nyuuzyou/wikis huggingface dataset:
Ordering on Natural Numbers is Compatible with Multiplication

== Theorem ==
Let $m, n, k \in \N$ where $\N$ is the set of natural numbers.
Let $k \ne 0$.
Then:
:$m < n \iff m \times k < n \times k$
=== Corollary ===
== Proof ==
Proof by induction:
First we note that if $k = 0$ then $m \times k = 0 = n \times k$ whatever $m$ and $n$ are, and the proposition clearly does not hold.
So, for all $k \in \N \setminus \set 0$, let $\map P k$ be the proposition:
:$m < n \iff m \times k < n \times k$
From Identity Element of Natural Number Multiplication is One:
:$m \times 1 = m < n = n \times 1$
and so $\map P 1$ is true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P j$ is true, where $j \ge 1$, then it logically follows that $\map P {j^+}$ is true.
So this is our induction hypothesis:
:$m < n \iff m \times j < n \times j$
Then we need to show:
:$m < n \iff m \times j^+ < n \times j^+$
=== Induction Step ===
This is our induction step:
Let $m < n$.
Then:
Similarly:
:$m < n \iff \paren {m \times j} + n < \paren {n \times j} + n$
But then from Ordering on Natural Numbers is Compatible with Addition, we also have:
:$m < n \iff \paren {m \times j} + m < \paren {m \times j} + n$
So from Natural Number Ordering is Transitive, we have:
:$m < n \iff \paren {m \times j} + m < \paren {n \times j} + n$
This gives:
So $\map P j \implies \map P {j^+}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall m, n, k \in \N, k \ne 0: m \times n \iff m \times k < n \times k$
== Sources ==
* : Introduction $\S 4$: The natural numbers
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.13: \ 2^\circ$
* : $\S 13$: Arithmetic
* : $\S 4$: Number systems $\text{I}$: Peano's Axioms
Category:Natural Number Multiplication
Category:Examples of Orderings
Ordering on Natural Numbers is Compatible with Multiplication/Corollary

== Corollary to Ordering on Natural Numbers is Compatible with Multiplication ==
Let $a, b, c, d \in \N$ where $\N$ is the set of natural numbers.
Then:
:$a > b, c > d \implies a c > b d$
== Proof ==
Finally:
== Sources ==
* : Introduction $\S 4$: The natural numbers: Exercise $1$
Category:Natural Number Multiplication
Category:Examples of Orderings
Ordering on Ordinal is Subset Relation

== Theorem ==
Let $\struct {S, \prec}$ be an ordinal.
Then $\forall x, y \in S:$
:$x \in y \iff x \prec y \iff S_x \subsetneqq S_y \iff x \subsetneqq y$
where $S_x$ and $S_y$ are the initial segments of $S$ determined by $x$ and $y$ respectively.
Thus there is no need to specify what the ordering on an ordinal is -- it is always the subset relation.
== Proof ==
The first equivalence is an immediate consequence of Equivalence of Definitions of Ordinal.
The second equivalence holds for any well-ordered set by Woset Isomorphic to Set of its Sections.
The third equivalence holds by definition of an ordinal.
It follows from Ordering is Equivalent to Subset Relation and Order Isomorphism between Wosets is Unique that this ordering is the only one.
== Also known as ==
Some sources refer to this result as Copi's identity for , from his statement of this in his $1954$ book .
However, it may have been known about earlier than that and may not be directly attributable to .
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals
Category:Ordinals
Category:Orderings
Category:Subset Relation
Ordering on Positive Integers is Equivalent to Ordering on Natural Numbers

== Theorem ==
Let $u, v \in \Z_{>0}$ be natural numbers.
Consider the mapping $\phi: \N_{>0} \to \Z_{>0}$ defined as:
:$\forall u \in \N_{>0}: \map \phi u = u'$
where $u' \in \Z$ denotes the (strictly) positive integer $\eqclass {b + u, b} {}$.
Let $u', v' \in \Z_{>0}$ be strictly positive integers.
Then:
:$u > v \iff u' > v'$
== Proof ==
Let $u' = \eqclass {b + u, b} {}$.
Let $v' = \eqclass {c + v, c} {}$.
Then:
== Sources ==
* : Introduction $\S 5$: The system of integers
Category:Orderings on Integers
Ordering on Real Numbers from Decimal Expansion

