nyuuzyou/wikis huggingface dataset:
Trace Sigma-Algebra of Generated Sigma-Algebra

== Theorem ==
Let $X$ be a Set, and let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Let $A \subseteq X$ be a subset of $X$.
Then the following equality holds:
:$A \cap \map \sigma \GG = \map \sigma {A \cap \GG}$
where
:$\map \sigma \GG$ denotes the smallest $\sigma$-algebra on $X$ that contains $\GG$
:$\map \sigma {A \cap \GG}$ denotes the smallest $\sigma$-algebra on $A$ that contains ${A \cap \GG}$
:$A \cap \map \sigma \GG$ denotes the trace $\sigma$-algebra on $A$
:$A \cap \GG$ is a shorthand for $\set {A \cap G: G \in \GG}$
== Proof ==
By definition of generated $\sigma$-algebra:
:$\GG \subseteq \map \sigma \GG$
whence from Set Intersection Preserves Subsets:
:$A \cap \GG \subseteq A \cap \map \sigma \GG$
and therefore, by definition of generated $\sigma$-algebra:
:$\map \sigma {A \cap \GG} \subseteq A \cap \map \sigma \GG$
For the reverse inclusion, define $\Sigma$ by:
:$\Sigma := \set {E \subseteq X: A \cap E \in \map \sigma {A \cap \GG} }$
We will show that $\Sigma$ is a $\sigma$-algebra on $X$.
Since $A \in \map \sigma {A \cap \GG}$:
:$A \cap X = A \in \map \sigma {A \cap \GG}$
and therefore $X \in \Sigma$.
Suppose that $E \in \Sigma$.
Then by Set Intersection Distributes over Set Difference and Intersection with Subset is Subset:
:$\paren {X \setminus E} \cap A = \paren {X \cap A} \setminus \paren {E \cap A} = A \setminus \paren {E \cap A}$
Since $E \cap A \in \map \sigma {A \cap \GG}$ and this is a $\sigma$-algebra on $A$:
:$A \setminus \paren {E \cap A} \in \map \sigma {A \cap \GG}$
Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\Sigma$.
Then by Intersection Distributes over Union:
:$\ds \paren {\bigcup_{n \mathop \in \N} E_n} \cap A = \bigcup_{n \mathop \in \N} \paren {E_n \cap A}$
The latter expression is a countable union of elements of $\map \sigma {A \cap \GG}$, hence again in $\map \sigma {A \cap \GG}$.
Therefore, $\Sigma$ is a $\sigma$-algebra.
It is also apparent that $\GG \subseteq \Sigma$ since:
:$A \cap \GG \subseteq \map \sigma {A \cap \GG}$
by definition of generated $\sigma$-algebra.
Thus, as $\Sigma$ is a $\sigma$-algebra:
:$\map \sigma \GG \subseteq \Sigma$
and therefore:
:$A \cap \map \sigma \GG \subseteq \map \sigma {A \cap \GG}$
Hence the result, by definition of set equality.
Category:Trace Sigma-Algebras
Category:Sigma-Algebras Generated by Collection of Subsets
Trace Sigma-Algebra of Measurable Set

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $A \in \Sigma$. 
Let $\Sigma_A$ be the trace $\sigma$-algebra of $A$ in $\Sigma$.
Then: 
:$\Sigma_A = \set {B \in \Sigma : B \subseteq A}$
That is, the elements of $\Sigma_A$ are precisely the $\Sigma$-measurable sets that are subsets of $A$. 
== Proof ==
Let: 
:$\Sigma' = \set {B \in \Sigma : B \subseteq A}$
We first show that:
:$\Sigma_A \subseteq \Sigma'$.
Let $B \in \Sigma_A$, then there exists $S \in \Sigma$ such that: 
:$B = A \cap S$
From Sigma-Algebra Closed under Finite Intersection, we have: 
:$B \in \Sigma$
while, from Intersection is Subset, we have: 
:$A \cap S \subseteq A$
so:
:$B \subseteq A$
giving: 
:$B \in \Sigma'$
So we have:
:$\Sigma_A \subseteq \Sigma'$
by the definition of set inclusion.
Now let $B \in \Sigma'$.
From Intersection with Subset is Subset, we have: 
:$B = B \cap A$ 
while $B \in \Sigma$, so: 
:$B \in \Sigma_A$
from the definition of trace $\sigma$-algebra.
So we have:
:$\Sigma' \subseteq \Sigma_A$
by the definition of set inclusion.
So, we have: 
:$\Sigma_A = \Sigma' = \set {B \in \Sigma : B \subseteq A}$
Category:Trace Sigma-Algebras
Trace in Terms of Dual Basis

