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-7600 | Denumerable Class is Set | Let $A$ be a class.
Let $\N$ be the natural numbers.
Suppose that $F: \N \to A$ is a bijection.
Then $A$ is a set. | By the Axiom of Infinity, $\N$ is a set.
Thus by the Axiom of Replacement, $A$ is also a set.
{{qed}}
Category:Set Theory
5hmyb1p303uf7bgmuq7hv5xmn9nuw12 | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Suppose that $F: \N \to A$ is a [[Definition:bijection|bijection]].
Then $A$ is a [[Definition:set|set]]. | By the [[Axiom:Axiom of Infinity|Axiom of Infinity]], $\N$ is a [[Definition:set|set]].
Thus by the [[Axiom:Axiom of Replacement|Axiom of Replacement]], $A$ is also a set.
{{qed}}
[[Category:Set Theory]]
5hmyb1p303uf7bgmuq7hv5xmn9nuw12 | Denumerable Class is Set | https://proofwiki.org/wiki/Denumerable_Class_is_Set | https://proofwiki.org/wiki/Denumerable_Class_is_Set | [
"Set Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Natural Numbers",
"Definition:bijection",
"Definition:set"
] | [
"Axiom:Axiom of Infinity",
"Definition:set",
"Axiom:Axiom of Replacement",
"Category:Set Theory"
] |
proofwiki-7601 | Relative Complement inverts Subsets | Let $S$ be a set.
Let $A \subseteq S, B \subseteq S$ be subsets of $S$.
Then:
:$A \subseteq B \iff \relcomp S B \subseteq \relcomp S A$
where $\complement_S$ denotes the complement relative to $S$. | {{begin-eqn}}
{{eqn | l = A
| o = \subseteq
| r = B
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = A \cap B
| r = A
| c = Intersection with Subset is Subset
}}
{{eqn | ll= \leadstoandfrom
| l = \relcomp S {A \cap B}
| r = \relcomp S A
| c = Relative Complement of Relat... | Let $S$ be a [[Definition:Set|set]].
Let $A \subseteq S, B \subseteq S$ be [[Definition:Subset|subsets]] of $S$.
Then:
:$A \subseteq B \iff \relcomp S B \subseteq \relcomp S A$
where $\complement_S$ denotes the [[Definition:Relative Complement|complement relative to $S$]]. | {{begin-eqn}}
{{eqn | l = A
| o = \subseteq
| r = B
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = A \cap B
| r = A
| c = [[Intersection with Subset is Subset]]
}}
{{eqn | ll= \leadstoandfrom
| l = \relcomp S {A \cap B}
| r = \relcomp S A
| c = [[Relative Complement of... | Relative Complement inverts Subsets/Proof 1 | https://proofwiki.org/wiki/Relative_Complement_inverts_Subsets | https://proofwiki.org/wiki/Relative_Complement_inverts_Subsets/Proof_1 | [
"Relative Complement inverts Subsets",
"Relative Complement",
"Subsets"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Relative Complement"
] | [
"Intersection with Subset is Subset",
"Relative Complement of Relative Complement",
"De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection",
"Union with Superset is Superset"
] |
proofwiki-7602 | Union of Subsets is Subset/Family of Sets | Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
where $\ds \bigcup_{i \mathop \in I} S_i$ is the union of $\family {S_i}$. | Suppose that $\forall i \in I: S_i \subseteq X$.
Consider any $\ds x \in \bigcup_{i \mathop \in I} S_i$.
By definition of set union:
:$\exists i \in I: x \in S_i$
But as $S_i \subseteq X$ it follows that $x \in X$.
Thus it follows that:
:$\ds \bigcup_{i \mathop \in I} S_i \subseteq X$
So:
:$\ds \paren {\forall i \in I:... | Let $\family {S_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets indexed by $I$]].
Then for all [[Definition:Set|sets]] $X$:
:$\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
where $\ds \bigcup_{i \mathop \in I} S_i$ is the [[Definition:Un... | Suppose that $\forall i \in I: S_i \subseteq X$.
Consider any $\ds x \in \bigcup_{i \mathop \in I} S_i$.
By definition of [[Definition:Union of Family|set union]]:
:$\exists i \in I: x \in S_i$
But as $S_i \subseteq X$ it follows that $x \in X$.
Thus it follows that:
:$\ds \bigcup_{i \mathop \in I} S_i \subseteq X$... | Union of Subsets is Subset/Family of Sets | https://proofwiki.org/wiki/Union_of_Subsets_is_Subset/Family_of_Sets | https://proofwiki.org/wiki/Union_of_Subsets_is_Subset/Family_of_Sets | [
"Union of Subsets is Subset"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set",
"Definition:Set Union/Family of Sets"
] | [
"Definition:Set Union/Family of Sets"
] |
proofwiki-7603 | Set Union is Self-Distributive/Sets of Sets | Let $A$ and $B$ denote sets of sets.
Then:
:$\ds \bigcup \paren {A \cup B} = \paren {\bigcup A} \cup \paren {\bigcup B}$
where $\ds \bigcup A$ denotes the union of $A$. | Let $\ds s \in \bigcup \paren {A \cup B}$.
Then by definition of union of set of sets:
:$\exists X \in A \cup B: s \in X$
By definition of set union, either:
:$X \in A$
or:
:$X \in B$
If $X \in A$, then:
:$s \in \set {x: \exists X \in A: x \in X}$
If $X \in B$, then:
:$s \in \set {x: \exists X \in B: x \in X}$
Thus by ... | Let $A$ and $B$ denote [[Definition:Set of Sets|sets of sets]].
Then:
:$\ds \bigcup \paren {A \cup B} = \paren {\bigcup A} \cup \paren {\bigcup B}$
where $\ds \bigcup A$ denotes the [[Definition:Union of Set of Sets|union of $A$]]. | Let $\ds s \in \bigcup \paren {A \cup B}$.
Then by definition of [[Definition:Union of Set of Sets|union of set of sets]]:
:$\exists X \in A \cup B: s \in X$
By definition of [[Definition:Set Union|set union]], either:
:$X \in A$
or:
:$X \in B$
If $X \in A$, then:
:$s \in \set {x: \exists X \in A: x \in X}$
If $X \... | Set Union is Self-Distributive/Sets of Sets | https://proofwiki.org/wiki/Set_Union_is_Self-Distributive/Sets_of_Sets | https://proofwiki.org/wiki/Set_Union_is_Self-Distributive/Sets_of_Sets | [
"Set Union is Self-Distributive"
] | [
"Definition:Set of Sets",
"Definition:Set Union/Set of Sets"
] | [
"Definition:Set Union/Set of Sets",
"Definition:Set Union",
"Definition:Set Union/Set of Sets",
"Definition:Set Union",
"Definition:Subset",
"Definition:Set Union",
"Definition:Set Union/Set of Sets",
"Set is Subset of Union",
"Definition:Subset",
"Definition:Set Equality",
"Category:Set Union i... |
proofwiki-7604 | Set Union is Self-Distributive/Families of Sets | Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Then:
:$\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cup \paren {\bigcup_{\a... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha}
| c =
}}
{{eqn | ll= \leadsto
| q = \exists \beta \in I
| l = x
| o = \in
| r = A_\beta \cup B_\beta
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadsto
... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be [[Definition:Indexed Family of Subsets|indexed families of subsets]] of a [[Definition:Set|set]] $S$.
Then:
:$\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cup B_\alpha}
| c =
}}
{{eqn | ll= \leadsto
| q = \exists \beta \in I
| l = x
| o = \in
| r = A_\beta \cup B_\beta
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadsto
... | Set Union is Self-Distributive/Families of Sets | https://proofwiki.org/wiki/Set_Union_is_Self-Distributive/Families_of_Sets | https://proofwiki.org/wiki/Set_Union_is_Self-Distributive/Families_of_Sets | [
"Set Union is Self-Distributive",
"Indexed Families"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Set",
"Definition:Set Union/Family of Sets"
] | [
"Set is Subset of Union/Family of Sets",
"Set is Subset of Union/Family of Sets",
"Definition:Subset",
"Definition:Subset",
"Definition:Set Equality/Definition 2"
] |
proofwiki-7605 | Set Intersection is Self-Distributive/Families of Sets | Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Then:
:$\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\a... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha}
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
| l = x
| o = \in
| r = A_\alpha \cap B_\alpha
| c = {{Defof|Intersection of Family}}
}}
{{eqn | ll= \lea... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be [[Definition:Indexed Family of Subsets|indexed families of subsets]] of a [[Definition:Set|set]] $S$.
Then:
:$\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha}
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
| l = x
| o = \in
| r = A_\alpha \cap B_\alpha
| c = {{Defof|Intersection of Family}}
}}
{{eqn | ll= \lea... | Set Intersection is Self-Distributive/Families of Sets | https://proofwiki.org/wiki/Set_Intersection_is_Self-Distributive/Families_of_Sets | https://proofwiki.org/wiki/Set_Intersection_is_Self-Distributive/Families_of_Sets | [
"Set Intersection is Self-Distributive",
"Indexed Families"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Set",
"Definition:Set Intersection/Family of Sets"
] | [
"Definition:Subset",
"Definition:Subset",
"Definition:Set Equality/Definition 2"
] |
proofwiki-7606 | Set Intersection is Self-Distributive/General Result | Let $\family {\mathbb S_i} _{i \mathop \in I}$ be an $I$-indexed family of sets of sets.
Then:
:$\ds \bigcap_{i \mathop \in I} \bigcap \mathbb S_i = \bigcap \bigcap_{i \mathop \in I} \mathbb S_i$ | {{proof wanted}}
Category:Set Intersection is Self-Distributive
f94t9xyxs3xx1tcjhcxqv6dbdajy8fj | Let $\family {\mathbb S_i} _{i \mathop \in I}$ be an [[Definition:Indexed Family of Sets|$I$-indexed family]] of [[Definition:Set of Sets|sets of sets]].
Then:
:$\ds \bigcap_{i \mathop \in I} \bigcap \mathbb S_i = \bigcap \bigcap_{i \mathop \in I} \mathbb S_i$ | {{proof wanted}}
[[Category:Set Intersection is Self-Distributive]]
f94t9xyxs3xx1tcjhcxqv6dbdajy8fj | Set Intersection is Self-Distributive/General Result | https://proofwiki.org/wiki/Set_Intersection_is_Self-Distributive/General_Result | https://proofwiki.org/wiki/Set_Intersection_is_Self-Distributive/General_Result | [
"Set Intersection is Self-Distributive"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set of Sets"
] | [
"Category:Set Intersection is Self-Distributive"
] |
proofwiki-7607 | Intersection Distributes over Union/Family of Sets | Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be a indexed family of subsets of a set $S$.
Let $B \subseteq S$.
Then:
:$\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$
where $\ds \bigcup_{\alpha \mathop \in I} A_\alpha... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap B}
| c =
}}
{{eqn | ll= \leadsto
| q = \exists \alpha \in I
| l = x
| o = \in
| r = A_\alpha \cap B
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadsto
| l = x
... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be a [[Definition:Indexed Family of Subsets|indexed family of subsets]] of a [[Definition:Set|set]] $S$.
Let $B \subseteq S$.
Then:
:$\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B} = \paren {\bigcup... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap B}
| c =
}}
{{eqn | ll= \leadsto
| q = \exists \alpha \in I
| l = x
| o = \in
| r = A_\alpha \cap B
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadsto
| l = x
... | Intersection Distributes over Union/Family of Sets | https://proofwiki.org/wiki/Intersection_Distributes_over_Union/Family_of_Sets | https://proofwiki.org/wiki/Intersection_Distributes_over_Union/Family_of_Sets | [
"Intersection Distributes over Union",
"Indexed Families"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Set",
"Definition:Set Union/Family of Sets"
] | [
"Set is Subset of Union/Family of Sets",
"Definition:Subset",
"Set is Subset of Union/Family of Sets",
"Definition:Subset",
"Definition:Set Equality/Definition 2"
] |
proofwiki-7608 | Union Distributes over Intersection/Family of Sets | Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $B \subseteq S$.
Then:
:$\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cup B$
where $\ds \bigcap_{\alpha \mathop \in I} A_\alph... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B}
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
| l = x
| o = \in
| r = A_\alpha \cup B
| c = Intersection is Subset
}}
{{eqn | ll= \leadsto
| q = \forall \... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family of subsets]] of a [[Definition:Set|set]] $S$.
Let $B \subseteq S$.
Then:
:$\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B} = \paren {\bigca... | {{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cup B}
| c =
}}
{{eqn | ll= \leadsto
| q = \forall \alpha \in I
| l = x
| o = \in
| r = A_\alpha \cup B
| c = [[Intersection is Subset/Family of Sets|Intersection is Subset]]
}}
... | Union Distributes over Intersection/Family of Sets | https://proofwiki.org/wiki/Union_Distributes_over_Intersection/Family_of_Sets | https://proofwiki.org/wiki/Union_Distributes_over_Intersection/Family_of_Sets | [
"Union Distributes over Intersection",
"Indexed Families"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family of Subsets",
"Definition:Set",
"Definition:Set Intersection/Family of Sets"
] | [
"Intersection is Subset/Family of Sets",
"Definition:Subset",
"Intersection is Subset/Family of Sets",
"Definition:Subset",
"Definition:Set Equality/Definition 2"
] |
proofwiki-7609 | Equivalence of Definitions of Well-Ordering | {{TFAE|def = Well-Ordering}}
Let $\struct {S, \preceq}$ be a ordered set. | === Definition 1 implies Definition 2 ===
{{:Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2}}
By hypothesis, every subset of $S$ has a smallest element.
By Smallest Element is Minimal it follows that every subset of $S$ has a minimal element.
Thus it follows that $\preceq$ is a well-order... | {{TFAE|def = Well-Ordering}}
Let $\struct {S, \preceq}$ be a [[Definition:Ordered Set|ordered set]]. | === [[Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2|Definition 1 implies Definition 2]] ===
{{:Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2}}
[[Definition:By Hypothesis|By hypothesis]], every [[Definition:Subset|subset]] of $S$ has a [[Definition:Smalles... | Equivalence of Definitions of Well-Ordering | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Well-Ordering | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Well-Ordering | [
"Well-Orderings",
"Equivalence of Definitions of Well-Ordering"
] | [
"Definition:Ordered Set"
] | [
"Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2",
"Definition:By Hypothesis",
"Definition:Subset",
"Definition:Smallest Element",
"Smallest Element is Minimal",
"Definition:Subset",
"Definition:Minimal/Element",
"Definition:Well-Ordering",
"Definition:Well-Ordering/Def... |
proofwiki-7610 | Equivalence of Definitions of Transitive Closure of Relation/Finite Chain is Smallest | Let $S$ be a set or class.
Let $\RR$ be a relation on $S$.
Let $\RR^+$ be the transitive closure of $\RR$ by the finite chain definition.
That is, for $x, y \in S$ let $x \mathrel {\RR^+} y$ {{iff}} for some natural number $n > 0$ there exist $s_0, s_1, \dots, s_n \in S$ such that $s_0 = x$, $s_n = y$, and:
:$\forall k... | ==== $\RR^+$ is transitive ====
Let $x, y, z \in S$.
Let $x \mathrel {\RR^+} y$ and $y \mathrel {\RR^+} z$.
Then for some $m, n \in \N_{>0}$ there are $s_0, s_1, \dots, s_m$ and $t_0, t_1, \dots, t_n$ such that $s_0 = x$, $s_m = y$, $t_0 = y$, $t_n = z$, and the following hold:
:$\forall k \in \N_m: s_k \mathrel {\RR^+... | Let $S$ be a [[Definition:set|set]] or [[Definition:Class (Class Theory)|class]].
Let $\RR$ be a [[Definition:Endorelation|relation]] on $S$.
Let $\RR^+$ be the [[Definition:Transitive Closure of Relation|transitive closure]] of $\RR$ by the [[Definition:Transitive Closure of Relation/Finite Chain|finite chain defini... | ==== $\RR^+$ is [[Definition:Transitive Relation|transitive]] ====
Let $x, y, z \in S$.
