nyuuzyou/wikis huggingface dataset:
Class Member of Class Builder

== Theorem ==
Let $A$ be a class.
Let $x$ be a set.
Let $\map P x$ be a well-formed formula in the language of set theory.
Let $\map P A$ denote the formula $\map P x$ with all free instances of $x$ replaced with instances of $A$.
Let $\set {x: \map P x}$ be a class specified using class builder notation.
Then:
:$A \in \set {x: \map P x} \iff \paren {\exists x: x = A \land \map P A}$
== Proof ==
== Also see ==
* Set Definition by Predicate
== Sources ==
* : $\S 4.12$
Category:Zermelo-Fraenkel Class Theory
Class Union Distributes over Class Intersection

== Theorem ==
Let $A$, $B$ and $C$ be classes.
Then:
:$A \cup \paren {B \cap C} = \paren {A \cup B} \cap \paren {A \cup C}$
where:
:$A \cup B$ denotes class union
:$B \cap C$ denotes class intersection.
== Proof ==
== Also see ==
* Union Distributes over Intersection
* Class Intersection Distributes over Class Union
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.6. \ \text {(b)}$
Category:Class Intersection
Category:Class Union
Category:Examples of Distributive Operations
Class Union Exists and is Unique

== Theorem ==
Let $V$ be a basic universe.
Let $A \subseteq V$ be a class.
Let $\bigcup A$ denote the union of $A$.
Then $\bigcup A$ is guaranteed to exist and is unique.
== Proof ==
By the Axiom of Specification the union of $A$ can be created:
:$\bigcup A := \set {x: \exists y: x \in y \land y \in A}$
Hence $\bigcup A$ exists.
Let $B$ and $C$ both be unions of $A$.
From the definition of union:
:$\forall A$:
::$x \in B \iff \exists y \in A: x \in y$
::$x \in C \iff \exists y \in A: x \in y$
From Biconditional is Commutative and Biconditional is Transitive:
:$x \in B \iff x \in C$
By the Axiom of Extension:
:$B = C$
Hence the union of $A$ is unique.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom
Category:Class Union
Class has Subclass which is not Element

== Theorem ==
Let $A$ be a class.
Then $A$ has at least one subclass $B$ which is not an element of $A$.
== Proof ==
Let a set $x$ be defined as ordinary $x \notin x$.
Let $\map \phi x$ be the set property defined as:
:$\map \pi x := \neg \paren {x \in x}$
Then by the there exists a class, which can be denoted $B$, such that:
:$B = \set {x \in A: \neg \paren {x \in x} }$
That is, $B$ contains as elements of all the ordinary sets of $A$.
By the $B$ is unique.
Thus $B$ is the class of all ordinary sets of $A$.
:$x \in B \iff \paren {x \in A \land x \notin x}$
Hence, for all $x$ which happen to be elements of $A$, we have:
:$x \in B \iff x \notin x$
 $B \in A$.
Then we could set $A$ for $x$, and so obtain:
:$B \in B \iff B \notin B$
This is a contradiction.
Hence by Proof by Contradiction $B$ cannot be an element of $A$.
That is:
:$B \notin V$
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 1$ Extensionality and separation: Theorem $1.2$
Category:Class Theory
Class is Extensional

== Theorem ==
Let $A$ be a class.
Then:
:$A = \set {x : x \in A}$
That is, $A$ is extensional.
== Proof ==
We have:
:$x \in \set {x : x \in A} \iff x \in A$
by Characterization of Class Membership (applied to $\set {x : x \in A}$).
By Universal Generalisation, it follows that:
:$\forall x: \paren {x \in A \iff x \in \set {x : x \in A} }$
Hence the result, by definition of class equality.
== Sources ==
* : $\S 4.9$
Category:Class Theory
Class is Not Element of Itself

== Theorem ==
Let $A$ be a class.
Then $A$ is not an element of itself:
:$A \notin A$
== Proof ==
$A \in A$ would create a relational loop for the class membership sign.
By No Membership Loops, this is a contradiction.
== Sources ==
* : $\S 5.21$
Category:Class Theory
Category:Axiom of Foundation
Class is Proper iff Bijection from Class to Proper Class

