id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-16800 | Box Topology may not be Coarsest Topology such that Projections are Continuous/Lemma | Let $\struct {X, \tau}$ be a topological space.
Let $U \in \tau$ such that $U \ne \O$ and $U \ne X$.
Let:
:$Y = \ds \prod_{n \mathop \in \N} X = X \times X \times X \times \ldots$
be the countable Cartesian product of $\family {X}_{n \mathop \in \N}$.
Let $\tau_T$ be the product topology on $Y$.
Let $\tau_b$ be the box... | By the definition of the box topology:
:$V \in \tau_b$
Let $W \in \tau_T$. | Let $\struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $U \in \tau$ such that $U \ne \O$ and $U \ne X$.
Let:
:$Y = \ds \prod_{n \mathop \in \N} X = X \times X \times X \times \ldots$
be the [[Definition:Countable Cartesian Product|countable Cartesian product]] of $\family {X}_{n \mathop... | By the definition of the [[Definition:Box Topology|box topology]]:
:$V \in \tau_b$
Let $W \in \tau_T$. | Box Topology may not be Coarsest Topology such that Projections are Continuous/Lemma | https://proofwiki.org/wiki/Box_Topology_may_not_be_Coarsest_Topology_such_that_Projections_are_Continuous/Lemma | https://proofwiki.org/wiki/Box_Topology_may_not_be_Coarsest_Topology_such_that_Projections_are_Continuous/Lemma | [
"Box Topology"
] | [
"Definition:Topological Space",
"Definition:Cartesian Product/Countable",
"Definition:Product Topology",
"Definition:Box Topology",
"Definition:Cartesian Product/Countable",
"Definition:Element",
"Definition:Box Topology",
"Definition:Element",
"Definition:Product Topology"
] | [
"Definition:Box Topology"
] |
proofwiki-16801 | P-adic Expansion Representative of P-adic Number is Unique | Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a$ be a $p$-adic number, that is left coset, in $\Q_p$.
Let $\ds \sum_{i \mathop = m}^\infty d_i p^i$ and $\ds \sum_{i \mathop = k}^\infty e_i p^i$ be $p$-adic expansions that represent $a$.
Then:
:$(1) \quad m = k$
:$(2)... | From P-adic Number times P-adic Norm is P-adic Unit there exists $n \in \Z$:
:$p^n a \in \Z^\times_p$
:$\norm a_p = p^n$
where $\Z^\times_p$ is the set of $p$-adic units.
Let $l = -n$.
From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient:
:$l = \min \set {i: i \ge m \land d_i \ne 0}$
and
:$... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:Valued Field of P-adic Numbers|$p$-adic numbers]].
Let $a$ be a [[Definition:P-adic Number|$p$-adic number]], that is [[Definition:Left Coset|left coset]], in $\Q_p$.
Let $\ds \sum_{i \mathop = m}^\in... | From [[P-adic Number times P-adic Norm is P-adic Unit]] there exists $n \in \Z$:
:$p^n a \in \Z^\times_p$
:$\norm a_p = p^n$
where $\Z^\times_p$ is the [[Definition:Set|set]] of [[Definition:P-adic Unit|$p$-adic units]].
Let $l = -n$.
From [[P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficien... | P-adic Expansion Representative of P-adic Number is Unique | https://proofwiki.org/wiki/P-adic_Expansion_Representative_of_P-adic_Number_is_Unique | https://proofwiki.org/wiki/P-adic_Expansion_Representative_of_P-adic_Number_is_Unique | [
"P-adic Number Theory"
] | [
"Definition:Prime Number",
"Definition:Valued Field of P-adic Numbers",
"Definition:P-adic Number",
"Definition:Coset/Left Coset",
"Definition:P-adic Expansion",
"Definition:P-adic Number/Representative",
"Definition:P-adic Expansion"
] | [
"P-adic Number times P-adic Norm is P-adic Unit",
"Definition:Set",
"Definition:P-adic Unit",
"P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient"
] |
proofwiki-16802 | Class is Subclass of Universal Class | Let $V$ denote the universal class.
Let $A$ be a class.
Then $A$ is a subclass of $V$. | 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.
{{qed}} | Let $V$ denote the [[Definition:Universal Class|universal class]].
Let $A$ be a [[Definition:Class (Class Theory)|class]].
Then $A$ is a [[Definition:Subclass|subclass]] of $V$. | By definition of [[Definition:Class (Class Theory)|class]], $A$ is a [[Definition:Collection|collection]] of [[Definition:Set|sets]].
Let $x \in A$ be a [[Definition:Set|set]].
By definition of [[Definition:Universal Class|universal class]], $V$ contains all [[Definition:Set|sets]] as [[Definition:Element of Class|el... | Class is Subclass of Universal Class | https://proofwiki.org/wiki/Class_is_Subclass_of_Universal_Class | https://proofwiki.org/wiki/Class_is_Subclass_of_Universal_Class | [
"Universal Class",
"Subclasses"
] | [
"Definition:Universal Class",
"Definition:Class (Class Theory)",
"Definition:Subclass"
] | [
"Definition:Class (Class Theory)",
"Definition:Collection",
"Definition:Set",
"Definition:Set",
"Definition:Universal Class",
"Definition:Set",
"Definition:Element/Class",
"Definition:Subclass"
] |
proofwiki-16803 | Not Every Class is a Set | Let $A$ be a class.
Then it is not necessarily the case that $A$ is also a set. | Let a set $x$ be defined as ordinary {{iff}} $x \notin x$.
Let $\map \phi x$ be the set property defined as:
:$\map \pi x := \neg \paren {x \in x}$
Then by the Axiom of Specification there exists a class, which can be denoted $A$, such that:
:$A = \set {x: \neg \paren {x \in x} }$
By the Axiom of Extension, $A$ is uniq... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Then it is not necessarily the case that $A$ is also a [[Definition:Set|set]]. | Let a [[Definition:Set|set]] $x$ be defined as [[Definition:Ordinary Set|ordinary]] {{iff}} $x \notin x$.
Let $\map \phi x$ be the [[Definition:First-Order Property of Sets|set property]] defined as:
:$\map \pi x := \neg \paren {x \in x}$
Then by the [[Axiom:Axiom of Specification (Classes)|Axiom of Specification]] t... | Not Every Class is a Set/Proof 1 | https://proofwiki.org/wiki/Not_Every_Class_is_a_Set | https://proofwiki.org/wiki/Not_Every_Class_is_a_Set/Proof_1 | [
"Class Theory",
"Not Every Class is a Set"
] | [
"Definition:Class (Class Theory)",
"Definition:Set"
] | [
"Definition:Set",
"Definition:Ordinary Set",
"Definition:First-Order Property of Sets",
"Axiom:Axiom of Specification/Class Theory",
"Definition:Class (Class Theory)",
"Axiom:Axiom of Extension/Class Theory",
"Definition:Unique",
"Definition:Class (Class Theory)",
"Definition:Ordinary Set",
"Defin... |
proofwiki-16804 | Not Every Class is a Set | Let $A$ be a class.
Then it is not necessarily the case that $A$ is also a set. | Suppose $A$ is the universal class, which is {{afortiori}} a class.
The result follows from Basic Universe is not Set.
{{qed}} | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Then it is not necessarily the case that $A$ is also a [[Definition:Set|set]]. | Suppose $A$ is the [[Definition:Universal Class|universal class]], which is {{afortiori}} a [[Definition:Class (Class Theory)|class]].
The result follows from [[Basic Universe is not Set]].
{{qed}} | Not Every Class is a Set/Proof 3 | https://proofwiki.org/wiki/Not_Every_Class_is_a_Set | https://proofwiki.org/wiki/Not_Every_Class_is_a_Set/Proof_3 | [
"Class Theory",
"Not Every Class is a Set"
] | [
"Definition:Class (Class Theory)",
"Definition:Set"
] | [
"Definition:Universal Class",
"Definition:Class (Class Theory)",
"Basic Universe is not Set"
] |
proofwiki-16805 | Class has Subclass which is not Element | Let $A$ be a class.
Then $A$ has at least one subclass $B$ which is not an element of $A$. | Let a set $x$ be defined as '''ordinary''' {{iff}} $x \notin x$.
Let $\map \phi x$ be the set property defined as:
:$\map \pi x := \neg \paren {x \in x}$
Then by the {{Axiom-link|Specification|Classes}} there exists a class, which can be denoted $B$, such that:
:$B = \set {x \in A: \neg \paren {x \in x} }$
That is, $B$... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Then $A$ has at least one [[Definition:Subclass|subclass]] $B$ which is not an [[Definition:Element of Class|element]] of $A$. | Let a [[Definition:Set|set]] $x$ be defined as '''ordinary''' {{iff}} $x \notin x$.
Let $\map \phi x$ be the [[Definition:First-Order Property of Sets|set property]] defined as:
:$\map \pi x := \neg \paren {x \in x}$
Then by the {{Axiom-link|Specification|Classes}} there exists a [[Definition:Class (Class Theory)|cla... | Class has Subclass which is not Element | https://proofwiki.org/wiki/Class_has_Subclass_which_is_not_Element | https://proofwiki.org/wiki/Class_has_Subclass_which_is_not_Element | [
"Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Subclass",
"Definition:Element/Class"
] | [
"Definition:Set",
"Definition:First-Order Property of Sets",
"Definition:Class (Class Theory)",
"Definition:Element/Class",
"Definition:Set",
"Definition:Unique",
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Element/Class",
"Definition:Contradiction",
"Proof by Contradiction"... |
proofwiki-16806 | Basic Universe is Supercomplete | Let $V$ be a basic universe.
Then $V$ is supercomplete. | By definition, a class $V$ is supercomplete {{iff}} $V$ is both transitive and swelled.
From the {{axiom-link|Transitivity}}, $V$ is transitive.
From the {{axiom-link|Swelledness}}, $V$ is swelled.
Hence the result.
{{qed}} | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Then $V$ is [[Definition:Supercomplete Class|supercomplete]]. | By definition, a [[Definition:Class (Class Theory)|class]] $V$ is [[Definition:Supercomplete Class|supercomplete]] {{iff}} $V$ is both [[Definition:Transitive Class|transitive]] and [[Definition:Swelled Class|swelled]].
From the {{axiom-link|Transitivity}}, $V$ is [[Definition:Transitive Class|transitive]].
From the ... | Basic Universe is Supercomplete | https://proofwiki.org/wiki/Basic_Universe_is_Supercomplete | https://proofwiki.org/wiki/Basic_Universe_is_Supercomplete | [
"Basic Universe",
"Supercomplete Classes"
] | [
"Definition:Basic Universe",
"Definition:Supercomplete Class"
] | [
"Definition:Class (Class Theory)",
"Definition:Supercomplete Class",
"Definition:Transitive Class",
"Definition:Swelled Class",
"Definition:Transitive Class",
"Definition:Swelled Class"
] |
proofwiki-16807 | Basic Universe is not Set | Let $V$ be a basic universe.
Then $V$ is not a set. | {{AimForCont}} $V$ were a set.
Then by the the {{axiom-link|Swelledness}}, $V$ is swelled.
That is, as $V$ is a set, every subclass of $V$ would also be a set.
From Class has Subclass which is not Element, $V$ has a subclass $S$ which is not an element of $V$.
That is:
:$\exists S \subseteq V: S \notin V$
But by defini... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Then $V$ is not a [[Definition:Set|set]]. | {{AimForCont}} $V$ were a [[Definition:Set|set]].
Then by the the {{axiom-link|Swelledness}}, $V$ is [[Definition:Swelled Class|swelled]].
That is, as $V$ is a [[Definition:Set|set]], every [[Definition:Subclass|subclass]] of $V$ would also be a [[Definition:Set|set]].
From [[Class has Subclass which is not Element]... | Basic Universe is not Set | https://proofwiki.org/wiki/Basic_Universe_is_not_Set | https://proofwiki.org/wiki/Basic_Universe_is_not_Set | [
"Basic Universe"
] | [
"Definition:Basic Universe",
"Definition:Set"
] | [
"Definition:Set",
"Definition:Swelled Class",
"Definition:Set",
"Definition:Subclass",
"Definition:Set",
"Class has Subclass which is not Element",
"Definition:Subclass",
"Definition:Element/Class",
"Definition:Basic Universe",
"Definition:Universal Class",
"Definition:Contradiction",
"Proof b... |
proofwiki-16808 | Empty Class Exists and is Unique | There is exactly one empty class. | Let $P$ be a property such that $\map P x$ is satisfied by no $x$ at all, for example:
:$\forall x: \map P x := \neg {x = x}$
Then by the Axiom of Specification we can create the class $A$ such that:
:$A := \set {x \in V \land \neg {x = x} }$
from which it is seen that $A$ has no elements.
Hence there exists an empty c... | There is [[Definition:Unique|exactly one]] [[Definition:Empty Class (Class Theory)|empty class]]. | Let $P$ be a [[Definition:Property|property]] such that $\map P x$ is satisfied by no $x$ at all, for example:
:$\forall x: \map P x := \neg {x = x}$
Then by the [[Axiom:Axiom of Specification (Classes)|Axiom of Specification]] we can create the [[Definition:Class (Class Theory)|class]] $A$ such that:
:$A := \set {x \... | Empty Class Exists and is Unique | https://proofwiki.org/wiki/Empty_Class_Exists_and_is_Unique | https://proofwiki.org/wiki/Empty_Class_Exists_and_is_Unique | [
"Empty Class"
] | [
"Definition:Unique",
"Definition:Empty Class (Class Theory)"
] | [
"Definition:Property",
"Axiom:Axiom of Specification/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Element/Class",
"Definition:Empty Class (Class Theory)",
"Definition:Empty Class (Class Theory)",
"Definition:Element/Class",
"Definition:Element/Class",
"Axiom:Axiom of Extension/Class... |
proofwiki-16809 | Empty Class is Subclass of All Classes | The empty class is a subclass of all classes. | Let $A$ be a class.
By definition of the empty class:
:$\forall x: \neg \paren {x \in \O}$
From False Statement implies Every Statement:
:$\forall x: \paren {x \in \O \implies x \in A}$
Hence the result by definition of subclass.
{{qed}} | The [[Definition:Empty Class (Class Theory)|empty class]] is a [[Definition:Subclass|subclass]] of all [[Definition:Class (Class Theory)|classes]]. | Let $A$ be a [[Definition:Class (Class Theory)|class]].
By definition of the [[Definition:Empty Class (Class Theory)|empty class]]:
:$\forall x: \neg \paren {x \in \O}$
From [[False Statement implies Every Statement]]:
:$\forall x: \paren {x \in \O \implies x \in A}$
Hence the result by definition of [[Definition:... | Empty Class is Subclass of All Classes | https://proofwiki.org/wiki/Empty_Class_is_Subclass_of_All_Classes | https://proofwiki.org/wiki/Empty_Class_is_Subclass_of_All_Classes | [
"Empty Class"
] | [
"Definition:Empty Class (Class Theory)",
"Definition:Subclass",
"Definition:Class (Class Theory)"
] | [
"Definition:Class (Class Theory)",
"Definition:Empty Class (Class Theory)",
"False Statement implies Every Statement",
"Definition:Subclass"
] |
proofwiki-16810 | Empty Class is Supercomplete | The empty class is supercomplete. | Vacuously, every element of $\O$ is also a subclass of $\O$.
Hence $\O$ is transitive by definition.
Vacuously, every subclass of every element of $\O$ is also an element of $\O$.
Hence $\O$ is swelled by definition.
The result follows by definition of supercomplete.
{{qed}} | The [[Definition:Empty Class (Class Theory)|empty class]] is [[Definition:Supercomplete Class|supercomplete]]. | [[Definition:Vacuous Truth|Vacuously]], every [[Definition:Element of Class|element]] of $\O$ is also a [[Definition:Subclass|subclass]] of $\O$.
Hence $\O$ is [[Definition:Transitive Class|transitive]] by definition.
[[Definition:Vacuous Truth|Vacuously]], every [[Definition:Subclass|subclass]] of every [[Definitio... | Empty Class is Supercomplete | https://proofwiki.org/wiki/Empty_Class_is_Supercomplete | https://proofwiki.org/wiki/Empty_Class_is_Supercomplete | [
"Empty Class",
"Supercomplete Classes"
] | [
"Definition:Empty Class (Class Theory)",
"Definition:Supercomplete Class"
] | [
"Definition:Vacuous Truth",
"Definition:Element/Class",
"Definition:Subclass",
"Definition:Transitive Class",
"Definition:Vacuous Truth",
"Definition:Subclass",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Swelled Class",
"Definition:Supercomplete Class"
] |
proofwiki-16811 | Basic Universe is not Empty | Let $V$ be a basic universe
Then $V$ is not the empty class. | The {{axiom-link|the Empty Set|Class Theory}} gives us:
{{:Axiom:Axiom of the Empty Set (Class Theory)}}
Hence the result by definition of empty class.
{{Qed}} | Let $V$ be a [[Definition:Basic Universe|basic universe]]
Then $V$ is not the [[Definition:Empty Class (Class Theory)|empty class]]. | The {{axiom-link|the Empty Set|Class Theory}} gives us:
{{:Axiom:Axiom of the Empty Set (Class Theory)}}
Hence the result by definition of [[Definition:Empty Class (Class Theory)|empty class]].
{{Qed}} | Basic Universe is not Empty | https://proofwiki.org/wiki/Basic_Universe_is_not_Empty | https://proofwiki.org/wiki/Basic_Universe_is_not_Empty | [
"Empty Class",
"Basic Universe"
] | [
"Definition:Basic Universe",
"Definition:Empty Class (Class Theory)"
] | [
"Definition:Empty Class (Class Theory)"
] |
proofwiki-16812 | Existence of Set is Equivalent to Existence of Empty Set | Let $V$ be a basic universe.
Let $P$ be the axiom:
:$V$ has at least one element.
Then $P$ is equivalent to the {{axiom-link|the Empty Set|Class Theory}}:
:The empty class $\O$ is a set. | === Necessary Condition ===
Let the {{axiom-link|the Empty Set|Class Theory}} hold.
That is:
:$\O$ is a set.
By definition of a basic universe, $V$ is a universal class.
Hence, by definition, every set is an element of $V$.
We have that $\O$ is a set {{hypothesis}}.
Thus:
:$\O \in V$
and by definition of empty class it... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $P$ be the [[Definition:Axiom|axiom]]:
:$V$ has at least one [[Definition:Element of Class|element]].
Then $P$ is [[Definition:Logical Equivalence|equivalent]] to the {{axiom-link|the Empty Set|Class Theory}}:
:The [[Definition:Empty Class (Class Theo... | === Necessary Condition ===
Let the {{axiom-link|the Empty Set|Class Theory}} hold.
That is:
:$\O$ is a [[Definition:Set|set]].
By definition of a [[Definition:Basic Universe|basic universe]], $V$ is a [[Definition:Universal Class|universal class]].
Hence, by definition, every [[Definition:Set|set]] is an [[Definit... | Existence of Set is Equivalent to Existence of Empty Set | https://proofwiki.org/wiki/Existence_of_Set_is_Equivalent_to_Existence_of_Empty_Set | https://proofwiki.org/wiki/Existence_of_Set_is_Equivalent_to_Existence_of_Empty_Set | [
"Empty Class",
"Basic Universe"
] | [
"Definition:Basic Universe",
"Definition:Axiom",
"Definition:Element/Class",
"Definition:Logical Equivalence",
"Definition:Empty Class (Class Theory)",
"Definition:Set"
] | [
"Definition:Set",
"Definition:Basic Universe",
"Definition:Universal Class",
"Definition:Set",
"Definition:Element/Class",
"Definition:Set",
"Definition:Empty Class (Class Theory)",
"Definition:Empty Class (Class Theory)",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Eleme... |
proofwiki-16813 | Singleton Class can be Formed from Set | Let $V$ be a basic universe.
Let $a \in V$ be a set.
Then the singleton class $\set a$ can be formed, which is a subclass of $V$. | Using the axiom of specification, let $A$ be the class defined as:
:$A := \set {x: x \in V \land x = a}$
That is:
:$A = \set a$
By the axiom of extension, $\set a$ is the only such class which has $a$ as an element. | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $a \in V$ be a [[Definition:Set|set]].
Then the [[Definition:Singleton Class|singleton class]] $\set a$ can be formed, which is a [[Definition:Subclass|subclass]] of $V$. | Using the [[Axiom:Axiom of Specification (Classes)|axiom of specification]], let $A$ be the [[Definition:Class (Class Theory)|class]] defined as:
:$A := \set {x: x \in V \land x = a}$
That is:
:$A = \set a$
By the [[Axiom:Axiom of Extension (Classes)|axiom of extension]], $\set a$ is the only such [[Definition:Class ... | Singleton Class can be Formed from Set | https://proofwiki.org/wiki/Singleton_Class_can_be_Formed_from_Set | https://proofwiki.org/wiki/Singleton_Class_can_be_Formed_from_Set | [
"Singleton Classes"
] | [
"Definition:Basic Universe",
"Definition:Set",
"Definition:Singleton Class",
"Definition:Subclass"
] | [
"Axiom:Axiom of Specification/Class Theory",
"Definition:Class (Class Theory)",
"Axiom:Axiom of Extension/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Element/Class"
] |
proofwiki-16814 | Singleton Class of Empty Set is Supercomplete | Let $\O$ denote the empty set.
Then the singleton $\set \O$ is supercomplete. | Let $x \in \set \O$ be any element of $\set \O$.
Then it has to be the case that $x = \O$.
