nyuuzyou/wikis huggingface dataset:
Ordinal Multiplication is Left Cancellable

== Theorem ==
Let $x, y, z$ be ordinals.
Let $z \ne 0$.
Then:
:$\ds \paren {z \cdot x} = \paren {z \cdot y} \implies x = y$
That is, ordinal multiplication is left cancellable.
== Proof ==
For the proof, $<$, $\in$, and $\subsetneq$ will be used interchangeably.
This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
Since $z \ne 0$ and $z \not < 0$, $0 < z$ by Ordinal Membership is Trichotomy.
Note that:
In addition:
This contradicts the consequents of the first two equations, so:
== Also see ==
* Natural Number Multiplication is Cancellable
== Sources ==
* : $\S 8.20$
Category:Ordinal Arithmetic
Ordinal Multiplication is Left Distributive

== Theorem ==
Let $x$, $y$, and $z$ be ordinals.
Let $\times$ denote ordinal multiplication.
Let $+$ denote ordinal addition.
Then:
:$x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$
== Proof ==
The proof shall proceed by Transfinite Induction, as follows:
=== Basis for the Induction ===
Let $0$ denote the ordinal zero.
This proves the basis for the induction.
=== Induction Step ===
This proves the induction step.
=== Limit Case ===
The inductive hypothesis for the limit case states that:
:$x \times \paren {y + w} = \paren {x \times y} + \paren {x \times w}$ for all $w \in z$ and $z$ is a limit ordinal.
The proof shall proceed by cases:
=== Case 1 ===
Suppose $x = 0$.
=== Case 2 ===
Suppose that $x \ne 0$.
Since $w$ is a limit ordinal, $y + w$ and $x \times w$ are limit ordinals by Limit Ordinals Preserved Under Ordinal Addition and Limit Ordinals Preserved Under Ordinal Multiplication.
Take any $w \in y + z$.
It follows that $w \in y \lor \paren {y \subseteq w \land w \in y + z}$ by Relation between Two Ordinals and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
Thus, $w \in y \lor w = y + u$ for some $u \in z$ by Ordinal Subtraction when Possible is Unique.
If $w < y$, then:
If $w = y + u$, then:
By Supremum Inequality for Ordinals:
: $x \times \paren {y + z} \subseteq \paren {x \times y} + \paren {x \times z}$
Conversely, if $v \in x \times z$, then:
By Supremum Inequality for Ordinals:
: $\paren {x \times y} + \paren {x \times z} \subseteq x \times \paren {y + z}$
By definition of set equality:
: $x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$
This proves the limit case.
== Sources ==
* : $\S 8.25$
Category:Ordinal Arithmetic
Ordinal Multiplication via Cantor Normal Form/Infinite Exponent

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $x > 1$.
Let $y \ge \omega$ where $\omega$ denotes the minimally inductive set.
Let $\sequence {a_i}$ be a sequence of ordinals that is strictly decreasing on $1 \le i \le n$.
Let $\sequence {b_i}$ be a sequence of ordinals such that $0 < b_i < x$ for all $1 \le i \le n$.
Then:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$
== Proof ==
It follows that:
By multiplying the inequalities by $x^y$ on the left:
Solving for both sides of the inequality:
Therefore:
== Also see ==
* Definition:Cantor Normal Form
== Sources ==
* : $\S 8.45$
Category:Ordinal Arithmetic
Ordinal Multiplication via Cantor Normal Form/Limit Base