== Theorem ==
Let $x, y \in \R$ be real numbers.
Let $x$ and $y$ be expressed by their decimal expansions:
Let $\preccurlyeq_l$ be the lexicographic ordering on $\R$ defined as:
:$x \preccurlyeq_l y$ :
::$m \prec n$
:or:
::$m = n$ and $\exists k \in \Z_{>0}: \paren {\forall j: 1 \le j < k: d_j = e_j} \land d_k < e_k$
:or:
::$m = n$ and $\forall j \in \Z_{>0}: d_j = e_j$.
Then:
:$x \le y$
where $\le$ denotes the usual ordering on $\R$.
== Proof ==
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $6$
Category:Real Numbers
Ordering on Singleton is Total Ordering

== Theorem ==
Let $S = \set s$ be a singleton.
Let $\RR$ be an ordering on $S$.
Then $\RR$ is a total ordering on $S$.
== Proof ==
By definition of ordering, $\RR$ is a reflexive relation.
Hence from Reflexive Relation on Singleton is Well-Ordering:
:$\struct {S, \RR}$ is a well-ordered set.
Hence by definition of well-ordered set:
:$\RR$ is a total ordering on $S$.
Hence the result.
Category:Singletons
Category:Total Orderings
Ordering where every Subclass has Smallest Element is Well-Ordering

== Theorem ==
Let $A$ be a class.
Let $\RR$ be an ordering on $A$.
Let $\RR$ be such that every subclass of $A$ has a smallest element under $\RR$.
Then $\RR$ is a well-ordering.
== Proof ==
We are given that $\RR$ is an ordering.
It will be shown that $\RR$ is a total ordering, from which the fact that it is a well-ordering follows directly .
Let $x$ and $y$ be elements of $A$.
Then $\set {x, y}$ is a subclass of $A$.
Hence $\set {x, y}$ has a smallest element.
That is, either $x \mathrel \RR y$ or $y \mathrel \RR x$.
As $x$ and $y$ are arbitrary, it follows that:
:$\forall x, y \in A: x \mathrel \RR y \lor y \mathrel \RR x$
Hence, by definition, $\RR$ is a total ordering.
So, we have that $\RR$ is a total ordering on $A$ such that every subclass of $A$ has a smallest element under $\RR$.
That is, $\RR$ is a well-ordering.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Exercise $1.1$
Category:Well-Orderings
Orderings on Set with 3 Elements

== Examples of Orderings ==
Let $S = \set {a, b, c}$ be an arbitrary set with $3$ elements.
The following are all the orderings that can be applied to $S$, grouped into isomorphism classes.
In the below, $\tuple {x, y}$ indicates that $x \preccurlyeq y$ for the ordering $\preccurlyeq$ under consideration.
=== Trivial Ordering ===
This is its Hasse diagram:
:
=== $2$ Elements Ordered ===
This is its Hasse diagram:
:
where the labels can be arbitrary.
=== $2$ Maximal Elements ===
This is its Hasse diagram:
:
where the labels can be arbitrary.
=== $2$ Minimal Elements ===
This is its Hasse diagram:
:
where the labels can be arbitrary.
=== Total Orderings ===
This is its Hasse diagram:
:
where the labels can be arbitrary.
These orderings are total orderings.
From Totally Ordered Set is Lattice, they are also lattice orderings.
== Summary ==
There are $19$ different orderings that can be applied to $S$, grouped into $5$ isomorphism classes.
Exactly one of those isomorphism classes is of a lattice ordering.
It contains $6$ such orderings.
Hence for a set with $3$ elements, there are $6$ possible lattice orderings that can be applied to that set.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.3$
Category:Order Isomorphisms
Orderings on Set with 4 Elements

== Examples of Orderings ==
Let $S$ be an arbitrary set with $4$ elements.
The following are the Hasse diagrams for the $15$ isomorphism classes of all the possible orderings on $S$:
:
Of the above, $\text{(l)}$ and $\text{(o)}$ are lattice orderings.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.4$
Category:Order Isomorphisms
Ordinal Addition/Examples/Ordinal Addition by Natural Number

== Theorem ==
Let $x$ be an ordinal.
Let $x^+$ denote the successor of $x$.
Let $n$ be a natural number.
Then:
:$x + \paren {n + 1} = \paren {x + n}^+$
where $+$ denotes ordinal addition.
== Proof ==
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 3$ Some ordinals
Category:Ordinal Addition
Category:Natural Numbers
Ordinal Addition/Examples/Ordinal Addition by One