Then every element of $\O$ is an element of $\set \O$ vacuously.
That is, $\set \O$ is swelled.
There is one element of $\set \O$, and that is $\O$.
This is a subclass of $\set \O$.
That is, $\set \O$ is transitive.
The result fo... | Let $\O$ denote the [[Definition:Empty Set|empty set]].
Then the [[Definition:Singleton Class|singleton]] $\set \O$ is [[Definition:Supercomplete Class|supercomplete]]. | Let $x \in \set \O$ be any [[Definition:Element|element]] of $\set \O$.
Then it has to be the case that $x = \O$.
Then every [[Definition:Element|element]] of $\O$ is an [[Definition:Element of Class|element]] of $\set \O$ [[Definition:Vacuous Truth|vacuously]].
That is, $\set \O$ is [[Definition:Swelled Class|swell... | Singleton Class of Empty Set is Supercomplete | https://proofwiki.org/wiki/Singleton_Class_of_Empty_Set_is_Supercomplete | https://proofwiki.org/wiki/Singleton_Class_of_Empty_Set_is_Supercomplete | [
"Singleton Classes",
"Empty Set",
"Supercomplete Classes"
] | [
"Definition:Empty Set",
"Definition:Singleton Class",
"Definition:Supercomplete Class"
] | [
"Definition:Element",
"Definition:Element",
"Definition:Element/Class",
"Definition:Vacuous Truth",
"Definition:Swelled Class",
"Definition:Element/Class",
"Definition:Subclass",
"Definition:Transitive Class",
"Definition:Supercomplete Class"
] |
proofwiki-16815 | Singleton Classes are Equal iff Sets are Equal | Let $a$ and $b$ be sets.
Let $\set a$ and $\set b$ denote the singleton classes of $a$ and $b$.
Then:
:$\set a = \set b \iff a = b$ | Let $a = b$.
Then $\set a$ and $\set b$ contain the same elements.
Hence by the axiom of extension it follows that $\set a = \set b$.
Let $\set a = \set b$.
We have that $a \in \set a$.
As $\set a = \set b$ we also have that $a \in \set b$.
But $b$ is the only element of $\set b$.
Hence it must be the case that $a = b$... | Let $a$ and $b$ be [[Definition:Set|sets]].
Let $\set a$ and $\set b$ denote the [[Definition:Singleton Class|singleton classes]] of $a$ and $b$.
Then:
:$\set a = \set b \iff a = b$ | Let $a = b$.
Then $\set a$ and $\set b$ contain the same [[Definition:Element of Class|elements]].
Hence by the [[Axiom:Axiom of Extension (Classes)|axiom of extension]] it follows that $\set a = \set b$.
Let $\set a = \set b$.
We have that $a \in \set a$.
As $\set a = \set b$ we also have that $a \in \set b$.
B... | Singleton Classes are Equal iff Sets are Equal | https://proofwiki.org/wiki/Singleton_Classes_are_Equal_iff_Sets_are_Equal | https://proofwiki.org/wiki/Singleton_Classes_are_Equal_iff_Sets_are_Equal | [
"Singleton Classes"
] | [
"Definition:Set",
"Definition:Singleton Class"
] | [
"Definition:Element/Class",
"Axiom:Axiom of Extension/Class Theory",
"Definition:Element/Class"
] |
proofwiki-16816 | Doubleton Class can be Formed from Two Sets | Let $V$ be a basic universe.
Let $a, b \in V$ be sets.
Then the doubleton class $\set {a, b}$ can be formed, which is a subclass of $V$. | Using the {{Axiom-link|Specification|Classes}}, let $A$ be the class defined as:
:$A := \set {x: x \in V \land \paren {x = a \lor x = b} }$
That is:
:$A = \set {a, b}$
By the {{Axiom-link|Extension|Classes}}, $\set {a, b}$ is the only such class which has $a$ and $b$ as elements. | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $a, b \in V$ be [[Definition:Set|sets]].
Then the [[Definition:Doubleton Class|doubleton class]] $\set {a, b}$ can be formed, which is a [[Definition:Subclass|subclass]] of $V$. | Using the {{Axiom-link|Specification|Classes}}, let $A$ be the [[Definition:Class (Class Theory)|class]] defined as:
:$A := \set {x: x \in V \land \paren {x = a \lor x = b} }$
That is:
:$A = \set {a, b}$
By the {{Axiom-link|Extension|Classes}}, $\set {a, b}$ is the only such [[Definition:Class (Class Theory)|class]] ... | Doubleton Class can be Formed from Two Sets | https://proofwiki.org/wiki/Doubleton_Class_can_be_Formed_from_Two_Sets | https://proofwiki.org/wiki/Doubleton_Class_can_be_Formed_from_Two_Sets | [
"Doubleton Classes"
] | [
"Definition:Basic Universe",
"Definition:Set",
"Definition:Doubleton/Class Theory",
"Definition:Subclass"
] | [
"Definition:Class (Class Theory)",
"Definition:Class (Class Theory)",
"Definition:Element/Class"
] |
proofwiki-16817 | Doubleton Class of Equal Sets is Singleton Class | Let $V$ be a basic universe.
Let $a, b \in V$ be sets.
Consider the doubleton class $\set {a, b}$.
Let $a = b$.
Then:
:$\set {a, b} = \set a$
where $\set a$ denotes the singleton class of $a$. | Let $A = \set {a, b}$
The existence of $A$ is shown in Doubleton Class can be Formed from Two Sets:
:$A := \set {x: \paren {x = a \lor x = b} }$
Let $a = b$.
Then:
{{begin-eqn}}
{{eqn | l = A
| r = \set {a, b}
| c = Definition of $A$
}}
{{eqn | r = \set {a, a}
| c = as $a = b$
}}
{{eqn | ll= \leadsto
... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $a, b \in V$ be [[Definition:Set|sets]].
Consider the [[Definition:Doubleton Class|doubleton class]] $\set {a, b}$.
Let $a = b$.
Then:
:$\set {a, b} = \set a$
where $\set a$ denotes the [[Definition:Singleton Class|singleton class]] of $a$. | Let $A = \set {a, b}$
The existence of $A$ is shown in [[Doubleton Class can be Formed from Two Sets]]:
:$A := \set {x: \paren {x = a \lor x = b} }$
Let $a = b$.
Then:
{{begin-eqn}}
{{eqn | l = A
| r = \set {a, b}
| c = Definition of $A$
}}
{{eqn | r = \set {a, a}
| c = as $a = b$
}}
{{eqn | ll=... | Doubleton Class of Equal Sets is Singleton Class | https://proofwiki.org/wiki/Doubleton_Class_of_Equal_Sets_is_Singleton_Class | https://proofwiki.org/wiki/Doubleton_Class_of_Equal_Sets_is_Singleton_Class | [
"Doubleton Classes"
] | [
"Definition:Basic Universe",
"Definition:Set",
"Definition:Doubleton/Class Theory",
"Definition:Singleton Class"
] | [
"Doubleton Class can be Formed from Two Sets",
"Rule of Idempotence"
] |
proofwiki-16818 | Equivalence of Formulations of Axiom of Pairing | The following formulations of the '''{{axiom-link|Pairing|Set Theory}}''' in the context of '''axiomatic set theory''' are equivalent: | === Strong Form implies Weak Form ===
Let the strong form of the Axiom of Pairing be assumed:
{{:Axiom:Axiom of Pairing/Set Theory/Strong Form}}
By definition of the biconditional, this can be expressed as:
:$\forall a: \forall b: \exists c: \forall z: \paren {\paren {z = a \lor z = b \implies z \in c} \land \paren {z ... | The following formulations of the '''{{axiom-link|Pairing|Set Theory}}''' in the context of '''[[Definition:Axiomatic Set Theory|axiomatic set theory]]''' are [[Definition:Logical Equivalence|equivalent]]: | === Strong Form implies Weak Form ===
Let [[Axiom:Axiom of Pairing/Set Theory/Strong Form|the strong form of the Axiom of Pairing]] be assumed:
{{:Axiom:Axiom of Pairing/Set Theory/Strong Form}}
By definition of the [[Definition:Biconditional|biconditional]], this can be expressed as:
:$\forall a: \forall b: \exists... | Equivalence of Formulations of Axiom of Pairing | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Pairing | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Pairing | [
"Axiom of Pairing",
"Definition Equivalences"
] | [
"Definition:Axiomatic Set Theory",
"Definition:Logical Equivalence"
] | [
"Axiom:Axiom of Pairing/Set Theory/Strong Form",
"Definition:Biconditional",
"Rule of Simplification",
"Axiom:Axiom of Pairing/Set Theory/Weak Form",
"Axiom:Axiom of Pairing/Set Theory/Weak Form",
"Axiom:Axiom of Pairing/Set Theory/Strong Form"
] |
proofwiki-16819 | Singleton Class of Set is Set | Let $x$ be a set.
Then the singleton class $\set x$ is likewise a set. | Let $x$ and $y$ be sets.
Let $x = y$.
From Doubleton Class of Equal Sets is Singleton Class, the doubleton class $\set {x, y}$ is the singleton class $\set x$.
From the axiom of pairing, the doubleton class $\set {x, y}$ is a set when $x$ and $y$ are sets.
Hence $\set x$ is a set.
{{qed}} | Let $x$ be a [[Definition:Set|set]].
Then the [[Definition:Singleton Class|singleton class]] $\set x$ is likewise a [[Definition:Set|set]]. | Let $x$ and $y$ be [[Definition:Set|sets]].
Let $x = y$.
From [[Doubleton Class of Equal Sets is Singleton Class]], the [[Definition:Doubleton Class|doubleton class]] $\set {x, y}$ is the [[Definition:Singleton Class|singleton class]] $\set x$.
From the [[Axiom:Axiom of Pairing (Class Theory)|axiom of pairing]], th... | Singleton Class of Set is Set | https://proofwiki.org/wiki/Singleton_Class_of_Set_is_Set | https://proofwiki.org/wiki/Singleton_Class_of_Set_is_Set | [
"Singleton Classes",
"Singletons"
] | [
"Definition:Set",
"Definition:Singleton Class",
"Definition:Set"
] | [
"Definition:Set",
"Doubleton Class of Equal Sets is Singleton Class",
"Definition:Doubleton/Class Theory",
"Definition:Singleton Class",
"Axiom:Axiom of Pairing/Class Theory",
"Definition:Doubleton/Class Theory",
"Definition:Set",
"Definition:Set",
"Definition:Set"
] |
proofwiki-16820 | Basic Universe has Infinite Number of Elements | Let $V$ be a basic universe.
Then $V$ has an infinite number of elements. | Let $a_n$ be the class defined as:
:$\forall n \in \N: a_n = \begin{cases} \O & : n = 0 \\ \set {a_{n - 1} } & : n > 0 \end {cases}$
It is shown by the Principle of Mathematical Induction that $a_n$ is a set for all $n \in \N$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$a_n$ is a set. | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Then $V$ has an [[Definition:Infinite Set|infinite number]] of [[Definition:Element of Class|elements]]. | Let $a_n$ be the [[Definition:Class (Class Theory)|class]] defined as:
:$\forall n \in \N: a_n = \begin{cases} \O & : n = 0 \\ \set {a_{n - 1} } & : n > 0 \end {cases}$
It is shown by the [[Principle of Mathematical Induction]] that $a_n$ is a [[Definition:Set|set]] for all $n \in \N$.
For all $n \in \Z_{\ge 0}$, le... | Basic Universe has Infinite Number of Elements | https://proofwiki.org/wiki/Basic_Universe_has_Infinite_Number_of_Elements | https://proofwiki.org/wiki/Basic_Universe_has_Infinite_Number_of_Elements | [
"Basic Universe"
] | [
"Definition:Basic Universe",
"Definition:Infinite Set",
"Definition:Element/Class"
] | [
"Definition:Class (Class Theory)",
"Principle of Mathematical Induction",
"Definition:Set",
"Definition:Proposition",
"Definition:Set",
"Definition:Set",
"Definition:Set",
"Definition:Set",
"Definition:Set",
"Definition:Set",
"Definition:Set",
"Definition:Set",
"Principle of Mathematical Ind... |
proofwiki-16821 | Equivalence of Formulations of Axiom of Pairing for Classes | The following formulations of the '''{{axiom-link|Pairing}}''' in the context of '''class theory''' are equivalent: | It is assumed that all classes are subclasses of a basic universe $V$. | The following formulations of the '''{{axiom-link|Pairing}}''' in the context of '''[[Definition:Class Theory|class theory]]''' are [[Definition:Logical Equivalence|equivalent]]: | It is assumed that all [[Definition:Class (Class Theory)|classes]] are [[Definition:Subclass|subclasses]] of a [[Definition:Basic Universe|basic universe]] $V$. | Equivalence of Formulations of Axiom of Pairing for Classes | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Pairing_for_Classes | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Pairing_for_Classes | [
"Axiom of Pairing",
"Definition Equivalences"
] | [
"Definition:Class Theory",
"Definition:Logical Equivalence"
] | [
"Definition:Class (Class Theory)",
"Definition:Subclass",
"Definition:Basic Universe",
"Definition:Class (Class Theory)",
"Definition:Subclass",
"Definition:Subclass",
"Definition:Basic Universe",
"Definition:Subclass"
] |
proofwiki-16822 | Equality of Ordered Pairs/Lemma | Let $\set {a, b}$ and $\set {a, d}$ be doubletons such that $\set {a, b} = \set {a, d}$.
Then:
:$b = d$ | We have that:
:$b \in \set {a, b}$
and so by definition of set equality:
:$b \in \set {a, d}$
So:
:$(1): \quad$ either $b = a$ or $b = d$.
First suppose that $b = a$.
Then:
:$\set {a, b} = \set {a, a} = \set a$
We have that:
:$d \in \set {a, d}$
and so by definition of set equality:
:$d \in \set {a, b}$
and so as $\set... | Let $\set {a, b}$ and $\set {a, d}$ be [[Definition:Doubleton|doubletons]] such that $\set {a, b} = \set {a, d}$.
Then:
:$b = d$ | We have that:
:$b \in \set {a, b}$
and so by definition of [[Definition:Set Equality|set equality]]:
:$b \in \set {a, d}$
So:
:$(1): \quad$ either $b = a$ or $b = d$.
First suppose that $b = a$.
Then:
:$\set {a, b} = \set {a, a} = \set a$
We have that:
:$d \in \set {a, d}$
and so by definition of [[Definition:Set ... | Equality of Ordered Pairs/Lemma | https://proofwiki.org/wiki/Equality_of_Ordered_Pairs/Lemma | https://proofwiki.org/wiki/Equality_of_Ordered_Pairs/Lemma | [
"Equality of Ordered Pairs"
] | [
"Definition:Doubleton"
] | [
"Definition:Set Equality",
"Definition:Set Equality"
] |
proofwiki-16823 | Equality of Ordered Pairs/Necessary Condition | Let $\tuple {a, b}$ and $\tuple {c, d}$ be ordered pairs such that $\tuple {a, b} = \tuple {c, d}$.
Then $a = c$ and $b = d$. | First a lemma:
{{:Equality of Ordered Pairs/Lemma}}{{qed|lemma}}
Let $\tuple {a, b} = \tuple {c, d}$.
From the empty set formalization:
:$\set {\set {\O, a}, \set {\set \O, b} } = \set {\set {\O, c}, \set {\set \O, d} }$
First we note the special case where $a = \set \O$ and $b = \O$.
Then we have:
{{begin-eqn}}
{{eqn ... | Let $\tuple {a, b}$ and $\tuple {c, d}$ be [[Definition:Ordered Pair|ordered pairs]] such that $\tuple {a, b} = \tuple {c, d}$.
Then $a = c$ and $b = d$. | First a [[Equality of Ordered Pairs/Lemma|lemma]]:
{{:Equality of Ordered Pairs/Lemma}}{{qed|lemma}}
Let $\tuple {a, b} = \tuple {c, d}$.
From the [[Definition:Empty Set Formalization of Ordered Pair|empty set formalization]]:
:$\set {\set {\O, a}, \set {\set \O, b} } = \set {\set {\O, c}, \set {\set \O, d} }$
Fir... | Equality of Ordered Pairs/Necessary Condition/Proof from Empty Set Formalization | https://proofwiki.org/wiki/Equality_of_Ordered_Pairs/Necessary_Condition | https://proofwiki.org/wiki/Equality_of_Ordered_Pairs/Necessary_Condition/Proof_from_Empty_Set_Formalization | [
"Equality of Ordered Pairs"
] | [
"Definition:Ordered Pair"
] | [
"Equality of Ordered Pairs/Lemma",
"Definition:Ordered Pair/Empty Set Formalization",
"Doubleton Class of Equal Sets is Singleton Class",
"Proof by Contradiction",
"Equality of Ordered Pairs/Lemma",
"Proof by Contradiction",
"Equality of Ordered Pairs/Lemma"
] |
proofwiki-16824 | Equality of Ordered Pairs/Necessary Condition | Let $\tuple {a, b}$ and $\tuple {c, d}$ be ordered pairs such that $\tuple {a, b} = \tuple {c, d}$.
Then $a = c$ and $b = d$. | First a lemma:
{{:Equality of Ordered Pairs/Lemma}}{{qed|lemma}}
Let $\tuple {a, b} = \tuple {c, d}$.
From the Kuratowski formalization:
:$\set {\set a, \set {a, b} } = \set {\set c, \set {c, d} }$
There are two cases: either $a = b$, or $a \ne b$.
==== Case 1 ====
Suppose $a = b$.
Then:
:$\set {\set a, \set {a, b} } ... | Let $\tuple {a, b}$ and $\tuple {c, d}$ be [[Definition:Ordered Pair|ordered pairs]] such that $\tuple {a, b} = \tuple {c, d}$.
Then $a = c$ and $b = d$. | First a [[Equality of Ordered Pairs/Lemma|lemma]]:
{{:Equality of Ordered Pairs/Lemma}}{{qed|lemma}}
Let $\tuple {a, b} = \tuple {c, d}$.
From the [[Definition:Kuratowski Formalization of Ordered Pair|Kuratowski formalization]]:
:$\set {\set a, \set {a, b} } = \set {\set c, \set {c, d} }$
There are two cases: eith... | Equality of Ordered Pairs/Necessary Condition/Proof from Kuratowski Formalization | https://proofwiki.org/wiki/Equality_of_Ordered_Pairs/Necessary_Condition | https://proofwiki.org/wiki/Equality_of_Ordered_Pairs/Necessary_Condition/Proof_from_Kuratowski_Formalization | [
"Equality of Ordered Pairs"
] | [
"Definition:Ordered Pair"
] | [
"Equality of Ordered Pairs/Lemma",
"Definition:Ordered Pair/Kuratowski Formalization",
"Definition:Element",
"Definition:Distinct/Plural",
"Equality of Ordered Pairs/Lemma"
] |
proofwiki-16825 | Equality of Ordered Pairs/Necessary Condition | Let $\tuple {a, b}$ and $\tuple {c, d}$ be ordered pairs such that $\tuple {a, b} = \tuple {c, d}$.
Then $a = c$ and $b = d$. | First a lemma:
{{:Equality of Ordered Pairs/Lemma}}{{qed|lemma}}
Let $\tuple {a, b} = \tuple {c, d}$.
From the Wiener formalization:
:$\set {\set {\O, \set a}, \set {\set b} } = \set {\set {\O, \set c}, \set {\set d} }$
Let $x \in \set {\set {\O, \set a}, \set {\set b} }$.
Then either:
:$x = \set {\O, \set a}$
or:
:$x ... | Let $\tuple {a, b}$ and $\tuple {c, d}$ be [[Definition:Ordered Pair|ordered pairs]] such that $\tuple {a, b} = \tuple {c, d}$.
Then $a = c$ and $b = d$. | First a [[Equality of Ordered Pairs/Lemma|lemma]]:
{{:Equality of Ordered Pairs/Lemma}}{{qed|lemma}}
Let $\tuple {a, b} = \tuple {c, d}$.
From the [[Definition:Wiener Formalization of Ordered Pair|Wiener formalization]]:
:$\set {\set {\O, \set a}, \set {\set b} } = \set {\set {\O, \set c}, \set {\set d} }$
Let $x ... | Equality of Ordered Pairs/Necessary Condition/Proof from Wiener Formalization | https://proofwiki.org/wiki/Equality_of_Ordered_Pairs/Necessary_Condition | https://proofwiki.org/wiki/Equality_of_Ordered_Pairs/Necessary_Condition/Proof_from_Wiener_Formalization | [
"Equality of Ordered Pairs"
] | [
"Definition:Ordered Pair"
] | [
"Equality of Ordered Pairs/Lemma",
"Definition:Ordered Pair/Wiener Formalization",
"Definition:Contradiction",
"Definition:Empty Set",
"Proof by Contradiction",
"Equality of Ordered Pairs/Lemma",
"Singleton Equality",
"Proof by Contradiction",
"Singleton Equality",
"Singleton Equality"
] |
proofwiki-16826 | Equality of Ordered Pairs/Sufficient Condition | Let $\tuple {a, b}$ and $\tuple {c, d}$ be ordered pairs.
Let $a = c$ and $b = d$.
Then:
:$\tuple {a, b} = \tuple {c, d}$ | Suppose $a = c$ and $b = d$.