Let $x \mathrel {\RR^+} y$ and $y \mathrel {\RR^+} z$.
Then for some $m, n \in \N_{>0}$ there are $s_0, s_1, \dots, s_m$ and $t_0, t_1, \dots, t_n$ such that $s_0 = x$, $s_m = y$, $t_0 = y$, $t_n = z$, and the following hold:
:$... | Equivalence of Definitions of Transitive Closure of Relation/Finite Chain is Smallest | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Transitive_Closure_of_Relation/Finite_Chain_is_Smallest | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Transitive_Closure_of_Relation/Finite_Chain_is_Smallest | [
"Equivalence of Definitions of Transitive Closure of Relation"
] | [
"Definition:set",
"Definition:Class (Class Theory)",
"Definition:Endorelation",
"Definition:Transitive Closure of Relation",
"Definition:Transitive Closure of Relation/Finite Chain",
"Definition:Natural Numbers",
"Definition:Transitive Relation",
"Definition:Transitive Relation"
] | [
"Definition:Transitive Relation",
"Definition:Transitive Relation",
"Definition:Transitive Relation",
"Definition:Transitive Relation"
] |
proofwiki-7611 | Order-Preserving Bijection on Wosets is Order Isomorphism | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be well-ordered sets.
Let $\phi: S \to T$ be a bijection such that $\phi: S \to T$ is order-preserving:
:$\forall x, y \in S: x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
Then:
:$\forall x, y \in S: \map \phi x \preceq_2 \map \phi y \implies x \prec... | A well-ordered set is a totally ordered set by definition.
A bijection is a surjection by definition.
The result follows from Order Isomorphism iff Strictly Increasing Surjection.
{{qed}}
Category:Well-Orderings
Category:Order Isomorphisms
bd9m2xcftcxo5aowtn8z62ncldgdsm8 | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Well-Ordered Set|well-ordered sets]].
Let $\phi: S \to T$ be a [[Definition:Bijection|bijection]] such that $\phi: S \to T$ is [[Definition:Order-Preserving|order-preserving]]:
:$\forall x, y \in S: x \preceq_1 y \implies \map \phi x \preceq_2 \... | A [[Definition:Well-Ordered Set|well-ordered set]] is a [[Definition:Totally Ordered Set|totally ordered set]] by definition.
A [[Definition:Bijection|bijection]] is a [[Definition:Surjection|surjection]] by definition.
The result follows from [[Order Isomorphism iff Strictly Increasing Surjection]].
{{qed}}
[[Categ... | Order-Preserving Bijection on Wosets is Order Isomorphism | https://proofwiki.org/wiki/Order-Preserving_Bijection_on_Wosets_is_Order_Isomorphism | https://proofwiki.org/wiki/Order-Preserving_Bijection_on_Wosets_is_Order_Isomorphism | [
"Well-Orderings",
"Order Isomorphisms"
] | [
"Definition:Well-Ordered Set",
"Definition:Bijection",
"Definition:Increasing",
"Definition:Order Isomorphism"
] | [
"Definition:Well-Ordered Set",
"Definition:Totally Ordered Set",
"Definition:Bijection",
"Definition:Surjection",
"Order Isomorphism iff Strictly Increasing Surjection",
"Category:Well-Orderings",
"Category:Order Isomorphisms"
] |
proofwiki-7612 | Inverse Image of Set under Set-Like Relation is Set | Let $A$ be a class.
Let $\RR$ be a set-like endorelation on $A$.
Let $B \subseteq A$ be a set.
Then $\map {\RR^{-1} } B$, the inverse image of $B$ under $\RR$, is also a set. | Since $\RR$ is set-like, $\map {\RR^{-1} } {\set x}$ is a set for each $x$ in $A$.
As $B \subseteq A$, this holds also for each $x \in B$.
{{explain|Explain better.}}
But then $\ds \map {\RR^{-1} } B = \bigcup_{x \mathop \in B} \map {\RR^{-1} } {\set x}$, which is a set by the Axiom of Unions.
{{qed}}
Category:Set Theo... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\RR$ be a [[Definition:Set-Like Relation|set-like]] [[Definition:Endorelation|endorelation]] on $A$.
Let $B \subseteq A$ be a [[Definition:set|set]].
Then $\map {\RR^{-1} } B$, the [[Definition:Inverse Image|inverse image]] of $B$ under $\RR$, is also a [... | Since $\RR$ is [[Definition:Set-Like Relation|set-like]], $\map {\RR^{-1} } {\set x}$ is a [[Definition:Set|set]] for each $x$ in $A$.
As $B \subseteq A$, this holds also for each $x \in B$.
{{explain|Explain better.}}
But then $\ds \map {\RR^{-1} } B = \bigcup_{x \mathop \in B} \map {\RR^{-1} } {\set x}$, which is a... | Inverse Image of Set under Set-Like Relation is Set | https://proofwiki.org/wiki/Inverse_Image_of_Set_under_Set-Like_Relation_is_Set | https://proofwiki.org/wiki/Inverse_Image_of_Set_under_Set-Like_Relation_is_Set | [
"Set Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Set-Like Relation",
"Definition:Endorelation",
"Definition:set",
"Definition:Inverse Image",
"Definition:set"
] | [
"Definition:Set-Like Relation",
"Definition:Set",
"Definition:Set",
"Axiom:Axiom of Unions/Class Theory",
"Category:Set Theory"
] |
proofwiki-7613 | Reciprocal of Holomorphic Function | Let $f: \C \to \C$ be a complex function.
Let $U \subseteq \C$ be an open set such that $f$ has no zeros in $U$.
Suppose further that $f$ is holomorphic in $U$.
Then the complex function
:$\dfrac 1 {f_{\restriction U} } : U \to \C$
is holomorphic. | Let $g: U \to \C$ be such that $\map g x = 1 / \map f x$.
Since $\map f x$ is nonzero for $x \in U$, $g$ is well-defined.
By Quotient Rule for Continuous Complex Functions, $g$ is continuous.
Let $z_0 \in U$.
As $g$ is continuous:
:$\ds \lim_{h \mathop \to 0} \frac 1 {\map f {z_0 + h} } = \frac 1 {\map f {z_0} }$
As $f... | Let $f: \C \to \C$ be a [[Definition:Complex Function|complex function]].
Let $U \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open set]] such that $f$ has no [[Definition:Root of Function|zeros]] in $U$.
Suppose further that $f$ is [[Definition:Holomorphic Function|holomorphic]] in $U$.
Then the [[D... | Let $g: U \to \C$ be such that $\map g x = 1 / \map f x$.
Since $\map f x$ is nonzero for $x \in U$, $g$ is well-defined.
By [[Quotient Rule for Continuous Complex Functions]], $g$ is [[Definition:Continuous Complex Function|continuous]].
Let $z_0 \in U$.
As $g$ is [[Definition:Continuous Complex Function|continuou... | Reciprocal of Holomorphic Function | https://proofwiki.org/wiki/Reciprocal_of_Holomorphic_Function | https://proofwiki.org/wiki/Reciprocal_of_Holomorphic_Function | [
"Holomorphic Functions",
"Reciprocals"
] | [
"Definition:Complex Function",
"Definition:Open Set/Complex Analysis",
"Definition:Root of Mapping",
"Definition:Holomorphic Function",
"Definition:Complex Function",
"Definition:Holomorphic Function"
] | [
"Combination Theorem for Continuous Functions/Complex/Quotient Rule",
"Definition:Continuous Complex Function",
"Definition:Continuous Complex Function",
"Definition:Holomorphic Function",
"Combination Theorem for Limits of Functions/Complex",
"Definition:Holomorphic Function",
"Category:Holomorphic Fun... |
proofwiki-7614 | Transitive Closure of Set-Like Relation is Set-Like | Let $A$ be a class.
Let $\RR$ be a set-like endorelation on $A$.
Let $\RR^+$ be the transitive closure of $\RR$.
Then $\RR^+$ is also a set-like relation. | Let $x \in A$.
Let $A'$ be the class of all subsets of $A$.
For each $s \in A'$, $\RR^{-1}$ is a subset of $A$.
Hence by Inverse Image of Set under Set-Like Relation is Set and the definition of endorelation:
:$\RR^{-1} \in A'$
Define a mapping $G: A' \to A'$ as:
:$\forall s \in A': \map G s = \map {\RR^{-1} } s$
Recur... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\RR$ be a [[Definition:Set-Like Relation|set-like]] [[Definition:Endorelation|endorelation]] on $A$.
Let $\RR^+$ be the [[Definition:Transitive Closure of Relation/Finite Chain|transitive closure]] of $\RR$.
Then $\RR^+$ is also a [[Definition:Set-Like Re... | Let $x \in A$.
Let $A'$ be the [[Definition:Class (Class Theory)|class]] of all [[Definition:Subset|subsets]] of $A$.
For each $s \in A'$, $\RR^{-1}$ is a [[Definition:Subset|subset]] of $A$.
Hence by [[Inverse Image of Set under Set-Like Relation is Set]] and the definition of [[Definition:Endorelation|endorelation... | Transitive Closure of Set-Like Relation is Set-Like | https://proofwiki.org/wiki/Transitive_Closure_of_Set-Like_Relation_is_Set-Like | https://proofwiki.org/wiki/Transitive_Closure_of_Set-Like_Relation_is_Set-Like | [
"Set Theory",
"Transitive Closures"
] | [
"Definition:Class (Class Theory)",
"Definition:Set-Like Relation",
"Definition:Endorelation",
"Definition:Transitive Closure of Relation/Finite Chain",
"Definition:Set-Like Relation"
] | [
"Definition:Class (Class Theory)",
"Definition:Subset",
"Definition:Subset",
"Inverse Image of Set under Set-Like Relation is Set",
"Definition:Endorelation",
"Definition:Mapping",
"Principle of Recursive Definition",
"Definition:Mapping",
"Definition:Set",
"Definition:Transitive Closure of Relati... |
proofwiki-7615 | Relational Closure from Transitive Closure | Let $A$ be a set or class.
Let $\RR$ be a relation on $A$.
Let $\RR^+$ be the transitive closure of $\RR$.
Let $B \subseteq A$.
Let $B' = B \cup \inv {\paren {\RR^+} } B$.
Let $C$ be an $\RR$-transitive subset or subclass of $A$ such that $B \subseteq C$.
Then:
:$B'$ is $\RR$-transitive
:$B' \subseteq C$
:If $B$ is a s... | === $B'$ is $\RR$-transitive ===
Let $x \in B'$ and $y \in A$, and let $y \mathrel \RR x$.
If $x \in B$, then by the definition of transitive closure:
:$y \mathrel {\RR^+} x$
so:
:$y \in B'$
Let $x \in \inv {\paren {\RR^+} } B$.
Then:
:$x \mathrel {\RR^+} b$
for some $b \in B$.
Since $\RR \subseteq \RR^+$, it follows t... | Let $A$ be a [[Definition:Set|set]] or [[Definition:Class (Class Theory)|class]].
Let $\RR$ be a [[Definition:Endorelation|relation]] on $A$.
Let $\RR^+$ be the [[Definition:Transitive Closure of Relation|transitive closure]] of $\RR$.
Let $B \subseteq A$.
Let $B' = B \cup \inv {\paren {\RR^+} } B$.
Let $C$ be an ... | === $B'$ is $\RR$-transitive ===
Let $x \in B'$ and $y \in A$, and let $y \mathrel \RR x$.
If $x \in B$, then by the definition of [[Definition:Transitive Closure of Relation|transitive closure]]:
:$y \mathrel {\RR^+} x$
so:
:$y \in B'$
Let $x \in \inv {\paren {\RR^+} } B$.
Then:
:$x \mathrel {\RR^+} b$
for some $b... | Relational Closure from Transitive Closure | https://proofwiki.org/wiki/Relational_Closure_from_Transitive_Closure | https://proofwiki.org/wiki/Relational_Closure_from_Transitive_Closure | [
"Relational Closures"
] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Endorelation",
"Definition:Transitive Closure of Relation",
"Definition:Transitive with Respect to a Relation",
"Definition:Subset",
"Definition:Subclass",
"Definition:Transitive with Respect to a Relation",
"Definition:Set",
"Defini... | [
"Definition:Transitive Closure of Relation",
"Definition:Transitive Relation",
"Definition:Transitive with Respect to a Relation"
] |
proofwiki-7616 | Minimal WRT Restriction | Let $A$ be a set or class.
Let $\RR$ be a relation on $A$.
Let $B$ be a subset or subclass of $A$.
Let $\RR'$ be the restriction of $\RR$ to $B$.
Let $m \in B$.
Then:
:$m$ is a strictly minimal element under $\RR$ in $B$
{{iff}}:
:$m$ is a strictly minimal element under $\RR'$ in $B$. | === Sufficient Condition ===
Let $m$ be a strictly minimal element under $\RR$ in $B$.
Let $x$ be any element of $B$.
{{AimForCont}} that $x \mathrel {\RR'} m$.
Then since $\RR' \subseteq \RR$:
:$x \mathrel \RR m$
contradicting the fact that $m$ is a strictly minimal element under $\RR$ in $B$.
Thus:
:$\lnot \paren {x ... | Let $A$ be a [[Definition:Set|set]] or [[Definition:Class (Class Theory)|class]].
Let $\RR$ be a [[Definition:Endorelation|relation]] on $A$.
Let $B$ be a [[Definition:Subset|subset]] or [[Definition:Subclass|subclass]] of $A$.
Let $\RR'$ be the [[Definition:Restriction|restriction]] of $\RR$ to $B$.
Let $m \in B$.... | === Sufficient Condition ===
Let $m$ be a [[Definition:Strictly Minimal Element|strictly minimal element]] under $\RR$ in $B$.
Let $x$ be any [[Definition:Element|element]] of $B$.
{{AimForCont}} that $x \mathrel {\RR'} m$.
Then since $\RR' \subseteq \RR$:
:$x \mathrel \RR m$
contradicting the fact that $m$ is a [... | Minimal WRT Restriction | https://proofwiki.org/wiki/Minimal_WRT_Restriction | https://proofwiki.org/wiki/Minimal_WRT_Restriction | [
"Restrictions"
] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Endorelation",
"Definition:Subset",
"Definition:Subclass",
"Definition:Restriction",
"Definition:Strictly Minimal Element",
"Definition:Strictly Minimal Element"
] | [
"Definition:Strictly Minimal Element",
"Definition:Element",
"Definition:Strictly Minimal Element",
"Definition:Strictly Minimal Element",
"Definition:Strictly Minimal Element",
"Definition:Strictly Minimal Element",
"Definition:Strictly Minimal Element"
] |
proofwiki-7617 | Intersection of Ordinals is Ordinal | Let $A$ be a non-empty class of ordinals.
Then its intersection $\bigcap A$ is an ordinal. | Let $\alpha = \bigcap A$.
It will be demonstrated that $\alpha$ is an ordinal according to Definition $2$:
{{:Definition:Ordinal/Definition 2}} | Let $A$ be a [[Definition:Non-Empty Class|non-empty]] [[Definition:Class (Class Theory)|class]] of [[Definition:Ordinal|ordinals]].
Then its [[Definition:Intersection of Class|intersection]] $\bigcap A$ is an [[Definition:Ordinal|ordinal]]. | Let $\alpha = \bigcap A$.
It will be demonstrated that $\alpha$ is an [[Definition:Ordinal/Definition 2|ordinal according to Definition $2$]]:
{{:Definition:Ordinal/Definition 2}} | Intersection of Ordinals is Ordinal | https://proofwiki.org/wiki/Intersection_of_Ordinals_is_Ordinal | https://proofwiki.org/wiki/Intersection_of_Ordinals_is_Ordinal | [
"Ordinals",
"Set Intersection"
] | [
"Definition:Non-Empty Set/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Ordinal",
"Definition:Class Intersection/Class of Sets",
"Definition:Ordinal"
] | [
"Definition:Ordinal/Definition 2",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Ordinal"
] |
proofwiki-7618 | Meet with Complement is Bottom | Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 2.