== Theorem ==
Let $A$ be a class.
Let $\mathrm P$ be a proper class.
Then $A$ is proper there exists a bijection from $A$ to $\mathrm P$.
=== Corollary ===
== Proof ==
=== Necessary Condition ===
Suppose that $A$ is a proper class.
Then by the Axiom of Limitation of Size there exists a bijection $f: A \to \mathrm U$, where $\mathrm U$ is the universal class.
There also exists a bijection $g: \mathrm P \to \mathrm U$.
By Inverse of Bijection is Bijection, there exists a bijection $g^{-1}: \mathrm U \to \mathrm P$
From Composite of Bijections is Bijection, it follows that $f \circ g^{-1}: A \to \mathrm P$ is a bijection.
=== Sufficient Condition ===
Suppose that there exists a bijection $f: A \to \mathrm P$.
Because $\mathrm P$ is a proper class, it follows from the Axiom of Limitation of Size that there exists a bijection $g: \mathrm P \to \mathrm U$.
Then by Composite of Bijections is Bijection, $f \circ g: A \to \mathrm U$ is a bijection from $A$ to the universe.
And so by the Axiom of Limitation of Size, $A$ is proper.
Hence the result.
Category:Gödel-Bernays Class Theory Category:Class Mappings
Class is Proper iff Bijection from Class to Proper Class/Corollary

== Theorem ==
Let $A$ be a class.
Let $P$ be a proper class.
Then $A$ is proper there exists a bijection from $P$ to $A$.
== Proof ==
From Biconditional is Transitive and Class is Proper iff Bijection from Class to Proper Class, it suffices to show that:
:There exists a bijection from $A$ to $P$ there exists a bijection from $P$ to $A$.
The rest follows from Inverse of Bijection is Bijection.
Category:Gödel-Bernays Class Theory
Category:Class Mappings
Class is Subclass of Universal Class

== Theorem ==
Let $V$ denote the universal class.
Let $A$ be a class.
Then $A$ is a subclass of $V$.
== Proof ==
By definition of class, $A$ is a collection of sets.
Let $x \in A$ be a set.
By definition of universal class, $V$ contains all sets as elements.
Hence $x \in V$.
So we have that:
:$x \in A \implies x \in V$
and the result follows by definition of subclass.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 1$ Extensionality and separation
Category:Universal Class
Category:Subclasses
Class is Transitive iff Union is Subclass

== Theorem ==
A class $A$ is transitive :
:$\bigcup A \subseteq A$
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity: Theorem $10.1$
Category:Class Theory
Category:Transitive Classes
Category:Class is Transitive iff Union is Subclass
Class of All Cardinals Contains Minimally Inductive Set

== Theorem ==
Let $\NN$ denote the class of all cardinal numbers.
Then:
:$\omega \subseteq \NN$
where $\omega$ denotes the minimally inductive set.
== Proof ==
Suppose $n \in \omega$.
By Cardinal of Finite Ordinal, $n = \card n$.
By Cardinal of Cardinal Equal to Cardinal/Corollary, $n \in \NN$.
== Sources ==
* : $\S 10.39$
Category:Class of All Cardinals
Category:Minimally Inductive Set
Class of All Cardinals is Proper Class

== Theorem ==
The class $\NN$ of all cardinal numbers is a proper class.
== Proof ==
 $\NN$ is a small class.
By Class of All Cardinals is Subclass of Class of All Ordinals:
:$\NN \subseteq \On$
Therefore, $\bigcup \NN$ is an ordinal by Union of Set of Ordinals is Ordinal.
Take $x = \set {y \in \On: y \preccurlyeq \bigcup \NN}$.
By Cardinal Equal to Collection of All Dominated Ordinals, $x$ is a cardinal number and there is no injection $f: x \to \bigcup \NN$.
But since $x$ is a cardinal number, $x \in \NN$, so $x \subseteq \bigcup \NN$ by Set is Subset of Union.
Therefore, the identity mapping $I_x: x \to \bigcup \NN$ is an injection.
This is a contradiction.
Therefore by Proof by Contradiction $\NN$ is not a small class.
== Sources ==
* : $\S 10.41$
Category:Class of All Cardinals
Category:Class Theory
Class of All Cardinals is Subclass of Class of All Ordinals