Then:
:$\set a = \set c$
and:
:$\set {a, b} = \set {c, d}$
Thus:
:$\set {\set a, \set {a, b} } = \set {\set c, \set {c, d} }$
and so by the Kuratowski formalization:
:$\tuple {a, b} = \tuple {c, d}$
{{qed}} | Let $\tuple {a, b}$ and $\tuple {c, d}$ be [[Definition:Ordered Pair|ordered pairs]].
Let $a = c$ and $b = d$.
Then:
:$\tuple {a, b} = \tuple {c, d}$ | Suppose $a = c$ and $b = d$.
Then:
:$\set a = \set c$
and:
:$\set {a, b} = \set {c, d}$
Thus:
:$\set {\set a, \set {a, b} } = \set {\set c, \set {c, d} }$
and so by the [[Definition:Kuratowski Formalization of Ordered Pair|Kuratowski formalization]]:
:$\tuple {a, b} = \tuple {c, d}$
{{qed}} | Equality of Ordered Pairs/Sufficient Condition | https://proofwiki.org/wiki/Equality_of_Ordered_Pairs/Sufficient_Condition | https://proofwiki.org/wiki/Equality_of_Ordered_Pairs/Sufficient_Condition | [
"Equality of Ordered Pairs"
] | [
"Definition:Ordered Pair"
] | [
"Definition:Ordered Pair/Kuratowski Formalization"
] |
proofwiki-16827 | Intersection of Class Exists and is Unique | Let $V$ be a basic universe.
Let $A \subseteq V$ be a class.
Let $\bigcap A$ denote the intersection of $A$.
Then $\bigcap A$ is guaranteed to exist and is unique. | By the Axiom of Specification we can create the subclass of $V$:
:$\bigcap A = \set {x \in V: \forall y \in A: x \in y}$
Hence $\bigcap A$ exists.
Suppose $\QQ \subseteq V$ such that $\QQ$ and $\bigcap A$ are both the intersection of $A$.
Then:
:$\QQ = \set {x \in V: \forall y \in A: x \in y}$
Thus:
:$x \in \QQ \implie... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $A \subseteq V$ be a [[Definition:Class (Class Theory)|class]].
Let $\bigcap A$ denote the [[Definition:Intersection of Class|intersection]] of $A$.
Then $\bigcap A$ is guaranteed to exist and is [[Definition:Unique|unique]]. | By the [[Axiom:Axiom of Specification (Classes)|Axiom of Specification]] we can create the [[Definition:Subclass|subclass]] of $V$:
:$\bigcap A = \set {x \in V: \forall y \in A: x \in y}$
Hence $\bigcap A$ exists.
Suppose $\QQ \subseteq V$ such that $\QQ$ and $\bigcap A$ are both the [[Definition:Intersection of Clas... | Intersection of Class Exists and is Unique | https://proofwiki.org/wiki/Intersection_of_Class_Exists_and_is_Unique | https://proofwiki.org/wiki/Intersection_of_Class_Exists_and_is_Unique | [
"Class Intersection"
] | [
"Definition:Basic Universe",
"Definition:Class (Class Theory)",
"Definition:Class Intersection/Class of Sets",
"Definition:Unique"
] | [
"Axiom:Axiom of Specification/Class Theory",
"Definition:Subclass",
"Definition:Class Intersection/Class of Sets",
"Axiom:Axiom of Extension/Class Theory",
"Definition:Unique"
] |
proofwiki-16828 | Intersection of Non-Empty Class is Set | Let $A$ be a non-empty class.
Let $\bigcap A$ denote the intersection of $A$.
Then $\bigcap A$ is a set. | Let $V$ denote the basic universe such that $A \subseteq V$.
We are given that $A$ is non-empty.
Then $\exists x \in A$ where $x$ is a set.
By definition of intersection of class, every element of $\bigcap A$ is an element of all elements of $A$.
Thus:
:$\bigcap A \subseteq x$
We are given that $A$ is a subclass of the... | Let $A$ be a [[Definition:Non-Empty Class|non-empty]] [[Definition:Class (Class Theory)|class]].
Let $\bigcap A$ denote the [[Definition:Intersection of Class|intersection]] of $A$.
Then $\bigcap A$ is a [[Definition:Set|set]]. | Let $V$ denote the [[Definition:Basic Universe|basic universe]] such that $A \subseteq V$.
We are given that $A$ is [[Definition:Non-Empty Class|non-empty]].
Then $\exists x \in A$ where $x$ is a [[Definition:Set|set]].
By definition of [[Definition:Intersection of Class|intersection of class]], every [[Definition:E... | Intersection of Non-Empty Class is Set/Proof 1 | https://proofwiki.org/wiki/Intersection_of_Non-Empty_Class_is_Set | https://proofwiki.org/wiki/Intersection_of_Non-Empty_Class_is_Set/Proof_1 | [
"Class Intersection",
"Intersection of Non-Empty Class is Set"
] | [
"Definition:Non-Empty Set/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Class Intersection/Class of Sets",
"Definition:Set"
] | [
"Definition:Basic Universe",
"Definition:Non-Empty Set/Class Theory",
"Definition:Set",
"Definition:Class Intersection/Class of Sets",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Subclass",
"Definition:Basic Universe",
"Definition:Basic Universe"... |
proofwiki-16829 | Intersection of Non-Empty Class is Set | Let $A$ be a non-empty class.
Let $\bigcap A$ denote the intersection of $A$.
Then $\bigcap A$ is a set. | Since $A$ is a non-empty class, there exists $S \in A$.
Since $S$ is an element of a class, it is not a proper class, and is thus a set.
By definition of class intersection:
:$x \in \bigcap A \implies x \in S$
By the subclass definition:
:$\bigcap A \subseteq S$
By Subclass of Set is Set, $\bigcap A$ is a set.
{{qed}} | Let $A$ be a [[Definition:Non-Empty Class|non-empty]] [[Definition:Class (Class Theory)|class]].
Let $\bigcap A$ denote the [[Definition:Intersection of Class|intersection]] of $A$.
Then $\bigcap A$ is a [[Definition:Set|set]]. | Since $A$ is a [[Definition:Non-Empty Class|non-empty class]], there exists $S \in A$.
Since $S$ is an [[Definition:Element of Class|element]] of a [[Definition:Class (Class Theory)|class]], it is not a [[Definition:Proper Class|proper class]], and is thus a [[Definition:Set|set]].
By definition of [[Definition:Inter... | Intersection of Non-Empty Class is Set/Proof 2 | https://proofwiki.org/wiki/Intersection_of_Non-Empty_Class_is_Set | https://proofwiki.org/wiki/Intersection_of_Non-Empty_Class_is_Set/Proof_2 | [
"Class Intersection",
"Intersection of Non-Empty Class is Set"
] | [
"Definition:Non-Empty Set/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Class Intersection/Class of Sets",
"Definition:Set"
] | [
"Definition:Non-Empty Set/Class Theory",
"Definition:Element/Class",
"Definition:Class (Class Theory)",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Set",
"Definition:Class Intersection/Class of Sets",
"Definition:Subclass",
"Subclass of Set is Set",
"Definition:Set"
] |
proofwiki-16830 | Intersection of Empty Set/Class Theory | Let $V$ be a basic universe.
Let $\O$ denote the empty class.
Then the intersection of $\O$ is $V$:
:$\bigcap \O = V$ | By definition of $V$, we have:
:$\bigcap \O \subseteq V$
By definition of empty class, there exists no set $x \in V$ which is an element of $\O$.
Hence it is vacuously true that every element of $V$ is an element of every element of $\O$.
Therefore every $x \in V$ is an element of $\bigcap \O$.
Therefore:
:$V \subseteq... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $\O$ denote the [[Definition:Empty Class (Class Theory)|empty class]].
Then the [[Definition:Intersection of Class|intersection]] of $\O$ is $V$:
:$\bigcap \O = V$ | By definition of $V$, we have:
:$\bigcap \O \subseteq V$
By definition of [[Definition:Empty Class (Class Theory)|empty class]], there exists no [[Definition:Set|set]] $x \in V$ which is an [[Definition:Element of Class|element]] of $\O$.
Hence it is [[Definition:Vacuous Truth|vacuously true]] that every [[Definitio... | Intersection of Empty Set/Class Theory | https://proofwiki.org/wiki/Intersection_of_Empty_Set/Class_Theory | https://proofwiki.org/wiki/Intersection_of_Empty_Set/Class_Theory | [
"Class Intersection",
"Intersection of Empty Set"
] | [
"Definition:Basic Universe",
"Definition:Empty Class (Class Theory)",
"Definition:Class Intersection/Class of Sets"
] | [
"Definition:Empty Class (Class Theory)",
"Definition:Set",
"Definition:Element/Class",
"Definition:Vacuous Truth",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Class Equality"
] |
proofwiki-16831 | Union of Subclass is Subclass of Union of Class | Let $A$ and $B$ be classes.
Let $\bigcup A$ and $\bigcup B$ denote the union of $A$ and union of $B$ respectively.
Let $A$ be a subclass of $B$:
:$A \subseteq B$
Then $\bigcup A$ is a subclass of $\bigcup B$:
:$\bigcup A \subseteq \bigcup B$ | Let $x \in \bigcup A$.
Then:
:$\exists y \in A: x \in y$
But as $A \subseteq B$ it follows that $y \in B$.
That is:
:$\exists y \in B: x \in y$
That is:
:$x \in \bigcup B$
Hence the result by definition of subclass.
{{Qed}} | Let $A$ and $B$ be [[Definition:Class (Class Theory)|classes]].
Let $\bigcup A$ and $\bigcup B$ denote the [[Definition:Union of Class|union]] of $A$ and [[Definition:Union of Class|union]] of $B$ respectively.
Let $A$ be a [[Definition:Subclass|subclass]] of $B$:
:$A \subseteq B$
Then $\bigcup A$ is a [[Definitio... | Let $x \in \bigcup A$.
Then:
:$\exists y \in A: x \in y$
But as $A \subseteq B$ it follows that $y \in B$.
That is:
:$\exists y \in B: x \in y$
That is:
:$x \in \bigcup B$
Hence the result by definition of [[Definition:Subclass|subclass]].
{{Qed}} | Union of Subclass is Subclass of Union of Class | https://proofwiki.org/wiki/Union_of_Subclass_is_Subclass_of_Union_of_Class | https://proofwiki.org/wiki/Union_of_Subclass_is_Subclass_of_Union_of_Class | [
"Class Union"
] | [
"Definition:Class (Class Theory)",
"Definition:Class Union/General Definition",
"Definition:Class Union/General Definition",
"Definition:Subclass",
"Definition:Subclass"
] | [
"Definition:Subclass"
] |
proofwiki-16832 | Intersection of Class is Subset of Intersection of Subclass | Let $V$ be a basic universe.
Let $A$ and $B$ be classes of $V$:
:$A \subseteq V, B \subseteq V$
such that it is not the case that $A = B = \O$.
Let $\bigcap A$ and $\bigcap B$ denote the intersection of $A$ and intersection of $B$ respectively.
Let $A$ be a subclass of $B$:
:$A \subseteq B$
Then $\bigcap B$ is a subset... | First we consider the degenerate case where $A = B = \O$.
By Intersection of Empty Class we have that:
:$\bigcap \O = V$
Thus we have:
:$\bigcap B = \bigcap A = V$
and neither $\bigcap B$ nor $\bigcap A$ are in fact sets.
So while in this case $\bigcap B \subseteq \bigcap A$, $\bigcap B$ is a subclass of $\bigcap A$ an... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $A$ and $B$ be [[Definition:Class (Class Theory)|classes]] of $V$:
:$A \subseteq V, B \subseteq V$
such that it is not the case that $A = B = \O$.
Let $\bigcap A$ and $\bigcap B$ denote the [[Definition:Intersection of Class|intersection]] of $A$ and [... | First we consider the [[Definition:Degenerate Case|degenerate case]] where $A = B = \O$.
By [[Intersection of Empty Class]] we have that:
:$\bigcap \O = V$
Thus we have:
:$\bigcap B = \bigcap A = V$
and neither $\bigcap B$ nor $\bigcap A$ are in fact [[Definition:Set|sets]].
So while in this case $\bigcap B \subsete... | Intersection of Class is Subset of Intersection of Subclass | https://proofwiki.org/wiki/Intersection_of_Class_is_Subset_of_Intersection_of_Subclass | https://proofwiki.org/wiki/Intersection_of_Class_is_Subset_of_Intersection_of_Subclass | [
"Class Intersection"
] | [
"Definition:Basic Universe",
"Definition:Class (Class Theory)",
"Definition:Class Intersection/Class of Sets",
"Definition:Class Intersection/Class of Sets",
"Definition:Subclass",
"Definition:Subset"
] | [
"Definition:Degenerate Case",
"Intersection of Empty Set/Class Theory",
"Definition:Set",
"Definition:Subclass",
"Definition:Subset",
"Definition:Empty Class (Class Theory)",
"Empty Class is Subclass of All Classes",
"Intersection of Empty Set/Class Theory",
"Definition:Basic Universe",
"Intersect... |
proofwiki-16833 | Union of Transitive Class is Subclass | Let $A$ be a transitive class.
Let $\bigcup A$ denote the union of $A$.
Then:
:$\bigcup A \subseteq A$ | Let $A$ be transitive.
Let $x \in \bigcup A$.
Then by definition:
:$\exists y \in A: x \in y$
By definition of transitive class:
:$x \in y \land y \in A \implies x \in A$
and so:
:$x \in A$
Hence the result by definition of subclass. | Let $A$ be a [[Definition:Transitive Class|transitive class]].
Let $\bigcup A$ denote the [[Definition:Union of Class|union]] of $A$.
Then:
:$\bigcup A \subseteq A$ | Let $A$ be [[Definition:Transitive Class|transitive]].
Let $x \in \bigcup A$.
Then by definition:
:$\exists y \in A: x \in y$
By definition of [[Definition:Transitive Class|transitive class]]:
:$x \in y \land y \in A \implies x \in A$
and so:
:$x \in A$
Hence the result by definition of [[Definition:Subclass|subcl... | Union of Transitive Class is Subclass | https://proofwiki.org/wiki/Union_of_Transitive_Class_is_Subclass | https://proofwiki.org/wiki/Union_of_Transitive_Class_is_Subclass | [
"Class is Transitive iff Union is Subclass"
] | [
"Definition:Transitive Class",
"Definition:Class Union/General Definition"
] | [
"Definition:Transitive Class",
"Definition:Transitive Class",
"Definition:Subclass"
] |
proofwiki-16834 | Union of Class is Subclass implies Class is Transitive | Let $A$ be a class.
Let $\bigcup A$ denote the union of $A$.
Let:
:$\bigcup A \subseteq A$
Then $A$ is transitive. | Let $\bigcup A \subseteq A$.
Let $x \in \bigcup A$.
Then by definition:
:$\exists y \in A: x \in y$
By definition of subclass:
:$x \in A$
Thus we have that:
:$x \in y \land y \in A \implies x \in A$
It follows by definition that $A$ is a transitive class. | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\bigcup A$ denote the [[Definition:Union of Class|union]] of $A$.
Let:
:$\bigcup A \subseteq A$
Then $A$ is [[Definition:Transitive Class|transitive]]. | Let $\bigcup A \subseteq A$.
Let $x \in \bigcup A$.
Then by definition:
:$\exists y \in A: x \in y$
By definition of [[Definition:Subclass|subclass]]:
:$x \in A$
Thus we have that:
:$x \in y \land y \in A \implies x \in A$
It follows by definition that $A$ is a [[Definition:Transitive Class|transitive class]]. | Union of Class is Subclass implies Class is Transitive | https://proofwiki.org/wiki/Union_of_Class_is_Subclass_implies_Class_is_Transitive | https://proofwiki.org/wiki/Union_of_Class_is_Subclass_implies_Class_is_Transitive | [
"Class is Transitive iff Union is Subclass"
] | [
"Definition:Class (Class Theory)",
"Definition:Class Union/General Definition",
"Definition:Transitive Class"
] | [
"Definition:Subclass",
"Definition:Transitive Class"
] |
proofwiki-16835 | Union of Transitive Class is Transitive | Let $A$ be a class.
Let $\bigcup A$ denote the union of $A$.
Let $A$ be transitive.
Then $\bigcup A$ is also transitive. | Let $A$ be transitive.
By Class is Transitive iff Union is Subclass:
:$\bigcup A \subseteq A$
By Union of Subclass is Subclass of Union of Class:
:$\map \bigcup {\bigcup A} \subseteq \bigcup A$
Then by Class is Transitive iff Union is Subclass:
:$\bigcup A$ is transitive.
{{qed}} | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\bigcup A$ denote the [[Definition:Union of Class|union]] of $A$.
Let $A$ be [[Definition:Transitive Class|transitive]].
Then $\bigcup A$ is also [[Definition:Transitive Class|transitive]]. | Let $A$ be [[Definition:Transitive Class|transitive]].
By [[Class is Transitive iff Union is Subclass]]:
:$\bigcup A \subseteq A$
By [[Union of Subclass is Subclass of Union of Class]]:
:$\map \bigcup {\bigcup A} \subseteq \bigcup A$
Then by [[Class is Transitive iff Union is Subclass]]:
:$\bigcup A$ is [[Definition... | Union of Transitive Class is Transitive/Proof 1 | https://proofwiki.org/wiki/Union_of_Transitive_Class_is_Transitive | https://proofwiki.org/wiki/Union_of_Transitive_Class_is_Transitive/Proof_1 | [
"Class Union",
"Transitive Classes",
"Union of Transitive Class is Transitive"
] | [
"Definition:Class (Class Theory)",
"Definition:Class Union/General Definition",
"Definition:Transitive Class",
"Definition:Transitive Class"
] | [
"Definition:Transitive Class",
"Class is Transitive iff Union is Subclass",
"Union of Subclass is Subclass of Union of Class",
"Class is Transitive iff Union is Subclass",
"Definition:Transitive Class"
] |
proofwiki-16836 | Union of Transitive Class is Transitive | Let $A$ be a class.
Let $\bigcup A$ denote the union of $A$.
Let $A$ be transitive.
Then $\bigcup A$ is also transitive. | Let $A$ be transitive.
Let $x \in \bigcup A$.
By Class is Transitive iff Union is Subclass we have that:
:$\bigcup A \subseteq A$
Thus by definition of subclass:
:$x \in A$
As $A$ is transitive:
:$x \subseteq A$
Let $z \in x$.
As $x \subseteq A$, it follows by definition of subclass that:
:$z \in A$
Thus we have that:
... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\bigcup A$ denote the [[Definition:Union of Class|union]] of $A$.
Let $A$ be [[Definition:Transitive Class|transitive]].
Then $\bigcup A$ is also [[Definition:Transitive Class|transitive]]. | Let $A$ be [[Definition:Transitive Class|transitive]].
Let $x \in \bigcup A$.
By [[Class is Transitive iff Union is Subclass]] we have that:
:$\bigcup A \subseteq A$
Thus by definition of [[Definition:Subclass|subclass]]:
:$x \in A$
As $A$ is [[Definition:Transitive Class|transitive]]:
:$x \subseteq A$
Let $z \in... | Union of Transitive Class is Transitive/Proof 2 | https://proofwiki.org/wiki/Union_of_Transitive_Class_is_Transitive | https://proofwiki.org/wiki/Union_of_Transitive_Class_is_Transitive/Proof_2 | [
"Class Union",
"Transitive Classes",
"Union of Transitive Class is Transitive"
] | [
"Definition:Class (Class Theory)",
"Definition:Class Union/General Definition",
"Definition:Transitive Class",
"Definition:Transitive Class"
] | [
"Definition:Transitive Class",
"Class is Transitive iff Union is Subclass",
"Definition:Subclass",
"Definition:Transitive Class",
"Definition:Subclass",
"Definition:Class Union/General Definition",
"Definition:Subclass",
"Definition:Transitive Class"
] |
proofwiki-16837 | Union of Class is Transitive if Every Element is Transitive | Let $A$ be a class.
Let $\bigcup A$ denote the union of $A$.
Let $A$ be such that every element of $A$ is transitive.
Then $\bigcup A$ is also transitive. | Let $A$ be such that every $y \in A$ is transitive.
Let $x \in \bigcup A$.
Then $x$ is an element of some element $y$ of $A$.
We have {{hypothesis}} that $y$ is transitive.
Hence, by definition of transitive class:
:$x \subseteq y$
Because $y \in A$, by definition of union of class:
:$y \subseteq \bigcup A$
So:
:$x \su... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\bigcup A$ denote the [[Definition:Union of Class|union]] of $A$.
Let $A$ be such that every [[Definition:Element of Class|element]] of $A$ is [[Definition:Transitive Set|transitive]].
Then $\bigcup A$ is also [[Definition:Transitive Class|transitive]]. | Let $A$ be such that every $y \in A$ is [[Definition:Transitive Set|transitive]].
Let $x \in \bigcup A$.
Then $x$ is an [[Definition:Element of Class|element]] of some [[Definition:Element of Class|element]] $y$ of $A$.
We have {{hypothesis}} that $y$ is [[Definition:Transitive Set|transitive]].