== Theorem ==
Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of dimension $n$.
Let $\tuple {e_1, \ldots, e_n}$ be a basis of $M$.
Let $\tuple {e_1^*,\ldots, e_n^*}$ be its dual basis
Let $f: M \to M$ be a linear operator.
Then its trace equals:
:$\map \tr f = \ds \sum_{i \mathop = 1}^n e_i^* \paren {\map f {e_i} }$
== Proof ==
Let $\ds \map f {e_i} = \sum_{j \mathop = 1}^n c_{ij} e_j$
Let $A$ be the matrix relative to the basis $\tuple {e_1, \ldots, e_n}$.
Then by the above assumption:
:$A_{ij} = c_{ij}$
Then:
Now it remains to show that $c_{ii} = e_i^* \paren {\map f {e_i} }$:
== Also see ==
* Trace in Terms of Orthonormal Basis
Category:Linear Algebra
Trace in Terms of Orthonormal Basis

== Theorem ==
Let $\mathbb K \subset \C$ be a field.
Let $\struct {V, \innerprod {\, \cdot \,} {\, \cdot \,} }$ be an inner product space over $\mathbb K$ of dimension $n$.
Let $\tuple {e_1, \ldots, e_n}$ be an orthonormal basis of $V$.
Let $f: V \to V$ be a linear operator.
Then its trace equals:
:$\map \tr f = \ds \sum_{i \mathop = 1}^n \innerprod {\map f {e_i} } {e_i}$
== Proof ==
Let $\ds \map f {e_i} = \sum_{j \mathop = 1}^n c_{ij} e_j$
Let $A$ be the matrix relative to the basis $\tuple {e_1, \ldots, e_n}$.
Then by the above assumption, $A_{ij} = c_{ij}$.
Then:
Now it remains to show that $c_{ii} = \innerprod {\map f {e_i} } {e_i}$:
== Also see ==
* Trace in Terms of Dual Basis
Category:Linear Algebra
Trace of Alternating Product of Matrices and Almost Zero Matrices

== Theorem ==
Let $R$ be a ring with unity.
Let $n, m$ be positive integers.
Let $E_{ij}$ denote the $n \times n$ matrix with only zeroes except a $1$ at the $\tuple {i, j}$th element.
Let $A_1, \ldots, A_m \in R^{n \times n}$.
Let $i_k, j_k \in \set {1, \ldots, n}$ for $k \in \set {1, \ldots, m}$.
Let $i_0 = i_m$ and $j_0 = j_m$.
Then:
:$\map \tr {A_1 E_{i_1, j_1} A_2 E_{i_2, j_2} \cdots A_m E_{i_m, j_m} } = \ds \prod_{k \mathop = 1}^m \sqbrk {A_k}_{j_{k - 1} i_k}$
== Proof ==
Use induction and the facts $E_{i j} A E_{k l} = A_{j k} E_{i l}$ and $\map \tr {A E_{i j} } = A_{j i}$ (induction basis).
Category:Matrix Theory
Trace of Matrix Product

== Theorem ==
Let $\mathbf A$ and $\mathbf B$ be square matrices of order $n$.
Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.
Then:
:$\ds \map \tr {\mathbf A \mathbf B} = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n a_{i j} b_{j i}$
where $\map \tr {\mathbf A \mathbf B}$ denotes the trace of $\mathbf A \mathbf B$.
Using the Einstein summation convention, this can be expressed as:
:$\map \tr {\mathbf A \mathbf B} = a_{i j} b_{j i}$
== Proof ==
Let $\mathbf C := \mathbf A \mathbf B$.
By definition of matrix product:
:$\ds c_{i k} = \sum_{j \mathop = 1}^n a_{i j} b_{j k}$
Thus for the diagonal elements:
:$\ds c_{i i} = \sum_{j \mathop = 1}^n a_{i j} b_{j i}$
By definition of trace:
:$\ds \map \tr {\mathbf C} = \sum_{i \mathop = 1}^n c_{i i}$
Hence the result.
== Sources ==
* : $2.2$: The summation convention: $(2.18)$
Category:Matrix Algebra
Trace of Matrix Product/General Result