== Theorem ==
Let $x$ be an ordinal.
Let $x^+$ denote the successor of $x$.
Let $1$ denote (ordinal) one, the successor of the zero ordinal $\O$.
Then:
:$x + 1 = x^+$
where $+$ denotes ordinal addition.
== Proof ==
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 3$ Some ordinals
Category:Ordinal Addition
Category:1
Ordinal Addition/Examples/Ordinal Addition by Two

== Theorem ==
Let $x$ be an ordinal.
Let $x^+$ denote the successor of $x$.
Let $2$ denote the successor of the ordinal $1$.
Then:
:$x + 2 = x^{++}$
where $+$ denotes ordinal addition.
== Proof ==
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 3$ Some ordinals
Category:Ordinal Addition
Category:2
Ordinal Addition by Zero

== Theorem ==
Let $x$ be an ordinal.
Let $\O$ be the zero ordinal.
Then:
:$x + \O = x = \O + x$
where $+$ denotes ordinal addition.
== Proof ==
By definition of ordinal addition, it is immediate that:
:$x + \O = x$
We shall use Transfinite Induction on $x$ to prove $\O + x = x$
=== Base Case ===
The induction basis $x = \O$ comes down to:
:$\O + \O = \O$
This follows by the above.
=== Inductive Case ===
For the induction step, suppose that $\O + x = x$.
Then, also:
=== Limit Case ===
Finally, the limit case.
So let $x$ be a limit ordinal, and suppose that:
:$\forall y \in x: \O + y = y$
Now we have:
Hence the result, by Transfinite Induction.
== Sources ==
* : $\S 8.3$
Category:Ordinal Arithmetic
Ordinal Addition is Associative

== Theorem ==
Ordinal addition is associative, i.e.:
:$\paren {x + y} + z = x + \paren {y + z}$
holds for all ordinals $x$, $y$ and $z$.
== Proof ==
By Transfinite Induction on $z$.
=== Basis for the Induction ===
This proves the basis for the induction.
=== Induction Step ===
Suppose that:
:$x + \paren {y + z} = \paren {x + y} + z$
Then:
This proves the induction step.
=== Limit Case ===
Let $z \in K_{II}$ be a limit ordinal, and suppose that:
:$\forall w \in z: x + \paren {y + w} = \paren {x + y} + w$
Then it follows that:
By the hypothesis, we have that $z \in K_{II}$.
From Limit Ordinals Preserved Under Ordinal Addition, also $y + z \in K_{II}$.
Therefore, by definition of ordinal addition:
:$x + \paren {y + z} = \ds \bigcup_{n \mathop \in y + z} \paren {x + n}$
So it is sufficient to prove that:
:$\ds \bigcup_{w \mathop \in z} \paren {x + \paren {y + w} } = \bigcup_{n \mathop \in y + z} \paren {x + n}$
Take any $w \in z$. 
Then by Membership is Left Compatible with Ordinal Addition, $y + w < y + z$.
Setting $n = y + w$, we now have that:
:$x + \paren {y + w} \le x + n$
By Supremum Inequality for Ordinals, it now follows that:
:$\ds \bigcup_{w \mathop \in z} \paren {x + \paren {y + w} } \le \bigcup_{n \mathop \in y + z} \paren {x + n}$
To prove the converse inequality, take any $n \in y + z$.
By definition of ordinal addition, this means:
:$n \in \ds \bigcup_{w \mathop \in z} \paren {y + w}$
and so, for some $w \in z$, we have $n \in y + w$.
By Membership is Left Compatible with Ordinal Addition, this yields:
:$x + n \in x + \paren {y + w}$
and whence by Supremum Inequality for Ordinals:
:$\ds \bigcup_{n \mathop \in y + z} \paren {x + n} \le \bigcup_{w \mathop \in z} \paren {x + \paren {y + w} }$
By definition of set equality:
:$\ds \bigcup_{n \mathop \in y + z} \paren {x + n} = \bigcup_{w \mathop \in z} \paren {x + \paren {y + w} }$
This proves the limit case.
Hence the result, by Transfinite Induction.
== Also see ==
* Natural Number Addition is Associative/Proof 2
== Sources ==
* : $\S 8.12$
Category:Ordinal Arithmetic
Ordinal Addition is Closed