Hence, by definiti... | Union of Class is Transitive if Every Element is Transitive/Proof | https://proofwiki.org/wiki/Union_of_Class_is_Transitive_if_Every_Element_is_Transitive | https://proofwiki.org/wiki/Union_of_Class_is_Transitive_if_Every_Element_is_Transitive/Proof | [
"Class Union",
"Transitive Classes",
"Union of Class is Transitive if Every Element is Transitive"
] | [
"Definition:Class (Class Theory)",
"Definition:Class Union/General Definition",
"Definition:Element/Class",
"Definition:Transitive Class",
"Definition:Transitive Class"
] | [
"Definition:Transitive Class",
"Definition:Element/Class",
"Definition:Element/Class",
"Definition:Transitive Class",
"Definition:Transitive Class",
"Definition:Class Union/General Definition",
"Definition:Transitive Class"
] |
proofwiki-16838 | Set Difference with Set Difference is Union of Set Difference with Intersection/Corollary | :$T \setminus \paren {S \setminus T} = T$ | From Set Difference with Set Difference is Union of Set Difference with Intersection:
:$R \setminus \paren {S \setminus T} = \paren {R \setminus S} \cup \paren {R \cap T}$
where $R, S, T$ are sets.
Hence:
{{begin-eqn}}
{{eqn | l = T \setminus \paren {S \setminus T}
| r = \paren {T \setminus S} \cup \paren {T \cap... | :$T \setminus \paren {S \setminus T} = T$ | From [[Set Difference with Set Difference is Union of Set Difference with Intersection]]:
:$R \setminus \paren {S \setminus T} = \paren {R \setminus S} \cup \paren {R \cap T}$
where $R, S, T$ are [[Definition:Set|sets]].
Hence:
{{begin-eqn}}
{{eqn | l = T \setminus \paren {S \setminus T}
| r = \paren {T \setmi... | Set Difference with Set Difference is Union of Set Difference with Intersection/Corollary | https://proofwiki.org/wiki/Set_Difference_with_Set_Difference_is_Union_of_Set_Difference_with_Intersection/Corollary | https://proofwiki.org/wiki/Set_Difference_with_Set_Difference_is_Union_of_Set_Difference_with_Intersection/Corollary | [
"Set Difference",
"Set Union",
"Set Difference with Set Difference is Union of Set Difference with Intersection"
] | [] | [
"Set Difference with Set Difference is Union of Set Difference with Intersection",
"Definition:Set",
"Set Difference with Set Difference is Union of Set Difference with Intersection",
"Set Intersection is Idempotent",
"Set Difference Union First Set is First Set",
"Category:Set Difference",
"Category:Se... |
proofwiki-16839 | Power Set Exists and is Unique | Let $V$ be a basic universe.
Let $x \in V$ be a set.
Let $\powerset x$ denote the power set of $x$.
Then $\powerset x$ is guaranteed to exist and is unique. | By the {{axiom-link|Specification|Classes}} an arbitrary subclass of $x$ can be created.
Hence we can create the class of all such subclasses.
Hence $\powerset x$ exists.
Let $\powerset x$, $\map \QQ x$ both be power sets of $x$.
From definition of power sets:
:$\forall T$:
::$T \in \powerset x \iff T \subseteq x$
::$T... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $x \in V$ be a [[Definition:Set|set]].
Let $\powerset x$ denote the [[Definition:Power Set (Class Theory)|power set]] of $x$.
Then $\powerset x$ is guaranteed to exist and is [[Definition:Unique|unique]]. | By the {{axiom-link|Specification|Classes}} an arbitrary [[Definition:Subclass|subclass]] of $x$ can be created.
Hence we can create the [[Definition:Class (Class Theory)|class]] of all such [[Definition:Subclass|subclasses]].
Hence $\powerset x$ exists.
Let $\powerset x$, $\map \QQ x$ both be [[Definition:Power Se... | Power Set Exists and is Unique | https://proofwiki.org/wiki/Power_Set_Exists_and_is_Unique | https://proofwiki.org/wiki/Power_Set_Exists_and_is_Unique | [
"Power Set"
] | [
"Definition:Basic Universe",
"Definition:Set",
"Definition:Power Set/Class Theory",
"Definition:Unique"
] | [
"Definition:Subclass",
"Definition:Class (Class Theory)",
"Definition:Subclass",
"Definition:Power Set/Class Theory",
"Definition:Power Set/Class Theory",
"Biconditional is Commutative",
"Biconditional is Transitive",
"Definition:Power Set/Class Theory",
"Definition:Unique"
] |
proofwiki-16840 | Element of Class is Subset of Union of Class | Let $A$ be a class.
Let $\ds \bigcup A$ denote the union of $A$.
Let $x \in A$.
Then:
:$x \subseteq \ds \bigcup A$ | Let $x \in A$.
By definition of class, $x$ is a set.
Let $y \in x$.
By definition of union of $A$:
:$\ds \bigcup A := \set {y: \exists x \in A: y \in x}$
It follows directly from that definition that:
:$y \in \ds \bigcup A$
The result follows by definition of subset.
{{qed}}
Category:Class Union
acmtwijny3dxqouvgoomb8v... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\ds \bigcup A$ denote the [[Definition:Union of Class|union]] of $A$.
Let $x \in A$.
Then:
:$x \subseteq \ds \bigcup A$ | Let $x \in A$.
By definition of [[Definition:Class (Class Theory)|class]], $x$ is a [[Definition:Set|set]].
Let $y \in x$.
By definition of [[Definition:Union of Class|union]] of $A$:
:$\ds \bigcup A := \set {y: \exists x \in A: y \in x}$
It follows directly from that definition that:
:$y \in \ds \bigcup A$
The r... | Element of Class is Subset of Union of Class | https://proofwiki.org/wiki/Element_of_Class_is_Subset_of_Union_of_Class | https://proofwiki.org/wiki/Element_of_Class_is_Subset_of_Union_of_Class | [
"Class Union"
] | [
"Definition:Class (Class Theory)",
"Definition:Class Union/General Definition"
] | [
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Class Union/General Definition",
"Definition:Subset",
"Category:Class Union"
] |
proofwiki-16841 | Set is Subset of Power Set of Union | Let $x$ be a set of sets.
Let $\bigcup x$ denote the union of $x$.
Let $\powerset {\bigcup x}$ denote the power set of $\bigcup x$.
Then:
:$x \subseteq \powerset {\bigcup x}$ | Let $z \in x$.
By Element of Class is Subset of Union of Class:
:$z \subseteq \bigcup x$
By definition of power set:
:$z \in \powerset {\bigcup x}$
The result follows by definition of subset.
{{qed}} | Let $x$ be a [[Definition:Set of Sets|set of sets]].
Let $\bigcup x$ denote the [[Definition:Union of Set of Sets|union]] of $x$.
Let $\powerset {\bigcup x}$ denote the [[Definition:Power Set|power set]] of $\bigcup x$.
Then:
:$x \subseteq \powerset {\bigcup x}$ | Let $z \in x$.
By [[Element of Class is Subset of Union of Class]]:
:$z \subseteq \bigcup x$
By definition of [[Definition:Power Set|power set]]:
:$z \in \powerset {\bigcup x}$
The result follows by definition of [[Definition:Subset|subset]].
{{qed}} | Set is Subset of Power Set of Union | https://proofwiki.org/wiki/Set_is_Subset_of_Power_Set_of_Union | https://proofwiki.org/wiki/Set_is_Subset_of_Power_Set_of_Union | [
"Power Set",
"Set Union"
] | [
"Definition:Set of Sets",
"Definition:Set Union/Set of Sets",
"Definition:Power Set"
] | [
"Element of Class is Subset of Union of Class",
"Definition:Power Set",
"Definition:Subset"
] |
proofwiki-16842 | Set equals Union of Power Set | Let $x$ be a set of sets.
Let $\powerset x$ denote the power set of $x$.
Let $\map \bigcup {\powerset x}$ denote the union of $\powerset x$.
Then:
:$x = \map \bigcup {\powerset x}$ | From Set is Element of its Power Set:
:$x \in \powerset x$
From Element of Class is Subset of Union of Class it follows that:
:$x \subseteq \map \bigcup {\powerset x}$
{{qed|lemma}}
Let $z \in \map \bigcup {\powerset x}$
Then by definition of unionunion:
:$\exists y \in \powerset x: z \in y$
By definition of $\powerset... | Let $x$ be a [[Definition:Set of Sets|set of sets]].
Let $\powerset x$ denote the [[Definition:Power Set|power set]] of $x$.
Let $\map \bigcup {\powerset x}$ denote the [[Definition:Union of Set of Sets|union]] of $\powerset x$.
Then:
:$x = \map \bigcup {\powerset x}$ | From [[Set is Element of its Power Set]]:
:$x \in \powerset x$
From [[Element of Class is Subset of Union of Class]] it follows that:
:$x \subseteq \map \bigcup {\powerset x}$
{{qed|lemma}}
Let $z \in \map \bigcup {\powerset x}$
Then by definition of [[Definition:Union of Set of Sets|union|union]]:
:$\exists y \in ... | Set equals Union of Power Set | https://proofwiki.org/wiki/Set_equals_Union_of_Power_Set | https://proofwiki.org/wiki/Set_equals_Union_of_Power_Set | [
"Power Set",
"Set Union"
] | [
"Definition:Set of Sets",
"Definition:Power Set",
"Definition:Set Union/Set of Sets"
] | [
"Set is Element of its Power Set",
"Element of Class is Subset of Union of Class",
"Definition:Set Union/Set of Sets",
"Definition:Subset",
"Definition:Subset",
"Definition:Set Equality"
] |
proofwiki-16843 | Cartesian Product Exists and is Unique | Let $A$ and $B$ be classes.
Let $A \times B$ be the '''cartesian product''' of $A$ and $B$.
Then $A \times B$ exists and is unique. | Let $A \times B$ be the cartesian product of $A$ and $B$.
Let $\tuple {x, y} \in A \times B$ such that $\tuple {x, y}$ satisfies the Kuratowski ordered pair formulation.
By Binary Cartesian Product in Kuratowski Formalization contained in Power Set of Power Set of Union:
:$A \times B \subseteq \powerset {\powerset {A \... | Let $A$ and $B$ be [[Definition:Class (Class Theory)|classes]].
Let $A \times B$ be the '''[[Definition:Cartesian Product (Class Theory)|cartesian product]]''' of $A$ and $B$.
Then $A \times B$ exists and is [[Definition:Unique|unique]]. | Let $A \times B$ be the [[Definition:Cartesian Product (Class Theory)|cartesian product]] of $A$ and $B$.
Let $\tuple {x, y} \in A \times B$ such that $\tuple {x, y}$ satisfies the [[Definition:Kuratowski Formalization of Ordered Pair|Kuratowski ordered pair formulation]].
By [[Binary Cartesian Product in Kuratowski ... | Cartesian Product Exists and is Unique | https://proofwiki.org/wiki/Cartesian_Product_Exists_and_is_Unique | https://proofwiki.org/wiki/Cartesian_Product_Exists_and_is_Unique | [
"Cartesian Product"
] | [
"Definition:Class (Class Theory)",
"Definition:Cartesian Product/Class Theory",
"Definition:Unique"
] | [
"Definition:Cartesian Product/Class Theory",
"Definition:Ordered Pair/Kuratowski Formalization",
"Binary Cartesian Product in Kuratowski Formalization contained in Power Set of Power Set of Union",
"Axiom:Axiom of Specification/Class Theory",
"Definition:Set",
"Definition:Cartesian Product/Class Theory",
... |
proofwiki-16844 | Cartesian Product of Sets is Set | Let $V$ be a basic universe.
Let $A$ and $B$ be sets in $V$.
Then their cartesian product $A \times B$ is also a set. | Let $A$ and $B$ be sets in $V$.
Because $V$ is a basic universe, the basic universe axioms apply.
Hence by the axiom of pairing $\set {A, B}$ is a set.
Then by the axiom of unions $\bigcup \set {A, B}$ is also a set.
We have that $A \cup B = \bigcup \set {A, B}$.
By the axiom of powers $\powerset {A \cup B}$ is a set.
... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $A$ and $B$ be [[Definition:Set|sets]] in $V$.
Then their [[Definition:Cartesian Product (Class Theory)|cartesian product]] $A \times B$ is also a [[Definition:Set|set]]. | Let $A$ and $B$ be [[Definition:Set|sets]] in $V$.
Because $V$ is a [[Definition:Basic Universe|basic universe]], the [[Definition:Basic Universe Axioms|basic universe axioms]] apply.
Hence by the [[Axiom:Axiom of Pairing (Class Theory)|axiom of pairing]] $\set {A, B}$ is a [[Definition:Set|set]].
Then by the [[Axi... | Cartesian Product of Sets is Set | https://proofwiki.org/wiki/Cartesian_Product_of_Sets_is_Set | https://proofwiki.org/wiki/Cartesian_Product_of_Sets_is_Set | [
"Cartesian Product"
] | [
"Definition:Basic Universe",
"Definition:Set",
"Definition:Cartesian Product/Class Theory",
"Definition:Set"
] | [
"Definition:Set",
"Definition:Basic Universe",
"Definition:Basic Universe Axioms",
"Axiom:Axiom of Pairing/Class Theory",
"Definition:Set",
"Axiom:Axiom of Unions/Class Theory",
"Definition:Set",
"Axiom:Axiom of Powers/Class Theory",
"Definition:Set",
"Axiom:Axiom of Powers/Class Theory",
"Defin... |
proofwiki-16845 | Jordan Curve Theorem/General Result | Let $M$ be a connected manifold of dimension $n - 1$ without boundary.
Let $M$ be embedded in Euclidean space $\R^n$.
Then $M$ divides $\R^n$ into an inside and an outside. | {{proof wanted}}
{{Namedfor|Marie Ennemond Camille Jordan}} | Let $M$ be a [[Definition:Connected Topological Space|connected]] [[Definition:Topological Manifold|manifold]] of [[Definition:Dimension of Topological Manifold|dimension]] $n - 1$ without [[Definition:Boundary (Topology)|boundary]].
Let $M$ be [[Definition:Embedding (Topology)|embedded]] in [[Definition:Real Euclidea... | {{proof wanted}}
{{Namedfor|Marie Ennemond Camille Jordan}} | Jordan Curve Theorem/General Result | https://proofwiki.org/wiki/Jordan_Curve_Theorem/General_Result | https://proofwiki.org/wiki/Jordan_Curve_Theorem/General_Result | [
"Jordan Curve Theorem"
] | [
"Definition:Connected Topological Space",
"Definition:Topological Manifold",
"Definition:Dimension (Topology)/Topological Manifold",
"Definition:Boundary (Topology)",
"Definition:Embedding (Topology)",
"Definition:Euclidean Space/Real"
] | [] |
proofwiki-16846 | Real Function with Positive Derivative is Increasing | Let:
:$\forall x \in \openint a b: \map {f'} x \ge 0$
Then $f$ is increasing on $\closedint a b$. | Let $c, d \in \closedint a b$ such that $c < d$.
Then $f$ satisfies the conditions of the Mean Value Theorem on $\closedint c d$.
Hence:
:$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$
Let $f$ be such that:
:$\forall x \in \openint a b: \map {f'} x \ge 0$
Then:
:$\map {f'} \xi \ge ... | Let:
:$\forall x \in \openint a b: \map {f'} x \ge 0$
Then $f$ is [[Definition:Increasing Real Function|increasing]] on $\closedint a b$. | Let $c, d \in \closedint a b$ such that $c < d$.
Then $f$ satisfies the conditions of the [[Mean Value Theorem]] on $\closedint c d$.
Hence:
:$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$
Let $f$ be such that:
:$\forall x \in \openint a b: \map {f'} x \ge 0$
Then:
:$\map {f'}... | Real Function with Positive Derivative is Increasing | https://proofwiki.org/wiki/Real_Function_with_Positive_Derivative_is_Increasing | https://proofwiki.org/wiki/Real_Function_with_Positive_Derivative_is_Increasing | [
"Increasing Real Functions",
"Real Functions",
"Differential Calculus"
] | [
"Definition:Increasing/Real Function"
] | [
"Mean Value Theorem",
"Definition:Increasing/Real Function"
] |
proofwiki-16847 | Real Function with Strictly Positive Derivative is Strictly Increasing | Let:
:$\forall x \in \openint a b: \map {f'} x > 0$
Then $f$ is strictly increasing on $\closedint a b$. | Let $c, d \in \closedint a b$ such that $c < d$.
Then $f$ satisfies the conditions of the Mean Value Theorem on $\closedint c d$.
Hence:
:$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$
Let $f$ be such that:
:$\forall x \in \openint a b: \map {f'} x > 0$
Then:
:$\map {f'} \xi > 0$
a... | Let:
:$\forall x \in \openint a b: \map {f'} x > 0$
Then $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $\closedint a b$. | Let $c, d \in \closedint a b$ such that $c < d$.
Then $f$ satisfies the conditions of the [[Mean Value Theorem]] on $\closedint c d$.
Hence:
:$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$
Let $f$ be such that:
:$\forall x \in \openint a b: \map {f'} x > 0$
Then:
:$\map {f'} \... | Real Function with Strictly Positive Derivative is Strictly Increasing | https://proofwiki.org/wiki/Real_Function_with_Strictly_Positive_Derivative_is_Strictly_Increasing | https://proofwiki.org/wiki/Real_Function_with_Strictly_Positive_Derivative_is_Strictly_Increasing | [
"Strictly Increasing Real Functions",
"Real Functions",
"Differential Calculus"
] | [
"Definition:Strictly Increasing/Real Function"
] | [
"Mean Value Theorem",
"Definition:Strictly Increasing/Real Function"
] |
proofwiki-16848 | Real Function with Strictly Negative Derivative is Strictly Decreasing | Let:
:$\forall x \in \openint a b: \map {f'} x < 0$
Then $f$ is strictly decreasing on $\closedint a b$. | Let $c, d \in \closedint a b$ such that $c < d$.
Then $f$ satisfies the conditions of the Mean Value Theorem on $\closedint c d$.
Hence:
:$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$
Let $f$ be such that:
:$\forall x \in \openint a b: \map {f'} x < 0$
Then:
:$\map {f'} \xi < 0$
a... | Let:
:$\forall x \in \openint a b: \map {f'} x < 0$
Then $f$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $\closedint a b$. | Let $c, d \in \closedint a b$ such that $c < d$.
Then $f$ satisfies the conditions of the [[Mean Value Theorem]] on $\closedint c d$.
Hence:
:$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$
Let $f$ be such that:
:$\forall x \in \openint a b: \map {f'} x < 0$
Then:
:$\map {f'} \... | Real Function with Strictly Negative Derivative is Strictly Decreasing | https://proofwiki.org/wiki/Real_Function_with_Strictly_Negative_Derivative_is_Strictly_Decreasing | https://proofwiki.org/wiki/Real_Function_with_Strictly_Negative_Derivative_is_Strictly_Decreasing | [
"Strictly Decreasing Real Functions",
"Real Functions",
"Differential Calculus"
] | [
"Definition:Strictly Decreasing/Real Function"
] | [
"Mean Value Theorem",
"Definition:Strictly Decreasing/Real Function"
] |
proofwiki-16849 | Real Function with Negative Derivative is Decreasing | Let:
:$\forall x \in \openint a b: \map {f'} x \le 0$
Then $f$ is decreasing on $\closedint a b$. | Let $c, d \in \closedint a b$ such that $c < d$.
Then $f$ satisfies the conditions of the Mean Value Theorem on $\closedint c d$.
Hence:
:$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$
Let $f$ be such that:
:$\forall x \in \openint a b: \map {f'} x \le 0$
Then:
:$\map {f'} \xi \le ... | Let:
:$\forall x \in \openint a b: \map {f'} x \le 0$
Then $f$ is [[Definition:Decreasing Real Function|decreasing]] on $\closedint a b$. | Let $c, d \in \closedint a b$ such that $c < d$.
Then $f$ satisfies the conditions of the [[Mean Value Theorem]] on $\closedint c d$.
Hence:
:$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$
Let $f$ be such that:
:$\forall x \in \openint a b: \map {f'} x \le 0$
Then:
:$\map {f'}... | Real Function with Negative Derivative is Decreasing | https://proofwiki.org/wiki/Real_Function_with_Negative_Derivative_is_Decreasing | https://proofwiki.org/wiki/Real_Function_with_Negative_Derivative_is_Decreasing | [
"Decreasing Real Functions",
"Real Functions",
"Differential Calculus"
] | [
"Definition:Decreasing/Real Function"
] | [
"Mean Value Theorem",
"Definition:Decreasing/Real Function"
] |
proofwiki-16850 | Continuous Real Function Differentiable on Borel Set | Let $\map \BB {\R, \size {\, \cdot \,} }$ be the Borel Sigma-Algebra on $\R$ with the usual topology.
Let $f: \R \to \R$ be a continuous real function.
Let $\map D f$ be the set of all points at which $f$ is differentiable.
Then $\map D f$ is a Borel Set with respect to $\map \BB {\R, \size {\, \cdot \,} }$. | By the definition of derivative:
:$\map {f'} x$ exists
{{iff}}:
:$\ds \lim_{h \mathop \to 0} \frac {\map f {x + h} - \map f x} h = L$ exists
where $L = \map {f'} x$.
For ease of presentation, denote:
:$\Delta f := \map f {x + h} - \map f x$
Since $f$ is continuous by hypothesis, so too is $\dfrac {\Delta f} h$ for $h ... | Let $\map \BB {\R, \size {\, \cdot \,} }$ be the [[Definition:Borel Sigma-Algebra|Borel Sigma-Algebra]] on $\R$ with the [[Definition:Usual Topology|usual topology]].
Let $f: \R \to \R$ be a [[Definition:Continuous Real Function|continuous real function]].
Let $\map D f$ be the set of all points at which $f$ is [[Def... | By the [[Definition:Derivative/Real Function/Derivative at Point/Definition 2|definition of derivative]]:
:$\map {f'} x$ exists
{{iff}}:
:$\ds \lim_{h \mathop \to 0} \frac {\map f {x + h} - \map f x} h = L$ exists
where $L = \map {f'} x$.