Then:
:$\exists \bot \in S: \forall a \in S: a \wedge \neg a = \bot$
where $\wedge$ denotes the meet operation in $S$.
This element $\bot$ is unique for any given $S$, and is named '''bottom'''. | Let $\exists r, s \in S: r \wedge \neg r = a, \ s \wedge \neg s = b$
Then:
{{begin-eqn}}
{{eqn | l = a
| r = r \wedge \neg r
| c = {{hypothesis}}
}}
{{eqn | r = \paren {s \wedge \neg s} \vee \paren {r \wedge \neg r}
| c = {{Boolean-algebra-axiom|2|5}}
}}
{{eqn | r = \paren {r \wedge \neg r} \vee \pare... | Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra/Definition 2|Boolean algebra, defined as in Definition 2]].
Then:
:$\exists \bot \in S: \forall a \in S: a \wedge \neg a = \bot$
where $\wedge$ denotes the [[Definition:Boolean Algebra|meet operation in $S$]].
This element $\bot$ is [[Definitio... | Let $\exists r, s \in S: r \wedge \neg r = a, \ s \wedge \neg s = b$
Then:
{{begin-eqn}}
{{eqn | l = a
| r = r \wedge \neg r
| c = {{hypothesis}}
}}
{{eqn | r = \paren {s \wedge \neg s} \vee \paren {r \wedge \neg r}
| c = {{Boolean-algebra-axiom|2|5}}
}}
{{eqn | r = \paren {r \wedge \neg r} \vee \pa... | Meet with Complement is Bottom | https://proofwiki.org/wiki/Meet_with_Complement_is_Bottom | https://proofwiki.org/wiki/Meet_with_Complement_is_Bottom | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra/Definition 2",
"Definition:Boolean Algebra",
"Definition:Unique"
] | [
"Definition:Unique",
"Definition:Symbol"
] |
proofwiki-7619 | Join with Complement is Top | Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 2.
Then:
:$\exists \top \in S: \forall a \in S: a \vee \neg a = \top$
where $\vee$ denotes the join operation in $S$.
This element $\top$ is unique for any given $S$, and is named '''top'''. | Let $\exists r, s \in S: r \vee \neg r = a, \ s \vee \neg s = b$
Then:
{{begin-eqn}}
{{eqn | l = a
| r = r \vee \neg r
| c = {{hypothesis}}
}}
{{eqn | r = \paren {s \vee \neg s} \wedge \paren {r \vee \neg r}
| c = {{Boolean-algebra-axiom|2|5}}
}}
{{eqn | r = \paren {r \vee \neg r} \wedge \paren {s \ve... | Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra/Definition 2|Boolean algebra, defined as in Definition 2]].
Then:
:$\exists \top \in S: \forall a \in S: a \vee \neg a = \top$
where $\vee$ denotes the [[Definition:Boolean Algebra|join operation in $S$]].
This element $\top$ is [[Definition:Un... | Let $\exists r, s \in S: r \vee \neg r = a, \ s \vee \neg s = b$
Then:
{{begin-eqn}}
{{eqn | l = a
| r = r \vee \neg r
| c = {{hypothesis}}
}}
{{eqn | r = \paren {s \vee \neg s} \wedge \paren {r \vee \neg r}
| c = {{Boolean-algebra-axiom|2|5}}
}}
{{eqn | r = \paren {r \vee \neg r} \wedge \paren {s \... | Join with Complement is Top | https://proofwiki.org/wiki/Join_with_Complement_is_Top | https://proofwiki.org/wiki/Join_with_Complement_is_Top | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra/Definition 2",
"Definition:Boolean Algebra",
"Definition:Unique"
] | [
"Definition:Unique",
"Definition:Symbol"
] |
proofwiki-7620 | Equivalence of Definitions of Order Isomorphism | {{TFAE|def = Order Isomorphism}}
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets. | === Definition 1 implies Definition 2 ===
Let $\phi: S \to T$ be an order isomorphism by Definition 1.
Then $\phi$ is bijective, and thus trivially surjective.
Let $x, y \in S$.
Then by Definition 1:
:$x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
Suppose that $\map \phi x \preceq_2 \map \phi y$.
Then by De... | {{TFAE|def = Order Isomorphism}}
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Ordered Set|ordered sets]]. | === Definition 1 implies Definition 2 ===
Let $\phi: S \to T$ be an order isomorphism by [[Definition:Order Isomorphism/Definition 1|Definition 1]].
Then $\phi$ is [[Definition:Bijection|bijective]], and thus trivially [[Definition:Surjection|surjective]].
Let $x, y \in S$.
Then by [[Definition:Order Isomorphism/De... | Equivalence of Definitions of Order Isomorphism | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Order_Isomorphism | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Order_Isomorphism | [
"Order Isomorphisms"
] | [
"Definition:Ordered Set"
] | [
"Definition:Order Isomorphism/Definition 1",
"Definition:Bijection",
"Definition:Surjection",
"Definition:Order Isomorphism/Definition 1",
"Definition:Order Isomorphism/Definition 1",
"Definition:Inverse of Mapping",
"Rule of Implication",
"Definition:Surjection",
"Definition:Order Isomorphism/Defin... |
proofwiki-7621 | Ordering is Equivalent to Subset Relation/Lemma | Let $\struct {S, \preceq}$ be an ordered set.
Then:
:$\forall a_1, a_2 \in S: \paren {a_1 \preceq a_2 \implies {a_1}^\preceq \subseteq {a_2}^\preceq}$
where ${a_1}^\preceq$ denotes the lower closure of $a_1$. | Let $a_1 \preceq a_2$.
Then by the definition of lower closure:
:$a_1 \in {a_2}^\preceq$
Let $a_3 \in {a_1}^\preceq$.
Then by definition:
:$a_3 \preceq a_1$
As an ordering is transitive, it follows that:
:$a_3 \preceq a_2$
and so:
:$a_3 \in {a_2}^\preceq$
This holds for all $a_3 \in {a_1}^\preceq$.
Thus by definition o... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then:
:$\forall a_1, a_2 \in S: \paren {a_1 \preceq a_2 \implies {a_1}^\preceq \subseteq {a_2}^\preceq}$
where ${a_1}^\preceq$ denotes the [[Definition:Lower Closure of Element|lower closure]] of $a_1$. | Let $a_1 \preceq a_2$.
Then by the definition of [[Definition:Lower Closure of Element|lower closure]]:
:$a_1 \in {a_2}^\preceq$
Let $a_3 \in {a_1}^\preceq$.
Then by definition:
:$a_3 \preceq a_1$
As an [[Definition:Ordering|ordering]] is [[Definition:Transitive Relation|transitive]], it follows that:
:$a_3 \prec... | Ordering is Equivalent to Subset Relation/Lemma | https://proofwiki.org/wiki/Ordering_is_Equivalent_to_Subset_Relation/Lemma | https://proofwiki.org/wiki/Ordering_is_Equivalent_to_Subset_Relation/Lemma | [
"Ordering is Equivalent to Subset Relation"
] | [
"Definition:Ordered Set",
"Definition:Lower Closure/Element"
] | [
"Definition:Lower Closure/Element",
"Definition:Ordering",
"Definition:Transitive Relation",
"Definition:Subset",
"Category:Ordering is Equivalent to Subset Relation"
] |
proofwiki-7622 | Smallest Element WRT Restricted Ordering | Let $S$ be a set or class.
Let $\preceq$ be an ordering on $S$.
Let $T$ be a subset or subclass of $S$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Let $m \in T$.
Then $m$ is the $\preceq$-smallest element of $T$ {{iff}} $m$ is the $\preceq'$-smallest element of $T$. | {{proof wanted|The same sort of utterly trivial thing as at Minimal WRT Restriction}}
Category:Smallest Elements
pk3bjqe63mj3zrgpqamspxwmg08vd65 | Let $S$ be a [[Definition:Set|set]] or [[Definition:Class (Class Theory)|class]].
Let $\preceq$ be an [[Definition:Ordering|ordering]] on $S$.
Let $T$ be a [[Definition:Subset|subset]] or [[Definition:Subclass|subclass]] of $S$.
Let $\preceq'$ be the [[Definition:Restriction|restriction]] of $\preceq$ to $T$.
Let $... | {{proof wanted|The same sort of utterly trivial thing as at Minimal WRT Restriction}}
[[Category:Smallest Elements]]
pk3bjqe63mj3zrgpqamspxwmg08vd65 | Smallest Element WRT Restricted Ordering | https://proofwiki.org/wiki/Smallest_Element_WRT_Restricted_Ordering | https://proofwiki.org/wiki/Smallest_Element_WRT_Restricted_Ordering | [
"Smallest Elements"
] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Ordering",
"Definition:Subset",
"Definition:Subclass",
"Definition:Restriction",
"Definition:Smallest Element"
] | [
"Category:Smallest Elements"
] |
proofwiki-7623 | Restriction to Subset of Strict Total Ordering is Strict Total Ordering | Let $S$ be a set or class.
Let $\prec$ be a strict total ordering on $A$.
Let $T$ be a subset or subclass of $A$.
Then the restriction of $\prec$ to $B$ is a strict total ordering of $B$. | Follows from:
:Restriction of Transitive Relation is Transitive
:Restriction of Antireflexive Relation is Antireflexive
:Restriction of Connected Relation is Connected
{{qed}}
Category:Total Orderings
i8o4agmo3vegz3a8i54osr3ngelqga6 | Let $S$ be a [[Definition:Set|set]] or [[Definition:Class (Class Theory)|class]].
Let $\prec$ be a [[Definition:Strict Total Ordering|strict total ordering]] on $A$.
Let $T$ be a [[Definition:Subset|subset]] or [[Definition:Subclass|subclass]] of $A$.
Then the [[Definition:Restriction of Relation|restriction]] of $... | Follows from:
:[[Restriction of Transitive Relation is Transitive]]
:[[Restriction of Antireflexive Relation is Antireflexive]]
:[[Restriction of Connected Relation is Connected]]
{{qed}}
[[Category:Total Orderings]]
i8o4agmo3vegz3a8i54osr3ngelqga6 | Restriction to Subset of Strict Total Ordering is Strict Total Ordering | https://proofwiki.org/wiki/Restriction_to_Subset_of_Strict_Total_Ordering_is_Strict_Total_Ordering | https://proofwiki.org/wiki/Restriction_to_Subset_of_Strict_Total_Ordering_is_Strict_Total_Ordering | [
"Total Orderings"
] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Strict Total Ordering",
"Definition:Subset",
"Definition:Subclass",
"Definition:Restriction/Relation",
"Definition:Strict Total Ordering"
] | [
"Restriction of Transitive Relation is Transitive",
"Restriction of Antireflexive Relation is Antireflexive",
"Restriction of Connected Relation is Connected",
"Category:Total Orderings"
] |
proofwiki-7624 | Restriction of Well-Founded Ordering is Well-Founded | Let $S$ be a set or class.
Let $T$ be a subset or subclass of $S$.
Let $\preceq$ be a well-founded ordering of $A$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Then $\preceq'$ is a well-founded ordering of $T$. | By Restriction of Ordering is Ordering, $\preceq'$ is an ordering.
By Restriction of Well-Founded Relation is Well-Founded, $\preceq'$ is a well-founded relation on $T$.
Hence the result by definition of well-founded ordering.
{{qed}}
Category:Order Theory
Category:Well-Founded Relations
gt3s509f9og72gfyqzi18hpu9x2ezcf | Let $S$ be a [[Definition:Set|set]] or [[Definition:Class (Class Theory)|class]].
Let $T$ be a [[Definition:Subset|subset]] or [[Definition:Subclass|subclass]] of $S$.
Let $\preceq$ be a [[Definition:Well-Founded Ordering|well-founded ordering]] of $A$.
Let $\preceq'$ be the [[Definition:Restriction of Ordering|rest... | By [[Restriction of Ordering is Ordering]], $\preceq'$ is an [[Definition:Ordering|ordering]].
By [[Restriction of Well-Founded Relation is Well-Founded]], $\preceq'$ is a [[Definition:Well-Founded Relation|well-founded relation]] on $T$.
Hence the result by definition of [[Definition:Well-Founded Ordering|well-found... | Restriction of Well-Founded Ordering is Well-Founded | https://proofwiki.org/wiki/Restriction_of_Well-Founded_Ordering_is_Well-Founded | https://proofwiki.org/wiki/Restriction_of_Well-Founded_Ordering_is_Well-Founded | [
"Order Theory",
"Well-Founded Relations"
] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Subset",
"Definition:Subclass",
"Definition:Well-Founded Ordered Set",
"Definition:Restriction of Ordering",
"Definition:Well-Founded Ordered Set"
] | [
"Restriction of Ordering is Ordering",
"Definition:Ordering",
"Restriction of Well-Founded Relation is Well-Founded",
"Definition:Well-Founded Relation",
"Definition:Well-Founded Ordered Set",
"Category:Order Theory",
"Category:Well-Founded Relations"
] |
proofwiki-7625 | Restriction of Well-Ordering is Well-Ordering | Let $S$ be a set or class.
Let $\preceq$ be a well-ordering of $S$.
Let $T$ be a subset or subclass of $S$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Then $\preceq'$ is a well-ordering of $T$. | By the definition of well-ordering, $\preceq$ is a well-founded total ordering.
By Restriction of Total Ordering is Total Ordering, $\preceq'$ is a total ordering.
By Restriction of Well-Founded Ordering is Well-Founded, $\preceq'$ is a well-founded ordering.
Thus $\preceq'$ is a well-ordering.
{{qed}}
Category:Well-Or... | Let $S$ be a [[Definition:set|set]] or [[Definition:Class (Class Theory)|class]].
Let $\preceq$ be a [[Definition:Well-Ordering|well-ordering]] of $S$.
Let $T$ be a [[Definition:Subset|subset]] or [[Definition:Subclass|subclass]] of $S$.
Let $\preceq'$ be the [[Definition:Restriction of Ordering|restriction]] of $\p... | By the definition of [[Definition:Well-Ordering|well-ordering]], $\preceq$ is a [[Definition:Well-Founded Relation|well-founded]] [[Definition:Total Ordering|total ordering]].
By [[Restriction of Total Ordering is Total Ordering]], $\preceq'$ is a [[Definition:Total Ordering|total ordering]].