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $x$ be a limit ordinal.
Let $y > 0$.
Let $\sequence {a_i}$ be a sequence of ordinals that is strictly decreasing on $1 \le i \le n$.
Let $\sequence {b_i}$ be a sequence of finite ordinals.
Then:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$
== Proof ==
The proof shall proceed by finite induction on $n$:
For all $n \in \N_{\le 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$
Since $x$ is a limit ordinal, it follows that $x^y$ is a limit ordinal by Limit Ordinals Closed under Ordinal Exponentiation.
=== Basis for the Induction ===
$\map P 1$ is the statement:
:$\ds \sum_{i \mathop = 1}^1 \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{i \mathop = 1}^k \paren {x^{a_i} b_i} \times x^y = x^{a_1 + y}$
Then we need to show:
:$\ds \sum_{i \mathop = 1}^{k \mathop + 1} \paren {x^{a_i} b_i} \times x^y = x^{a_1 \mathop + y}$
=== Induction Step ===
This is our induction step:
By Upper Bound of Ordinal Sum, it follows that:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_{i \mathop + 1} } b_{i \mathop + 1} } < x^{a_1}$
By Membership is Left Compatible with Ordinal Multiplication:
But:
Therefore:
== Also see ==
* Definition:Cantor Normal Form
== Sources ==
* : $\S 8.46$
Category:Ordinal Arithmetic
Ordinal Number Equivalent to Cardinal Number

== Theorem ==
Let $x$ be an ordinal.
Let $\card x$ denote the cardinal number of $x$.
Then:
:$x \sim \card x$
where $\sim$ denotes set equivalence.
== Proof ==
From Set is Equivalent to Itself:
:$x \sim x$
Therefore, $x$ is equivalent to some ordinal.
By Condition for Set Equivalent to Cardinal Number:
:$x \sim \card x$
== Sources ==
* : $\S 10.11$
Category:Cardinals
Ordinal Power of Power

== Theorem ==
Let $x$, $y$, and $z$ be ordinals.
Then:
:$\paren {x^y}^z = x^{y \mathop \times z}$
== Proof ==
The proof shall proceed by Transfinite Induction on $z$.
=== Basis for the Induction ===
This proves the basis for the induction.
=== Induction Step ===
Suppose $\paren {x^y}^z = x^{y \times z}$.
Then:
This proves the induction step.
=== Limit Case ===
Suppose $\paren {x^y}^w = x^{y \times w}$ for all $w \in z$ where $z$ is a limit ordinal.
Then:
To prove the statement, it is necessary and sufficient to prove that:
:$\ds \bigcup_{w \mathop \in z} x^{y \times w} = x^{y \times z}$
The proof of this shall proceed by cases:
==== Case 1 ====
If $y = 0$, it follows that:
==== Case 2 ====
If $y \ne 0$:
Therefore, by Supremum Inequality for Ordinals, it follows that:
:$\ds \bigcup_{w \mathop \in z} x^{y \times w} \le x^{y \times z}$
Conversely:
By Supremum Inequality for Ordinals, it follows that:
:$\ds \bigcup_{v \mathop \in y \times z} x^v \le \bigcup_{w \mathop \in z} x^{y \times w}$
Therefore, by definition of ordinal exponentiation:
This proves the limit case.
== Sources ==
* : $\S 8.42$
Category:Ordinal Arithmetic
Ordinal Space is Completely Normal

== Theorem ==
Let $\Gamma$ denote a limit ordinal.
Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\hointr 0 \Gamma$ and $\closedint 0 \Gamma$ are both completely normal.
== Proof ==
By definition, $\hointr 0 \Gamma$ and $\closedint 0 \Gamma$ are both linearly ordered spaces.
The result follows from Linearly Ordered Space is Completely Normal.
== Sources ==
* : Part $\text {II}$: Counterexamples: $40 \text { - } 43$. Ordinal Space: $4$
Category:Ordinal Spaces
Category:Completely Normal Spaces
Ordinal Space is Strongly Locally Compact

== Theorem ==
Let $T$ denote an ordinal space on a limit ordinal $\Gamma$.
Then $T$ is a strongly locally compact space.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $40 \text { - } 43$. Ordinal Space: $6$
Category:Ordinal Spaces
Category:Locally Compact Spaces
Ordinal Subset is Well-Ordered