For ease of presentation, denote:
:$\Delta f := \map f {x + h} - \map f x... | Continuous Real Function Differentiable on Borel Set | https://proofwiki.org/wiki/Continuous_Real_Function_Differentiable_on_Borel_Set | https://proofwiki.org/wiki/Continuous_Real_Function_Differentiable_on_Borel_Set | [
"Sigma-Algebras",
"Borel Sets",
"Continuity",
"Borel Sets"
] | [
"Definition:Borel Sigma-Algebra",
"Definition:Euclidean Space/Euclidean Topology",
"Definition:Continuous Real Function",
"Definition:Differentiable Mapping/Real Function",
"Definition:Borel Sigma-Algebra/Borel Set"
] | [
"Definition:Derivative/Real Function/Derivative at Point/Definition 2",
"Definition:Continuous Real Function",
"Definition:By Hypothesis",
"Definition:Rational Number",
"Definition:Real Number",
"Rational Numbers are Everywhere Dense in Set of Real Numbers/Topology",
"Definition:Limit of Real Function",... |
proofwiki-16851 | Limit with Rational Epsilon and Delta | Let $\openint a b$ be an open real interval.
Let $c \in \openint a b$.
Let $f: \openint a b \setminus \set c \to \R$ be a real function.
Let $L \in \R$.
Suppose that:
:$\forall \epsilon > 0 \in \Q_{>0}: \exists \delta \in \Q_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$
Th... | Recall the definition of a limit of a real function:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$
Denote by $\map P {\epsilon,\delta}$ the proposition that the above statement holds.
Suppose that $\map P {\epsilon,\del... | Let $\openint a b$ be an [[Definition:Open Real Interval|open real interval]].
Let $c \in \openint a b$.
Let $f: \openint a b \setminus \set c \to \R$ be a [[Definition:Real Function|real function]].
Let $L \in \R$.
Suppose that:
:$\forall \epsilon > 0 \in \Q_{>0}: \exists \delta \in \Q_{>0}: \forall x \in \R: 0 <... | Recall the definition of a [[Definition:Limit of Real Function|limit of a real function]]:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$
Denote by $\map P {\epsilon,\delta}$ the [[Definition:Proposition|proposition]] ... | Limit with Rational Epsilon and Delta | https://proofwiki.org/wiki/Limit_with_Rational_Epsilon_and_Delta | https://proofwiki.org/wiki/Limit_with_Rational_Epsilon_and_Delta | [
"Limits of Real Functions"
] | [
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Limit of Real Function"
] | [
"Definition:Limit of Real Function",
"Definition:Proposition",
"Between two Real Numbers exists Rational Number",
"Between two Real Numbers exists Rational Number",
"Definition:Conditional/Language of Conditional/Strong"
] |
proofwiki-16852 | Limit with Epsilon Powers of 2 | Let $\openint a b$ be an open real interval.
Let $c \in \openint a b$.
Let $f: \openint a b \setminus \set c \to \R$ be a real function.
Let $L \in \R$.
Suppose that:
:$\forall n > 0 \in \N: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < 2^{-n} $
Then the limit ... | Denote by $\map P {2^{-n},\delta}$ the proposition considered in the theorem exposition:
:$\forall \in \N: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < 2^{-n} $
Let the limit of $f$ as $x \to c$ exist and equal $L$, as described in the definition of limit:
:$\... | Let $\openint a b$ be an [[Definition:Open Real Interval|open real interval]].
Let $c \in \openint a b$.
Let $f: \openint a b \setminus \set c \to \R$ be a [[Definition:Real Function|real function]].
Let $L \in \R$.
Suppose that:
:$\forall n > 0 \in \N: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - ... | Denote by $\map P {2^{-n},\delta}$ the [[Definition:Proposition|proposition]] considered in the theorem exposition:
:$\forall \in \N: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < 2^{-n} $
Let the limit of $f$ as $x \to c$ exist and equal $L$, as described in... | Limit with Epsilon Powers of 2 | https://proofwiki.org/wiki/Limit_with_Epsilon_Powers_of_2 | https://proofwiki.org/wiki/Limit_with_Epsilon_Powers_of_2 | [
"Limits of Real Functions"
] | [
"Definition:Real Interval/Open",
"Definition:Real Function",
"Definition:Limit of Real Function"
] | [
"Definition:Proposition",
"Definition:Limit of Real Function",
"Definition:Proposition",
"Axiom of Archimedes",
"Ordering of Reciprocals",
"Definition:Conditional/Language of Conditional/Weak"
] |
proofwiki-16853 | Characterization of Probability Density Function | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X: \Omega \to \R$ be a continuous random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $\Omega_X = \Img X$, the image of $X$.
Let the '''probability density function''' of $X$ is the mapping $f_X: \R \to \closedint 0 1$ be defined as:
:$\forall x \in \... | {{begin-eqn}}
{{eqn | l = \map \Pr {x - \frac \epsilon 2 \le X \le x + \frac \epsilon 2}
| r = \map {F_X} {x + \frac \epsilon 2} - \map {F_X} {x - \frac \epsilon 2} + \map \Pr {X = x -\frac \epsilon 2}
}}
{{eqn | r = \map {F_X} {x + \frac \epsilon 2} - \map {F_X} {x - \frac \epsilon 2}
| c = Probability ... | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $X: \Omega \to \R$ be a [[Definition:Continuous Random Variable|continuous random variable]] on $\struct {\Omega, \Sigma, \Pr}$.
Let $\Omega_X = \Img X$, the [[Definition:Image of Mapping|image]] of $X$.
Let the '''prob... | {{begin-eqn}}
{{eqn | l = \map \Pr {x - \frac \epsilon 2 \le X \le x + \frac \epsilon 2}
| r = \map {F_X} {x + \frac \epsilon 2} - \map {F_X} {x - \frac \epsilon 2} + \map \Pr {X = x -\frac \epsilon 2}
}}
{{eqn | r = \map {F_X} {x + \frac \epsilon 2} - \map {F_X} {x - \frac \epsilon 2}
| c = [[Probabilit... | Characterization of Probability Density Function | https://proofwiki.org/wiki/Characterization_of_Probability_Density_Function | https://proofwiki.org/wiki/Characterization_of_Probability_Density_Function | [
"Probability"
] | [
"Definition:Probability Space",
"Definition:Random Variable/Continuous",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping",
"Definition:Cumulative Distribution Function",
"Definition:Continuously Differentiable",
"Definition:Real Function",
"Definition:Mapping"
] | [
"Probability of Continuous Random Variable at Single Point is Zero",
"Continuous Derivative as Average of One-Sided Derivatives"
] |
proofwiki-16854 | Abel's Test | Let $\ds \sum b_n$ be a convergent real series.
Let $\sequence {a_n}$ be a decreasing sequence of positive real numbers.
Then the series $\ds \sum a_n b_n$ is also convergent. | {{MissingLinks|throughout}}
Let $b_0 = 0$.
Let $B_N = \ds \sum_{k \mathop = 0}^N b_k$.
Then:
:$\forall n \ge 1: b_n = B_n − B_{n − 1}$
From Abel's Lemma:
$\ds \sum_{k \mathop = 1}^N a_k b_k = \sum_{k \mathop = 1}^{N - 1} B_k \paren {a_k - a_{k + 1} } + a_N B_N$
By the Monotone Convergence Theorem:
:$\sequence {a_n}$ c... | Let $\ds \sum b_n$ be a [[Definition:Convergent Real Series|convergent real series]].
Let $\sequence {a_n}$ be a [[Definition:Decreasing Real Sequence|decreasing sequence]] of [[Definition:Positive Real Number|positive real numbers]].
Then the [[Definition:Real Series|series]] $\ds \sum a_n b_n$ is also [[Definition... | {{MissingLinks|throughout}}
Let $b_0 = 0$.
Let $B_N = \ds \sum_{k \mathop = 0}^N b_k$.
Then:
:$\forall n \ge 1: b_n = B_n − B_{n − 1}$
From [[Abel's Lemma/Formulation 2|Abel's Lemma]]:
$\ds \sum_{k \mathop = 1}^N a_k b_k = \sum_{k \mathop = 1}^{N - 1} B_k \paren {a_k - a_{k + 1} } + a_N B_N$
By the [[Monotone C... | Abel's Test | https://proofwiki.org/wiki/Abel's_Test | https://proofwiki.org/wiki/Abel's_Test | [
"Abel's Test",
"Convergence Tests",
"Convergence"
] | [
"Definition:Convergent Series/Number Field",
"Definition:Decreasing/Sequence/Real Sequence",
"Definition:Positive/Real Number",
"Definition:Series/Real",
"Definition:Convergent Series/Number Field"
] | [
"Abel's Lemma/Formulation 2",
"Monotone Convergence Theorem (Real Analysis)/Decreasing Sequence"
] |
proofwiki-16855 | Abel's Test for Uniform Convergence | Let $\sequence {\map {a_n} z}$ and $\sequence {\map {b_n} z}$ be sequences of complex functions on a compact set $K$.
Let $\sequence {\map {a_n} z}$ be such that:
:$\sequence {\map {a_n} z}$ is bounded in $K$
:$\ds \sum \size {\map {a_n} z - \map {a_{n + 1} } z}$ is convergent with a sum which is bounded in $K$
:$\ds \... | {{tidy|Use {{ProofWiki}} templates in order to make this page readable}}
{{MissingLinks|throughout}}
{{ProofWanted}}
First we modify the statement of Abel's Lemma | Let $\sequence {\map {a_n} z}$ and $\sequence {\map {b_n} z}$ be [[Definition:Sequence|sequences]] of [[Definition:Complex Function|complex functions]] on a [[Definition:Compact Subset of Complex Plane|compact set]] $K$.
Let $\sequence {\map {a_n} z}$ be such that:
:$\sequence {\map {a_n} z}$ is [[Definition:Bounded C... | {{tidy|Use {{ProofWiki}} templates in order to make this page readable}}
{{MissingLinks|throughout}}
{{ProofWanted}}
First we modify the statement of [[Abel's Lemma]] | Abel's Test for Uniform Convergence | https://proofwiki.org/wiki/Abel's_Test_for_Uniform_Convergence | https://proofwiki.org/wiki/Abel's_Test_for_Uniform_Convergence | [
"Abel's Test",
"Complex Analysis",
"Uniform Convergence"
] | [
"Definition:Sequence",
"Definition:Complex Function",
"Definition:Compact Space/Metric Space/Complex",
"Definition:Bounded Sequence/Complex",
"Definition:Convergent Sequence/Complex Numbers",
"Definition:Bounded Sequence/Complex",
"Definition:Uniform Convergence",
"Definition:Uniform Convergence"
] | [
"Abel's Lemma"
] |
proofwiki-16856 | Equivalence of Definitions of Topology Generated by Synthetic Basis/Definition 1 iff Definition 2 | Let $S$ be a set.
Let $\BB$ be a synthetic basis on $S$.
Let $\tau$ be the topology on $S$ generated by the synthetic basis $\BB$:
:$\ds \tau = \set {\bigcup \AA: \AA \subseteq \BB}$
Then:
:$\ds \forall U \subseteq S: U \in \tau \iff U = \bigcup \set {B \in \BB: B \subseteq U}$ | Trivially, the reverse implication holds, as $\set {B \in \BB: B \subseteq U} \subseteq \BB$.
We now show that the forward implication holds.
Suppose $U \in \tau$.
Then, by definition:
:$\ds \exists \AA \subseteq \BB: U = \bigcup \AA$
By Union is Smallest Superset: General Result:
:$\forall B \in \AA: B \subseteq U$
By... | Let $S$ be a [[Definition:Set|set]].
Let $\BB$ be a [[Definition:Synthetic Basis|synthetic basis]] on $S$.
Let $\tau$ be the [[Definition:Topology Generated by Synthetic Basis|topology on $S$ generated by the synthetic basis $\BB$]]:
:$\ds \tau = \set {\bigcup \AA: \AA \subseteq \BB}$
Then:
:$\ds \forall U \subsete... | Trivially, the [[Definition:Reverse Implication|reverse implication]] holds, as $\set {B \in \BB: B \subseteq U} \subseteq \BB$.
We now show that the [[Definition:Forward Implication|forward implication]] holds.
Suppose $U \in \tau$.
Then, by definition:
:$\ds \exists \AA \subseteq \BB: U = \bigcup \AA$
By [[Union... | Equivalence of Definitions of Topology Generated by Synthetic Basis/Definition 1 iff Definition 2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Topology_Generated_by_Synthetic_Basis/Definition_1_iff_Definition_2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Topology_Generated_by_Synthetic_Basis/Definition_1_iff_Definition_2 | [
"Equivalence of Definitions of Topology Generated by Synthetic Basis"
] | [
"Definition:Set",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Topology Generated by Synthetic Basis"
] | [
"Definition:Biconditional",
"Definition:Biconditional",
"Union is Smallest Superset/General Result",
"Definition:Subset",
"Union of Subset of Family is Subset of Union of Family",
"Union is Smallest Superset/General Result",
"Definition:Set Equality/Definition 2"
] |
proofwiki-16857 | Equivalence of Definitions of Topology Generated by Synthetic Basis/Definition 1 iff Definition 3 | Let $S$ be a set.
Let $\BB$ be a synthetic basis on $S$.
Let $\tau$ be the topology on $S$ generated by the synthetic basis $\BB$:
:$\ds \tau = \set {\bigcup \AA: \AA \subseteq \BB}$
Then:
:$\forall U \subseteq S: U \in \tau \iff \forall x \in U: \exists B \in \BB: x \in B \subseteq U$ | From Set is Subset of Union: General Result, the forward implication directly follows.
We now show that the reverse implication holds.
By hypothesis, we have that:
:$\ds U \subseteq \bigcup \set {B \in \BB: B \subseteq U}$
By Union is Smallest Superset: General Result:
:$\ds \bigcup \set {B \in \BB: B \subseteq U} \sub... | Let $S$ be a [[Definition:Set|set]].
Let $\BB$ be a [[Definition:Synthetic Basis|synthetic basis]] on $S$.
Let $\tau$ be the [[Definition:Topology Generated by Synthetic Basis|topology on $S$ generated by the synthetic basis $\BB$]]:
:$\ds \tau = \set {\bigcup \AA: \AA \subseteq \BB}$
Then:
:$\forall U \subseteq S:... | From [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]], the [[Definition:Forward Implication|forward implication]] directly follows.
We now show that the [[Definition:Reverse Implication|reverse implication]] holds.
[[Definition:By Hypothesis|By hypothesis]], we have that:
:$\ds U \sub... | Equivalence of Definitions of Topology Generated by Synthetic Basis/Definition 1 iff Definition 3 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Topology_Generated_by_Synthetic_Basis/Definition_1_iff_Definition_3 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Topology_Generated_by_Synthetic_Basis/Definition_1_iff_Definition_3 | [
"Equivalence of Definitions of Topology Generated by Synthetic Basis"
] | [
"Definition:Set",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Topology Generated by Synthetic Basis"
] | [
"Set is Subset of Union/General Result",
"Definition:Biconditional",
"Definition:Biconditional",
"Definition:By Hypothesis",
"Union is Smallest Superset/General Result",
"Definition:Set Equality/Definition 2"
] |
proofwiki-16858 | Projection from Product Topology is Open and Continuous | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its fa... | === Projection is Continuous ===
{{:Projection from Product Topology is Continuous}}{{qed|lemma}} | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $T = \struct {T_1 \times T_2, \tau}$ be the [[Definition:Product Space (Topology) of Two Factor Spaces|product space]] of $T_1$ and $T_2$, where $\tau$ is the [[Definition:Product Topology on... | === [[Projection from Product Topology is Continuous|Projection is Continuous]] ===
{{:Projection from Product Topology is Continuous}}{{qed|lemma}} | Projection from Product Topology is Open and Continuous | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Open_and_Continuous | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Open_and_Continuous | [
"Product Topology",
"Continuous Mappings",
"Open Mappings",
"Projections",
"Projection from Product Topology is Open and Continuous"
] | [
"Definition:Topological Space",
"Definition:Product Space (Topology)/Two Factor Spaces",
"Definition:Product Topology/Two Factor Spaces",
"Definition:Projection (Mapping Theory)",
"Definition:Product Topology/Factor Space",
"Definition:Open Mapping",
"Definition:Continuous Mapping (Topology)"
] | [
"Projection from Product Topology is Continuous"
] |
proofwiki-16859 | Projection from Product Topology is Open and Continuous/General Result | Let $\family {T_i}_{i \mathop \in I} = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the corresponding product space.
Let $\tau$ denote the product topology on $S$.
Let $\pr_i: S \to S_i$ b... | === Projection is Continuous ===
{{:Projection from Product Topology is Continuous/General Result/Proof}}{{qed|lemma}} | Let $\family {T_i}_{i \mathop \in I} = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the correspo... | === [[Projection from Product Topology is Continuous/General Result/Proof|Projection is Continuous]] ===
{{:Projection from Product Topology is Continuous/General Result/Proof}}{{qed|lemma}} | Projection from Product Topology is Open and Continuous/General Result | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Open_and_Continuous/General_Result | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Open_and_Continuous/General_Result | [
"Projection from Product Topology is Open and Continuous"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space of Topological Spaces",
"Definition:Product Topology",
"Definition:Projection (Mapping Theory)",
"Definition:Open Mapping",
"Definition:Continuous Mapping (Topology)"
] | [
"Projection from Product Topology is Continuous/General Result/Proof"
] |
proofwiki-16860 | Projection from Product Topology is Continuous/General Result | Let $\family {T_i}_{i \mathop \in I} = \family {\struct{S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the corresponding product space.
Let $\tau$ denote the Tychonoff topology on $S$.
Let $\pr_i: S \to S_i$ ... | By definition of the product topology on $S$:
:$\tau$ is the initial topology on $S$ with respect to $\family {\pr_i}_{i \mathop \in I}$
By definition of the Initial Topoplogy: Definition 2:
:$\tau$ is the coarsest topology on $S$ such that each $\pr_i: S \to S_i$ is a $\struct{\tau, \tau_i}$-continuous. | Let $\family {T_i}_{i \mathop \in I} = \family {\struct{S_i, \tau_i} }_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the correspon... | By definition of the [[Definition:Product Topology|product topology]] on $S$:
:$\tau$ is the [[Definition:Initial Topology|initial topology]] on $S$ with respect to $\family {\pr_i}_{i \mathop \in I}$
By definition of the [[Definition:Initial Topology/Definition 2|Initial Topoplogy: Definition 2]]:
:$\tau$ is the [[De... | Projection from Product Topology is Continuous/General Result/Proof | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Continuous/General_Result | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Continuous/General_Result/Proof | [
"Projection from Product Topology is Continuous"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space of Topological Spaces",
"Definition:Product Topology",
"Definition:Projection (Mapping Theory)",
"Definition:Continuous Mapping (Topology)"
] | [
"Definition:Product Topology",
"Definition:Initial Topology",
"Definition:Initial Topology/Definition 2",
"Definition:Coarser Topology",
"Definition:Continuous Mapping (Topology)"
] |
proofwiki-16861 | Projection from Product Topology is Open/General Result | Let $\family {T_i}_{i \mathop \in I} = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the corresponding product space.
Let $\tau$ denote the product topology on $S$.
Let $\pr_i: S \to S_i$ b... | Let $U \in \tau$.
It follows from the definition of product topology that $U$ can be expressed as:
:$\ds U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k, j} }^{-1} } {U_{k, j} }$
where:
:$J$ is an arbitrary index set
:$n_j \in \N$
:$i_{k, j} \in I$
:$U_{k, j} \in \tau_{i_{k, j} }$.
For all $... | Let $\family {T_i}_{i \mathop \in I} = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the correspo... | Let $U \in \tau$.
It follows from the definition of [[Definition:Product Topology|product topology]] that $U$ can be expressed as:
:$\ds U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k, j} }^{-1} } {U_{k, j} }$
where:
:$J$ is an arbitrary [[Definition:Indexing Set|index set]]
:$n_j \in \N... | Projection from Product Topology is Open/General Result/Proof | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Open/General_Result | https://proofwiki.org/wiki/Projection_from_Product_Topology_is_Open/General_Result/Proof | [
"Projection from Product Topology is Open"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space of Topological Spaces",
"Definition:Product Topology",
"Definition:Projection (Mapping Theory)",
"Definition:Open Mapping"
] | [
"Definition:Product Topology",
"Definition:Indexing Set",
"Image of Union under Relation/Family of Sets",
"Cartesian Product of Intersections/General Case",
"Definition:Open Mapping"
] |
proofwiki-16862 | Product Space Basis Induced from Factor Space Bases | Let $\family {\struct{S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.
Let $\BB_\alpha$ be a basis for the topology $\tau_\alpha$ for each $\alpha \in I$.
Let $\struct {S, \tau} = \ds \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\... | Let $x = \family {x_\alpha}_{\alpha \mathop \in I} \in S$.
Let $W \in \tau$ such that $x \in W$.
From Natural Basis of Product Topology, the set $\BB'$ of cartesian products of the form $\ds \prod_{\alpha \mathop \in I} V_\alpha$ where:
:for all $\alpha \in I : V_\alpha \in \tau_\alpha$
:for all but finitely many indic... | Let $\family {\struct{S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] for $\alpha$ in some [[Definition:Indexing Set|indexing set]] $I$.
