element: $\O$.
But $\O$ is not itself an element of $S$.
Thus not all elements of elements of $S$ are themselves elements of $S$.
Hence by definition $S$ is not transitive.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity: Exercise $10.1 \ \text {(a)}$
Category:Examples of Transitive Classes
Transitive Class of Ordinals is Subset of Ordinal not in it

== Theorem ==
Let $A$ be a transitive class of ordinals.
Let $\alpha$ be an ordinal which is not an element of $A$.
Then:
:$A \subseteq \alpha$
== Proof ==
Let $A$ and $\alpha$ be .
Let $\beta \in A$ be arbitrary.
Because $\beta \in A$ and $\alpha \notin A$ we have that:
:$\beta \ne \alpha$
Because $\beta \in A$ and $A$ is transitive:
:all elements of $\beta$ are in $A$
But because $\alpha \notin A$:
:$\alpha \notin \beta$
Thus we have:
:$\alpha \notin \beta$
and:
:$\alpha \ne \beta$
Hence from Ordinal Membership is Trichotomy:
:$\beta \in \alpha$
As $\beta$ is arbitrary, it follows that:
:$A \subseteq \alpha$
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 2$ Ordinals and transitivity: Lemma $2.1$
Category:Ordinals
Category:Transitive Classes
Transitive Closure is Closure Operator

== Definition ==
Let $S$ be a set.
Let $\TT$ be the set of all endorelations on $S$.
Then the transitive closure operator is a closure operator on $\TT$.
== Proof ==
Let $\QQ$ be the set of transitive relations on $S$.
By Intersection of Transitive Relations is Transitive, the intersection of any subset of $\QQ$ is in $\QQ$.
Recall the definition of transitive closure as the intersection of transitive supersets:
:The transitive closure of a relation $\RR$ on $S$ is the intersection of elements of $\QQ$ that contain $S$.
From Closure Operator from Closed Sets we conclude that transitive closure is a closure operator.
Category:Transitive Closures
Category:Closure Operators