== Theorem ==
Let $\On$ be the class of all ordinals.
Then:
:$\forall x, y \in \On: x + y \in \On$
That is: the sum $x+y$ is an ordinal.
== Proof ==
Using Transfinite Induction on $y$:
=== Base Case ===
=== Inductive Case ===
=== Limit Case ===
== Sources ==
* : $\S 8.2$
Category:Ordinal Arithmetic
Ordinal Addition is Left Cancellable

== Theorem ==
Let $x, y, z$ be ordinals.
Then:
:$\paren {z + x} = \paren {z + y} \implies x = y$
That is, ordinal addition 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.
Note that:
However:
This contradicts the consequents of the first two equations, so:
== Warning ==
Ordinal addition is not right cancellable.
By Finite Ordinal Plus Transfinite Ordinal
:$\paren {1 + \omega} = \paren {0 + \omega}$
but $1 \ne 0$.
== Also see ==
* Natural Number Addition is Cancellable: natural number addition is both left and right cancellable.
== Sources ==
* : $\S 8.5$
Category:Ordinal Arithmetic
Ordinal Equal to Rank

== 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 $x$ equals the rank of $S$ $S \in \map V {x + 1} \land S \notin \map V x$.
== Proof ==
=== Sufficient Condition ===
Suppose that $x$ is equal to the rank of $S$.
Then $S \in \map V {x + 1}$ by the definition of rank.
Suppose $S \in \map V x$.
Then $x \ne 0$, so $x = y^+$ for some ordinal $y$ or $x$ is a limit ordinal.
Suppose $x = y^+$.
Then, $S \in \map V {y + 1}$.
This contradicts the fact that the rank is the smallest set $x$ such that $S \in V\left({x+1}\right)$.
Suppose $x$ is a limit ordinal.
Then, $S \in \map V y$ for some $y$, so $S \in \map V {y + 1}$ and $y+1 < x$.
This contradicts the fact that the rank is the smallest set $x$ such that $S \in V \left({x+1}\right)$.
Therefore, $S \notin \map V x$.
=== Necessary Condition ===
Suppose $S \notin \map V x$ and $S \in \map V {x + 1}$.
Then, if $y < x$, then $S \notin \map V {y + 1}$ by Von Neumann Hierarchy Comparison.
If $x < y$, then $S \in \map V y$.
Therefore, $x$ is the unique ordinal that satisfies $S \notin \map V x$ and $S \in \map V {x + 1}$.
Moreover, the rank of $S$ also satisfies $S \notin \map V x$ and $S \in \map V {x + 1}$.
Therefore, $x = \map {\operatorname{rank} } S$.
== Sources ==
* : $\S 9.15(2)$
Category:Von Neumann Hierarchy
Ordinal Exponentiation is Closed

== Theorem ==
Let $x$ and $y$ be ordinals.
Then:
:$x^y \in \On$
That is, ordinal exponentiation is closed.
== Proof ==
Let $x = 0$ and $y = 0$.
Then by the definition of ordinal exponentiation:
:$x^y = 1$
Let $x = 0$ and $y \ne 0$.
Then by the definition of ordinal exponentiation:
:$x^y = 0$
In either case, $x^y$ is an ordinal.
Now suppose that $x \ne 0$.
The proof shall proceed by Transfinite Induction on $y$.
=== Basis for the Induction ===
By definition of ordinal exponentiation:
:$x^0 = 1$
which is an ordinal.
This proves the basis for the induction.
=== Induction Step ===
The inductive hypothesis states that $x^y \in \On$.
Suppose the inductive hypothesis holds.
Then:
This proves the induction step.
=== Limit Case ===
The inductive hypothesis states that $x^z \in \On$ for all $z \in y$.
This proves the limit case.
Category:Ordinal Arithmetic
Ordinal Exponentiation of Terms

== Theorem ==
Let $x, y, z$ be ordinals.
Let $n$ be a finite ordinal.
Let $x$ be a limit ordinal.
Let $y, z, n$ all be greater than $0$.
Then:
:$\paren {x^y \times n}^z = x^{y \mathop \times z} \times n$ if $z$ is not a limit ordinal
:$\paren {x^y \times n}^z = x^{y \mathop \times z}$ if $z$ is a limit ordinal.
== Proof ==
The proof shall proceed by Transfinite Induction on $z$.
=== Basis for the Induction ===
The hypothesis requires that $z \ne 0$, so the induction starts at $z = 1$.
This proves the basis for the induction.
=== Induction Step ===
This proves the induction step.
=== Limit Case ===
Suppose that this statement holds for all $w \in z$ where $z$ is a limit ordinal.
Then:
This proves the limit case.
== Sources ==
* : $\S 8.47$
Category:Ordinal Arithmetic
Ordinal Exponentiation via Cantor Normal Form/Corollary