== Theorem ==
Let $S$ be a class.
Let every element of $S$ be an ordinal.
Then $\struct {S, \in}$ is a strict well-ordering.
== Proof ==
Let $\On$ denote the class of all ordinals.
By definition of subset, $S \subseteq \On$.
But by Class of All Ordinals is Ordinal $\On$ is an ordinal.
Therefore $\On$ is well-ordered by $\in$.
This means that $S$ is also well-ordered by $\in$.
Category:Ordinals
Ordinal Subset of Ordinal is Initial Segment

== Theorem ==
Let $S$ be an ordinal.
Let $T \subset S$ also be an ordinal.
Then $\exists a \in S: T = S_a$, where $S_a$ is the initial segment of $S$ determined by $a$.
That is, $T = S_a = a \in S$.
== Proof ==
Recall that the Ordering on Ordinal is Subset Relation.
Let $a$ be the minimal element of $S \setminus T$.
By definition of minimal element:
:$\forall x \in S: x \subset a$
Hence $x \notin S \setminus T$.
By definition of set difference:
:$x \in T$
Thus by definition of initial segment:
:$S_a \subseteq T$
Now let $b \in T$.
Then by definition of ordinal:
:$T_b = b = S_b$
Now, if $a \subset b$ then $a \in S_b$.
So $a \in T_b$ and hence $a \in T$.
But $a \in S \setminus T$, so by definition of set difference, this is not the case.
So $a \not \subset b$.
We have that an ordinal is well-ordered by $\subseteq$.
So by the definition of well-ordering, $\subseteq$ is a total ordering on $S$
So it follows that $b \subseteq a$.
But $b \ne a$ since $b \in T$.
Hence $b \subset a$, and so $b \in S_a$.
This proves that $T \subseteq S_a$.
Thus $T = S_a$.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.7$
Category:Ordinals
Ordinal Subtraction when Possible is Unique

== Theorem ==
Let $x$ and $y$ be ordinals such that $x \le y$.
Then there exists a unique ordinal $z$ such that $x + z = y$.
That is:
:$x \le y \implies \exists! z \in \On: x + z = y$
== Proof ==
By transfinite induction on $y$.
=== Base Case ===
=== Inductive Case ===
The last step is justified because of:
:$x + z^+ = y^+$
which guarantees existence, and by Ordinal Addition is Left Cancellable:
:$x + w = x + z^+ \implies w = z^+$
which guarantees uniqueness.
=== Limit Case ===
Set $A = \set {w : x < w \land w < y}$.
The above statement shows that $A$ is nonempty.
Then:
:$\forall w \in A: \exists ! z: x + z = w$
Create a mapping $F$, that sends each $w \in A$ to the unique $z$ that satisfies $x + z = w$.
Finally, we must prove that:
:$\ds \bigcup_{w \mathop \in A} \paren {x + \map F w} = x + \bigcup_{w \mathop \in A} \map F w$
It suffices to prove that $\ds \bigcup_{w \mathop \in A} \map F w$ is a limit ordinal.
Let $w = x^+$.
Then $\map F w = 1$, and Union of Ordinals is Least Upper Bound.
Thus:
:$\ds \bigcup_{w \mathop \in A} \map F w \ne \O$
Thus:
:$\ds \bigcup_{w \mathop \in A} \map F w \ne z^+$
Therefore $\ds \bigcup_{w \mathop \in A} \map F w$ must be a limit ordinal.
To prove uniqueness, assume $x + z = y$.
Then by Ordinal Addition is Left Cancellable:
:$\ds x + z = x + \bigcup_{w \mathop \in A} \map F w \implies z = \bigcup_{w \mathop \in A} \map F w$
== Also see ==
* Definition:Ordinal Subtraction
== Sources ==
* : $\S 8.8$
Category:Ordinal Arithmetic
Ordinal Sum of Powers