By [[Restriction of Well... | Restriction of Well-Ordering is Well-Ordering | https://proofwiki.org/wiki/Restriction_of_Well-Ordering_is_Well-Ordering | https://proofwiki.org/wiki/Restriction_of_Well-Ordering_is_Well-Ordering | [
"Well-Orderings"
] | [
"Definition:set",
"Definition:Class (Class Theory)",
"Definition:Well-Ordering",
"Definition:Subset",
"Definition:Subclass",
"Definition:Restriction of Ordering",
"Definition:Well-Ordering"
] | [
"Definition:Well-Ordering",
"Definition:Well-Founded Relation",
"Definition:Total Ordering",
"Restriction of Total Ordering is Total Ordering",
"Definition:Total Ordering",
"Restriction of Well-Founded Ordering is Well-Founded",
"Definition:Well-Founded Ordered Set",
"Definition:Well-Ordering",
"Cat... |
proofwiki-7626 | Event Space contains Sample Space | :$\Omega \in \Sigma$ | {{begin-eqn}}
{{eqn | l = \Sigma
| o = \ne
| r = \O
| c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 1)$}}
}}
{{eqn | ll= \leadsto
| q = \exists A
| l = A
| o = \in
| r = \Sigma
| c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadsto
| l = \Omega \setminus A
... | :$\Omega \in \Sigma$ | {{begin-eqn}}
{{eqn | l = \Sigma
| o = \ne
| r = \O
| c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 1)$}}
}}
{{eqn | ll= \leadsto
| q = \exists A
| l = A
| o = \in
| r = \Sigma
| c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadsto
| l = \Omega \setminus A
... | Event Space contains Sample Space | https://proofwiki.org/wiki/Event_Space_contains_Sample_Space | https://proofwiki.org/wiki/Event_Space_contains_Sample_Space | [
"Event Spaces"
] | [] | [
"Union with Relative Complement"
] |
proofwiki-7627 | Event Space contains Empty Set | :$\O \in \Sigma$ | {{begin-eqn}}
{{eqn | l = \Sigma
| o = \ne
| r = \O
| c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 1)$}}
}}
{{eqn | ll= \leadsto
| q = \exists A
| l = A
| o = \in
| r = \Sigma
| c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadsto
| l = A \setminus A
| ... | :$\O \in \Sigma$ | {{begin-eqn}}
{{eqn | l = \Sigma
| o = \ne
| r = \O
| c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 1)$}}
}}
{{eqn | ll= \leadsto
| q = \exists A
| l = A
| o = \in
| r = \Sigma
| c = {{Defof|Empty Set}}
}}
{{eqn | ll= \leadsto
| l = A \setminus A
| ... | Event Space contains Empty Set | https://proofwiki.org/wiki/Event_Space_contains_Empty_Set | https://proofwiki.org/wiki/Event_Space_contains_Empty_Set | [
"Event Spaces"
] | [] | [
"Set Difference with Self is Empty Set"
] |
proofwiki-7628 | Power Set of Sample Space is Event Space | Let $\EE$ be an experiment whose sample space is $\Omega$.
Let $\powerset \Omega$ be the power set of $\Omega$.
Then $\powerset \Omega$ is an event space of $\EE$. | Let $\powerset \Omega := \Sigma$.
;Event Space Axiom $(\text {ES} 1)$:
From Empty Set is Subset of All Sets we have that $\O \subseteq \Omega$.
By the definition of power set:
:$\O \in \Sigma$
thus fulfilling axiom $(\text {ES} 1)$.
{{qed|lemma}}
;Event Space Axiom $(\text {ES} 2)$:
Let $A \in \Sigma$.
Then by the defi... | Let $\EE$ be an [[Definition:Experiment|experiment]] whose [[Definition:Sample Space|sample space]] is $\Omega$.
Let $\powerset \Omega$ be the [[Definition:Power Set|power set]] of $\Omega$.
Then $\powerset \Omega$ is an [[Definition:Event Space|event space]] of $\EE$. | Let $\powerset \Omega := \Sigma$.
;[[Definition:Event Space|Event Space Axiom $(\text {ES} 1)$]]:
From [[Empty Set is Subset of All Sets]] we have that $\O \subseteq \Omega$.
By the definition of [[Definition:Power Set|power set]]:
:$\O \in \Sigma$
thus fulfilling [[Definition:Event Space|axiom $(\text {ES} 1)$]].
... | Power Set of Sample Space is Event Space/Proof 1 | https://proofwiki.org/wiki/Power_Set_of_Sample_Space_is_Event_Space | https://proofwiki.org/wiki/Power_Set_of_Sample_Space_is_Event_Space/Proof_1 | [
"Event Spaces",
"Power Set of Sample Space is Event Space"
] | [
"Definition:Experiment",
"Definition:Sample Space",
"Definition:Power Set",
"Definition:Event Space"
] | [
"Definition:Event Space",
"Empty Set is Subset of All Sets",
"Definition:Power Set",
"Definition:Event Space",
"Definition:Event Space",
"Definition:Power Set",
"Set Difference and Intersection form Partition/Corollary 2",
"Definition:Power Set",
"Definition:Event Space",
"Definition:Event Space",... |
proofwiki-7629 | Power Set of Sample Space is Event Space | Let $\EE$ be an experiment whose sample space is $\Omega$.
Let $\powerset \Omega$ be the power set of $\Omega$.
Then $\powerset \Omega$ is an event space of $\EE$. | For $\powerset \Omega$ to be an event space of $\EE$, it needs to fulfil the following properties:
:$(1): \quad \powerset \Omega \ne \O$, that is, an event space can not be empty.
:$(2): \quad$ If $A \in \powerset \Omega$, then $\relcomp \Omega A \in \powerset \Omega$, that is, the complement of $A$ relative to $\Omega... | Let $\EE$ be an [[Definition:Experiment|experiment]] whose [[Definition:Sample Space|sample space]] is $\Omega$.
Let $\powerset \Omega$ be the [[Definition:Power Set|power set]] of $\Omega$.
Then $\powerset \Omega$ is an [[Definition:Event Space|event space]] of $\EE$. | For $\powerset \Omega$ to be an [[Definition:Event Space|event space]] of $\EE$, it needs to fulfil the following properties:
:$(1): \quad \powerset \Omega \ne \O$, that is, an event space can not be [[Definition:Empty Set|empty]].
:$(2): \quad$ If $A \in \powerset \Omega$, then $\relcomp \Omega A \in \powerset \Omeg... | Power Set of Sample Space is Event Space/Proof 2 | https://proofwiki.org/wiki/Power_Set_of_Sample_Space_is_Event_Space | https://proofwiki.org/wiki/Power_Set_of_Sample_Space_is_Event_Space/Proof_2 | [
"Event Spaces",
"Power Set of Sample Space is Event Space"
] | [
"Definition:Experiment",
"Definition:Sample Space",
"Definition:Power Set",
"Definition:Event Space"
] | [
"Definition:Event Space",
"Definition:Empty Set",
"Definition:Relative Complement",
"Definition:Set Union",
"Definition:Countable Set",
"Definition:Element",
"Power Set is Sigma-Algebra"
] |
proofwiki-7630 | Event Space from Single Subset of Sample Space | Let $\EE$ be an experiment whose sample space is $\Omega$.
Let $\O \subsetneqq A \subsetneqq \Omega$.
Then $\Sigma := \set {\O, A, \Omega \setminus A, \Omega}$ is an event space of $\EE$. | ;Event Space Axiom $(\text {ES} 1)$:
From its definition:
:$\Sigma \ne \O$
thus fulfilling axiom $(\text {ES} 1)$.
{{qed|lemma}}
;Event Space Axiom $(\text {ES} 2)$:
From Set Difference with Empty Set is Self:
:$\Omega \setminus \O = \Omega \in \Sigma$
From Set Difference with Self is Empty Set:
:$\Omega \setminus \Ome... | Let $\EE$ be an [[Definition:Experiment|experiment]] whose [[Definition:Sample Space|sample space]] is $\Omega$.
Let $\O \subsetneqq A \subsetneqq \Omega$.
Then $\Sigma := \set {\O, A, \Omega \setminus A, \Omega}$ is an [[Definition:Event Space|event space]] of $\EE$. | ;[[Definition:Event Space|Event Space Axiom $(\text {ES} 1)$]]:
From its definition:
:$\Sigma \ne \O$
thus fulfilling [[Definition:Event Space|axiom $(\text {ES} 1)$]].
{{qed|lemma}}
;[[Definition:Event Space|Event Space Axiom $(\text {ES} 2)$]]:
From [[Set Difference with Empty Set is Self]]:
:$\Omega \setminus \O ... | Event Space from Single Subset of Sample Space | https://proofwiki.org/wiki/Event_Space_from_Single_Subset_of_Sample_Space | https://proofwiki.org/wiki/Event_Space_from_Single_Subset_of_Sample_Space | [
"Probability Theory"
] | [
"Definition:Experiment",
"Definition:Sample Space",
"Definition:Event Space"
] | [
"Definition:Event Space",
"Definition:Event Space",
"Definition:Event Space",
"Set Difference with Empty Set is Self",
"Set Difference with Self is Empty Set",
"Relative Complement of Relative Complement",
"Definition:Event Space",
"Definition:Event Space",
"Union with Empty Set",
"Union with Supe... |
proofwiki-7631 | Intersection of Events is Event | :$A, B \in \Sigma \implies A \cap B \in \Sigma$ | {{begin-eqn}}
{{eqn | l = A, B
| o = \in
| r = \Sigma
| c=
}}
{{eqn | ll= \leadsto
| l = \Omega \setminus A, \ \Omega \setminus B
| o = \in
| r = \Sigma
| c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 2)$}}
}}
{{eqn | ll= \leadsto
| l = \paren {\Omega \setminus ... | :$A, B \in \Sigma \implies A \cap B \in \Sigma$ | {{begin-eqn}}
{{eqn | l = A, B
| o = \in
| r = \Sigma
| c=
}}
{{eqn | ll= \leadsto
| l = \Omega \setminus A, \ \Omega \setminus B
| o = \in
| r = \Sigma
| c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 2)$}}
}}
{{eqn | ll= \leadsto
| l = \paren {\Omega \setminus ... | Intersection of Events is Event | https://proofwiki.org/wiki/Intersection_of_Events_is_Event | https://proofwiki.org/wiki/Intersection_of_Events_is_Event | [
"Intersections of Events",
"Event Spaces"
] | [] | [
"De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection",
"Relative Complement of Relative Complement"
] |
proofwiki-7632 | Set Difference of Events is Event | :$A, B \in \Sigma \implies A \setminus B \in \Sigma$ | {{begin-eqn}}
{{eqn | l = A, B
| o = \in
| r = \Sigma
| c=
}}
{{eqn | ll= \leadsto
| l = A, \Omega \setminus B
| o = \in
| r = \Sigma
| c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 2)$}}
}}
{{eqn | ll= \leadsto
| l = A \cap \paren {\Omega \setminus B}
| o... | :$A, B \in \Sigma \implies A \setminus B \in \Sigma$ | {{begin-eqn}}
{{eqn | l = A, B
| o = \in
| r = \Sigma
| c=
}}
{{eqn | ll= \leadsto
| l = A, \Omega \setminus B
| o = \in
| r = \Sigma
| c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 2)$}}
}}
{{eqn | ll= \leadsto
| l = A \cap \paren {\Omega \setminus B}
| o... | Set Difference of Events is Event | https://proofwiki.org/wiki/Set_Difference_of_Events_is_Event | https://proofwiki.org/wiki/Set_Difference_of_Events_is_Event | [
"Event Spaces"
] | [] | [
"Intersection of Events is Event",
"Set Difference as Intersection with Relative Complement"
] |
proofwiki-7633 | Composition of Affine Transformations is Affine Transformation | Let $\EE$, $\FF$ and $\GG$ be affine spaces with difference spaces $E$, $F$ and $G$ respectively.
Let $\LL: \EE \to \FF$ and $\MM: \FF \to \GG$ be affine transformations.
Let $L$ and $M$ be the tangent maps of $\LL$ and $\MM$ respectively.
Then the composition $\MM \circ \LL: \EE \to \FF$ is an affine transformation wi... | Let $\NN = \MM \circ \LL : \EE \to \GG$ be the composition.
We want to show that for any $p, q \in \EE$
:$\map \GG Q = \map \GG p + \map {M \circ L} {\vec {p q} }$
We find that:
{{begin-eqn}}
{{eqn | l = \map \GG q
| r = \map {\MM \circ \LL} q
}}
{{eqn | r = \map \MM {\map \LL p} + \map L {\vec{p q} }
| c =... | Let $\EE$, $\FF$ and $\GG$ be [[Definition:Affine Space|affine spaces]] with [[Definition:Difference Space|difference spaces]] $E$, $F$ and $G$ respectively.
Let $\LL: \EE \to \FF$ and $\MM: \FF \to \GG$ be [[Definition:Affine Transformation|affine transformations]].
Let $L$ and $M$ be the [[Definition:Tangent Map of... | Let $\NN = \MM \circ \LL : \EE \to \GG$ be the [[Definition:Composition of Mappings|composition]].
We want to show that for any $p, q \in \EE$
:$\map \GG Q = \map \GG p + \map {M \circ L} {\vec {p q} }$
We find that:
{{begin-eqn}}
{{eqn | l = \map \GG q
| r = \map {\MM \circ \LL} q
}}
{{eqn | r = \map \MM {\map... | Composition of Affine Transformations is Affine Transformation | https://proofwiki.org/wiki/Composition_of_Affine_Transformations_is_Affine_Transformation | https://proofwiki.org/wiki/Composition_of_Affine_Transformations_is_Affine_Transformation | [
"Affine Geometry",
"Composite Mappings"
] | [
"Definition:Affine Space",
"Definition:Tangent Space (Affine Geometry)",
"Definition:Affine Transformation",
"Definition:Tangent Map/Affine Transformation",
"Definition:Composition of Mappings",
"Definition:Affine Transformation",
"Definition:Tangent Map/Affine Transformation"
] | [
"Definition:Composition of Mappings",
"Definition:Affine Transformation",
"Definition:Affine Transformation",
"Category:Affine Geometry",
"Category:Composite Mappings"
] |
proofwiki-7634 | Symmetric Difference of Events is Event | :$A, B \in \Sigma \implies A \ast B \in \Sigma$ | {{begin-eqn}}
{{eqn | l = A, B
| o = \in
| r = \Sigma
| c=
}}
{{eqn | ll= \leadsto
| l = A \cup B
| o = \in
| r = \Sigma
| c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 3)$}}
}}
{{eqn | lo= \land
| l = A \cap B
| o = \in
| r = \Sigma
| c = Inte... | :$A, B \in \Sigma \implies A \ast B \in \Sigma$ | {{begin-eqn}}
{{eqn | l = A, B
| o = \in
| r = \Sigma
| c=
}}
{{eqn | ll= \leadsto
| l = A \cup B
| o = \in
| r = \Sigma
| c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 3)$}}
}}
{{eqn | lo= \land
| l = A \cap B
| o = \in
| r = \Sigma
| c = [[In... | Symmetric Difference of Events is Event | https://proofwiki.org/wiki/Symmetric_Difference_of_Events_is_Event | https://proofwiki.org/wiki/Symmetric_Difference_of_Events_is_Event | [
"Event Spaces"
] | [] | [
"Intersection of Events is Event",
"Set Difference of Events is Event"
] |
proofwiki-7635 | Characterization of Affine Transformations | Let $\EE$ and $\FF$ be affine spaces over a field $K$.
Let $\LL: \EE \to \FF$ be a mapping.
Then $\LL$ is an affine transformation {{iff}}:
:$\forall p, q \in \EE: \forall \lambda \in K: \map \LL {\lambda p + \paren {1 - \lambda} q} = \lambda \map \LL p + \paren {1 - \lambda} \map \LL q$
where $\lambda p + \paren {1 - ... | === Sufficient Condition ===
Let $\LL$ be an affine transformation.
Let $L$ be the tangent map.
Let $r \in \EE$ be any point.
Then by definition we have:
:$\lambda p + \paren {1 - \lambda} q = r + \lambda \vec{r p} + \paren {1 - \lambda} \vec{r q}$
Thus we find:
{{begin-eqn}}
{{eqn | l = \map \LL {\lambda p + \paren {1... | Let $\EE$ and $\FF$ be [[Definition:Affine Space|affine spaces]] over a [[Definition:Field (Abstract Algebra)|field]] $K$.
Let $\LL: \EE \to \FF$ be a [[Definition:Mapping|mapping]].
Then $\LL$ is an [[Definition:Affine Transformation|affine transformation]] {{iff}}:
:$\forall p, q \in \EE: \forall \lambda \in K: \... | === Sufficient Condition ===
Let $\LL$ be an [[Definition:Affine Transformation|affine transformation]].
Let $L$ be the [[Definition:Tangent Map of Affine Transformation|tangent map]].
Let $r \in \EE$ be any point.