== Theorem ==
Let $\NN$ denote the class of all cardinals.
Let $\On$ denote the class of all ordinals.
Then:
:$\NN \subseteq \On$
== Proof ==
By definition of the class of all cardinals:
:$\NN = \set {x \in \On: \exists y: x = \card y}$
Every element of $\NN$ is thus an element of $\On$.
== Sources ==
* : $\S 10.37$
Category:Class of All Cardinals
Category:Class of All Ordinals
Class of All Ordinals is G-Tower

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $g$ be the successor mapping:
:$\forall x \in \On: \map g x = x \cup \set x$
Then $\On$ is a $g$-tower.
== Proof ==
From Successor Mapping is Progressing, $g$ is a progressing mapping.
From Class of All Ordinals is Minimally Superinductive over Successor Mapping, $\On$ is superinductive over $g$.
Hence the result by definition of $g$-tower.
Category:G-Towers
Category:Class of All Ordinals
Class of All Ordinals is Minimally Superinductive over Successor Mapping

== Theorem ==
The class of all ordinals $\On$ is the unique class which is minimally superinductive under the successor mapping.
== Proof ==
We need to show that:
:$\On$ is a superinductive class under the successor mapping
and:
:no proper subclass of $\On$ is superinductive class under the successor mapping.
We recall immediately that Successor Mapping is Progressing.
This validates the definition of superinductive class under the successor mapping.
By definition of ordinal:
:$S \in \On$
:
:$S$ is an element of every superinductive class under the successor mapping.
Hence $\On$ is a subclass of every superinductive class under the successor mapping.
Let $\cap S$ denote the intersection of every superinductive class under the successor mapping.
Thus by definition of class intersection:
:$\On$ is a subclass of $\cap S$.
Let $M$ be an arbitrary superinductive class under the successor mapping.
We have by definition of superinductive class:
:$\O \in M$
Thus as $M$ is arbitrary:
:$\O \in \cap S$
Thus $\O$ is an element of every superinductive class.
Hence:
:$\O \in \On$
Now let $\alpha \in \On$.
Then by definition of ordinal:
:$\alpha$ is an element of every superinductive class under the successor mapping.
Hence by definition of superinductive class:
:$\alpha^+$ is an element of every superinductive class under the successor mapping
where $\alpha^+$ denotes the successor set of $\alpha$.
Hence by definition:
:$\alpha^+ \in \On$
Thus $\On$ is closed under the successor mapping.
Let $C$ be a chain in $\On$.
Then $C$ is a chain in every superinductive class under the successor mapping.
By definition of superinductive class:
:$\cup C$ is an element of every superinductive class under the successor mapping
where $\cup C$ denotes the union of $C$.
That is:
:$\cup C \in \On$
demonstrating that $\On$ is closed under chain unions.
Thus $\On$ is a superinductive class.
 $A$ is a superinductive class which is a proper subclass of $\On$.
Then:
:$\exists x \in \On: x \notin A$
Hence $\On \not \subseteq A$
But this contradicts the statement that $\On$ is a subclass of every superinductive class.
Hence by Proof by Contradiction no proper subclass of $\On$ is superinductive class under the successor mapping.
Hence by definition $\On$ is a minimally superinductive under the successor mapping.
It remains to prove uniqueness.
Suppose $\On'$ is another minimally superinductive class under the successor mapping.
Then we have:
:$\On' \subseteq \On$
but also we have:
:$\On \subseteq \On'$
and so:
:$\On' = \On$
Thus $\On$ is the unique minimally superinductive class under the successor mapping.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers
Category:Class of All Ordinals
Category:Minimally Superinductive Classes
Category:Successor Mapping
Class of All Ordinals is Only Proper Class of Ordinals

== Theorem ==
Let $A$ be a transitive proper class of ordinals.
Then $A$ is the class of all ordinals $\On$.
== Proof ==
Let $A$ be a transitive class of ordinals.
Let there exist $\alpha \in \On$ such that $\alpha \notin A$.
Then by Transitive Class of Ordinals is Subset of Ordinal not in it:
:$A \subseteq \alpha$
But that makes $A$ a set.
So if $A$ is a proper class, it must contain all ordinals.
That is:
:$A = \On$
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 2$ Ordinals and transitivity: Theorem $2.2 \ (2)$
Category:Class of All Ordinals
Class of All Ordinals is Ordinal