== Theorem ==
Let $x$, $y$, and $z$ be ordinals.
Then:
:$x^y \times x^z = x^{y + z}$
== Proof ==
The proof shall proceed by Transfinite Induction on $z$.
=== Basis for the Induction ===
:$x^0 = 1$ for all $x$
This proves the basis for the induction.
=== Induction Step ===
Suppose that $x^y \times x^z = x^{y + z}$.
Then:
This proves the induction step.
=== Limit Case ===
Suppose that $\forall w \in z: x^y \times x^w = x^{y + w}$ for limit ordinal $z$.
Conversely:
But this means that $u$ is bounded above by $x^v$ for some $v \in z$.
Thus there exists a $v \in z$ such that:
:$w \le x^y \times x^v$
By Supremum Inequality for Ordinals, it follows that:
:$\ds x^y \times x^z \le \bigcup_{w \mathop \in z} \paren {x^y \times x^w}$
Conversely:
Thus, $u$ is bounded above by $y + v$ for some $v \in z$.
Therefore:
:$x^u \le x^{y + v}$
By Supremum Inequality for Ordinals, it follows that:
:$\ds x^{y + z} \le \bigcup_{w \mathop \in z} x^{y + w}$
Thus, by definition of set equality:
:$\ds x^{y + z} = \bigcup_{w \mathop \in z} x^{y + w}$
Combining the results of the first and second lemmas for the limit case:
:$x^{y + z} = x^y \times x^z$
This proves the limit case.
== Sources ==
* : $\S 8.41$
Category:Ordinal Arithmetic
Ordinal equals Successor of its Union

== Theorem ==
Let $\alpha$ be an ordinal.
Then:
:$\bigcup \alpha^+ = \alpha$
where:
:$\alpha^+$ denotes the successor set of $\alpha$
:$\bigcup \alpha^+$ denotes the union of $\alpha^+$.
== Proof ==
Let $x \in \bigcup \alpha^+$.
Then there exists an ordinal $\beta$ such that:
:$x \in \beta$
and:
:$\beta \in \alpha^+$
By definition of the usual ordering of ordinals:
:$\beta < \alpha^+$
Thus:
:$\beta \le \alpha$
Hence because $x < \beta$:
:$x < \alpha$
and thus:
:$x \in \alpha$
That is:
:$\bigcup \alpha^+ \subseteq \alpha$
Now suppose $x \in \alpha$.
Since we have:
:$\alpha \in \alpha^+$
we have:
:$x \in \bigcup \alpha^+$
Thus:
:$a \subseteq \bigcup \alpha^+$
Hence by set equality:
:$a = \bigcup \alpha^+$
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.20 \ (2)$
Category:Ordinals
Category:Set Union
Category:Successor Mapping
Ordinal equals its Initial Segment

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $<$ denote the (strict) usual ordering of $\On$.
Let $\alpha$ be an ordinal.
Then $\alpha$ is equal to its own initial segment:
:$\alpha = \set {\beta \in \On: \beta < \alpha}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
Category:Ordinal equals its Initial Segment
Category:Ordinals
Ordinal equals its Initial Segment/Proof 4

== Theorem ==
== Proof ==
Let $\alpha$ be an ordinal by Definition 4.
From Strict Ordering of Ordinals is Equivalent to Membership Relation:
:$\forall \alpha, \beta \in \On: \beta < \alpha \iff \alpha \in \beta$
Hence the statement of the result is equivalent to:
:$\alpha = \set {\beta \in \On: \beta \in \alpha}$
which is trivially true by definition of a set.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Corollary $1.13$
Category:Ordinal equals its Initial Segment
Ordinal in Aleph iff Cardinal in Aleph

== Theorem ==
Let $x$ and $y$ be ordinals.
Then:
:$x \in \aleph_y \iff \card x \in \aleph_y$
where $\aleph$ denotes the aleph mapping.
== Proof ==
By the definition of the aleph mapping, $\aleph_y$ is an element of the class of infinite cardinals.
By Cardinal Inequality implies Ordinal Inequality, it follows that:
:$x \in \aleph_y \iff \card x \in \aleph_y$
== Sources ==
* : $\S 10.46 \ (1)$
Category:Aleph Mapping
Ordinal is Finite iff Natural Number