Then by definition we have:
:$\lambda p + \paren {1 - \lambda} q = r + \lambda \vec{r p} + \paren {1 ... | Characterization of Affine Transformations | https://proofwiki.org/wiki/Characterization_of_Affine_Transformations | https://proofwiki.org/wiki/Characterization_of_Affine_Transformations | [
"Affine Geometry"
] | [
"Definition:Affine Space",
"Definition:Field (Abstract Algebra)",
"Definition:Mapping",
"Definition:Affine Transformation",
"Definition:Barycenter"
] | [
"Definition:Affine Transformation",
"Definition:Tangent Map/Affine Transformation",
"Definition:Linear Transformation",
"Definition:Linear Transformation"
] |
proofwiki-7636 | Probability of Empty Event is Zero | :$\map \Pr \O = 0$ | From the conditions for $\Pr$ to be a probability measure, we have:
:$(1): \quad \forall A \in \Sigma: 0 \le \map \Pr A$
:$(2): \quad \map \Pr \Omega = 1$
:$(3): \quad \ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$ where all $A_i$ are pairwise disjoint.
From the definition of even... | :$\map \Pr \O = 0$ | From the conditions for $\Pr$ to be a [[Definition:Probability Measure|probability measure]], we have:
:$(1): \quad \forall A \in \Sigma: 0 \le \map \Pr A$
:$(2): \quad \map \Pr \Omega = 1$
:$(3): \quad \ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$ where all $A_i$ are [[Defini... | Probability of Empty Event is Zero | https://proofwiki.org/wiki/Probability_of_Empty_Event_is_Zero | https://proofwiki.org/wiki/Probability_of_Empty_Event_is_Zero | [
"Probability Theory"
] | [] | [
"Definition:Probability Measure",
"Definition:Pairwise Disjoint",
"Definition:Event Space",
"Intersection with Empty Set",
"Definition:Pairwise Disjoint",
"Union with Empty Set"
] |
proofwiki-7637 | Ordinal Membership is Asymmetric | Let $m$ and $n$ be ordinals.
Then it is not the case that $m \in n$ and $n \in m$. | {{AimForCont}} $m \in n$ and $n \in m$.
Since $m$ is an ordinal, it is transitive.
Thus since $m \in n$ and $n \in m$, it follows that $m \in m$.
But this contradicts Ordinal is not Element of Itself.
{{qed}}
Category:Ordinals
oapnhc6j5eyfppc6gujiamnal6c9qm9 | Let $m$ and $n$ be [[Definition:Ordinal|ordinals]].
Then it is not the case that $m \in n$ and $n \in m$. | {{AimForCont}} $m \in n$ and $n \in m$.
Since $m$ is an [[Definition:Ordinal/Definition 1|ordinal]], it is [[Definition:Transitive Set|transitive]].
Thus since $m \in n$ and $n \in m$, it follows that $m \in m$.
But this contradicts [[Ordinal is not Element of Itself]].
{{qed}}
[[Category:Ordinals]]
oapnhc6j5eyfppc... | Ordinal Membership is Asymmetric | https://proofwiki.org/wiki/Ordinal_Membership_is_Asymmetric | https://proofwiki.org/wiki/Ordinal_Membership_is_Asymmetric | [
"Ordinals"
] | [
"Definition:Ordinal"
] | [
"Definition:Ordinal/Definition 1",
"Definition:Transitive Class",
"Ordinal is not Element of Itself",
"Category:Ordinals"
] |
proofwiki-7638 | Intersection of Ordinals is Smallest | Let $A$ be a non-empty set or class of ordinals.
Let $m = \bigcap A$ be the intersection of all the elements of $A$.
Then $m$ is the smallest element of $A$. | By Intersection of Ordinals is Ordinal, $m$ is an ordinal.
By Intersection is Largest Subset, $m \subseteq a$ for each $a \in A$.
It remains to show that $m \in A$.
Let $m^+ = m \cup \set m$ be the successor of $m$.
By Relation between Two Ordinals:
:for each $a \in A$, either $m^+ \subseteq a$ or $a \in m^+$.
Let $m^+... | Let $A$ be a non-empty [[Definition:Set|set]] or [[Definition:Class (Class Theory)|class]] of [[Definition:Ordinal|ordinals]].
Let $m = \bigcap A$ be the [[Definition:Intersection of Set of Sets|intersection]] of all the [[Definition:Element|elements]] of $A$.
Then $m$ is the [[Definition:Smallest Set by Set Inclusi... | By [[Intersection of Ordinals is Ordinal]], $m$ is an [[Definition:Ordinal|ordinal]].
By [[Intersection is Largest Subset]], $m \subseteq a$ for each $a \in A$.
It remains to show that $m \in A$.
Let $m^+ = m \cup \set m$ be the [[Definition:Successor Set|successor]] of $m$.
By [[Relation between Two Ordinals]]:
:f... | Intersection of Ordinals is Smallest | https://proofwiki.org/wiki/Intersection_of_Ordinals_is_Smallest | https://proofwiki.org/wiki/Intersection_of_Ordinals_is_Smallest | [
"Ordinals"
] | [
"Definition:Set",
"Definition:Class (Class Theory)",
"Definition:Ordinal",
"Definition:Set Intersection/Set of Sets",
"Definition:Element",
"Definition:Smallest Set by Set Inclusion"
] | [
"Intersection of Ordinals is Ordinal",
"Definition:Ordinal",
"Intersection is Largest Subset",
"Definition:Successor Mapping/Successor Set",
"Relation between Two Ordinals",
"Intersection is Largest Subset",
"Definition:Contradiction",
"Ordinal is not Element of Itself",
"Definition:Contradiction",
... |
proofwiki-7639 | Class of All Ordinals is Well-Ordered by Subset Relation | Let $\On$ be the class of all ordinals.
Then the restriction of the subset relation, $\subseteq$, to $\On$ is a well-ordering.
That is:
:$\subseteq$ is an ordering on $\On$.
:If $A$ is a non-empty subclass of $\On$, then $A$ has a smallest element under the subset relation. | By Subset Relation on Class is Ordering, $\subseteq$ is an ordering of any class.
Let $A$ be a subclass of $\On$.
By Intersection of Ordinals is Smallest, $A$ has a smallest element under the subset relation.
{{qed}} | Let $\On$ be the [[Definition:Class of All Ordinals|class of all ordinals]].
Then the [[Definition:Restriction of Relation (Class Theory)|restriction]] of the [[Definition:Subset Relation|subset relation]], $\subseteq$, to $\On$ is a [[Definition:Well-Ordering (Class Theory)|well-ordering]].
That is:
:$\subseteq$ is... | By [[Subset Relation on Class is Ordering]], $\subseteq$ is an [[Definition:Ordering (Class Theory)|ordering]] of any [[Definition:Class (Class Theory)|class]].
Let $A$ be a [[Definition:Subclass|subclass]] of $\On$.
By [[Intersection of Ordinals is Smallest]], $A$ has a [[Definition:Smallest Element (Class Theory)|s... | Class of All Ordinals is Well-Ordered by Subset Relation/Proof 1 | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_Well-Ordered_by_Subset_Relation | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_Well-Ordered_by_Subset_Relation/Proof_1 | [
"Class of All Ordinals is Well-Ordered by Subset Relation",
"Class of All Ordinals",
"Subset Relation"
] | [
"Definition:Class of All Ordinals",
"Definition:Restriction/Relation/Class Theory",
"Definition:Subset Relation",
"Definition:Well-Ordering/Class Theory",
"Definition:Ordering",
"Definition:Non-Empty Set/Class Theory",
"Definition:Subclass",
"Definition:Smallest Element/Class Theory",
"Definition:Su... | [
"Subset Relation is Ordering/Class Theory",
"Definition:Ordering/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Subclass",
"Intersection of Ordinals is Smallest",
"Definition:Smallest Element/Class Theory",
"Definition:Subset Relation"
] |
proofwiki-7640 | Class of All Ordinals is Well-Ordered by Subset Relation | Let $\On$ be the class of all ordinals.
Then the restriction of the subset relation, $\subseteq$, to $\On$ is a well-ordering.
That is:
:$\subseteq$ is an ordering on $\On$.
:If $A$ is a non-empty subclass of $\On$, then $A$ has a smallest element under the subset relation. | From Class of All Ordinals is $g$-Tower, $\On$ is a $g$-tower.
The result follows from $g$-Tower is Well-Ordered under Subset Relation.
{{qed}} | Let $\On$ be the [[Definition:Class of All Ordinals|class of all ordinals]].
Then the [[Definition:Restriction of Relation (Class Theory)|restriction]] of the [[Definition:Subset Relation|subset relation]], $\subseteq$, to $\On$ is a [[Definition:Well-Ordering (Class Theory)|well-ordering]].
That is:
:$\subseteq$ is... | From [[Class of All Ordinals is G-Tower|Class of All Ordinals is $g$-Tower]], $\On$ is a [[Definition:G-Tower|$g$-tower]].
The result follows from [[G-Tower is Well-Ordered under Subset Relation|$g$-Tower is Well-Ordered under Subset Relation]].
{{qed}} | Class of All Ordinals is Well-Ordered by Subset Relation/Proof 2 | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_Well-Ordered_by_Subset_Relation | https://proofwiki.org/wiki/Class_of_All_Ordinals_is_Well-Ordered_by_Subset_Relation/Proof_2 | [
"Class of All Ordinals is Well-Ordered by Subset Relation",
"Class of All Ordinals",
"Subset Relation"
] | [
"Definition:Class of All Ordinals",
"Definition:Restriction/Relation/Class Theory",
"Definition:Subset Relation",
"Definition:Well-Ordering/Class Theory",
"Definition:Ordering",
"Definition:Non-Empty Set/Class Theory",
"Definition:Subclass",
"Definition:Smallest Element/Class Theory",
"Definition:Su... | [
"Class of All Ordinals is G-Tower",
"Definition:G-Tower",
"G-Tower is Well-Ordered under Subset Relation"
] |
proofwiki-7641 | Subset is Compatible with Ordinal Successor | Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x \in y$.
Then:
:$x^+ \in y^+$ | {{begin-eqn}}
{{eqn|l = x \in y
|o = \implies
|r = x \ne y
|c = No Membership Loops
}}
{{eqn|o = \implies
|r = x^+ \ne y^+
|c = Equality of Successors
}}
{{eqn|l = x \in y
|o = \implies
|r = y \notin x
|c = No Membership Loops
}}
{{eqn|o = \implies
|r = y \notin x^+
|c ... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] and let $x^+$ denote the [[Definition:Successor Set|successor set]] of $x$.
Let $x \in y$.
Then:
:$x^+ \in y^+$ | {{begin-eqn}}
{{eqn|l = x \in y
|o = \implies
|r = x \ne y
|c = [[No Membership Loops]]
}}
{{eqn|o = \implies
|r = x^+ \ne y^+
|c = [[Equality of Successors]]
}}
{{eqn|l = x \in y
|o = \implies
|r = y \notin x
|c = [[No Membership Loops]]
}}
{{eqn|o = \implies
|r = y \notin ... | Subset is Compatible with Ordinal Successor/Proof 1 | https://proofwiki.org/wiki/Subset_is_Compatible_with_Ordinal_Successor | https://proofwiki.org/wiki/Subset_is_Compatible_with_Ordinal_Successor/Proof_1 | [
"Ordinals",
"Subset is Compatible with Ordinal Successor"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"No Membership Loops",
"Equality of Successors",
"No Membership Loops",
"Successor Set of Ordinal is Ordinal",
"Set is Element of Successor",
"Ordinal Membership is Trichotomy"
] |
proofwiki-7642 | Subset is Compatible with Ordinal Successor | Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x \in y$.
Then:
:$x^+ \in y^+$ | First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x \in y$.
We wish to show that $x^+ \in y^+$.
By Ordinal Membership is Trichotomy:
:Either $x^+ \in y^+$, $y^+ \in x^+$, or $x^+ = y^+$.
{{AimForCont}} $y^+ = x^+$.
Then $y \in x$ or $y = x$ by the definition of successor set.
If... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] and let $x^+$ denote the [[Definition:Successor Set|successor set]] of $x$.
Let $x \in y$.
Then:
:$x^+ \in y^+$ | First note that by [[Successor Set of Ordinal is Ordinal]], $x^+$ and $y^+$ are [[Definition:Ordinal|ordinals]].
Let $x \in y$.
We wish to show that $x^+ \in y^+$.
By [[Ordinal Membership is Trichotomy]]:
:Either $x^+ \in y^+$, $y^+ \in x^+$, or $x^+ = y^+$.
{{AimForCont}} $y^+ = x^+$.
Then $y \in x$ or $y = x$ ... | Subset is Compatible with Ordinal Successor/Proof 2 | https://proofwiki.org/wiki/Subset_is_Compatible_with_Ordinal_Successor | https://proofwiki.org/wiki/Subset_is_Compatible_with_Ordinal_Successor/Proof_2 | [
"Ordinals",
"Subset is Compatible with Ordinal Successor"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Successor Set of Ordinal is Ordinal",
"Definition:Ordinal",
"Ordinal Membership is Trichotomy",
"Definition:Successor Mapping/Successor Set",
"Ordinal is not Element of Itself",
"Definition:Transitive Class",
"Ordinal is not Element of Itself",
"Definition:Successor Mapping/Successor Set",
"Ordinal... |
proofwiki-7643 | Subset is Compatible with Ordinal Successor | Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x \in y$.
Then:
:$x^+ \in y^+$ | First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
By Ordinal Membership is Trichotomy, one of the following must be true:
{{begin-eqn}}
{{eqn | l = x^+
| r = y^+
}}
{{eqn | l = y^+
| o = \in
| r = x^+
}}
{{eqn | l = x^+
| o = \in
| r = y^+
}}
{{end-eqn}}... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] and let $x^+$ denote the [[Definition:Successor Set|successor set]] of $x$.
Let $x \in y$.
Then:
:$x^+ \in y^+$ | First note that by [[Successor Set of Ordinal is Ordinal]], $x^+$ and $y^+$ are [[Definition:Ordinal|ordinals]].
By [[Ordinal Membership is Trichotomy]], one of the following must be true:
{{begin-eqn}}
{{eqn | l = x^+
| r = y^+
}}
{{eqn | l = y^+
| o = \in
| r = x^+
}}
{{eqn | l = x^+
| o = \i... | Subset is Compatible with Ordinal Successor/Proof 3 | https://proofwiki.org/wiki/Subset_is_Compatible_with_Ordinal_Successor | https://proofwiki.org/wiki/Subset_is_Compatible_with_Ordinal_Successor/Proof_3 | [
"Ordinals",
"Subset is Compatible with Ordinal Successor"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Successor Set of Ordinal is Ordinal",
"Definition:Ordinal",
"Ordinal Membership is Trichotomy",
"Ordinal is not Element of Itself",
"Ordinal Membership is Asymmetric",
"Equality of Successors",
"Definition:Ordinal/Definition 1",
"Definition:Transitive Class",
"Definition:Contradiction"
] |
proofwiki-7644 | Successor is Less than Successor/Sufficient Condition/Proof 1 | Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x^+ \in y^+$.
Then:
: $x \in y$ | Suppose $y^+ \in x^+$.
By the definition of successor, $y^+ \in x \lor y^+ = x$.
Suppose $y^+ = x$.
By Ordinal is Less than Successor, $y \in x$.
Suppose $y^+ \in x$.
By Ordinal is Less than Successor, $y \in y^+$.
By Ordinal is Transitive, $y \in x$. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] and let $x^+$ denote the [[Definition:Successor Set|successor set]] of $x$.
Let $x^+ \in y^+$.
Then:
: $x \in y$ | Suppose $y^+ \in x^+$.
By the definition of [[Definition:Successor Set|successor]], $y^+ \in x \lor y^+ = x$.
Suppose $y^+ = x$.
By [[Ordinal is Less than Successor]], $y \in x$.
Suppose $y^+ \in x$.
By [[Ordinal is Less than Successor]], $y \in y^+$.
By [[Ordinal is Transitive]], $y \in x$. | Successor is Less than Successor/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Successor_is_Less_than_Successor/Sufficient_Condition/Proof_1 | https://proofwiki.org/wiki/Successor_is_Less_than_Successor/Sufficient_Condition/Proof_1 | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Definition:Successor Mapping/Successor Set",
"Ordinal is Less than Successor",
"Ordinal is Less than Successor",
"Ordinal is Transitive"
] |
proofwiki-7645 | Successor is Less than Successor/Sufficient Condition/Proof 2 | Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x^+ \in y^+$.