== Theorem ==
The class of all ordinals $\On$ is an ordinal.
== Proof ==
The epsilon relation is equivalent to the strict subset relation when restricted to ordinals by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
It follows that:
:$\forall x \in \On: x \subset \On$
The initial segment of the class of all ordinals is:
:$\set {x \in \On : x \subset \On}$
This class is equal to $\On$.
Therefore, by the definition of ordinal, $\On$ is an ordinal.
== Sources ==
* : $7.12$
Category:Class of All Ordinals
Class of All Ordinals is Proper Class/Proof 1

== Theorem ==
== Proof ==
We have that Successor Mapping on Ordinals is Strictly Progressing.
The result follows from Superinductive Class under Strictly Progressing Mapping is Proper Class.
 
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.10$
Category:Class of All Ordinals is Proper Class
Class of All Ordinals is Proper Class/Proof 2

== Theorem ==
== Proof ==
 $\On$ is a set.
Then from the Burali-Forti Paradox, a contradiction could be deduced.
Hence by Proof by Contradiction, $\On$ cannot be a set.
Thus $\On$ is a class that is not a set.
Hence $\On$ is a proper class.
 
Category:Class of All Ordinals is Proper Class
Class of All Ordinals is Well-Ordered by Subset Relation

== Theorem ==
Let $\On$ be the class of all ordinals.
Then the restriction of the subset relation, $\subseteq$, to $\On$ is a well-ordering.
That is:
:$\subseteq$ is an ordering on $\On$.
:If $A$ is a non-empty subclass of $\On$, then $A$ has a smallest element under the subset relation.
== Proof 1 ==
== Proof 2 ==
Category:Class of All Ordinals is Well-Ordered by Subset Relation
Category:Class of All Ordinals
Category:Subset Relation
Class of All Ordinals is Well-Ordered by Subset Relation/Proof 1

== Theorem ==
== Proof ==
By Subset Relation on Class is Ordering, $\subseteq$ is an ordering of any class.
Let $A$ be a subclass of $\On$.
By Intersection of Ordinals is Smallest, $A$ has a smallest element under the subset relation.
Category:Class of All Ordinals is Well-Ordered by Subset Relation
Class of All Ordinals is Well-Ordered by Subset Relation/Proof 2

== Theorem ==
== Proof ==
From Class of All Ordinals is $g$-Tower, $\On$ is a $g$-tower.
The result follows from $g$-Tower is Well-Ordered under Subset Relation.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.3$
Category:Class of All Ordinals is Well-Ordered by Subset Relation
Class of Finite Character is Closed under Chain Unions

== Theorem ==
Let $A$ be a class which has finite character.
Then $A$ is closed under chain unions.
== Proof ==
Let $C$ be a chain of elements of $A$.
$C$ is a set.
To show that $\bigcup C \in A$ it is sufficient to show that every finite subset of $\bigcup C$ is also an element of $A$.
Let $\set {y_1, y_2, \ldots, y_n}$ be an arbitrary subset of $\bigcup C$.
For each $i \le n$, we have that $y_i$ is an element of some $c_i$ of $C$.
Let $c$ be the greatest element by the subset relation of the sets $\set {c_1, c_2, \ldots, c_n}$.
Then each of the sets $y_1, y_2, \ldots, y_n$ is an element of $c$.
Hence $\set {y_1, y_2, \ldots, y_n} \subseteq c$.
Thus every finite subset of $\bigcup C$ is the subset of some element of $C$,
Hence every finite subset of $\bigcup C$ is the subset of some element of $A$.
Hence by Class of Finite Character is Swelled, every finite subset of $\bigcup C$ is an element of $A$.
So, as $A$ has finite character:
:$\bigcup C \in A$
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Lemma $5.4 \ (2)$
Category:Closure under Chain Unions
Category:Finite Character
Class of Finite Character is Swelled