== Theorem ==
Let $x$ be an ordinal.
Then $x$ is a finite set $x$ is an element of the minimally inductive set.
== Proof ==
$x$ is finite $x \sim \N_n$ for some $n \in \N$, by definition.
But $x$ is an ordinal, and by definition, it is equal to its initial segment.
By definition of the von Neumann construction of natural numbers, it follows that $x \sim n$ for some $n$.
By Finite Ordinal is equal to Natural Number, it follows that $x$ is equal to $n$.
Thus, $x$ is an element of the minimally inductive set.
== Sources ==
* : $\S 10.30$
Category:Ordinals
Category:Minimally Inductive Set
Ordinal is Less than Ordinal to Limit Power

== Theorem ==
Let $x$, $y$, and $z$ be ordinals.
Let $z$ be a limit ordinal.
Then:
:$x < y^z \iff \exists w \in z: x < y^w$
== Proof ==
== Sources ==
* : $\S 8.40$
Category:Ordinal Arithmetic
Ordinal is Less than Successor

== Theorem ==
Let $x$ be an ordinal.
Let $x^+$ denote the successor of $x$.
Then:
:$x \in x^+$
:$x \subset x^+$
== Proof ==
$x$ is an ordinal and so by definition is also a set.
:$x \in \left({x \cup \left\{ {x}\right\} }\right) \land x \subset \left({x \cup \left\{ {x}\right\} }\right)$
so by applying the definition of a successor set:
:$x \in x^+ \land x \subset x^+$
== Sources ==
* : $\S 7.23$
Category:Ordinals
Ordinal is Less than Sum

== Theorem ==
Let $x$ and $y$ be ordinals.
Then:
:$x \le \paren {x + y}$
:$x \le \paren {y + x}$
== Proof ==
By Proof by Cases, one of the following holds by Empty Set is Subset of All Sets:
:$\O < y$
:$y = \O$
By Ordinal Addition by Zero:
:$x = \paren {x + \O} = \paren {\O + x}$
By Membership is Left Compatible with Ordinal Addition:
:$\O < y \implies x < \paren {x + y}$
But if $y = \O$, then it is clear the inequality $x \le \paren {x + y}$ holds as well.
So in either case:
:$x \le \paren {x + y}$
Similarly, by Subset is Right Compatible with Ordinal Addition:
:$\O \le y \implies x \le \paren {y + x}$
The fact that $\O \le y$ is clear from Empty Set is Subset of All Sets.
Therefore:
:$x \le \paren {y + x}$
Category:Ordinal Arithmetic
Ordinal is Member of Class of All Ordinals

== Theorem ==
Let $A$ be an ordinal.
Then:
:$A \in \On \lor A = \On$
where $\On$ denotes the class of all ordinals.
== Proof ==
We have that $A$ is an ordinal
From Class of All Ordinals is Ordinal and Ordinal Membership is Trichotomy:
:$A \in \On \lor A = \On \lor \On \in A$
But by the Burali-Forti Paradox $\On$ is a proper class.
Therefore:
:$A \in \On \lor A = \On$
== Sources ==
* : $7.14$
Category:Class of All Ordinals
Ordinal is Proper Subset of Successor

== Theorem ==
Let $\alpha$ be an ordinal.
Then:
:$\alpha \subsetneqq \alpha^+$
where $\alpha^+$ denotes the successor set of $\alpha$.
That is:
:$\alpha$ is a proper subset of $\alpha^+$.
== Proof ==
 $\alpha = \alpha^+$.
By definition:
:$\alpha^+ = \alpha \cup \set \alpha$
and so:
:$\alpha \subseteq \alpha^+$
and:
:$\alpha \in \alpha^+$
which leads to:
:$\alpha \in \alpha$
But from Ordinal is not Element of Itself:
:$\alpha \notin \alpha$
Hence by Proof by Contradiction:
:$\alpha \ne \alpha^+$
But we have:
:$\alpha \subseteq \alpha^+$
Hence:
:$\alpha \subsetneqq \alpha^+$
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Corollary $1.9$
Category:Ordinals
Category:Successor Mapping
Ordinal is Subset of Class of All Ordinals