Then:
: $x \in y$ | First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting Ordinal is not Element of Itself.
Thus $x \in y$. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] and let $x^+$ denote the [[Definition:Successor Set|successor set]] of $x$.
Let $x^+ \in y^+$.
Then:
: $x \in y$ | First note that by [[Successor Set of Ordinal is Ordinal]], $x^+$ and $y^+$ are [[Definition:Ordinal|ordinals]].
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting [[Ordinal is not Element of Itself]].
Thus $x \in y$. | Successor is Less than Successor/Sufficient Condition/Proof 2 | https://proofwiki.org/wiki/Successor_is_Less_than_Successor/Sufficient_Condition/Proof_2 | https://proofwiki.org/wiki/Successor_is_Less_than_Successor/Sufficient_Condition/Proof_2 | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Successor Mapping/Successor Set"
] | [
"Successor Set of Ordinal is Ordinal",
"Definition:Ordinal",
"Ordinal is not Element of Itself"
] |
proofwiki-7646 | Successor is Less than Successor/Sufficient Condition | == Proof 1 ==
{{:Successor is Less than Successor/Sufficient Condition/Proof 1}} | Suppose $y^+ \in x^+$.
By the definition of successor, $y^+ \in x \lor y^+ = x$.
Suppose $y^+ = x$.
By Ordinal is Less than Successor, $y \in x$.
Suppose $y^+ \in x$.
By Ordinal is Less than Successor, $y \in y^+$.
By Ordinal is Transitive, $y \in x$. | == [[Successor is Less than Successor/Sufficient Condition/Proof 1|Proof 1]] ==
{{:Successor is Less than Successor/Sufficient Condition/Proof 1}} | Suppose $y^+ \in x^+$.
By the definition of [[Definition:Successor Set|successor]], $y^+ \in x \lor y^+ = x$.
Suppose $y^+ = x$.
By [[Ordinal is Less than Successor]], $y \in x$.
Suppose $y^+ \in x$.
By [[Ordinal is Less than Successor]], $y \in y^+$.
By [[Ordinal is Transitive]], $y \in x$. | Successor is Less than Successor/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Successor_is_Less_than_Successor/Sufficient_Condition | https://proofwiki.org/wiki/Successor_is_Less_than_Successor/Sufficient_Condition/Proof_1 | [
"Ordinals"
] | [
"Successor is Less than Successor/Sufficient Condition/Proof 1"
] | [
"Definition:Successor Mapping/Successor Set",
"Ordinal is Less than Successor",
"Ordinal is Less than Successor",
"Ordinal is Transitive"
] |
proofwiki-7647 | Successor is Less than Successor/Sufficient Condition | == Proof 1 ==
{{:Successor is Less than Successor/Sufficient Condition/Proof 1}} | First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting Ordinal is not Element of Itself.
Thus $x \in y$. | == [[Successor is Less than Successor/Sufficient Condition/Proof 1|Proof 1]] ==
{{:Successor is Less than Successor/Sufficient Condition/Proof 1}} | First note that by [[Successor Set of Ordinal is Ordinal]], $x^+$ and $y^+$ are [[Definition:Ordinal|ordinals]].
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting [[Ordinal is not Element of Itself]].
Thus $x \in y$. | Successor is Less than Successor/Sufficient Condition/Proof 2 | https://proofwiki.org/wiki/Successor_is_Less_than_Successor/Sufficient_Condition | https://proofwiki.org/wiki/Successor_is_Less_than_Successor/Sufficient_Condition/Proof_2 | [
"Ordinals"
] | [
"Successor is Less than Successor/Sufficient Condition/Proof 1"
] | [
"Successor Set of Ordinal is Ordinal",
"Definition:Ordinal",
"Ordinal is not Element of Itself"
] |
proofwiki-7648 | Probability of Event not Occurring | :$\forall A \in \Sigma: \map \Pr {\Omega \setminus A} = 1 - \map \Pr A$ | From the conditions for $\Pr$ to be a probability measure, we have:
:$(1): \quad \forall A \in \Sigma: 0 \le \map \Pr A$
:$(2): \quad \map \Pr \Omega = 1$
:$(3): \quad \ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$ where all $A_i$ are pairwise disjoint.
Let $A \in \Sigma$ be an ev... | :$\forall A \in \Sigma: \map \Pr {\Omega \setminus A} = 1 - \map \Pr A$ | From the conditions for $\Pr$ to be a [[Definition:Probability Measure|probability measure]], we have:
:$(1): \quad \forall A \in \Sigma: 0 \le \map \Pr A$
:$(2): \quad \map \Pr \Omega = 1$
:$(3): \quad \ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$ where all $A_i$ are [[Defini... | Probability of Event not Occurring | https://proofwiki.org/wiki/Probability_of_Event_not_Occurring | https://proofwiki.org/wiki/Probability_of_Event_not_Occurring | [
"Probability Theory"
] | [] | [
"Definition:Probability Measure",
"Definition:Pairwise Disjoint",
"Definition:Event",
"Definition:Event Space",
"Intersection with Relative Complement is Empty",
"Union with Relative Complement"
] |
proofwiki-7649 | Delta-Algebra is Sigma-Algebra | Every $\delta$-algebra is a $\sigma$-algebra. | Let $\DD$ be a $\delta$-algebra whose unit is $\mathbb U$.
Let $A_1, A_2, \ldots$ be a countably infinite collection of elements of $\DD$.
Then:
{{begin-eqn}}
{{eqn | q = \forall i
| l = \mathbb U \setminus A_i
| o = \in
| r = \DD
| c = $\DD$ is closed under relative complement with $\mathbb U$
... | Every [[Definition:Delta-Algebra|$\delta$-algebra]] is a [[Definition:Sigma-Algebra|$\sigma$-algebra]]. | Let $\DD$ be a [[Definition:Delta-Algebra|$\delta$-algebra]] whose [[Definition:Unit of System of Sets|unit]] is $\mathbb U$.
Let $A_1, A_2, \ldots$ be a [[Definition:Countable|countably infinite]] collection of [[Definition:Element|elements]] of $\DD$.
Then:
{{begin-eqn}}
{{eqn | q = \forall i
| l = \mathbb ... | Delta-Algebra is Sigma-Algebra | https://proofwiki.org/wiki/Delta-Algebra_is_Sigma-Algebra | https://proofwiki.org/wiki/Delta-Algebra_is_Sigma-Algebra | [
"Sigma-Algebras"
] | [
"Definition:Delta-Algebra",
"Definition:Sigma-Algebra"
] | [
"Definition:Delta-Algebra",
"Definition:Unit of System of Sets",
"Definition:Countable Set",
"Definition:Element",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Relative Complement",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Countable Set",
"... |
proofwiki-7650 | Countable Intersection of Events is Event | :$\quad A_1, A_2, \ldots \in \Sigma \implies \ds \bigcap_{i \mathop = 1}^\infty A_i \in \Sigma$ | By definition, a probability space $\struct {\Omega, \Sigma, \Pr}$ is a measure space.
So, again by definition, an event space $\Sigma$ is a $\sigma$-algebra on $\Omega$.
From Sigma-Algebra is Delta-Algebra:
:$\ds A_1, A_2, \ldots \in \Sigma \implies \bigcap_{i \mathop = 1}^\infty A_i \in \Sigma$
by definition of $\del... | :$\quad A_1, A_2, \ldots \in \Sigma \implies \ds \bigcap_{i \mathop = 1}^\infty A_i \in \Sigma$ | By definition, a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$ is a [[Definition:Measure Space|measure space]].
So, again by definition, an [[Definition:Event Space|event space]] $\Sigma$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $\Omega$.
From [[Sigma-Algebra is Delta... | Countable Intersection of Events is Event | https://proofwiki.org/wiki/Countable_Intersection_of_Events_is_Event | https://proofwiki.org/wiki/Countable_Intersection_of_Events_is_Event | [
"Event Spaces"
] | [] | [
"Definition:Probability Space",
"Definition:Measure Space",
"Definition:Event Space",
"Definition:Sigma-Algebra",
"Sigma-Algebra is Delta-Algebra",
"Definition:Delta-Algebra"
] |
proofwiki-7651 | Subset Relation is Ordering/General Result | Let $\mathbb S$ be a set of sets.
Then $\subseteq$ is an ordering on $\mathbb S$.
In other words, let $\struct {\mathbb S, \subseteq}$ be the relational structure defined on $\mathbb S$ by the subset relation $\subseteq$.
Then $\struct {\mathbb S, \subseteq}$ is an ordered set. | To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.
So, checking in turn each of the criteria for an ordering: | Let $\mathbb S$ be a [[Definition:Set of Sets|set of sets]].
Then $\subseteq$ is an [[Definition:Ordering|ordering]] on $\mathbb S$.
In other words, let $\struct {\mathbb S, \subseteq}$ be the [[Definition:Relational Structure|relational structure]] defined on $\mathbb S$ by the [[Definition:Subset Relation|subset ... | To establish that $\subseteq$ is an [[Definition:Ordering|ordering]], we need to show that it is [[Definition:Reflexive Relation|reflexive]], [[Definition:Antisymmetric Relation|antisymmetric]] and [[Definition:Transitive Relation|transitive]].
So, checking in turn each of the criteria for an [[Definition:Ordering|ord... | Subset Relation is Ordering/General Result | https://proofwiki.org/wiki/Subset_Relation_is_Ordering/General_Result | https://proofwiki.org/wiki/Subset_Relation_is_Ordering/General_Result | [
"Subset Relation is Ordering"
] | [
"Definition:Set of Sets",
"Definition:Ordering",
"Definition:Relational Structure",
"Definition:Subset Relation",
"Definition:Ordered Set"
] | [
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Antisymmetric Relation",
"Definition:Transitive Relation",
"Definition:Ordering",
"Definition:Reflexive Relation",
"Definition:Antisymmetric Relation",
"Definition:Transitive Relation",
"Definition:Ordering"
] |
proofwiki-7652 | Noetherian Domain is Factorization Domain | Let $R$ be a noetherian integral domain.
Then $R$ is a factorization domain. | Let $\FF$ be the set of ideals of $R$ of the form $x R$, with $x$ not a unit and such that $x$ cannot be decomposed in the form:
:$x = u p_1 \dotsm p_r$
where $u$ is a unit and $p_1, \dotsc, p_r$ irreducible.
We show by contradiction that $\FF = \O$.
{{AimForCont}} $\FF \ne \O$.
Since $R$ is noetherian, we can choose a... | Let $R$ be a [[Definition:Noetherian Ring|noetherian]] [[Definition:Integral Domain|integral domain]].
Then $R$ is a [[Definition:Factorization Domain|factorization domain]]. | Let $\FF$ be the set of [[Definition:Ideal of Ring|ideals]] of $R$ of the form $x R$, with $x$ not a [[Definition:Unit of Ring|unit]] and such that $x$ cannot be [[Definition:Decomposable Element|decomposed]] in the form:
:$x = u p_1 \dotsm p_r$
where $u$ is a [[Definition:Unit of Ring|unit]] and $p_1, \dotsc, p_r$ [[D... | Noetherian Domain is Factorization Domain | https://proofwiki.org/wiki/Noetherian_Domain_is_Factorization_Domain | https://proofwiki.org/wiki/Noetherian_Domain_is_Factorization_Domain | [
"Ring Theory",
"Factorization"
] | [
"Definition:Noetherian Ring",
"Definition:Integral Domain",
"Definition:Factorization Domain"
] | [
"Definition:Ideal of Ring",
"Definition:Unit of Ring",
"Definition:Decomposable Element",
"Definition:Unit of Ring",
"Definition:Irreducible Element of Ring",
"Proof by Contradiction",
"Definition:Noetherian Ring",
"Definition:Maximal/Element",
"Definition:Irreducible Element of Ring",
"Definition... |
proofwiki-7653 | Probability Measure is Monotone | Let $A, B \in \Sigma$ such that $A \subseteq B$.
Then:
:$\map \Pr A \le \map \Pr B$ | From Set Difference Union Second Set is Union:
:$A \cup B = \paren {B \setminus A} \cup A$
From Set Difference Intersection with Second Set is Empty Set:
:$\paren {B \setminus A} \cap A = \O$
From the Addition Law of Probability:
:$\map \Pr {A \cup B} = \map \Pr {B \setminus A} + \map \Pr A$
From Union with Superset is... | Let $A, B \in \Sigma$ such that $A \subseteq B$.
Then:
:$\map \Pr A \le \map \Pr B$ | From [[Set Difference Union Second Set is Union]]:
:$A \cup B = \paren {B \setminus A} \cup A$
From [[Set Difference Intersection with Second Set is Empty Set]]:
:$\paren {B \setminus A} \cap A = \O$
From the [[Addition Law of Probability]]:
:$\map \Pr {A \cup B} = \map \Pr {B \setminus A} + \map \Pr A$
From [[Uni... | Probability Measure is Monotone/Proof 1 | https://proofwiki.org/wiki/Probability_Measure_is_Monotone | https://proofwiki.org/wiki/Probability_Measure_is_Monotone/Proof_1 | [
"Probability Theory",
"Probability Measure is Monotone"
] | [] | [
"Set Difference Union Second Set is Union",
"Set Difference Intersection with Second Set is Empty Set",
"Addition Law of Probability",
"Union with Superset is Superset",
"Definition:Probability Measure"
] |
proofwiki-7654 | Probability Measure is Monotone | Let $A, B \in \Sigma$ such that $A \subseteq B$.
Then:
:$\map \Pr A \le \map \Pr B$ | As by definition a probability measure is a measure, we can directly use the result Measure is Monotone.
{{qed}} | Let $A, B \in \Sigma$ such that $A \subseteq B$.
Then:
:$\map \Pr A \le \map \Pr B$ | As by definition a [[Definition:Probability Measure|probability measure]] is a [[Definition:Measure (Measure Theory)|measure]], we can directly use the result [[Measure is Monotone]].
{{qed}} | Probability Measure is Monotone/Proof 2 | https://proofwiki.org/wiki/Probability_Measure_is_Monotone | https://proofwiki.org/wiki/Probability_Measure_is_Monotone/Proof_2 | [
"Probability Theory",
"Probability Measure is Monotone"
] | [] | [
"Definition:Probability Measure",
"Definition:Measure (Measure Theory)",
"Measure is Monotone"
] |
proofwiki-7655 | Transfinite Induction/Principle 1/Proof 2 | Let $\On$ denote the class of all ordinals.
Let $A$ denote a class.
Suppose that:
:For each element $x$ of $\On$, if $\forall y \in \On: \paren {y < x \implies y \in A}$ then $x$ is an element of $A$.
Then $\On \subseteq A$. | <onlyinclude>
{{AimForCont}} $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.
By Class of All Ordinals is Well-Ordered by Subset Relation, $\On \setminus A$ must have a smallest element $y$.
Then every strict predecessor of ... | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $A$ denote a [[Definition:Class (Class Theory)|class]].
Suppose that:
:For each [[Definition:Element of Class|element]] $x$ of $\On$, if $\forall y \in \On: \paren {y < x \implies y \in A}$ then $x$ is an [[Definition:Element of Cla... | <onlyinclude>
{{AimForCont}} $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From [[Set Difference is Subset]], $\On \setminus A$ is a [[Definition:Subclass|subclass]] of the [[Definition:Ordinal|ordinals]].