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $x$ be a limit ordinal and 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 natural numbers.
Then:
:$\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^{x^y} = x^{a_1 \mathop \times x^y}$
== Proof ==
By the hypothesis, $x^y$ is a limit ordinal by Limit Ordinals Closed under Ordinal Exponentiation.
The result follows from Ordinal Exponentiation via Cantor Normal Form/Limit Exponents.
== Also see ==
* Cantor Normal Form
== Sources ==
* : $\S 8.50$
Category:Ordinal Arithmetic
Ordinal Exponentiation via Cantor Normal Form/Limit Exponents

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $x$ and $y$ be limit ordinals.
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 natural numbers.
Then:
:$\ds \paren {\sum_{i \mathop = 1}^n x^{a_i} \times b_i}^y = x^{a_1 \mathop \times y}$
== Proof ==
By Upper Bound of Ordinal Sum:
:$\ds \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i} \le x^{a_1} \times \paren {b_1 + 1}$
Furthermore:
:$\ds x^{a_1} \le \sum_{i \mathop = 1}^n \paren {x^{a_i} \times b_i}$
It follows that:
It follows that:
== Also see ==
* Definition:Cantor Normal Form
== Sources ==
* : $\S 8.49$
Category:Ordinal Arithmetic
Ordinal Less than Successor Aleph

== Theorem ==
Let $x$ and $y$ be ordinals.
Then:
:$y < \aleph_{x + 1} \iff y < \aleph_x \lor y \sim \aleph_x$
== Proof ==
=== Sufficient Condition ===
But $\card y$ is a cardinal number, so it is either finite or an element of the class of infinite cardinala.
It is a finite set $y \in \omega$ by Ordinal is Finite iff Natural Number.
If $y \in \omega$, then $y \in \aleph_x$ since $\aleph_x$ is an infinite set by Aleph is Infinite.
If it is an element of the infinite cardinal class, then $y = \aleph_z$ for some ordinal $z$ by the definition of the aleph mapping.
This means that:
==== Case 1: $z < x$ ====
In the first case $z < x$:
==== Case 2: $z = x$ ====
Therefore:
:$y < \aleph_x \lor y \sim \aleph_x$
=== Necessary Condition ===
Suppose $y < \aleph_x \lor y \sim \aleph_x$.
==== Case 1: $y < \aleph_x$ ====
Suppose $y < \aleph_x$.
Since $\aleph$ is strictly monotone by the definition of the aleph mapping:
:$y < \aleph_{x + 1}$
==== Case 2: $y \sim \aleph_x$ ====
Suppose $y \sim \aleph_x$.
Then by Equivalent Sets have Equal Cardinal Numbers:
:$\card y = \aleph_x$
But since $\aleph$ is strictly monotone:
:$\aleph_x < \aleph_{x + 1}$
Therefore, by Ordinal in Aleph iff Cardinal in Aleph:
:$\card y < \aleph_{x + 1}$ and $y < \aleph_{x + 1}$
Category:Aleph Mapping
Ordinal Membership is Asymmetric

== Theorem ==
Let $m$ and $n$ be ordinals.
Then it is not the case that $m \in n$ and $n \in m$.
== Proof ==
 $m \in n$ and $n \in m$.
Since $m$ is an ordinal, it is transitive.
Thus since $m \in n$ and $n \in m$, it follows that $m \in m$.
But this contradicts Ordinal is not Element of Itself.
Category:Ordinals
Ordinal Membership is Transitive

== Theorem ==
Let $\On$ denote the class of all ordinals.
Then:
:$\forall \alpha, \beta, \gamma \in \On: \paren {\alpha \in \beta} \land \paren {\beta \in \gamma} \implies \alpha \in \gamma$
== Proof ==
By Strict Ordering of Ordinals is Equivalent to Membership Relation the statement to be proved is equivalent to:
:$\forall \alpha, \beta, \gamma \in \On: \paren {\alpha \subsetneqq \beta} \land \paren {\beta \subsetneqq \gamma} \implies \alpha \subsetneqq \gamma$
which follows (indirectly) from Subset Relation is Transitive.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Corollary $1.16$
Category:Ordinals
Category:Transitive Relations
Ordinal Membership is Trichotomy