== Theorem ==
Suppose $A$ is an ordinal.
Then:
:$A \subseteq \On$
where $\On$ represents the class of all ordinals.
== Proof ==
By Ordinal is Member of Class of All Ordinals:
:$A \in \On \lor A = \On$
In either case:
:$A \subseteq \On$
since $\On$ is transitive.
== Sources ==
* : $7.15$
Category:Class of All Ordinals
Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy

== Theorem ==
Let $x$ be an ordinal.
Let $S$ be a small class.
Let $\map V x$ denote the von Neumann hierarchy on the ordinal $x$.
Then $x$ is a subset of the rank of $S$ $S \notin \map V x$.
== Proof ==
=== Necessary Condition ===
Let $x \subseteq \map {\operatorname{rank} } S$.
Then by Von Neumann Hierarchy Comparison:
:$S \in \map V x \implies S \in \map V {\map {\operatorname{rank} } S}$
But by Ordinal Equal to Rank:
:$S \notin \map V {\map {\operatorname{rank} } S}$
By contraposition:
:$S \notin \map V x$
=== Sufficient Condition ===
Let $S \notin \map V x$.
Then:
== Sources ==
* : $\S 9.15 \ (3)$
Category:Von Neumann Hierarchy
Ordinal is Subset of Successor

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $y^+$ denote the successor of $y$.
Then:
: $x \subseteq y^+ \iff \left({x \subseteq y \lor x = y^+}\right)$
== Proof ==
Let $A \subset B$ denote that $A$ is a proper subset of $B$.
Let $A \in B$ denote that $A$ is an element of $B$.
From Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, $\subset$ and $\in$ can be used interchangeably.
Thus:
Conversely:
Category:Ordinals
Ordinal is Transitive/Proof 4

== Theorem ==
== Proof ==
Let $\alpha$ be an ordinal by Definition 4.
The proof proceeds by the Principle of Superinduction.
From Empty Class is Transitive we start with the fact that $0$ is transitive.
Let $x$ be transitive.
From Successor Set of Transitive Set is Transitive:
:$x^+$ is transitive.
We have that Class is Transitive iff Union is Subclass.
Hence the union of a chain of transitive sets is transitive.
Hence the result by the Principle of Superinduction.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.7$
Category:Ordinal is Transitive
Ordinal is not Element of Itself/Proof 1

== Theorem ==
== Proof ==
By Successor Set of Ordinal is Ordinal, the successor of $x$ is an ordinal.
That is, $x^+ = x \cup \set x$ is an ordinal.
By Set is Element of Successor, $x \in x^+$.
Because $x^+$ is an ordinal, it is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_{x^+} }$.
Because a strict ordering is antireflexive and $x \in x^+$, we conclude that $x \notin x$.
Category:Ordinal is not Element of Itself
Ordinal is not Element of Itself/Proof 3