By [[Class of All Ordinals is Well-Ordered by Subset Relation]], $\On \setminus A$ must have... | Transfinite Induction/Principle 1/Proof 2 | https://proofwiki.org/wiki/Transfinite_Induction/Principle_1/Proof_2 | https://proofwiki.org/wiki/Transfinite_Induction/Principle_1/Proof_2 | [
"Transfinite Induction"
] | [
"Definition:Class of All Ordinals",
"Definition:Class (Class Theory)",
"Definition:Element/Class",
"Definition:Element/Class"
] | [
"Set Difference is Subset",
"Definition:Subclass",
"Definition:Ordinal",
"Class of All Ordinals is Well-Ordered by Subset Relation",
"Definition:Smallest Element",
"Definition:Strictly Precede",
"Definition:Element/Class",
"Definition:Element/Class",
"Category:Transfinite Induction"
] |
proofwiki-7656 | Immediate Successor is Unique in Toset | Let $(S, \preceq)$ be a totally ordered set.
Let $x, y \in S$.
Suppose that $y$ is an immediate successor of $x$.
Then $y$ is the ''only'' immediate successor of $x$. | {{AimForCont}} that $x$ has another immediate successor, $z ≠ y$.
Then by definition:
:$x \prec y$
:$x \prec z$
Since $\preceq$ is a total ordering and $y ≠ z$:
:$y \prec z$ or $z \prec y$.
If $y \prec z$ then $x \prec y \prec z$, contradicting the fact that $z$ is an immediate successor of $x$.
If $z \prec y$ then $x ... | Let $(S, \preceq)$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $x, y \in S$.
Suppose that $y$ is an [[Definition:Immediate Successor Element|immediate successor]] of $x$.
Then $y$ is the ''only'' immediate successor of $x$. | {{AimForCont}} that $x$ has another [[Definition:Immediate Successor Element|immediate successor]], $z ≠ y$.
Then by definition:
:$x \prec y$
:$x \prec z$
Since $\preceq$ is a [[Definition:Total Ordering|total ordering]] and $y ≠ z$:
:$y \prec z$ or $z \prec y$.
If $y \prec z$ then $x \prec y \prec z$, contradicti... | Immediate Successor is Unique in Toset | https://proofwiki.org/wiki/Immediate_Successor_is_Unique_in_Toset | https://proofwiki.org/wiki/Immediate_Successor_is_Unique_in_Toset | [
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Immediate Successor Element"
] | [
"Definition:Immediate Successor Element",
"Definition:Total Ordering"
] |
proofwiki-7657 | Mapping from Set to Class of All Ordinals is Bounded Above | Let $x$ be a set.
Let $\On$ be the class of all ordinals.
Let $f: x \to \On$ be a mapping.
Then $f$ has an upper bound. | Let $I$ be the image of $f$.
By {{Corollary|Union of Set of Ordinals is Ordinal}}, $\bigcup I$ is an ordinal.
But by Union is Smallest Superset, each element of $I$ is a subset of $\bigcup I$.
Thus $\bigcup I$ is an upper bound of $f$.
{{qed}}
Category:Class of All Ordinals
Category:Class Mappings
dn066mvr3jb0qbefutzup... | Let $x$ be a [[Definition:Set|set]].
Let $\On$ be the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $f: x \to \On$ be a [[Definition:Mapping|mapping]].
Then $f$ has an [[Definition:Upper Bound of Mapping|upper bound]]. | Let $I$ be the [[Definition:Image of Mapping|image]] of $f$.
By {{Corollary|Union of Set of Ordinals is Ordinal}}, $\bigcup I$ is an [[Definition:Ordinal|ordinal]].
But by [[Union is Smallest Superset]], each element of $I$ is a [[Definition:Subset|subset]] of $\bigcup I$.
Thus $\bigcup I$ is an [[Definition:Upper B... | Mapping from Set to Class of All Ordinals is Bounded Above | https://proofwiki.org/wiki/Mapping_from_Set_to_Class_of_All_Ordinals_is_Bounded_Above | https://proofwiki.org/wiki/Mapping_from_Set_to_Class_of_All_Ordinals_is_Bounded_Above | [
"Class of All Ordinals",
"Class Mappings"
] | [
"Definition:Set",
"Definition:Class of All Ordinals",
"Definition:Mapping",
"Definition:Upper Bound of Mapping"
] | [
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Ordinal",
"Union is Smallest Superset",
"Definition:Subset",
"Definition:Upper Bound of Mapping",
"Category:Class of All Ordinals",
"Category:Class Mappings"
] |
proofwiki-7658 | Equivalence of Definitions of Independent Events | Let $\EE$ be an experiment with probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be events of $\EE$ such that $\map \Pr A > 0$ and $\map \Pr B > 0$.
{{TFAE|def = Independent Events}} | {{begin-eqn}}
{{eqn | l = \condprob A B
| r = \map \Pr A
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}
| r = \map \Pr A
| c = {{Defof|Conditional Probability}}
}}
{{eqn | ll= \leadstoandfrom
| l = \map \Pr {A \cap B}
| r = \map \Pr A \, ... | Let $\EE$ be an [[Definition:Experiment|experiment]] with [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
Let $A, B \in \Sigma$ be [[Definition:Event|events]] of $\EE$ such that $\map \Pr A > 0$ and $\map \Pr B > 0$.
{{TFAE|def = Independent Events}} | {{begin-eqn}}
{{eqn | l = \condprob A B
| r = \map \Pr A
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}
| r = \map \Pr A
| c = {{Defof|Conditional Probability}}
}}
{{eqn | ll= \leadstoandfrom
| l = \map \Pr {A \cap B}
| r = \map \Pr A \, ... | Equivalence of Definitions of Independent Events | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Independent_Events | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Independent_Events | [
"Independent Events"
] | [
"Definition:Experiment",
"Definition:Probability Space",
"Definition:Event"
] | [] |
proofwiki-7659 | Independent Events are Independent of Complement/General Result | Let $A_1, A_2, \ldots, A_m$ be events in a probability space $\struct {\Omega, \Sigma, \Pr}$.
Then $A_1, A_2, \ldots, A_m$ are independent {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m$ are also independent. | Proof by induction:
For all $n \in \N: n \ge 2$, let $\map P n$ be the proposition:
:$A_1, A_2, \ldots, A_n$ are independent {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$ are independent. | Let $A_1, A_2, \ldots, A_m$ be [[Definition:Event|events]] in a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
Then $A_1, A_2, \ldots, A_m$ are [[Definition:Independent Events/General Definition|independent]] {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setm... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N: n \ge 2$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$A_1, A_2, \ldots, A_n$ are [[Definition:Independent Events|independent]] {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$ are [[Defi... | Independent Events are Independent of Complement/General Result | https://proofwiki.org/wiki/Independent_Events_are_Independent_of_Complement/General_Result | https://proofwiki.org/wiki/Independent_Events_are_Independent_of_Complement/General_Result | [
"Independent Events"
] | [
"Definition:Event",
"Definition:Probability Space",
"Definition:Independent Events/General Definition",
"Definition:Independent Events/General Definition"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Independent Events",
"Definition:Independent Events/General Definition",
"Definition:Independent Events",
"Definition:Independent Events",
"Definition:Independent Events/General Definition",
"Definition:Independent Events/Gen... |
proofwiki-7660 | Probability of Independent Events Not Happening/Corollary | Let $A$ be an event in an event space of an experiment $\EE$ whose probability space is $\struct {\Omega, \Sigma, \Pr}$.
Let $\map \Pr A = p$.
Suppose that the nature of $\EE$ is that its outcome is independent of previous trials of $\EE$.
Then the probability that $A$ does not occur during the course of $m$ trials of ... | This is an instance of Probability of Independent Events Not Happening with all of $A_1, A_2, \ldots, A_m$ being instances of $A$.
The result follows directly.
{{qed}}
Category:Probability Theory
jfor173v5hwwv46csmqgmacah73w0e4 | Let $A$ be an [[Definition:Event|event]] in an [[Definition:Event Space|event space]] of an [[Definition:Experiment|experiment]] $\EE$ whose [[Definition:Probability Space|probability space]] is $\struct {\Omega, \Sigma, \Pr}$.
Let $\map \Pr A = p$.
Suppose that the nature of $\EE$ is that its [[Definition:Outcome|ou... | This is an instance of [[Probability of Independent Events Not Happening]] with all of $A_1, A_2, \ldots, A_m$ being instances of $A$.
The result follows directly.
{{qed}}
[[Category:Probability Theory]]
jfor173v5hwwv46csmqgmacah73w0e4 | Probability of Independent Events Not Happening/Corollary | https://proofwiki.org/wiki/Probability_of_Independent_Events_Not_Happening/Corollary | https://proofwiki.org/wiki/Probability_of_Independent_Events_Not_Happening/Corollary | [
"Probability Theory"
] | [
"Definition:Event",
"Definition:Event Space",
"Definition:Experiment",
"Definition:Probability Space",
"Definition:Elementary Event",
"Definition:Independent Events",
"Definition:Probability",
"Definition:Event/Occurrence"
] | [
"Probability of Independent Events Not Happening",
"Category:Probability Theory"
] |
proofwiki-7661 | Probability of Limit of Sequence of Events/Increasing | Let $\sequence {A_n}_{n \mathop \in \N}$ be an increasing sequence of events.
Let $\ds A = \bigcup_{i \mathop \in \N} A_i$ be the limit of $\sequence {A_n}_{n \mathop \in \N}$.
Then:
:$\ds \map \Pr A = \lim_{n \mathop \to \infty} \map \Pr {A_n}$ | Let $\ds B_i = A_i \setminus A_{i - 1}$ for $i \in \N: i > 0$.
Then:
:$A = A_0 \cup B_1 \cup B_2 \cup \cdots$
is the union of disjoint events in $\Sigma$.
By definition of probability measure:
{{begin-eqn}}
{{eqn | l = \map \Pr A
| r = \map \Pr {A_0} + \map \Pr {B_1} + \map \Pr {B_2} + \cdots
| c =
}}
{{eq... | Let $\sequence {A_n}_{n \mathop \in \N}$ be an [[Definition:Increasing Sequence of Events|increasing sequence of events]].
Let $\ds A = \bigcup_{i \mathop \in \N} A_i$ be the [[Definition:Limit of Sequence of Events|limit of $\sequence {A_n}_{n \mathop \in \N}$]].
Then:
:$\ds \map \Pr A = \lim_{n \mathop \to \infty}... | Let $\ds B_i = A_i \setminus A_{i - 1}$ for $i \in \N: i > 0$.
Then:
:$A = A_0 \cup B_1 \cup B_2 \cup \cdots$
is the [[Definition:Set Union|union]] of [[Definition:Disjoint Events|disjoint events]] in $\Sigma$.
By definition of [[Definition:Probability Measure|probability measure]]:
{{begin-eqn}}
{{eqn | l = \map \P... | Probability of Limit of Sequence of Events/Increasing | https://proofwiki.org/wiki/Probability_of_Limit_of_Sequence_of_Events/Increasing | https://proofwiki.org/wiki/Probability_of_Limit_of_Sequence_of_Events/Increasing | [
"Probability Theory"
] | [
"Definition:Increasing Sequence of Events",
"Definition:Limit of Sequence of Events"
] | [
"Definition:Set Union",
"Definition:Disjoint Events",
"Definition:Probability Measure",
"Telescoping Series/Example 2"
] |
proofwiki-7662 | Probability of Limit of Sequence of Events/Decreasing | Let $\sequence {B_n}_{n \mathop \in \N}$ be a decreasing sequence of events.
Let $\ds B = \bigcap_{i \mathop \in \N} B_i$ be the limit of $\sequence {B_n}_{n \mathop \in \N}$.
Then:
:$\ds \map \Pr B = \lim_{n \mathop \to \infty} \map \Pr {B_n}$ | {{expand|The below needs to be done properly.}}
Set $A_i = \Omega \setminus B_i$ and then apply De Morgan's laws and the result for an increasing sequence of events.
{{qed}} | Let $\sequence {B_n}_{n \mathop \in \N}$ be a [[Definition:Decreasing Sequence of Events|decreasing sequence of events]].
Let $\ds B = \bigcap_{i \mathop \in \N} B_i$ be the [[Definition:Limit of Sequence of Events|limit of $\sequence {B_n}_{n \mathop \in \N}$]].
Then:
:$\ds \map \Pr B = \lim_{n \mathop \to \infty} ... | {{expand|The below needs to be done properly.}}
Set $A_i = \Omega \setminus B_i$ and then apply [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union|De Morgan's laws]] and the [[Probability of Limit of Sequence of Events/Increasing|result for an increasing sequence of events]].
{{qed}} | Probability of Limit of Sequence of Events/Decreasing | https://proofwiki.org/wiki/Probability_of_Limit_of_Sequence_of_Events/Decreasing | https://proofwiki.org/wiki/Probability_of_Limit_of_Sequence_of_Events/Decreasing | [
"Probability Theory"
] | [
"Definition:Decreasing Sequence of Events",
"Definition:Limit of Sequence of Events"
] | [
"De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union",
"Probability of Limit of Sequence of Events/Increasing"
] |
proofwiki-7663 | Sum of Discrete Random Variables | Let $U: \Omega \to \R$ be defined as:
:$\forall \omega \in \Omega: \map U \omega = \map X \omega + \map Y \omega$
Then $U$ is also a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$. | To show that $U$ is discrete random variable on $\struct {\Omega, \Sigma, \Pr}$, we need to show that:
:$(1): \quad$ The image of $U$ is a countable subset of $\R$;
:$(2): \quad \forall x \in \R: \set {\omega \in \Omega: \map U \omega = x} \in \Sigma$.
First we consider any $U_u = \set {\omega \in \Omega: \map U \omega... | Let $U: \Omega \to \R$ be defined as:
:$\forall \omega \in \Omega: \map U \omega = \map X \omega + \map Y \omega$
Then $U$ is also a [[Definition:Discrete Random Variable|discrete random variable]] on $\struct {\Omega, \Sigma, \Pr}$. | To show that $U$ is [[Definition:Discrete Random Variable|discrete random variable]] on $\struct {\Omega, \Sigma, \Pr}$, we need to show that:
:$(1): \quad$ The [[Definition:Image of Mapping|image]] of $U$ is a [[Definition:Countable|countable]] [[Definition:Subset|subset]] of $\R$;
:$(2): \quad \forall x \in \R: \set... | Sum of Discrete Random Variables | https://proofwiki.org/wiki/Sum_of_Discrete_Random_Variables | https://proofwiki.org/wiki/Sum_of_Discrete_Random_Variables | [
"Discrete Random Variables"
] | [
"Definition:Random Variable/Discrete"
] | [
"Definition:Random Variable/Discrete",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Countable Set",
"Definition:Subset",
"Definition:Random Variable/Discrete",
"Definition:Probability Space",
"Definition:Set Union",
"Definition:Set Intersection",
"Definition:Random Variable/Discrete"... |
proofwiki-7664 | Product of Discrete Random Variables | Let $V: \Omega \to \R$ be defined as:
:$\forall \omega \in \Omega: \map V \omega = \map X \omega \map Y \omega$
Then $V$ is also a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$. | To show that $V$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$, we need to show that:
:$(1): \quad $The image of $U$ and $V$ are countable subsets of $\R$;
:$(2): \quad \forall x \in \R: \set {\omega \in \Omega: \map V \omega = x} \in \Sigma$.
First we consider any $V_V = \set {\omega \in \Omega: \ma... | Let $V: \Omega \to \R$ be defined as:
:$\forall \omega \in \Omega: \map V \omega = \map X \omega \map Y \omega$
Then $V$ is also a [[Definition:Discrete Random Variable|discrete random variable]] on $\struct {\Omega, \Sigma, \Pr}$. | To show that $V$ is a [[Definition:Discrete Random Variable|discrete random variable]] on $\struct {\Omega, \Sigma, \Pr}$, we need to show that:
:$(1): \quad $The [[Definition:Image of Mapping|image]] of $U$ and $V$ are [[Definition:Countable|countable]] [[Definition:Subset|subsets]] of $\R$;
:$(2): \quad \forall x \i... | Product of Discrete Random Variables | https://proofwiki.org/wiki/Product_of_Discrete_Random_Variables | https://proofwiki.org/wiki/Product_of_Discrete_Random_Variables | [
"Probability Theory"
] | [
"Definition:Random Variable/Discrete"
] | [
"Definition:Random Variable/Discrete",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Countable Set",
"Definition:Subset",
"Definition:Random Variable/Discrete",
"Definition:Probability Space",
"Definition:Set Union",
"Definition:Set Intersection",
"Definition:Random Variable/Discrete"... |
proofwiki-7665 | Bernoulli Process as Geometric Distribution/Shifted | Let $\sequence {Y_i}$ be a binomial experiment with parameter $p$.