== Theorem ==
Let $\mathbf A_1, \mathbf A_2, \ldots, \mathbf A_m$ be square matrices of order $n$.
Let $\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m$ be the (conventional) matrix product of $\mathbf A_1, \mathbf A_2, \ldots, \mathbf A_m$.
Then:
:$(1): \quad \ds \map \tr {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, i_1}$
where:
:$\map {a_1} {i_1, i_2}$ (for example) denotes the element of $\mathbf A_1$ whose indices are $i_1$ and $i_2$
:$\map \tr {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m}$ denotes the trace of $\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m$.
In $(1)$, the Einstein summation convention is used, with the implicit understanding that a summation is performed over each of the indices $i_1$ to $i_m$.
== Proof ==
Let $\mathbf C = \mathbf A_1 \mathbf A_2 \cdots \mathbf A_m$
From Product of Finite Sequence of Matrices, the general element of $\mathbf C$ is given in the Einstein summation convention by:
:$\map c {i_1, j} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, j}$
Thus for the diagonal elements:
:$\ds \map c {i_1, i_1} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, i_1}$
which is the Einstein summation convention for the trace of $\mathbf C$.
== Sources ==
* : $2.2$: The summation convention
Category:Matrix Algebra
Trace of Product of Matrices

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix over $R$.
Let $\mathbf B = \sqbrk b_{n m}$ be an $n \times m$ matrix over $R$.
Then:
:$\map \tr {\mathbf A \mathbf B} = \map \tr {\mathbf B \mathbf A}$
where $\map \tr {\mathbf A}$ denotes the trace of $\mathbf A$.
== Proof ==
Let $\mathbf A \mathbf B = \mathbf C = \sqbrk c_m$.
Let $\mathbf B \mathbf A = \mathbf D = \sqbrk d_n$.
Then by definition of matrix products:
Therefore:
Category:Traces of Matrices
Category:Conventional Matrix Multiplication
Trace of Sum of Matrices is Sum of Traces

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be square matrices of order $n$.
let $\mathbf A + \mathbf B$ denote the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
Then:
:$\map \tr {\mathbf A + \mathbf B} = \map \tr {\mathbf A} + \map \tr {\mathbf B}$
where $\map \tr {\mathbf A}$ denotes the trace of $\mathbf A$.
== Proof ==
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace
Category:Traces of Matrices
Trace of Unit Matrix

== Theorem ==
Let $\mathbf I_n$ be the unit matrix of order $n$.
Then:
:$\map \tr {\mathbf I_n} = n$
where $\map \tr {\mathbf I_n}$ denotes the trace of $\mathbf I_n$.
== Proof ==
By definition:
:$\mathbf I_n := \sqbrk a_n: a_{i j} = \delta_{i j}$
That is: each of the elements on the main diagonal is equal to $1$.
There are $n$ such elements.
Hence the result.
== Sources ==
* : $2.1$: Matrices: $(2.8)$
Category:Matrix Algebra
Tractrix has Asymptote Perpendicular to Axis

== Theorem ==
Let $\TT$ be a tractrix embedded in a Cartesian plane whose axis is the $y$-axis generated by the line segment beign dragged along the $x$-axis.
Then the $x$-axis is an asymptote to $\TT$ which is perpendicular to the axis of $\TT$.
Thus the asymptote of $\TT$ is indeed an asymptote.
== Proof ==
== Sources ==
* 
* 
Category:Tractrix
Tractrix has Cusp on Axis

== Theorem ==
Let $\TT$ be a tractrix embedded in a Cartesian plane whose axis is the $y$-axis and whose asymptote is the $x$-axis.
The point $\paren {0, a}$ is a single cusp of $\TT$ of the first kind.
== Proof ==
== Sources ==
* 
* 
Category:Tractrix
Transcendence of Sum or Product of Transcendentals

== Theorem ==
Let $a$ and $b$ be two transcendental numbers.
Then at least one of $a + b$ and $a \times b$ is transcendental.
== Proof ==
Proof by Contradiction:
 $a + b$ and $a \times b$ are both not transcendental.
Hence by definition, they are both algebraic.
Hence, $\left({z - a}\right) \left({z - b}\right)$ is a polynomial with algebraic coefficients.
Therefore, $a$ and $b$ must both be algebraic.
However, this contradicts with the assumption that $a$ and $b$ are both transcendental.
Hence by Proof by Contradiction it must follow that at least one of $a + b$ and $a \times b$ is transcendental.
Category:Transcendental Numbers
Transcendental Function/Examples/Exponential Functions