== Theorem ==
Let $A$ be a class which has finite character.
Then $A$ is a swelled class.
== Proof ==
Let $x \in A$ and $y \subseteq x$.
Then every finite subset of $y$ is also a finite subset of $x$.
Hence every finite subset of $y$ is in $A$.
Hence again $y \in A$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Lemma $5.4 \ (1)$
Category:Swelled Classes
Category:Finite Character
Class of Infinite Cardinals is Proper Class

== Theorem ==
The class of infinite cardinals $\NN’$ is a proper class.
== Proof ==
 $\NN'$ is a small class.
By Union of Small Classes is Small, $\NN’ \cup \omega$ is a small class.
By definition of the class of infinite cardinals, $\NN \subseteq \NN' \cup \omega$.
But by Axiom of Subsets Equivalents, this means that $\NN$ is a small class.
This contradicts Class of All Cardinals is Proper Class.
Therefore, $\NN’$ is not a small class.
== Sources ==
* : $\S 10.43$
Category:Class of All Cardinals
Class such that Every Transitive Subset is Element of it Contains All Ordinals

== Theorem ==
Let $K$ be a class.
Let $K$ be such that every transitive subset of $K$ is an element of $K$.
Then every ordinal is an element of $K$.
== Proof ==
Let us assume the hypothesis.
Let $\On$ denote the class of all ordinals.
From Well-Ordering of Class of All Ordinals under Subset Relation, $\On$ is well-ordered under $\subseteq$.
Hence we can use the First Principle of Transfinite Induction.
Let $\alpha$ be an ordinal such that every ordinal less than $\alpha$ is an element of $K$.
Thus $\alpha \subseteq K$.
As Ordinal is Transitive, it follows that:
:$\alpha \in K$
So if all ordinals less than $\alpha$ are elements of $K$, then so is $\alpha$.
Hence by the First Principle of Transfinite Induction, all ordinals are in $K$.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 2$ Ordinals and transitivity: Theorem $2.4$
Category:Ordinals
Category:Transitive Classes
Class under Progressing Mapping such that Elements are Sandwiched is Nest

== Theorem ==
Let $A$ be a class.
Let $g: A \to A$ be a progressing mapping on $A$ such that:
:$\forall x, y \in A: \map g x \subseteq y \lor y \subseteq x$
Then $A$ is a nest:
:$\forall x, y \in A: x \subseteq y \lor y \subseteq x$
== Proof ==
By definition of progressing mapping:
:$\forall x \in A: x \subseteq \map g x$
Thus by Subset Relation is Transitive:
:$\map g x \subseteq y \implies x \subseteq y$
and it follows that:
:$\forall x, y \in A: x \subseteq y \lor y \subseteq x$
Hence the result by definition of nest.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Lemma $4.9$
Category:Progressing Mappings
Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable

== Theorem ==
Let $B$ be a well-orderable class.
Let $A$ be a class such that there exists an injection $f: A \to C$, where $C$ is a subclass of $B$.
Then $A$ is a well-orderable class.
== Proof ==
Let $\RR$ be a well-ordering that can be established on $B$.
This can always be done, as $B$ is a well-orderable class.
Let $F$ be an injection that maps each element $x$ of $A$ to an element $\map F x$ of $B$.
Let $\preccurlyeq$ be the class of all ordered pairs $\tuple {x, y}$ of elements of $A$ such that $\tuple {\map F x, \map F y} \in \RR$.
Thus:
:$(1): \quad x \preccurlyeq y \iff \tuple {\map F x, \map F y} \in \RR$
It will be demonstrated that $A$ is well-ordered under $\preccurlyeq$.
Let $\Img F$ denote the image of $F$.
By Subclass of Well-Ordered Class is Well-Ordered, $\Img F$ is well-ordered under $\RR$.
We have that $\RR$ is a well-ordering
Hence :
:$\RR$ is reflexive, transitive and antisymmetric
:Every pair of elements of $B$ is comparable under $\RR$
From reflexivity:
:$\forall a \in A: \forall a \in A: \tuple {\map F a, \map F a} \in \RR$
and so from $(1)$:
:$\forall a \in A: a \preccurlyeq a$
Thus $\preccurlyeq$ is reflexive on $A$.
Let $a, b, c, \in A$ such that $a \preccurlyeq b$ and $b \preccurlyeq c$.
Then from $(1)$:
:$\tuple {\map F a, \map F b} \in \RR$
and:
:$\tuple {\map F b, \map F c} \in \RR$
As $\RR$ is transitive:
:$\tuple {\map F a, \map F c} \in \RR$
Hence from $(1)$:
:$a \preccurlyeq c$
Thus $\preccurlyeq$ is transitive on $A$.
Let $a, b \in A$ such that $a \preccurlyeq b$ and $b \preccurlyeq a$.
Then from $(1)$:
:$\tuple {\map F a, \map F b} \in \RR$
and:
:$\tuple {\map F b, \map F a} \in \RR$
As $\RR$ is antisymmetric:
:$\map F a = \map F b$
Because $F$ is an injection:
:$a = b$
Thus $\preccurlyeq$ is antisymmetric on $A$.
Let $a, b \in A$ be arbitrary.
As every pair of elements of $B$ is comparable under $\RR$:
:$\tuple {\map F a, \map F b} \in \RR$ or $\tuple {\map F b, \map F a} \in \RR$
Hence from $(1)$:
:$a \preccurlyeq b$ or $b \preccurlyeq a$
Hence every pair of elements of $A$ is comparable under $\preccurlyeq$.
Thus $\preccurlyeq$ is a total ordering on $A$.
Let $C$ be an arbitrary non-empty subclass of $A$.
Let $C'$ be the class of all $\map F x$ such that $x \in C$.
We have that $C'$ is a subclass of $B$.
Then as $B$ is well-ordered by $\RR$, it follows that $C'$ has a smallest element.
This smallest element is $\map F b$ for some $b \in C$.
Let $x \in C$.
Then:
:$\map F x \in C'$
and so:
:$\tuple {\map F b, \map F x} \in \RR$
Hence by $(1)$:
:$b \preccurlyeq x$
As $x$ is arbitrary:
:$\forall x \in C: b \preccurlyeq x$
and so $b$ is the smallest element of $C$ $\preccurlyeq$.
It follows by definition that $\preccurlyeq$ is a well-ordering.
Thus $A$ is well-ordered by $\preccurlyeq$.
Hence the result.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Proposition $1.2$
Category:Well-Orderings
Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable/Mistake

== Source Work ==
:
:Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering:
::$\S 1$ Introduction to well ordering:
:::Proposition $1.2$: Proof
== Mistake ==
:Finally, suppose $C$ is any non-empty subclass of $A$. Let $C'$ be the class of all elements $x'$ such that $x \in C$. Then $C'$ contains a least element ( $R$), and this element is $b'$ for some $b \in C$. Then for any $x \in C$, $x' \in C'$, and so $x' R b'$, and therefore $x \le b$. Thus $b$ is the least element of $C$ (with respect to $\le$).
== Correction ==
As $b'$ is the least element of $C'$, it follows that for arbitrary $x' \in C'$ that $b' R x'$, not $x' R b'$.
Hence similarly $b \le x$, not $x \le b$.
However, the conclusion is correct.
== Also see ==
* Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable: 's rendition of this result
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Proposition $1.2$
Category:Mistakes/Set Theory and the Continuum Problem
Classical Probability is Probability Measure

== Theorem ==
The classical probability model is a probability measure.
== Proof ==
We check all the Kolmogorov axioms in turn:
=== First Axiom ===
From Empty Set is Subset of All Sets and from the definitions of the event space and sample space:
:$\O \subseteq \Sigma \subseteq \Omega$
From Cardinality of Empty Set and Cardinality of Subset of Finite Set:
:$0 \le \card \Sigma \le \card \Omega$
Dividing all terms by $\card \Omega$:
:$0 \le \dfrac {\card \Sigma} {\card \Omega} \le 1$
The middle term is the asserted definition of $\map \Pr {\, \cdot \,}$.
=== Second Axiom ===
By hypothesis:
:$\map \Pr \Omega = \dfrac {\card \Omega} {\card \Omega} = 1$
=== Third Axiom ===
Follows from Cardinality is Additive Function and the corollary to the Inclusion-Exclusion Principle.
Category:Probability Theory