== Theorem ==
Let $\alpha$ and $\beta$ be ordinals.
Then:
:$\paren {\alpha = \beta} \lor \paren {\alpha \in \beta} \lor \paren {\beta \in \alpha}$
where $\lor$ denotes logical or.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
Category:Ordinal Membership is Trichotomy
Category:Ordinals
Ordinal Membership is Trichotomy/Corollary

== Corollary to Ordinal Membership is Trichotomy ==
Let $\alpha$ be an ordinal.
Let $x, y \in \alpha$ such that $x \ne y$.
Then either:
:$x \in y$
or:
:$y \in x$
== Proof ==
We have that Element of Ordinal is Ordinal.
The result then follows directly from Ordinal Membership is Trichotomy.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Corollary $1.15$
Category:Ordinal Membership is Trichotomy
Ordinal Membership is Trichotomy/Proof 1

== Theorem ==
== Proof ==
From Class of All Ordinals is Well-Ordered by Subset Relation, $\On$ is a nest.
Hence:
:$\forall \alpha, \beta \in \On: \paren {\alpha \subsetneqq \beta} \lor \paren {\beta \subsetneqq \alpha} \lor \paren {\alpha = \beta}$
From Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, this is equivalent to:
:$\forall \alpha, \beta \in \On: \paren {\alpha \in \beta} \lor \paren {\beta \in \alpha} \lor \paren {\alpha = \beta}$
Hence the result.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Corollary $1.14$
Category:Ordinal Membership is Trichotomy
Ordinal Membership is Trichotomy/Proof 2

== Theorem ==
== Proof ==
By Relation between Two Ordinals, it follows that:
:$\paren {\alpha = \beta} \lor \paren {\alpha \subset \beta} \lor \paren {\beta \subset \alpha}$
By Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, the result follows.
== Sources ==
* : $\S 7.10$
Category:Ordinal Membership is Trichotomy
Ordinal Multiplication by One

== Theorem ==
Let $x$ be an ordinal.
Let $1$ denote the ordinal one.
== Proof ==
The proof of the other equality shall proceed 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 ==
* : $\S 8.18$
Category:Ordinal Arithmetic
Ordinal Multiplication by Zero

== Theorem ==
Let $x$ be an ordinal.
== Proof ==
For $\O \cdot x = \O$, the proof shall proceed by Transfinite Induction on $x$.
=== 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 ==
* : $\S 8.18$
Category:Ordinal Arithmetic
Ordinal Multiplication is Associative

== Theorem ==
Let $x, y, z$ be ordinals.
Let $\times$ denote ordinal multiplication.
Then:
:$x \times \paren {y \times z} = \paren {x \times y} \times z$
== Proof ==
The proof shall proceed by Transfinite Induction on $z$:
=== Basis for the Induction ===
Let $0$ denote the zero ordinal.
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:
:$\forall w \in Z: x \times \paren {y \times w} = \paren {x \times y} \times w$
where $z$ is a limit ordinal.
The proof shall proceed by cases:
=== Case 1 ===
If $y = 0$, then:
=== Case 2 ===
If $y \ne 0$, then $y \times z$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Multiplication.
It follows that:
If $u < \paren {y \times z}$, then $u < \paren {y \times w}$ for some $w \in z$ by Ordinal is Less than Ordinal times Limit.
Generalizing, the result follows for all $u \in \paren {y \times z}$.
Therefore by Supremum Inequality for Ordinals:
:$x \times \paren {y \times z} \le \paren {x \times y} \times z$
Conversely, take any $w < z$.
It follows by Supremum Inequality for Ordinals that:
:$\paren {x \times y} \times z \le x \times \paren {y \times z}$
By definition of set equality:
: $x \times \paren {y \times z} = \paren {x \times y} \times z$
This proves the limit case.
== Sources ==
* : $\S 8.26$
Category:Ordinal Arithmetic
Ordinal Multiplication is Closed

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $\On$ denote the class of all ordinals.
:$x \cdot y \in \On$
== Proof ==
The proof proceeds by transfinite induction on $y$.
=== 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 ==
* : $\S 8.16$
Category:Ordinal Arithmetic