== Theorem ==
== Proof ==
Let $\On$ denote the class of all ordinals.
From Class of All Ordinals is Minimally Superinductive over Successor Mapping, $\On$ is superinductive.
Hence we can use the Principle of Superinduction.
By Zero is Smallest Ordinal, $0$ is the smallest element of $\On$.
We identify the natural number $0$ via the von Neumann construction of the natural numbers as:
:$0 := \O$
From Empty Set is Ordinary:
:$\O \notin \O$
and so:
:$0 \notin 0$
Let $x$ be an ordinal for which it has been established that:
:$x \notin x$
Then by Successor Set of Ordinary Transitive Set is Ordinary‎:
:$x^+ \notin x^+$
It remains to be shown that if $C$ is a chain of ordinals all of which are ordinary, then $\bigcup C$ is ordinary.
Let $C$ be a chain of ordinals.
From Union of Set of Ordinals is Ordinal, $\bigcup C$ is itself an ordinal.
Let $\alpha = \bigcup C$.
 $\bigcup C$ is not ordinary.
That is, suppose that $\alpha \in \alpha$.
Then:
:$\alpha \in \bigcup C$
Hence for some $\beta \in C$:
:$\alpha \in \beta$
But:
:$\beta \in C$
and so:
:$\beta \subseteq \bigcup C$
That is:
:$\beta \subseteq \alpha$
Because $\alpha \in \beta$ and Ordinal is Transitive:
:$\alpha \subseteq \beta$
Hence by definition of set equality:
:$\alpha = \beta$
That is:
:$\beta \in \beta$
Thus some element of $C$, that is, $\beta$, is an element of itself.
That is, not all elements of $C$ are ordinary.
This contradicts our assertion that all elements of $C$ are ordinary.
Hence by Proof by Contradiction we have that if $C$ is a chain of ordinals all of which are ordinary, then $\bigcup C$ is ordinary:
:$\bigcup C \notin \bigcup C$
The result then follows by the Principle of Superinduction.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.8$
Category:Ordinal is not Element of Itself
Ordinals Isomorphic to the Same Well-Ordered Set

== Theorem ==
Let $A$ and $B$ be ordinals.
Let $\left({\prec, S}\right)$ be a strict well-ordering.
Let $\left({\in, A}\right)$ and $\left({\prec, S}\right)$ be order isomorphic.
Let $\left({\in, B}\right)$ and $\left({\prec, S}\right)$ be order isomorphic.
Then:
: $A = B$
== Proof ==
Let $\phi_1$ denote the mapping creating the order isomorphism between $\left({\in, A}\right)$ and $\left({\prec, S}\right)$.
Let $\phi_2$ denote the mapping creating the order isomorphism between $\left({\in, B}\right)$ and $\left({\prec, S}\right)$.
Then $\phi_2^{-1}: S \to B$ is an order isomorphism by Inverse of Order Isomorphism is Order Isomorphism.
$\phi_1 \circ \phi_2^{-1}: A \to B$ is an order isomorphism by Composite of Order Isomorphisms is Order Isomorphism.
So $\left({\in, A}\right)$ and $\left({\in, B}\right)$ are order isomorphic ordinals.
By Isomorphic Ordinals are Equal:
:$A = B$
== Sources ==
* : $\S 7.39$
Category:Ordinals
Category:Order Isomorphisms
Ordinals are Well-Ordered/Corollary

== Theorem ==
Let $A$ be a set of ordinals.
Let $\Epsilon {\restriction_A}$ denote the epsilon restriction on $A$.
Then $A$ is strictly well-ordered by $\Epsilon {\restriction_A}$.
== Proof ==
Let $A$ be a set of ordinals.
Let $\Epsilon {\restriction_A}$ denote the epsilon restriction on $A$.
It is to be shown that $\Epsilon {\restriction_A}$ is antireflexive.
Suppose there is a $a \in A$ such that $a \in a$.
Then by ordinal is element of ordinal iff proper subset of ordinal, we have $a \subsetneq a$.
And so:
:$\exists x \in a: x \notin a$
But this is a contradiction, so $\Epsilon {\restriction_A}$ must be antireflexive.
By Strict Subset Relation is Transitive, it follows that $\Epsilon {\restriction_A}$ is transitive.
By Ordinals are Well-Ordered, it follows that for any non-empty subset $B \subseteq A$ there is a smallest element $x \in B$ such that:
:$\forall y \in B: x \in y \lor x = y$
It has been shown that $\Epsilon {\restriction_A}$ is a strict ordering such that every non-empty subset of $A$ has a smallest element.
Hence the result.
Category:Ordinals
Category:Ordinals are Well-Ordered