== Examples of Transcendental Functions ==
The exponential functions are transcendental.
== Proof ==
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(g)}$ Transcendental Functions
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $10$
* 
* 
Category:Examples of Transcendental Functions
Transcendental Function/Examples/Logarithm Functions

== Examples of Transcendental Functions ==
The logarithm functions are transcendental.
=== Real Logarithm Function is Transcendental ===
=== Complex Logarithm Function is Transcendental ===
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(g)}$ Transcendental Functions
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $10$
* 
* 
Category:Examples of Transcendental Functions
Transcendental Function/Examples/Trigonometric Functions

== Examples of Transcendental Functions ==
The trigonometric functions are transcendental.
== Proof ==
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(g)}$ Transcendental Functions
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $10$
* 
* 
Category:Examples of Transcendental Functions
Transcendental Numbers are Uncountable

== Theorem ==
The set of transcendental real numbers is uncountable.
== Proof ==
By definition, a transcendental number (in this context) is a real number which is not an algebraic number.
Recall that the Real Numbers are Uncountable.
Also recall that the Algebraic Numbers are Countable.
The result follows from Uncountable Set less Countable Set is Uncountable.
== Sources ==
* : $0 \cdotp 11000 10000 00000 00000 00010 00000 00000 00000 0 \ldots$
* : $0 \cdotp 11000 \, 10000 \, 00000 \, 00000 \, 00010 \, 00000 \, 00000 \, 00000 \, 0 \ldots$
Category:Transcendental Numbers
Transcendental Slope

== Theorem ==
The slope of a line may be transcendental.
== Proof ==
The slope form of any number $x$ may be produced by: 
If $x$ is transcendental, then the slope of a line $\mathrm m$ is transcendental.
== Example ==
$\mathrm \pi$ is proven to be transcendental by the Lindemann-Weiersrass Theorem
Slope $m$ is transcendental.
Category:Transcendental Number Theory‏‎
Transfer Operator with respect to One-Sided Shift Space of Finite Type is Linear Bounded Operator

== Theorem ==
Let $\struct {X ^+, \sigma}$ be a one-sided shift of finite type.
Let $\struct {B, \norm \cdot }$ be either:
:$\struct {\map C {X ^+, \C}, \norm \cdot_\infty}$ the continuous mapping space with the supremum norm
:or:
:$\struct {\map {F_\theta^+} {X^+}, \norm \cdot_\theta}$ the space of Lipschitz functions with the Lipschitz norm
Let $f \in B$.
Then the transfer operator $\LL_f : B \to B$ is a bounded linear operator.
== Proof ==
Category:Ergodic Theory
Category:Functional Analysis
Transfinite Induction/Principle 1

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $A$ denote a class.
Suppose that:
:For all elements $x$ of $\On$, if $x$ is a subset of $A$, then $x$ is an element of $A$.
Then $\On \subseteq A$.
== Proof ==
 that $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.
By Epsilon Relation is Strongly Well-Founded on Ordinal Class, $\On \setminus A$ must have a strictly minimal element $y$ under $\in$.
By Element of Ordinal is Ordinal, $y$ must be a subset of $\On$, the class of all ordinals.
However, from the fact that $y$ is a strictly minimal element under $\in$ of $\On \setminus A$:
:$\paren {\On \setminus A} \cap y = \O$
So by its subsethood of $\On$:
:$\paren {\On \cap y} \setminus A = \paren {y \setminus A} = \O$
Therefore $y \subseteq A$.
However, by the hypothesis, $y$ must also be an element of $A$.
This contradicts the fact that $y$ is an element of $\On \setminus A$.
Therefore $\On \subseteq A$.
== Sources ==
* : $\S 17$: Well Ordering
* : $7.17$
Category:Transfinite Induction
Transfinite Induction/Principle 1/Proof 2

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $A$ denote a class.
Suppose that:
:For each element $x$ of $\On$, if $\forall y \in \On: \paren {y < x \implies y \in A}$ then $x$ is an element of $A$.
Then $\On \subseteq A$.
== Proof ==
<onlyinclude>
 $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.
By Class of All Ordinals is Well-Ordered by Subset Relation, $\On \setminus A$ must have a smallest element $y$.
Then every strict predecessor of $y$ must lie in $A$, so by the premise, $y$ must also be an element of $A$.
This contradicts the fact that $y$ is an element of $\On \setminus A$.
Therefore $\On \subseteq A$.
Category:Transfinite Induction
Transfinite Induction/Principle 2