Let $\BB_\alpha$ be a [[Definition:Analytic Basis|basis]] for the [[Definit... | Let $x = \family {x_\alpha}_{\alpha \mathop \in I} \in S$.
Let $W \in \tau$ such that $x \in W$.
From [[Natural Basis of Product Topology]], the [[Definition:Set|set]] $\BB'$ of [[Definition:Cartesian Product|cartesian products]] of the form $\ds \prod_{\alpha \mathop \in I} V_\alpha$ where:
:for all $\alpha \in I :... | Product Space Basis Induced from Factor Space Bases | https://proofwiki.org/wiki/Product_Space_Basis_Induced_from_Factor_Space_Bases | https://proofwiki.org/wiki/Product_Space_Basis_Induced_from_Factor_Space_Bases | [
"Topological Bases",
"Product Topology"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Basis (Topology)/Analytic Basis",
"Definition:Topology",
"Definition:Product Space (Topology)",
"Definition:Set",
"Definition:Cartesian Product",
"Definition:Finite Set",
"Definition:Index",
... | [
"Natural Basis of Product Topology",
"Definition:Set",
"Definition:Cartesian Product",
"Definition:Finite Set",
"Definition:Index",
"Definition:Basis (Topology)/Analytic Basis",
"Definition:Finite Set",
"Definition:Basis (Topology)/Analytic Basis",
"Definition:Finite Set",
"Category:Topological Ba... |
proofwiki-16863 | Existence and Uniqueness of Domain of Relation | Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Then the domain $\Dom \RR$ of $\RR$ exists and is unique. | By the Axiom of Specification we can create the subclass of $V$:
:$\Dom \RR = \set {x \in V: \exists y \in V: \tuple {x, y} \in \RR}$
Hence $\Dom \RR$ exists.
Suppose $\QQ \subseteq V$ such that $\QQ$ and $\Dom \RR$ are both the domain of $\RR$.
Then:
:$\QQ = \set {x \in V: \exists y \in V: \tuple {x, y} \in \RR}$
Thus... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $\RR \subseteq V \times V$ be a [[Definition:Relation (Class Theory)|relation]].
Then the [[Definition:Domain of Relation (Class Theory)|domain]] $\Dom \RR$ of $\RR$ exists and is [[Definition:Unique|unique]]. | By the [[Axiom:Axiom of Specification (Classes)|Axiom of Specification]] we can create the [[Definition:Subclass|subclass]] of $V$:
:$\Dom \RR = \set {x \in V: \exists y \in V: \tuple {x, y} \in \RR}$
Hence $\Dom \RR$ exists.
Suppose $\QQ \subseteq V$ such that $\QQ$ and $\Dom \RR$ are both the [[Definition:Domain of... | Existence and Uniqueness of Domain of Relation | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Domain_of_Relation | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Domain_of_Relation | [
"Relation Theory"
] | [
"Definition:Basic Universe",
"Definition:Relation/Class Theory",
"Definition:Domain (Set Theory)/Relation/Class Theory",
"Definition:Unique"
] | [
"Axiom:Axiom of Specification/Class Theory",
"Definition:Subclass",
"Definition:Domain (Set Theory)/Relation/Class Theory",
"Axiom:Axiom of Extension/Class Theory",
"Definition:Unique"
] |
proofwiki-16864 | Existence and Uniqueness of Image of Relation | Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Then the image $\Img \RR$ of $\RR$ exists and is unique. | By the Axiom of Specification we can create the subclass of $V$:
:$\Img \RR = \set {y \in V: \exists x \in V: \tuple {x, y} \in \RR}$
Hence $\Img \RR$ exists.
Suppose $\QQ \subseteq V$ such that $\QQ$ and $\Img \RR$ are both the image of $\RR$.
Then:
:$\QQ = \set {y \in V: \exists x \in V: \tuple {x, y} \in \RR}$
Thus:... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $\RR \subseteq V \times V$ be a [[Definition:Relation (Class Theory)|relation]].
Then the [[Definition:Image of Relation (Class Theory)|image]] $\Img \RR$ of $\RR$ exists and is [[Definition:Unique|unique]]. | By the [[Axiom:Axiom of Specification (Classes)|Axiom of Specification]] we can create the [[Definition:Subclass|subclass]] of $V$:
:$\Img \RR = \set {y \in V: \exists x \in V: \tuple {x, y} \in \RR}$
Hence $\Img \RR$ exists.
Suppose $\QQ \subseteq V$ such that $\QQ$ and $\Img \RR$ are both the [[Definition:Image of ... | Existence and Uniqueness of Image of Relation | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Image_of_Relation | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Image_of_Relation | [
"Relation Theory"
] | [
"Definition:Basic Universe",
"Definition:Relation/Class Theory",
"Definition:Image (Set Theory)/Relation/Relation/Class Theory",
"Definition:Unique"
] | [
"Axiom:Axiom of Specification/Class Theory",
"Definition:Subclass",
"Definition:Image (Set Theory)/Relation/Relation/Class Theory",
"Axiom:Axiom of Extension/Class Theory",
"Definition:Unique"
] |
proofwiki-16865 | Union of Union of Relation is Union of Domain with Image | Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\Dom \RR$ denote the domain of $\RR$.
Then:
:$\map \bigcup {\bigcup \RR} = \Dom \RR \cup \Img \RR$
where:
:$\bigcup \RR$ denotes the union of $\RR$
:$\Dom \RR$ denotes the domain of $\RR$
:$\Img \RR$ denotes the image of $\RR$. | {{begin-eqn}}
{{eqn | l = \bigcup \RR
| r = \set {z: \exists \tuple {x, y} \in \RR: z \in \tuple {x, y} }
| c = {{Defof|Union of Class}}
}}
{{eqn | r = \set {z: \exists \set {\set x, \set {x, y} } \in \RR: z \in \tuple {x, y} }
| c = {{Defof|Kuratowski Formalization of Ordered Pair}}
}}
{{eqn | r = \s... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $\RR \subseteq V \times V$ be a [[Definition:Relation|relation]].
Let $\Dom \RR$ denote the [[Definition:Domain of Relation|domain]] of $\RR$.
Then:
:$\map \bigcup {\bigcup \RR} = \Dom \RR \cup \Img \RR$
where:
:$\bigcup \RR$ denotes the [[Definition:Un... | {{begin-eqn}}
{{eqn | l = \bigcup \RR
| r = \set {z: \exists \tuple {x, y} \in \RR: z \in \tuple {x, y} }
| c = {{Defof|Union of Class}}
}}
{{eqn | r = \set {z: \exists \set {\set x, \set {x, y} } \in \RR: z \in \tuple {x, y} }
| c = {{Defof|Kuratowski Formalization of Ordered Pair}}
}}
{{eqn | r = \s... | Union of Union of Relation is Union of Domain with Image | https://proofwiki.org/wiki/Union_of_Union_of_Relation_is_Union_of_Domain_with_Image | https://proofwiki.org/wiki/Union_of_Union_of_Relation_is_Union_of_Domain_with_Image | [
"Class Union",
"Relation Theory"
] | [
"Definition:Basic Universe",
"Definition:Relation",
"Definition:Domain (Set Theory)/Relation",
"Definition:Class Union/General Definition",
"Definition:Domain (Set Theory)/Relation",
"Definition:Image (Set Theory)/Relation/Relation"
] | [
"Category:Class Union",
"Category:Relation Theory"
] |
proofwiki-16866 | Domain of Relation is Subclass of Union of Union of Relation | Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\Dom \RR$ denote the domain of $\RR$.
Then:
:$\Dom \RR \subseteq \map \bigcup {\bigcup \RR}$
where $\bigcup \RR$ denotes the union of $\RR$. | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \Dom \RR
| c = {{Defof|Domain of Relation (Class Theory)}}
}}
{{eqn | ll= \leadsto
| q = \exists x
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = {{Defof|Domain of Relation (Class Theory)}}
}}
{{eqn | ll= \leadsto
| l = \se... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $\RR \subseteq V \times V$ be a [[Definition:Relation (Class Theory)|relation]].
Let $\Dom \RR$ denote the [[Definition:Domain of Relation (Class Theory)|domain]] of $\RR$.
Then:
:$\Dom \RR \subseteq \map \bigcup {\bigcup \RR}$
where $\bigcup \RR$ denot... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \Dom \RR
| c = {{Defof|Domain of Relation (Class Theory)}}
}}
{{eqn | ll= \leadsto
| q = \exists x
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = {{Defof|Domain of Relation (Class Theory)}}
}}
{{eqn | ll= \leadsto
| l = \se... | Domain of Relation is Subclass of Union of Union of Relation/Proof | https://proofwiki.org/wiki/Domain_of_Relation_is_Subclass_of_Union_of_Union_of_Relation | https://proofwiki.org/wiki/Domain_of_Relation_is_Subclass_of_Union_of_Union_of_Relation/Proof | [
"Class Union",
"Relations",
"Domain of Relation is Subclass of Union of Union of Relation"
] | [
"Definition:Basic Universe",
"Definition:Relation/Class Theory",
"Definition:Domain (Set Theory)/Relation/Class Theory",
"Definition:Class Union/General Definition"
] | [] |
proofwiki-16867 | Image of Relation is Subclass of Union of Union of Relation | Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\Img \RR$ denote the image of $\RR$.
Then:
:$\Img \RR \subseteq \map \bigcup {\bigcup \RR}$
where $\bigcup \RR$ denotes the union of $\RR$. | {{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \Img \RR
| c = {{Defof|Image of Relation (Class Theory)}}
}}
{{eqn | ll= \leadsto
| q = \exists x
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = {{Defof|Image of Relation (Class Theory)}}
}}
{{eqn | ll= \leadsto
| l = \set ... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $\RR \subseteq V \times V$ be a [[Definition:Relation (Class Theory)|relation]].
Let $\Img \RR$ denote the [[Definition:Image of Relation (Class Theory)|image]] of $\RR$.
Then:
:$\Img \RR \subseteq \map \bigcup {\bigcup \RR}$
where $\bigcup \RR$ denotes... | {{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \Img \RR
| c = {{Defof|Image of Relation (Class Theory)}}
}}
{{eqn | ll= \leadsto
| q = \exists x
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = {{Defof|Image of Relation (Class Theory)}}
}}
{{eqn | ll= \leadsto
| l = \set ... | Image of Relation is Subclass of Union of Union of Relation | https://proofwiki.org/wiki/Image_of_Relation_is_Subclass_of_Union_of_Union_of_Relation | https://proofwiki.org/wiki/Image_of_Relation_is_Subclass_of_Union_of_Union_of_Relation | [
"Class Union"
] | [
"Definition:Basic Universe",
"Definition:Relation/Class Theory",
"Definition:Image (Set Theory)/Relation/Relation/Class Theory",
"Definition:Class Union/General Definition"
] | [] |
proofwiki-16868 | Relation is Set implies Domain and Image are Sets | Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\RR$ be a set.
Then $\Dom \RR$ and $\Img \RR$ are also sets. | From Domain of Relation is Subclass of Union of Union of Relation:
:$\Dom \RR \subseteq \map \bigcup {\bigcup \RR}$
From Image of Relation is Subclass of Union of Union of Relation:
:$\Img \RR \subseteq \map \bigcup {\bigcup \RR}$
We are given that $\RR$ is a set.
From the {{axiom-link|Unions|Class Theory}}:
:$\bigcup ... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $\RR \subseteq V \times V$ be a [[Definition:Relation (Class Theory)|relation]].
Let $\RR$ be a [[Definition:Set|set]].
Then $\Dom \RR$ and $\Img \RR$ are also [[Definition:Set|sets]]. | From [[Domain of Relation is Subclass of Union of Union of Relation]]:
:$\Dom \RR \subseteq \map \bigcup {\bigcup \RR}$
From [[Image of Relation is Subclass of Union of Union of Relation]]:
:$\Img \RR \subseteq \map \bigcup {\bigcup \RR}$
We are given that $\RR$ is a [[Definition:Set|set]].
From the {{axiom-link|Un... | Relation is Set implies Domain and Image are Sets | https://proofwiki.org/wiki/Relation_is_Set_implies_Domain_and_Image_are_Sets | https://proofwiki.org/wiki/Relation_is_Set_implies_Domain_and_Image_are_Sets | [
"Relations"
] | [
"Definition:Basic Universe",
"Definition:Relation/Class Theory",
"Definition:Set",
"Definition:Set"
] | [
"Domain of Relation is Subclass of Union of Union of Relation",
"Image of Relation is Subclass of Union of Union of Relation",
"Definition:Set",
"Definition:Set",
"Definition:Set",
"Definition:Basic Universe",
"Definition:Subclass",
"Definition:Set"
] |
proofwiki-16869 | Set is Transitive iff Subset of Power Set | A set $a$ is transitive {{iff}}:
:$a \subseteq \powerset a$
where $\powerset a$ denotes the power set of $a$. | === Necessary Condition ===
Let $a$ be transitive.
Let $x \in a$.
By definition of transitive set:
:$x \subseteq a$
Then by definition of power set:
:$x \in \powerset a$
Hence, by definition of subset:
:$a \subseteq \powerset a$
{{qed|lemma}} | A [[Definition:Set|set]] $a$ is [[Definition:Transitive Set|transitive]] {{iff}}:
:$a \subseteq \powerset a$
where $\powerset a$ denotes the [[Definition:Power Set|power set]] of $a$. | === Necessary Condition ===
Let $a$ be [[Definition:Transitive Set|transitive]].
Let $x \in a$.
By definition of [[Definition:Transitive Set|transitive set]]:
:$x \subseteq a$
Then by definition of [[Definition:Power Set|power set]]:
:$x \in \powerset a$
Hence, by definition of [[Definition:Subset|subset]]:
:$a \s... | Set is Transitive iff Subset of Power Set | https://proofwiki.org/wiki/Set_is_Transitive_iff_Subset_of_Power_Set | https://proofwiki.org/wiki/Set_is_Transitive_iff_Subset_of_Power_Set | [
"Power Set",
"Transitive Classes"
] | [
"Definition:Set",
"Definition:Transitive Class",
"Definition:Power Set"
] | [
"Definition:Transitive Class",
"Definition:Transitive Class",
"Definition:Power Set",
"Definition:Subset",
"Definition:Subset",
"Definition:Power Set",
"Definition:Transitive Class"
] |
proofwiki-16870 | Power Set of Transitive Set is Transitive | Let $x$ be a transitive set.
Then its power set $\powerset x$ is also a transitive set. | Let $x$ be transitive.
By Set is Transitive iff Subset of Power Set:
:$x \subseteq \powerset x$
Then by Power Set of Subset:
:$\powerset x \subseteq \powerset {\powerset x}$
Thus by Set is Transitive iff Subset of Power Set:
:$\powerset x$ is a transitive set.
{{qed}} | Let $x$ be a [[Definition:Transitive Set|transitive set]].
Then its [[Definition:Power Set|power set]] $\powerset x$ is also a [[Definition:Transitive Set|transitive set]]. | Let $x$ be [[Definition:Transitive Set|transitive]].
By [[Set is Transitive iff Subset of Power Set]]:
:$x \subseteq \powerset x$
Then by [[Power Set of Subset]]:
:$\powerset x \subseteq \powerset {\powerset x}$
Thus by [[Set is Transitive iff Subset of Power Set]]:
:$\powerset x$ is a [[Definition:Transitive Set|tr... | Power Set of Transitive Set is Transitive | https://proofwiki.org/wiki/Power_Set_of_Transitive_Set_is_Transitive | https://proofwiki.org/wiki/Power_Set_of_Transitive_Set_is_Transitive | [
"Power Set",
"Transitive Classes"
] | [
"Definition:Transitive Class",
"Definition:Power Set",
"Definition:Transitive Class"
] | [
"Definition:Transitive Class",
"Set is Transitive iff Subset of Power Set",
"Power Set of Subset",
"Set is Transitive iff Subset of Power Set",
"Definition:Transitive Class"
] |
proofwiki-16871 | Universal Class less Set is not Transitive | Let $V$ be a basic universe.
Let $a \in V$ be a set.
Then:
:$V \setminus \set a$ is not a transitive class
where $\setminus$ denotes class difference. | By definition, $V$ is the class of all sets.
As $a \in V$, by definition of $V$ it follows that $a$ is a set.
Consider the power set $\powerset a$ of $a$.
From the axiom of powers:
:$\powerset a$ is also a set
and:
:$\powerset {\powerset a}$ is also a set.
By definition:
:$a \in \powerset a$
and so:
:$\set a \in \power... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Let $a \in V$ be a [[Definition:Set|set]].
Then:
:$V \setminus \set a$ is not a [[Definition:Transitive Class|transitive class]]
where $\setminus$ denotes [[Definition:Class Difference|class difference]]. | By definition, $V$ is the [[Definition:Class (Class Theory)|class]] of all [[Definition:Set|sets]].
As $a \in V$, by definition of $V$ it follows that $a$ is a [[Definition:Set|set]].
Consider the [[Definition:Power Set|power set]] $\powerset a$ of $a$.
From the [[Axiom:Axiom of Powers (Class Theory)|axiom of powers... | Universal Class less Set is not Transitive | https://proofwiki.org/wiki/Universal_Class_less_Set_is_not_Transitive | https://proofwiki.org/wiki/Universal_Class_less_Set_is_not_Transitive | [
"Examples of Transitive Classes"
] | [
"Definition:Basic Universe",
"Definition:Set",
"Definition:Transitive Class",
"Definition:Class Difference"
] | [
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Set",
"Definition:Power Set",
"Axiom:Axiom of Powers/Class Theory",
"Definition:Set",
"Definition:Set",
"Definition:Set",
"Definition:Class Difference",
"Definition:Element",
"Definition:Element",
"Definition:Transitive Class"
] |
proofwiki-16872 | Basic Universe is Inductive | Let $V$ be a basic universe.
Then $V$ is an inductive class. | By definition of basic universe, $V$ is a class containing all sets as elements.
By the {{axiom-link|the Empty Set|Class Theory}}:
:the empty class $\O$ is a set.
Hence $\O$ is an element of $V$.
By the {{axiom-link|Powers|Class Theory}}, if $x$ is a set, then $\powerset x$ is a set.
By definition of power set:
:$\set ... | Let $V$ be a [[Definition:Basic Universe|basic universe]].
Then $V$ is an [[Definition:Inductive Class|inductive class]]. | By definition of [[Definition:Basic Universe|basic universe]], $V$ is a [[Definition:Class (Class Theory)|class]] containing all [[Definition:Set|sets]] as [[Definition:Element of Class|elements]].
By the {{axiom-link|the Empty Set|Class Theory}}:
:the [[Definition:Empty Class (Class Theory)|empty class]] $\O$ is a [[... | Basic Universe is Inductive | https://proofwiki.org/wiki/Basic_Universe_is_Inductive | https://proofwiki.org/wiki/Basic_Universe_is_Inductive | [
"Basic Universe",
"Inductive Classes"
] | [
"Definition:Basic Universe",
"Definition:Inductive Class"
] | [
"Definition:Basic Universe",
"Definition:Class (Class Theory)",
"Definition:Set",
"Definition:Element/Class",
"Definition:Empty Class (Class Theory)",
"Definition:Set",
"Definition:Element/Class",
"Definition:Set",
"Definition:Set",
"Definition:Power Set",
"Definition:Set",
"Definition:Set",
... |
proofwiki-16873 | Equivalence of Formulations of Axiom of Infinity for Zermelo Universe | The following formulations of the '''{{axiom-link|Infinity|Class Theory}}''' in the context of a '''Zermelo universe''' are equivalent: | Let the {{axiom-link|Infinity|Class Theory}}, in each of its formulations, be applied to a basic universe $V$ separately, as follows. | The following formulations of the '''{{axiom-link|Infinity|Class Theory}}''' in the context of a '''[[Definition:Zermelo Universe|Zermelo universe]]''' are [[Definition:Logical Equivalence|equivalent]]: | Let the {{axiom-link|Infinity|Class Theory}}, in each of its formulations, be applied to a [[Definition:Basic Universe|basic universe]] $V$ separately, as follows. | Equivalence of Formulations of Axiom of Infinity for Zermelo Universe | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Infinity_for_Zermelo_Universe | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Infinity_for_Zermelo_Universe | [
"Axiom of Infinity",
"Definition Equivalences"
] | [
"Definition:Zermelo Universe",
"Definition:Logical Equivalence"
] | [
"Definition:Basic Universe",
"Definition:Basic Universe"
] |
proofwiki-16874 | Inductive Construction of Natural Numbers fulfils Peano's Axioms | Let $P$ denote the set of natural numbers by definition as an inductive set.
Then $P$ fulfils Peano's axioms. | By definition of inductive set:
:$\O \in P$
By definition of the natural numbers, $\O$ is identified with $0$ (zero).
Thus {{PeanoAxiom|1}} holds.
{{qed|lemma}}
Let $x$ be a natural number.
By definition, $x$ is an element of every inductive set.
Thus if $x \in P$ it follows that $x^+ \in P$.
Thus {{PeanoAxiom|2}} hold... | Let $P$ denote the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] by [[Definition:Inductive Set Definition for Natural Numbers|definition as an inductive set]].
Then $P$ fulfils [[Axiom:Peano's Axioms|Peano's axioms]]. | By definition of [[Definition:Inductive Set|inductive set]]:
:$\O \in P$
By definition of the [[Definition:Natural Number|natural numbers]], $\O$ is identified with [[Definition:Zero (Number)|$0$ (zero)]].
Thus {{PeanoAxiom|1}} holds.
{{qed|lemma}}
Let $x$ be a [[Definition:Natural Number|natural number]].
By defi... | Inductive Construction of Natural Numbers fulfils Peano's Axioms | https://proofwiki.org/wiki/Inductive_Construction_of_Natural_Numbers_fulfils_Peano's_Axioms | https://proofwiki.org/wiki/Inductive_Construction_of_Natural_Numbers_fulfils_Peano's_Axioms | [
"Peano's Axioms",
"Inductive Sets",
"Natural Numbers"
] | [
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Natural Numbers/Inductive Set Definition",
"Axiom:Peano's Axioms"
] | [
"Definition:Inductive Set",
"Definition:Natural Numbers",
"Definition:Zero (Number)",
"Definition:Natural Numbers",
"Definition:Element",
"Definition:Inductive Set",
"Inductive Construction of Natural Numbers fulfils Peano's Axiom of Injectivity",
"Definition:Element",
"Definition:Empty Set",
"Def... |
proofwiki-16875 | Natural Number is Transitive Set | Let $n$ be a natural number.
Then $n$ is a transitive set. | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$n$ is a transitive set. | Let $n$ be a [[Definition:Natural Number|natural number]].
Then $n$ is a [[Definition:Transitive Set|transitive set]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$n$ is a [[Definition:Transitive Set|transitive set]]. | Natural Number is Transitive Set | https://proofwiki.org/wiki/Natural_Number_is_Transitive_Set | https://proofwiki.org/wiki/Natural_Number_is_Transitive_Set | [
"Transitive Classes",
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Transitive Class"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Transitive Class",
"Definition:Transitive Class",
"Definition:Transitive Class",
"Definition:Transitive Class",
"Definition:Transitive Class",
"Definition:Transitive Class",
"Principle of Mathematical Induction",
"Definit... |
proofwiki-16876 | Natural Number is Ordinary Set | Let $n$ be a natural number.
Then $n$ is an ordinary set.
That is:
:$n \notin n$ | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$n$ is an ordinary set. | Let $n$ be a [[Definition:Natural Number|natural number]].
Then $n$ is an [[Definition:Ordinary Set|ordinary set]].
That is:
:$n \notin n$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$n$ is an [[Definition:Ordinary Set|ordinary set]]. | Natural Number is Ordinary Set | https://proofwiki.org/wiki/Natural_Number_is_Ordinary_Set | https://proofwiki.org/wiki/Natural_Number_is_Ordinary_Set | [
"Ordinary Sets",
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Ordinary Set"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Ordinary Set",
"Definition:Ordinary Set",
"Definition:Ordinary Set",
"Principle of Mathematical Induction",
"Definition:Ordinary Set"
] |
proofwiki-16877 | Product Space Local Basis Induced from Factor Spaces Local Bases | Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an indexed family of topological spaces for $\alpha$ in some indexing set $I$.
Let $\struct {S, \tau} = \ds\prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha ... | Let $W \in \tau$ such that $x \in W$.
From Natural Basis of Product Topology, the set $\BB$ of cartesian products of the form $\ds \prod_{\alpha \mathop \in I} V_\alpha$ where:
:for all $\alpha \in I : V_\alpha \in \tau_\alpha$
:for all but finitely many indices $\alpha : V_\alpha = X_\alpha$
is a basis for $\tau$.
The... | Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] for $\alpha$ in some [[Definition:Indexing Set|indexing set]] $I$.
Let $\struct {S, \tau} = \ds\prod_{\alpha \mathop \in I} \struct {S_\alpha... | Let $W \in \tau$ such that $x \in W$.
From [[Natural Basis of Product Topology]], the [[Definition:Set|set]] $\BB$ of [[Definition:Cartesian Product|cartesian products]] of the form $\ds \prod_{\alpha \mathop \in I} V_\alpha$ where:
:for all $\alpha \in I : V_\alpha \in \tau_\alpha$
:for all but finitely many [[Defin... | Product Space Local Basis Induced from Factor Spaces Local Bases | https://proofwiki.org/wiki/Product_Space_Local_Basis_Induced_from_Factor_Spaces_Local_Bases | https://proofwiki.org/wiki/Product_Space_Local_Basis_Induced_from_Factor_Spaces_Local_Bases | [
"Topological Bases",
"Product Spaces",
"Local Bases"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Local Basis",
"Definition:Topological Space",
"Definition:Set",
"Definition:Cartesian Product",
"Definition:Finite Set",
"Definition:Index",
"Definition... | [
"Natural Basis of Product Topology",
"Definition:Set",
"Definition:Cartesian Product",
"Definition:Index",
"Definition:Basis (Topology)/Analytic Basis",
"Definition:Finite Set",
"Definition:Local Basis",
"Definition:Finite Set",
"Category:Topological Bases",
"Category:Product Spaces",
"Category:... |
proofwiki-16878 | Natural Numbers cannot be Elements of Each Other | Let $m$ and $n$ be natural numbers.
Then it cannot be the case that both $m \in n$ and $n \in m$. | {{AimForCont}} both $m \in n$ and $n \in m$.
We have $m \in n$
From Natural Number is Transitive Set:
:$m \subseteq n$
by definition of transitive.
Thus:
:$n \in m \subseteq n$
and so:
:$n \in n$
But from Natural Number is Ordinary Set:
:$n \notin n$
The result follows by Proof by Contradiction.
{{qed}} | Let $m$ and $n$ be [[Definition:Natural Number|natural numbers]].
Then it cannot be the case that both $m \in n$ and $n \in m$. | {{AimForCont}} both $m \in n$ and $n \in m$.
We have $m \in n$
From [[Natural Number is Transitive Set]]:
:$m \subseteq n$
by definition of [[Definition:Transitive Set|transitive]].
Thus:
:$n \in m \subseteq n$
and so:
:$n \in n$
But from [[Natural Number is Ordinary Set]]:
:$n \notin n$
The result follows by [[Pr... | Natural Numbers cannot be Elements of Each Other | https://proofwiki.org/wiki/Natural_Numbers_cannot_be_Elements_of_Each_Other | https://proofwiki.org/wiki/Natural_Numbers_cannot_be_Elements_of_Each_Other | [
"Ordinary Sets",
"Natural Numbers"
] | [
"Definition:Natural Numbers"
] | [
"Natural Number is Transitive Set",
"Definition:Transitive Class",
"Natural Number is Ordinary Set",
"Proof by Contradiction"
] |
proofwiki-16879 | Inductive Construction of Natural Numbers fulfils Peano's Axiom of Injectivity | Let $P$ denote the set of natural numbers by definition as an inductive set.
Then $P$ fulfils:
:{{PeanoAxiom|3}}
where $s$ denotes the successor mapping. | Let $m$ and $n$ be natural numbers such that $n^+ = m^+$.
By construction:
:$n \in n^+$
and:
:$m \in m^+$
Thus as $n^+ = m^+$ we have:
:$n \in m^+$
and:
:$m \in n^+$
This gives us:
:$n \in m \lor n = m$
and:
:$m \in n \lor m = n$
{{AimForCont}} that $n \ne m$.
Then from $n \in m \lor n = m$ we have:
:$n \in m$
and from... | Let $P$ denote the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] by [[Definition:Inductive Set Definition for Natural Numbers|definition as an inductive set]].
Then $P$ fulfils:
:{{PeanoAxiom|3}}
where $s$ denotes the [[Definition:Successor Mapping|successor mapping]]. | Let $m$ and $n$ be [[Definition:Natural Numbers|natural numbers]] such that $n^+ = m^+$.
By construction:
:$n \in n^+$
and:
:$m \in m^+$
Thus as $n^+ = m^+$ we have:
:$n \in m^+$
and:
:$m \in n^+$
This gives us:
:$n \in m \lor n = m$
and:
:$m \in n \lor m = n$
{{AimForCont}} that $n \ne m$.
Then from $n \in m \lor... | Inductive Construction of Natural Numbers fulfils Peano's Axiom of Injectivity | https://proofwiki.org/wiki/Inductive_Construction_of_Natural_Numbers_fulfils_Peano's_Axiom_of_Injectivity | https://proofwiki.org/wiki/Inductive_Construction_of_Natural_Numbers_fulfils_Peano's_Axiom_of_Injectivity | [
"Peano's Axioms",
"Inductive Sets",
"Natural Numbers"
] | [
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Natural Numbers/Inductive Set Definition",
"Definition:Successor Mapping"
] | [
"Definition:Natural Numbers",
"Natural Numbers cannot be Elements of Each Other",
"Proof by Contradiction",
"Definition:Injection"
] |
proofwiki-16880 | Element of Natural Number is Natural Number | Let $n$ be a natural number.
Let $m \in n$.
Then $m$ is also a natural number. | The proof proceeds by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
:for all $m \in n$: $m$ is a natural number. | Let $n$ be a [[Definition:Natural Number|natural number]].
Let $m \in n$.
Then $m$ is also a [[Definition:Natural Number|natural number]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:for all $m \in n$: $m$ is a [[Definition:Natural Number|natural number]]. | Element of Natural Number is Natural Number | https://proofwiki.org/wiki/Element_of_Natural_Number_is_Natural_Number | https://proofwiki.org/wiki/Element_of_Natural_Number_is_Natural_Number | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Natural Numbers"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Principle of Mathemati... |
proofwiki-16881 | Natural Number is not Subset of its Union | Let $n \in \N$ be a natural number as defined by the von Neumann construction.
Then, except in the degenerate case where $n = 0$, it is not the case that:
:$n \subseteq \bigcup n$ | First we note that from Union of Empty Set we have:
:$\bigcup \O = \O$
leading to:
:$\O \subseteq \bigcup \O$
thus disposing of the degenerate case.
Let $n \in \N$ such that $n \ne \O$.
By definition of the von Neumann construction:
:$n = \set {0, 1, 2, \ldots, n - 1}$
Thus, by definition, $m \in n$ for $m = 0, 1, 2, \... | Let $n \in \N$ be a [[Definition:Natural Number|natural number]] as defined by the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]].
Then, except in the [[Definition:Degenerate Case|degenerate case]] where $n = 0$, it is not the case that:
:$n \subseteq \bigcup n$ | First we note that from [[Union of Empty Set]] we have:
:$\bigcup \O = \O$
leading to:
:$\O \subseteq \bigcup \O$
thus disposing of the [[Definition:Degenerate Case|degenerate case]].
Let $n \in \N$ such that $n \ne \O$.
By definition of the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann const... | Natural Number is not Subset of its Union | https://proofwiki.org/wiki/Natural_Number_is_not_Subset_of_its_Union | https://proofwiki.org/wiki/Natural_Number_is_not_Subset_of_its_Union | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Degenerate Case"
] | [
"Union of Empty Set",
"Definition:Degenerate Case",
"Definition:Natural Numbers/Von Neumann Construction",
"Natural Numbers cannot be Elements of Each Other",
"Definition:Class Union/General Definition"
] |
proofwiki-16882 | Natural Number is Superset of its Union | Let $n \in \N$ be a natural number as defined by the von Neumann construction.
Then:
:$\bigcup n \subseteq n$ | Let $n \in \N$.
From Natural Number is Transitive Set, $n$ is transitive.
From Class is Transitive iff Union is Subclass it follows directly that:
:$\bigcup n \subseteq n$
{{qed}} | Let $n \in \N$ be a [[Definition:Natural Number|natural number]] as defined by the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]].
Then:
:$\bigcup n \subseteq n$ | Let $n \in \N$.
From [[Natural Number is Transitive Set]], $n$ is [[Definition:Transitive Set|transitive]].
From [[Class is Transitive iff Union is Subclass]] it follows directly that:
:$\bigcup n \subseteq n$
{{qed}} | Natural Number is Superset of its Union | https://proofwiki.org/wiki/Natural_Number_is_Superset_of_its_Union | https://proofwiki.org/wiki/Natural_Number_is_Superset_of_its_Union | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Natural Numbers/Von Neumann Construction"
] | [
"Natural Number is Transitive Set",
"Definition:Transitive Class",
"Class is Transitive iff Union is Subclass"
] |
proofwiki-16883 | Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton | Let $T_1$ and $T_2$ be non-empty topological spaces.
Let $b \in T_2$.
Let $T_1 \times T_2$ be the product space of $T_1$ and $T_2$.
Let $T_2 \times T_1$ be the product space of $T_2$ and $T_1$.
Then:
:$T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$
:$T_1$ is homeomorphic to the subspace $\... | From Finite Cartesian Product of Non-Empty Sets is Non-Empty both $T_1 \times T_2$ and $T_2 \times T_1$ are both non-empty.
The conclusions follow immediately from Subspace of Product Space is Homeomorphic to Factor Space.
{{qed}} | Let $T_1$ and $T_2$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Topological Space|topological spaces]].
Let $b \in T_2$.
Let $T_1 \times T_2$ be the [[Definition:Product Space (Topology) of Two Factor Spaces|product space]] of $T_1$ and $T_2$.
Let $T_2 \times T_1$ be the [[Definition:Product Space (Topolo... | From [[Finite Cartesian Product of Non-Empty Sets is Non-Empty]] both $T_1 \times T_2$ and $T_2 \times T_1$ are both [[Definition:Non-Empty Set|non-empty]].
The conclusions follow immediately from [[Subspace of Product Space is Homeomorphic to Factor Space]].
{{qed}} | Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Proof 1 | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space/Product_with_Singleton | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space/Product_with_Singleton/Proof_1 | [
"Subspace of Product Space is Homeomorphic to Factor Space"
] | [
"Definition:Non-Empty Set",
"Definition:Topological Space",
"Definition:Product Space (Topology)/Two Factor Spaces",
"Definition:Product Space (Topology)/Two Factor Spaces",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Topological Subspace",
"Definition:Homeomorphism/Topological Spaces",
... | [
"Finite Cartesian Product of Non-Empty Sets is Non-Empty",
"Definition:Non-Empty Set",
"Subspace of Product Space is Homeomorphic to Factor Space"
] |
proofwiki-16884 | Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton | Let $T_1$ and $T_2$ be non-empty topological spaces.
Let $b \in T_2$.
Let $T_1 \times T_2$ be the product space of $T_1$ and $T_2$.
Let $T_2 \times T_1$ be the product space of $T_2$ and $T_1$.
Then:
:$T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$
:$T_1$ is homeomorphic to the subspace $\... | The conclusions are symmetrical.
{{WLOG}}, therefore, it will be shown that $T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$.
Let $f: T_1 \to T_1 \times \set b$ be defined as:
:$\map f x = \tuple {x, b}$
=== Lemma ===
{{:Subspace of Product Space is Homeomorphic to Factor Space/Product with... | Let $T_1$ and $T_2$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Topological Space|topological spaces]].
Let $b \in T_2$.
Let $T_1 \times T_2$ be the [[Definition:Product Space (Topology) of Two Factor Spaces|product space]] of $T_1$ and $T_2$.
Let $T_2 \times T_1$ be the [[Definition:Product Space (Topolo... | The conclusions are symmetrical.
{{WLOG}}, therefore, it will be shown that $T_1$ is [[Definition:Homeomorphism|homeomorphic]] to the [[Definition:Topological Subspace|subspace]] $T_1 \times \set b$ of $T_1 \times T_2$.
Let $f: T_1 \to T_1 \times \set b$ be defined as:
:$\map f x = \tuple {x, b}$
=== [[Subspace of ... | Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Proof 2 | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space/Product_with_Singleton | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space/Product_with_Singleton/Proof_2 | [
"Subspace of Product Space is Homeomorphic to Factor Space"
] | [
"Definition:Non-Empty Set",
"Definition:Topological Space",
"Definition:Product Space (Topology)/Two Factor Spaces",
"Definition:Product Space (Topology)/Two Factor Spaces",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Topological Subspace",
"Definition:Homeomorphism/Topological Spaces",
... | [
"Definition:Homeomorphism",
"Definition:Topological Subspace",
"Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Lemma",
"Definition:Restriction/Mapping",
"Definition:Subspace",
"Definition:Projection",
"Projection from Product Topology is Continuous ",
"Definition:Cont... |
proofwiki-16885 | Natural Number is Union of its Successor | Let $n \in \N$ be a natural number as defined by the von Neumann construction.
Then:
:$\map \bigcup {n^+} = n$ | {{begin-eqn}}
{{eqn | o =
| r = \map \bigcup {n^+}
| c =
}}
{{eqn | r = \map \bigcup {\set n \cup n}
| c = {{Defof|Von Neumann Construction of Natural Numbers}}
}}
{{eqn | r = \bigcup \set n \cup \bigcup n
| c = Set Union is Self-Distributive
}}
{{eqn | r = n \cup \bigcup n
| c = Union o... | Let $n \in \N$ be a [[Definition:Natural Number|natural number]] as defined by the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]].
Then:
:$\map \bigcup {n^+} = n$ | {{begin-eqn}}
{{eqn | o =
| r = \map \bigcup {n^+}
| c =
}}
{{eqn | r = \map \bigcup {\set n \cup n}
| c = {{Defof|Von Neumann Construction of Natural Numbers}}
}}
{{eqn | r = \bigcup \set n \cup \bigcup n
| c = [[Set Union is Self-Distributive]]
}}
{{eqn | r = n \cup \bigcup n
| c = [[U... | Natural Number is Union of its Successor | https://proofwiki.org/wiki/Natural_Number_is_Union_of_its_Successor | https://proofwiki.org/wiki/Natural_Number_is_Union_of_its_Successor | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Natural Numbers/Von Neumann Construction"
] | [
"Set Union is Self-Distributive",
"Union of Singleton",
"Natural Number is Superset of its Union",
"Union with Superset is Superset"
] |
proofwiki-16886 | Set of Natural Numbers Equals its Union | Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction on a Zermelo universe $V$.
Then:
:$\bigcup \omega = \omega$ | We have {{hypothesis}} that:
:$\omega \in V$
where $V$ is a Zermelo universe.
By the {{axiom-link|Infinity|Class Theory}} we have that $\omega$ is a set.
By the {{axiom-link|Transitivity}}, $\omega$ is transitive.
Hence by Class is Transitive iff Union is Subclass:
:$\bigcup \omega \subseteq \omega$
{{qed|lemma}}
Let $... | Let $\omega$ denote the [[Definition:Natural Numbers|set of natural numbers]] as defined by the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]] on a [[Definition:Zermelo Universe|Zermelo universe]] $V$.
Then:
:$\bigcup \omega = \omega$ | We have {{hypothesis}} that:
:$\omega \in V$
where $V$ is a [[Definition:Zermelo Universe|Zermelo universe]].
By the {{axiom-link|Infinity|Class Theory}} we have that $\omega$ is a [[Definition:Set|set]].
By the {{axiom-link|Transitivity}}, $\omega$ is [[Definition:Transitive Set|transitive]].
Hence by [[Class is Tr... | Set of Natural Numbers Equals its Union | https://proofwiki.org/wiki/Set_of_Natural_Numbers_Equals_its_Union | https://proofwiki.org/wiki/Set_of_Natural_Numbers_Equals_its_Union | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Zermelo Universe"
] | [
"Definition:Zermelo Universe",
"Definition:Set",
"Definition:Transitive Class",
"Class is Transitive iff Union is Subclass",
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Class Union/General Definition",
"Definition:Set Equality"
] |
proofwiki-16887 | Set of Natural Numbers Equals Union of its Successor | Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction on a Zermelo universe $V$.
Then:
:$\bigcup \omega^+ = \omega$ | We have that:
:$\omega^+ = \omega \cup \set \omega$
and so:
:$\omega \subseteq \omega^+$
{{qed|lemma}}
By definition:
:$\bigcup \omega^+ = \set {x: \exists y \in \omega^+: x \in y}$
Thus:
:$x \in \bigcup \omega^+ \implies x \in \omega$
{{qed|lemma}}
So by definition of set equality:
:$\bigcup \omega^+ = \omega$
{{qed}} | Let $\omega$ denote the [[Definition:Natural Numbers|set of natural numbers]] as defined by the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]] on a [[Definition:Zermelo Universe|Zermelo universe]] $V$.
Then:
:$\bigcup \omega^+ = \omega$ | We have that:
:$\omega^+ = \omega \cup \set \omega$
and so:
:$\omega \subseteq \omega^+$
{{qed|lemma}}
By definition:
:$\bigcup \omega^+ = \set {x: \exists y \in \omega^+: x \in y}$
Thus:
:$x \in \bigcup \omega^+ \implies x \in \omega$
{{qed|lemma}}
So by definition of [[Definition:Set Equality|set equality]]:
:... | Set of Natural Numbers Equals Union of its Successor | https://proofwiki.org/wiki/Set_of_Natural_Numbers_Equals_Union_of_its_Successor | https://proofwiki.org/wiki/Set_of_Natural_Numbers_Equals_Union_of_its_Successor | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Zermelo Universe"
] | [
"Definition:Set Equality"
] |
proofwiki-16888 | Equivalence of Definitions of Minimally Inductive Class | Let $A$ be a class.
Let $g$ be a mapping on $A$.
{{TFAE|def = Minimally Inductive Class under General Mapping|view = minimally inductive class under $g$}} | === $(1)$ implies $(2)$ ===
Let it be given that $A$ is inductive under $g$.
Let $A$ be a minimally inductive class under $g$ by definition 1.
Then by definition:
:$A$ has no proper subclass $B$ such that $B$ is inductive under $g$.
Let $A$ have a subclass $C$ which is inductive under $g$.
Then by definition, $C$ is no... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $g$ be a [[Definition:Class Mapping|mapping]] on $A$.