== Theorem ==
Let $A$ be a class satisfying the following conditions:
* $\O \in A$
* $\forall x \in A: x^+ \in A$
* If $y$ is a limit ordinal, then $\paren {\forall x < y: x \in A} \implies y \in A$
where $x^+$ denotes the successor of $x$.
Then $\On \subseteq A$.
== Proof ==
We shall prove this using the first principle of transfinite induction.
Assume $y$ is an ordinal such that $y \subseteq A$.
This implies $\forall x < y: x \in A$.
By definition, $y$ is a limit ordinal, $y = \O$, or $\exists x: y = x^+$.
If $y$ is a limit ordinal, then, since:
:$\forall x < y: x \in A$
it follows that $y \in A$.
If $y = \O$, then $y \in A$ .
Finally, assume $y = z^+$ for some ordinal $z$.
Since $z < z^+ = y$ and $\forall x < y: x \in A$, it follows that $z \in A$.
By hypothesis, we conclude $y = z^+ \in A$.
In any case, $y \in A$.
Therefore, for all ordinals $y$, if $y \subseteq A$, then $y \in A$.
By the first principle of transfinite induction, $\On \subseteq A$.
Category:Transfinite Induction
Transfinite Induction/Schema 1

== Theorem ==
Let $\map P x$ be a property 
Suppose that:
:If $\map P x$ holds for all ordinals $x$ less than $y$, then $\map P y$ also holds.
Then $\map P x$ holds for all ordinals $x$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.9$
Category:Transfinite Induction
Transfinite Induction/Schema 1/Proof 1

== Theorem ==
== Proof ==
It should be noted that for any two ordinals, $x < y \iff x \in y$.
Let $\map P x$ be a property that satisfies the above conditions.
 $y$ is an ordinal such that $\neg \map P y$.
By the Rule of Transposition, it follows that if $y$ is an ordinal such that $\neg \map P y$, then there exists an ordinal $x < y \iff x \in y$ such that $\neg \map P x$.
Let $\alpha$ be the smallest ordinal in $y$ such that $\neg \map P \alpha$.
We can choose such an element because an ordinal is strictly well-ordered by definition.
Since $\neg \map P \alpha$, let $\beta$ be the smallest ordinal in $\alpha$ such that $\neg \map P \beta$.
Since $y$ is transitive by definition:
:$\beta \in \alpha \land \alpha \in y \implies \beta \in y$
And so $\beta$ is an ordinal in $y$ such that $\neg \map P \beta$.
But since $\beta < \alpha$, this contradicts the fact that $\alpha$ is the smallest ordinal of $y$ such that $\neg \map P \alpha$.
Therefore $\map P x$ holds for all ordinals.
Hence the result.
Category:Transfinite Induction
Transfinite Induction/Schema 1/Proof 2

== Theorem ==
== Proof ==
It should be noted that for any two ordinals, $x < y \iff x \in y$.
Let $\map P x$ be a property that satisfies the above conditions.
 $y$ is an ordinal such that $\neg \map P y$.
Let $S$ be the set defined as:
:$S = \set {x \in y^+: \neg \map P x}$
It is seen that this set is non-empty because by Set is Element of Successor:
:$y \in y^+$
$y^+$ is an ordinal because of Successor Set of Ordinal is Ordinal.
By Element of Ordinal is Ordinal, $S$ is a set of ordinals.
By the Rule of Transposition, it follows that for every $x \in S$, there exists an ordinal $\alpha < x \iff \alpha \in x$ such that $\neg \map P \alpha$.
But by transitivity:
:$\alpha \in x \land x \in y^+ \implies \alpha \in y^+$
So the epsilon restriction $\in_S$ is a right-total relation on $S$.
Therefore by the Axiom of Dependent Choice there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that:
:$\forall n \in \N: x_{n + 1} \in x_n$
But this contradicts No Infinitely Descending Membership Chains.
Thus $\map P x$ holds for all ordinals.
Hence the result.
Category:Transfinite Induction
Transfinite Induction/Schema 2