{{TFAE|def = Minimally Inductive Class under General Mapping|view = minimally inductive class under $g$}} | === $(1)$ implies $(2)$ ===
Let it be given that $A$ is [[Definition:Inductive Class under General Mapping|inductive under $g$]].
Let $A$ be a [[Definition:Minimally Inductive Class under General Mapping/Definition 1|minimally inductive class under $g$ by definition 1]].
Then by definition:
:$A$ has no [[Definition:... | Equivalence of Definitions of Minimally Inductive Class | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Minimally_Inductive_Class | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Minimally_Inductive_Class | [
"Minimally Inductive Classes"
] | [
"Definition:Class (Class Theory)",
"Definition:Mapping/Class Theory"
] | [
"Definition:Inductive Class/General",
"Definition:Minimally Inductive Class under General Mapping/Definition 1",
"Definition:Proper Subclass",
"Definition:Inductive Class/General",
"Definition:Subclass",
"Definition:Inductive Class/General",
"Definition:Proper Subclass",
"Definition:Proper Subclass",
... |
proofwiki-16889 | Principle of General Induction | Let $M$ be a class.
Let $g: M \to M$ be a mapping on $M$.
Let $M$ be minimally inductive under $g$.
Let $P: M \to \set {\T, \F}$ be a propositional function on $M$.
Suppose that:
{{begin-axiom}}
{{axiom | n = 1
| ml= \map P \O
| mo= =
| mr= \T
}}
{{axiom | n = 2
| q = \forall x \in M
... | We are given that $M$ is a minimally inductive class under $g$.
That is, $M$ is an inductive class under $g$ with the extra property that $M$ has no proper class which is also an inductive class under $g$.
Let $P$ be a propositional function on $M$ which has the properties specified:
:$(1): \quad \map P \O = \T$
:$(2):... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Class Mapping|mapping]] on $M$.
Let $M$ be [[Definition:Minimally Inductive Class under General Mapping|minimally inductive under $g$]].
Let $P: M \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional func... | We are given that $M$ is a [[Definition:Minimally Inductive Class under General Mapping|minimally inductive class under $g$]].
That is, $M$ is an [[Definition:Inductive Class under General Mapping|inductive class under $g$]] with the extra property that $M$ has no [[Definition:Proper Class|proper class]] which is also... | Principle of General Induction | https://proofwiki.org/wiki/Principle_of_General_Induction | https://proofwiki.org/wiki/Principle_of_General_Induction | [
"Mathematical Induction",
"Proof Techniques",
"Principle of General Induction"
] | [
"Definition:Class (Class Theory)",
"Definition:Mapping/Class Theory",
"Definition:Minimally Inductive Class under General Mapping",
"Definition:Propositional Function"
] | [
"Definition:Minimally Inductive Class under General Mapping",
"Definition:Inductive Class/General",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Inductive Class/General",
"Definition:Propositional Function",
"Definition:Class (Class Theory)",
"Definition:Element/Class",
"Definition:Indu... |
proofwiki-16890 | Von Neumann Construction of Natural Numbers is Minimally Inductive | Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction.
$\omega$ is a minimally inductive class under the successor mapping. | Consider Peano's axioms:
{{:Axiom:Peano's Axioms}}
From Inductive Construction of Natural Numbers fulfils Peano's Axioms, $\omega$ fulfils Peano's axioms.
We note that from {{PeanoAxiom|1}}:
:$\O \in \omega$
We acknowledge from {{PeanoAxiom|2}}:
:the successor mapping defines that $n^+ := n \cup \set n$
and from {{Pean... | Let $\omega$ denote the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] as defined by the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]].
$\omega$ is a [[Definition:Minimally Inductive Class under General Mapping|minimally inductive class]] under the [[Def... | Consider [[Axiom:Peano's Axioms|Peano's axioms]]:
{{:Axiom:Peano's Axioms}}
From [[Inductive Construction of Natural Numbers fulfils Peano's Axioms]], $\omega$ fulfils [[Axiom:Peano's Axioms|Peano's axioms]].
We note that from {{PeanoAxiom|1}}:
:$\O \in \omega$
We acknowledge from {{PeanoAxiom|2}}:
:the [[Definition... | Von Neumann Construction of Natural Numbers is Minimally Inductive | https://proofwiki.org/wiki/Von_Neumann_Construction_of_Natural_Numbers_is_Minimally_Inductive | https://proofwiki.org/wiki/Von_Neumann_Construction_of_Natural_Numbers_is_Minimally_Inductive | [
"Natural Numbers",
"Minimally Inductive Classes"
] | [
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Minimally Inductive Class under General Mapping",
"Definition:Natural Numbers/Von Neumann Construction/Successor Mapping"
] | [
"Axiom:Peano's Axioms",
"Inductive Construction of Natural Numbers fulfils Peano's Axioms",
"Axiom:Peano's Axioms",
"Definition:Natural Numbers/Von Neumann Construction/Successor Mapping"
] |
proofwiki-16891 | Cartesian Product of Subsets/Family of Subsets | Let $\family {S_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.
Let $S = \ds \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.
Let $\family {T_i}_{i \mathop \in I}$ be a family of sets.
Let $T = \ds \prod_{i \mathop \in I} T_i$ be the Cartesian pr... | Let $T_i \subseteq S_i$ for all $i \in I$.
Then:
{{begin-eqn}}
{{eqn | l = \family {x_i}
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| q = \forall i \in I
| l = x_i
| o = \in
| r = T_i
| c = {{Defof|Cartesian Product of Family}}
}}
{{eqn | ll= \leadsto
| q = \f... | Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $S = \ds \prod_{i \mathop \in I} S_i$ be the [[Definition:Cartesian Product of Family|Cartesian product]] of $\family {S_i}_{i \mathop \in I}$.
Let $\fa... | Let $T_i \subseteq S_i$ for all $i \in I$.
Then:
{{begin-eqn}}
{{eqn | l = \family {x_i}
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| q = \forall i \in I
| l = x_i
| o = \in
| r = T_i
| c = {{Defof|Cartesian Product of Family}}
}}
{{eqn | ll= \leadsto
| q = \... | Cartesian Product of Subsets/Family of Subsets | https://proofwiki.org/wiki/Cartesian_Product_of_Subsets/Family_of_Subsets | https://proofwiki.org/wiki/Cartesian_Product_of_Subsets/Family_of_Subsets | [
"Cartesian Product of Subsets",
"Indexed Families"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Indexing Set",
"Definition:Cartesian Product/Family of Sets",
"Definition:Indexing Set/Family of Sets",
"Definition:Cartesian Product of Family "
] | [
"Definition:Subset",
"Category:Cartesian Product of Subsets",
"Category:Indexed Families"
] |
proofwiki-16892 | Double Induction Principle | Let $M$ be a class.
Let $g: M \to M$ be a mapping on $M$.
Let $M$ be a minimally inductive class under $g$.
Let $\RR$ be a relation on $M$ which satisfies:
{{begin-axiom}}
{{axiom | n = \text D_1
| q = \forall x \in M
| m = \map \RR {x, \O}
}}
{{axiom | n = \text D_2
| q = \forall x, y \in M
... | The proof proceeds by general induction.
Let an element $x$ of $M$ be defined as:
:'''left normal''' with respect to $\RR$ {{iff}} $\map \RR {x, y}$ for all $y \in M$
:'''right normal''' with respect to $\RR$ {{iff}} $\map \RR {y, x}$ for all $y \in M$.
Let the hypothesis be assumed.
First we demonstrate a lemma:
=== L... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Class Mapping|mapping]] on $M$.
Let $M$ be a [[Definition:Minimally Inductive Class under General Mapping|minimally inductive class]] under $g$.
Let $\RR$ be a [[Definition:Relation (Class Theory)|relation]] on $M$ which satis... | The proof proceeds by [[Principle of General Induction|general induction]].
Let an [[Definition:Element of Class|element]] $x$ of $M$ be defined as:
:'''[[Definition:Left Normal Element of Relation|left normal]]''' with respect to $\RR$ {{iff}} $\map \RR {x, y}$ for all $y \in M$
:'''[[Definition:Right Normal Element ... | Double Induction Principle/Proof 1 | https://proofwiki.org/wiki/Double_Induction_Principle | https://proofwiki.org/wiki/Double_Induction_Principle/Proof_1 | [
"Mathematical Induction",
"Named Theorems",
"Proof Techniques",
"Minimally Closed Classes",
"Double Induction Principle"
] | [
"Definition:Class (Class Theory)",
"Definition:Mapping/Class Theory",
"Definition:Minimally Inductive Class under General Mapping",
"Definition:Relation/Class Theory"
] | [
"Principle of General Induction",
"Definition:Element/Class",
"Definition:Left Normal Element of Relation",
"Definition:Right Normal Element of Relation",
"Definition:Lemma",
"Double Induction Principle/Lemma",
"Principle of General Induction",
"Definition:Right Normal Element of Relation",
"Double ... |
proofwiki-16893 | Double Induction Principle | Let $M$ be a class.
Let $g: M \to M$ be a mapping on $M$.
Let $M$ be a minimally inductive class under $g$.
Let $\RR$ be a relation on $M$ which satisfies:
{{begin-axiom}}
{{axiom | n = \text D_1
| q = \forall x \in M
| m = \map \RR {x, \O}
}}
{{axiom | n = \text D_2
| q = \forall x, y \in M
... | By definition, a minimally inductive class under $g$ is a minimally closed class under $g$ with respect to $\O$.
Recall the Double Induction Principle for Minimally Closed Class:
Let $\RR$ be a relation on $M$ which satisfies:
{{begin-axiom}}
{{axiom | n = \text D_1
| q = \forall x \in M
| m = \map \RR ... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Class Mapping|mapping]] on $M$.
Let $M$ be a [[Definition:Minimally Inductive Class under General Mapping|minimally inductive class]] under $g$.
Let $\RR$ be a [[Definition:Relation (Class Theory)|relation]] on $M$ which satis... | By definition, a [[Definition:Minimally Inductive Class under General Mapping|minimally inductive class]] under $g$ is a [[Definition:Minimally Closed Class|minimally closed class under $g$ with respect to $\O$]].
Recall the [[Double Induction Principle for Minimally Closed Class]]:
Let $\RR$ be a [[Definition:Relati... | Double Induction Principle/Proof 2 | https://proofwiki.org/wiki/Double_Induction_Principle | https://proofwiki.org/wiki/Double_Induction_Principle/Proof_2 | [
"Mathematical Induction",
"Named Theorems",
"Proof Techniques",
"Minimally Closed Classes",
"Double Induction Principle"
] | [
"Definition:Class (Class Theory)",
"Definition:Mapping/Class Theory",
"Definition:Minimally Inductive Class under General Mapping",
"Definition:Relation/Class Theory"
] | [
"Definition:Minimally Inductive Class under General Mapping",
"Definition:Minimally Closed Class",
"Double Induction Principle/Minimally Closed Class",
"Definition:Relation/Class Theory"
] |
proofwiki-16894 | Successor Mapping on Natural Numbers is Progressing | Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction.
Let $s: \omega \to \omega$ denote the successor mapping on $\omega$.
Then $s$ is a progressing mapping. | By definition of the von Neumann construction:
:$n^+ = n \cup \set n$
from which it follows that:
:$n \subseteq n^+$
Hence the result by definition of progressing mapping.
{{qed}} | Let $\omega$ denote the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] as defined by the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]].
Let $s: \omega \to \omega$ denote the [[Definition:Successor Mapping on Von Neumann Construction|successor mapping]] on... | By definition of the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]]:
:$n^+ = n \cup \set n$
from which it follows that:
:$n \subseteq n^+$
Hence the result by definition of [[Definition:Progressing Mapping|progressing mapping]].
{{qed}} | Successor Mapping on Natural Numbers is Progressing/Proof 1 | https://proofwiki.org/wiki/Successor_Mapping_on_Natural_Numbers_is_Progressing | https://proofwiki.org/wiki/Successor_Mapping_on_Natural_Numbers_is_Progressing/Proof_1 | [
"Progressing Mappings",
"Successor Mapping on Natural Numbers is Progressing"
] | [
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Natural Numbers/Von Neumann Construction/Successor Mapping",
"Definition:Progressing Mapping"
] | [
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Progressing Mapping"
] |
proofwiki-16895 | Successor Mapping on Natural Numbers is Progressing | Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction.
Let $s: \omega \to \omega$ denote the successor mapping on $\omega$.
Then $s$ is a progressing mapping. | By definition, the successor mapping on $\omega$ is indeed an example of a successor mapping.
The result follows from Successor Mapping is Progressing.
{{qed}} | Let $\omega$ denote the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] as defined by the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]].
Let $s: \omega \to \omega$ denote the [[Definition:Successor Mapping on Von Neumann Construction|successor mapping]] on... | By definition, the [[Definition:Successor Mapping on Von Neumann Construction|successor mapping]] on $\omega$ is indeed an example of a [[Definition:Successor Mapping|successor mapping]].
The result follows from [[Successor Mapping is Progressing]].
{{qed}} | Successor Mapping on Natural Numbers is Progressing/Proof 2 | https://proofwiki.org/wiki/Successor_Mapping_on_Natural_Numbers_is_Progressing | https://proofwiki.org/wiki/Successor_Mapping_on_Natural_Numbers_is_Progressing/Proof_2 | [
"Progressing Mappings",
"Successor Mapping on Natural Numbers is Progressing"
] | [
"Definition:Set",
"Definition:Natural Numbers",
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Natural Numbers/Von Neumann Construction/Successor Mapping",
"Definition:Progressing Mapping"
] | [
"Definition:Natural Numbers/Von Neumann Construction/Successor Mapping",
"Definition:Successor Mapping",
"Successor Mapping is Progressing"
] |
proofwiki-16896 | Cartesian Product of Subsets/Family of Nonempty Subsets | Let $T_i \ne \O$ for all $i \in I$.
Then:
:$T \subseteq S \iff \forall i \in I: T_i \subseteq S_i$. | From Cartesian Product of Family of Subsets:
:$\paren {\forall i \in I: T_i \subseteq S_i} \implies T \subseteq S$
So it remains to show that:
:$T \subseteq S \implies \forall i \in I : T_i \subseteq S_i$.
Let $T \subseteq S$.
Let $x_j \in T_j$ for some $j \in I$.
Let $\map x j = x_j$
Suppose $k \in I: k \ne j$.
As $T_... | Let $T_i \ne \O$ for all $i \in I$.
Then:
:$T \subseteq S \iff \forall i \in I: T_i \subseteq S_i$. | From [[Cartesian Product of Family of Subsets]]:
:$\paren {\forall i \in I: T_i \subseteq S_i} \implies T \subseteq S$
So it remains to show that:
:$T \subseteq S \implies \forall i \in I : T_i \subseteq S_i$.
Let $T \subseteq S$.
Let $x_j \in T_j$ for some $j \in I$.
Let $\map x j = x_j$
Suppose $k \in I: k \ne ... | Cartesian Product of Subsets/Family of Nonempty Subsets | https://proofwiki.org/wiki/Cartesian_Product_of_Subsets/Family_of_Nonempty_Subsets | https://proofwiki.org/wiki/Cartesian_Product_of_Subsets/Family_of_Nonempty_Subsets | [
"Cartesian Product of Subsets",
"Indexed Families"
] | [] | [
"Cartesian Product of Subsets/Family of Subsets",
"Axiom:Axiom of Choice",
"Definition:Cartesian Product",
"Definition:Element",
"Definition:Element",
"Category:Cartesian Product of Subsets",
"Category:Indexed Families"
] |
proofwiki-16897 | Projection is Injection iff Factor is Singleton/Family of Sets | Let $\family {S_i}_{i \mathop \in I}$ be a non-empty family of non-empty sets where $I$ is an arbitrary index set.
Let $S = \ds \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.
Let $\pr_j: S \to S_j$ be the $j$th projection on $S$.
Then $\pr_j$ is an injection {{iff}} $S_i$ is... | === Sufficient Condition ===
{{:Projection is Injection iff Factor is Singleton/Family of Sets/Sufficient Condition}}{{qed|lemma}} | Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Indexed Family of Sets|family]] of [[Definition:Non-Empty Set|non-empty sets]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $S = \ds \prod_{i \mathop \in I} S_i$ be the [[Definition:Cartesian Product ... | === [[Projection is Injection iff Factor is Singleton/Family of Sets/Sufficient Condition|Sufficient Condition]] ===
{{:Projection is Injection iff Factor is Singleton/Family of Sets/Sufficient Condition}}{{qed|lemma}} | Projection is Injection iff Factor is Singleton/Family of Sets | https://proofwiki.org/wiki/Projection_is_Injection_iff_Factor_is_Singleton/Family_of_Sets | https://proofwiki.org/wiki/Projection_is_Injection_iff_Factor_is_Singleton/Family_of_Sets | [
"Projection is Injection iff Factor is Singleton"
] | [
"Definition:Non-Empty Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Non-Empty Set",
"Definition:Indexing Set",
"Definition:Cartesian Product of Family ",
"Definition:Projection (Mapping Theory)",
"Definition:Injection",
"Definition:Singleton"
] | [
"Projection is Injection iff Factor is Singleton/Family of Sets/Sufficient Condition"
] |
proofwiki-16898 | Progressing Function Lemma | Let $A$ be a class.
Let $g$ be a progressing mapping on $A$.
Let $\RR$ be the relation defined as:
:$\map \RR {x, y} \iff \map g x \subseteq y \lor y \subseteq x$
where $\lor$ denotes disjunction (inclusive "or").
Then:
{{begin-axiom}}
{{axiom | n = 1
| q = \forall y \in \Dom g
| ml= \map \RR {y, \O}
}}... | In the following, let $x$ and $y$ be arbitrary elements of $A$ in the domain of $g$.
From Empty Set is Subset of All Sets:
:$\O \subseteq y$
Thus by the Rule of Addition:
:$\map g y \subseteq \O \lor \O \subseteq y$
and so it is seen that $\map \RR {y, \O}$.
That is, $(1)$ holds.
{{qed|lemma}}
Let $\map \RR {x, y}$ and... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $g$ be a [[Definition:Progressing Mapping|progressing mapping]] on $A$.
Let $\RR$ be the [[Definition:Relation (Class Theory)|relation]] defined as:
:$\map \RR {x, y} \iff \map g x \subseteq y \lor y \subseteq x$
where $\lor$ denotes [[Definition:Disjunctio... | In the following, let $x$ and $y$ be arbitrary [[Definition:Element of Class|elements]] of $A$ in the [[Definition:Domain of Mapping|domain]] of $g$.
From [[Empty Set is Subset of All Sets]]:
:$\O \subseteq y$
Thus by the [[Rule of Addition]]:
:$\map g y \subseteq \O \lor \O \subseteq y$
and so it is seen that $\map... | Progressing Function Lemma | https://proofwiki.org/wiki/Progressing_Function_Lemma | https://proofwiki.org/wiki/Progressing_Function_Lemma | [
"Progressing Mappings",
"Named Theorems"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:Relation/Class Theory",
"Definition:Disjunction",
"Definition:Conjunction"
] | [
"Definition:Element/Class",
"Definition:Domain (Set Theory)/Mapping",
"Empty Set is Subset of All Sets",
"Rule of Addition",
"Definition:Progressing Mapping",
"Definition:Set Equality",
"Definition:Set Equality"
] |
proofwiki-16899 | Minimally Inductive Class under Progressing Mapping induces Nest | Let $M$ be a class which is minimally inductive under a progressing mapping $g$.
Then $M$ is a nest in which:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$ | Let $\RR$ be the relation on $M$ defined as:
:$\forall x, y \in M: \map \RR {x, y} \iff \map g x \subseteq y \lor y \subseteq x$
We are given that $g$ is a progressing mapping.
From the Progressing Function Lemma, we have that:
{{begin-axiom}}
{{axiom | n = 1
| q = \forall y \in \Dom g
| ml= \map \RR {y... | Let $M$ be a [[Definition:Class (Class Theory)|class]] which is [[Definition:Minimally Inductive Class under General Mapping|minimally inductive]] under a [[Definition:Progressing Mapping|progressing mapping]] $g$.
Then $M$ is a [[Definition:Nest (Class Theory)|nest]] in which:
:$\forall x, y \in M: \map g x \subseteq... | Let $\RR$ be the [[Definition:Relation (Class Theory)|relation]] on $M$ defined as:
:$\forall x, y \in M: \map \RR {x, y} \iff \map g x \subseteq y \lor y \subseteq x$
We are given that $g$ is a [[Definition:Progressing Mapping|progressing mapping]].
From the [[Progressing Function Lemma]], we have that:
{{begin-a... | Minimally Inductive Class under Progressing Mapping induces Nest/Proof 1 | https://proofwiki.org/wiki/Minimally_Inductive_Class_under_Progressing_Mapping_induces_Nest | https://proofwiki.org/wiki/Minimally_Inductive_Class_under_Progressing_Mapping_induces_Nest/Proof_1 | [
"Progressing Mappings",
"Minimally Inductive Classes",
"Minimally Inductive Class under Progressing Mapping induces Nest"
] | [
"Definition:Class (Class Theory)",
"Definition:Minimally Inductive Class under General Mapping",
"Definition:Progressing Mapping",
"Definition:Nest/Class Theory"
] | [
"Definition:Relation/Class Theory",
"Definition:Progressing Mapping",
"Progressing Function Lemma",
"Double Induction Principle",
"Definition:Relation/Class Theory",
"Definition:Nest/Class Theory"
] |
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