== Theorem ==
Let $\map \phi x$ be a property satisfying the following conditions:
:$(1): \quad \map \phi \O$ is true
:$(2): \quad$ If $x$ is an ordinal, then $\map \phi x \implies \map \phi {x^+}$
:$(3): \quad$ If $y$ is a limit ordinal, then $\paren {\forall x < y: \map \phi x} \implies \map \phi y$
where $x^+$ denotes the successor of $x$.
Then, $\map \phi x$ is true for all ordinals $x$.
== Proof 1 ==
== Proof 2 ==
In proofs by transfinite induction using this particular schema, the following terms are used.
=== Basis for the Induction ===
The proposition $\map \phi \O$ is called the basis for the induction.
=== Induction Step ===
The proposition $\map \phi x = \T \implies \map \phi {x^+} = \T$ is called the inductive step.
It says that the propositional function will pass from an ordinal number to its successor.
=== Limit Case ===
$\paren {\forall x < y: \map \phi x = \T} \implies \map \phi y = \T$ is called the limit case.
It states that if $\phi$ is true for every ordinal strictly less than $y$, then $\phi$ is true for $y$.
It is essentially proving that the proposition will hold for limit ordinals.
Then, this formulation of transfinite induction says that if the basis for the induction, inductive step, and limit case are all satisfied, then the statement holds for all ordinals.
== Sources ==
* : Appendix $\text{A}$: Set Theory: Induction
Category:Transfinite Induction
Transfinite Induction/Schema 2/Proof 1

== Theorem ==
== Proof ==
It should be noted that for any two ordinals, $x \lt y \iff x \in y$.
Let $\map \phi x$ be a property that satisfies the above conditions.
 $y$ be an ordinal such that $\neg \map \phi y$.
It is noted that $y \ne \O$.
Therefore $y$ must be either a successor ordinal or a limit ordinal.
If $y$ is a successor ordinal, then let $x$ be the ordinal such that $x^+ = y$.
By the Rule of Transposition, it is seen that $\neg \map \phi x$.
By Set is Element of Successor, it follows that $x \in y$, and so $x \lt y$.
If $y$ is a limit ordinal, then by the Rule of Transposition there exists an ordinal $x \lt y \iff x \in y$ such that $\neg \map \phi x$.
It has been shown that if $y$ is an ordinal such that $\neg \map \phi y$, then there exists an ordinal $x \in y$ such that $\neg \map \phi x$.
The rest follows from the proof of Schema 1.
Category:Transfinite Induction
Transfinite Induction/Schema 2/Proof 2

== Theorem ==
== Proof ==
Define the class:
:$A := \set {x \in \On: \map \phi x = \T}$.
Then $\map \phi x = \T$ is equivalent to the statement:
:that $x \in A$
The three conditions in the hypothesis become:
:$(1a): \quad \O \in A$
:$(2a): \quad x \in A \implies x^+ \in A$
:$(3a): \quad \paren {\forall x < y: x \in A} \implies y \in A$
These are precisely the conditions for the class $A$ in the second principle of transfinite induction.
Therefore, $\On \subseteq A$.
Thus, $\map \phi x$ holds for all $x \in \On$.
Category:Transfinite Induction
Transfinite Recursion Theorem

== Theorem ==
=== Formulation 1 ===
=== Formulation 2 ===
=== Formulation 3 ===
=== Formulation 4 ===
=== Formulation 5 ===
== Also presented as ==
There are variants in how this result is presented, as the topic can be approached from a number of different directions.
Here are examples:
=== First Principle of Transfinite Recursion ===
=== Second Principle of Transfinite Recursion ===
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Theorem $5.1$
Category:Ordinals
Category:Transfinite Recursion Theorem
Category:Named Theorems
Transfinite Recursion Theorem/Corollary

== Theorem ==
Let $x$ be an ordinal.
Let $G$ be a mapping
There exists a unique mapping $f$ that satisfies the following properties:
:The domain of $f$ is $x$
:$\forall y \in x: \map f y = \map G {f \restriction y}$
== Proof ==
Construct $K$ and $F$ as in the First Principle of Transfinite Recursion.
Set $f = \paren {F \restriction x}$.
Then since $x \subseteq \Dom F$, the domain of $f$ is $x$.
Thus such a mapping $f$ exists.
Suppose there are two mappings $f$ and $g$ that satisfy these conditions.
We will use the first principle of transfinite induction to show that $f = g$.
Take $y \in x$.
Suppose $\forall z \in y: \map f z = \map g z$.
Then:
:$\paren {f \restriction y} = \paren {g \restriction y}$
and so:
So $\map f y = \map g y$ for all $y \in x$ by transfinite induction.
Since $x$ is the domain of $f$ and $g$, it follows that $f = g$.
Thus the mapping $f$ is unique.
== Sources ==
* : $\S 7.41$
Category:Transfinite Recursion Theorem