Let $\EE$ be the experiment which consists of performing the Bernoulli trial $Y_i$ as many times as it takes to achieve a success, and then stop.
Let $k$ be the number of Bernoulli trials to achieve a success.
Then $k$ is modelled by a shifted geometric... | Follows directly from the definition of shifted geometric distribution.
Let $Y$ be the discrete random variable defined as the number of trials for the first success to be achieved.
Thus the last trial (and the last trial only) will be a success, and the others will be failures.
The probability that $k-1$ failures are ... | Let $\sequence {Y_i}$ be a [[Definition:Binomial Experiment|binomial experiment with parameter $p$]].
Let $\EE$ be the experiment which consists of performing the [[Definition:Bernoulli Trial|Bernoulli trial]] $Y_i$ as many times as it takes to achieve a [[Definition:Success|success]], and then stop.
Let $k$ be the n... | Follows directly from the definition of [[Definition:Shifted Geometric Distribution|shifted geometric distribution]].
Let $Y$ be the [[Definition:Discrete Random Variable|discrete random variable]] defined as the number of trials for the first [[Definition:Success|success]] to be achieved.
Thus the last trial (and th... | Bernoulli Process as Geometric Distribution/Shifted | https://proofwiki.org/wiki/Bernoulli_Process_as_Geometric_Distribution/Shifted | https://proofwiki.org/wiki/Bernoulli_Process_as_Geometric_Distribution/Shifted | [
"Bernoulli Distribution",
"Geometric Distribution"
] | [
"Definition:Binomial Experiment",
"Definition:Bernoulli Trial",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Trial",
"Definition:Bernoulli Distribution",
"Definition:Geometric Distribution/Shifted"
] | [
"Definition:Geometric Distribution/Shifted",
"Definition:Random Variable/Discrete",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Distribution"
] |
proofwiki-7666 | Probability Mass Function of Negative Binomial Distribution/Type 2 | Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the '''type $2$ negative binomial distribution with parameters $r$ and $p$'''.
Then the probability mass function of $X$ is given by:
:$\map \Pr {X = k} = \dbinom {k + r - 1} {r - 1} p^k \paren {1 - p}^r$
where:
:... | {{Recall|Negative Binomial Distribution (Type 2)}}
{{:Definition:Negative Binomial Distribution (Type 2)}}
The number of Bernoulli trials may be as few as $0$, so the image is correct:
:$\Img X = \set {0, 1, 2, \ldots}$
If $X$ takes the value $k$, then there must have been $k + r$ trials altogether.
So, after $r + k - ... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] on a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the '''[[Definition:Negative Binomial Distribution (Type 2)|type $2$ negative binomial distribution]] with parameters $r$ and $p$'''.
The... | {{Recall|Negative Binomial Distribution (Type 2)}}
{{:Definition:Negative Binomial Distribution (Type 2)}}
The number of [[Definition:Bernoulli Trial|Bernoulli trials]] may be as few as $0$, so the [[Definition:Image of Mapping|image]] is correct:
:$\Img X = \set {0, 1, 2, \ldots}$
If $X$ takes the value $k$, then t... | Probability Mass Function of Negative Binomial Distribution/Type 2 | https://proofwiki.org/wiki/Probability_Mass_Function_of_Negative_Binomial_Distribution/Type_2 | https://proofwiki.org/wiki/Probability_Mass_Function_of_Negative_Binomial_Distribution/Type_2 | [
"Probability Mass Function of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 2)",
"Probability Mass Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Space",
"Definition:Negative Binomial Distribution/Type 2",
"Definition:Probability Mass Function"
] | [
"Definition:Bernoulli Trial",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Experiment",
"Definition:Bernoulli Distribution",
"Definition:Experiment",
"Definition:Bernoulli Distribution",
"Definition:Probability",
"Definition:Event/Occurrence",
"Definition:Event",
"Definition:Binomi... |
proofwiki-7667 | Geometric Distribution Gives Rise to Probability Mass Function/Shifted | Let $Y$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $Y$ have the shifted geometric distribution with parameter $p$ (where $0 < p < 1$).
Then $Y$ gives rise to a probability mass function. | By definition:
:$\map \Omega Y = \N_{>0} = \set {1, 2, 3, \ldots}$
:$\map \Pr {Y = k} = p \paren {1 - p}^{k - 1}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
| r = \sum_{k \mathop \ge 1} p \paren {1 - p}^{k - 1}
| c = {{Defof|Shifted Geometric Distribution}}
}}
{{eqn | r = \sum_{j \mathop \ge 0} p \pare... | Let $Y$ be a [[Definition:Discrete Random Variable|discrete random variable]] on a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
Let $Y$ have the [[Definition:Shifted Geometric Distribution|shifted geometric distribution with parameter $p$]] (where $0 < p < 1$).
Then $Y$ gives r... | By [[Definition:Shifted Geometric Distribution|definition]]:
:$\map \Omega Y = \N_{>0} = \set {1, 2, 3, \ldots}$
:$\map \Pr {Y = k} = p \paren {1 - p}^{k - 1}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
| r = \sum_{k \mathop \ge 1} p \paren {1 - p}^{k - 1}
| c = {{Defof|Shifted Geometric Distributi... | Geometric Distribution Gives Rise to Probability Mass Function/Shifted | https://proofwiki.org/wiki/Geometric_Distribution_Gives_Rise_to_Probability_Mass_Function/Shifted | https://proofwiki.org/wiki/Geometric_Distribution_Gives_Rise_to_Probability_Mass_Function/Shifted | [
"Shifted Geometric Distribution",
"Probability Mass Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Space",
"Definition:Geometric Distribution/Shifted",
"Definition:Probability Mass Function"
] | [
"Definition:Geometric Distribution/Shifted",
"Geometric Distribution Gives Rise to Probability Mass Function"
] |
proofwiki-7668 | Probability Mass Function of Negative Binomial Distribution/Type 1 | Let $X$ be a discrete random variable on a probability space $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the '''type $1$ negative binomial distribution with parameters $r$ and $p$'''..
Then the probability mass function of $X$ is given by:
:$\map \Pr {X = k} = \dbinom {k - 1} {r - 1} p^r \paren {1 - p}^{k - r}$
where... | {{Recall|Negative Binomial Distribution (Type 1)}}
{{:Definition:Negative Binomial Distribution (Type 1)}}
First note that the number of Bernoulli trials has to be at least $r$, so the image is correct: $\Img X = \set {r, r + 1, r + 2, \ldots}$.
Now, note that if $X$ takes the value $k$, then in the first $k - 1$ trial... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] on a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
Let $X$ have the '''[[Definition:Negative Binomial Distribution (Type 1)|type $1$ negative binomial distribution]] with parameters $r$ and $p$'''..
Th... | {{Recall|Negative Binomial Distribution (Type 1)}}
{{:Definition:Negative Binomial Distribution (Type 1)}}
First note that the number of [[Definition:Bernoulli Trial|Bernoulli trials]] has to be at least $r$, so the [[Definition:Image of Mapping|image]] is correct: $\Img X = \set {r, r + 1, r + 2, \ldots}$.
Now, note... | Probability Mass Function of Negative Binomial Distribution/Type 1 | https://proofwiki.org/wiki/Probability_Mass_Function_of_Negative_Binomial_Distribution/Type_1 | https://proofwiki.org/wiki/Probability_Mass_Function_of_Negative_Binomial_Distribution/Type_1 | [
"Probability Mass Function of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 1)",
"Probability Mass Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Space",
"Definition:Negative Binomial Distribution/Type 1",
"Definition:Probability Mass Function"
] | [
"Definition:Bernoulli Trial",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Distribution",
"Definition:Experiment",
"Definition:Probability",
"Definition:Event/Occurrence",
"Definition:Event",
"Defin... |
proofwiki-7669 | Negative Binomial Distribution Gives Rise to Probability Mass Function/Type 2 | Let $X$ have the type $2$ negative binomial distribution with parameters $n$ and $p$ ($0 < p < 1$).
Then $X$ gives rise to a probability mass function. | By definition:
:$\Img X = \set {0, 1, 2, \ldots}$
:$\map \Pr {X = k} = \dbinom {n + k - 1} {n - 1} p^k \paren {1 - p}^n$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
| r = \sum_{k \mathop \ge n} \binom {n + k - 1} {n - 1} p^k \paren {1 - p}^n
| c =
}}
{{eqn | r = \paren {1 - p}^n \sum_{k \mathop \ge n} ... | Let $X$ have the [[Definition:Negative Binomial Distribution (Type 2)|type $2$ negative binomial distribution with parameters $n$ and $p$]] ($0 < p < 1$).
Then $X$ gives rise to a [[Definition:Probability Mass Function|probability mass function]]. | By [[Definition:Negative Binomial Distribution (Type 2)|definition]]:
:$\Img X = \set {0, 1, 2, \ldots}$
:$\map \Pr {X = k} = \dbinom {n + k - 1} {n - 1} p^k \paren {1 - p}^n$
Then:
{{begin-eqn}}
{{eqn | l = \map \Pr \Omega
| r = \sum_{k \mathop \ge n} \binom {n + k - 1} {n - 1} p^k \paren {1 - p}^n
| c... | Negative Binomial Distribution Gives Rise to Probability Mass Function/Type 2 | https://proofwiki.org/wiki/Negative_Binomial_Distribution_Gives_Rise_to_Probability_Mass_Function/Type_2 | https://proofwiki.org/wiki/Negative_Binomial_Distribution_Gives_Rise_to_Probability_Mass_Function/Type_2 | [
"Negative Binomial Distribution (Type 2)"
] | [
"Definition:Negative Binomial Distribution/Type 2",
"Definition:Probability Mass Function"
] | [
"Definition:Negative Binomial Distribution/Type 2",
"Symmetry Rule for Binomial Coefficients",
"Negated Upper Index of Binomial Coefficient",
"Binomial Theorem",
"Category:Negative Binomial Distribution (Type 2)"
] |
proofwiki-7670 | Negative Binomial Distribution as Generalized Geometric Distribution/Type 2 | The type $2$ negative binomial distribution is a generalization of the geometric distribution:
Let $\sequence {X_i}$ be a binomial experiment with parameter $p$.
Let $\EE$ be the experiment which consists of:
:Perform the Bernoulli trial $X_i$ until $n$ failures occur, and then stop.
Let $k$ be the number of successes ... | Consider the experiment $\EE$ as described.
By definition, $\EE$ is modelled by a type $2$ negative binomial distribution with parameters $n$ and $p$:
:$\forall k \in \Z, k \ge 0: \map \Pr {X = k} = \dbinom {n + k - 1} {n - 1} p^k q^n $
where $q = 1 - p$.
Now consider the experiment $\EE'$ as described.
By Bernoulli Pr... | The [[Definition:Negative Binomial Distribution (Type 2)|type $2$ negative binomial distribution]] is a generalization of the [[Definition:Geometric Distribution|geometric distribution]]:
Let $\sequence {X_i}$ be a [[Definition:Binomial Experiment|binomial experiment with parameter $p$]].
Let $\EE$ be the [[Definiti... | Consider the [[Definition:Experiment|experiment]] $\EE$ as described.
By definition, $\EE$ is modelled by a [[Definition:Negative Binomial Distribution (Type 2)|type $2$ negative binomial distribution with parameters $n$ and $p$]]:
:$\forall k \in \Z, k \ge 0: \map \Pr {X = k} = \dbinom {n + k - 1} {n - 1} p^k q^n $
w... | Negative Binomial Distribution as Generalized Geometric Distribution/Type 2 | https://proofwiki.org/wiki/Negative_Binomial_Distribution_as_Generalized_Geometric_Distribution/Type_2 | https://proofwiki.org/wiki/Negative_Binomial_Distribution_as_Generalized_Geometric_Distribution/Type_2 | [
"Negative Binomial Distribution as Generalized Geometric Distribution",
"Negative Binomial Distribution (Type 2)",
"Geometric Distribution"
] | [
"Definition:Negative Binomial Distribution/Type 2",
"Definition:Geometric Distribution",
"Definition:Binomial Experiment",
"Definition:Experiment",
"Definition:Bernoulli Trial",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Distribution",
"Definition... | [
"Definition:Experiment",
"Definition:Negative Binomial Distribution/Type 2",
"Definition:Experiment",
"Bernoulli Process as Geometric Distribution",
"Definition:Geometric Distribution"
] |
proofwiki-7671 | Negative Binomial Distribution as Generalized Geometric Distribution/Type 1 | The type $1$ negative binomial distribution is a generalization of the shifted geometric distribution:
Let $\sequence {Y_i}$ be a binomial experiment with parameter $p$.
Let $\FF$ be the experiment which consists of:
:Perform the Bernoulli trial $Y_i$ as many times as it takes to achieve $n$ successes, and then stop.
L... | Consider the experiment $\FF$ as described.
By definition, $\FF$ is modelled by a negative binomial distribution with parameters $n$ and $p$:
:$\ds \forall k \in \Z, k \ge n: \map \Pr {Y = k} = \binom {k - 1} {n - 1} q^{k - n} p^n$
where $q = 1 - p$.
Now consider the experiment $\FF'$ as described.
By Bernoulli Process... | The [[Definition:Negative Binomial Distribution (Type 1)|type $1$ negative binomial distribution]] is a generalization of the [[Definition:Shifted Geometric Distribution|shifted geometric distribution]]:
Let $\sequence {Y_i}$ be a [[Definition:Binomial Experiment|binomial experiment with parameter $p$]].
Let $\FF$ b... | Consider the experiment $\FF$ as described.
By definition, $\FF$ is modelled by a [[Definition:Negative Binomial Distribution (Type 1)|negative binomial distribution with parameters $n$ and $p$]]:
:$\ds \forall k \in \Z, k \ge n: \map \Pr {Y = k} = \binom {k - 1} {n - 1} q^{k - n} p^n$
where $q = 1 - p$.
Now conside... | Negative Binomial Distribution as Generalized Geometric Distribution/Type 1 | https://proofwiki.org/wiki/Negative_Binomial_Distribution_as_Generalized_Geometric_Distribution/Type_1 | https://proofwiki.org/wiki/Negative_Binomial_Distribution_as_Generalized_Geometric_Distribution/Type_1 | [
"Negative Binomial Distribution as Generalized Geometric Distribution",
"Negative Binomial Distribution (Type 1)",
"Geometric Distribution"
] | [
"Definition:Negative Binomial Distribution/Type 1",
"Definition:Geometric Distribution/Shifted",
"Definition:Binomial Experiment",
"Definition:Experiment",
"Definition:Bernoulli Trial",
"Definition:Bernoulli Distribution",
"Definition:Bernoulli Trial",
"Definition:Bernoulli Trial",
"Definition:Berno... | [
"Definition:Negative Binomial Distribution/Type 1",
"Bernoulli Process as Geometric Distribution/Shifted",
"Definition:Geometric Distribution/Shifted",
"Category:Negative Binomial Distribution as Generalized Geometric Distribution",
"Category:Negative Binomial Distribution (Type 1)",
"Category:Geometric D... |
proofwiki-7672 | Equivalence of Definitions of Variance of Discrete Random Variable | Let $X$ be a discrete random variable.
Let $\mu = \expect X$ be the expectation of $X$.
{{TFAE|def = Variance of Discrete Random Variable}} | === Definition 1 equivalent to Definition 2 ===
Let $\var X$ be defined as:
:$\var X := \expect {\paren {X - \expect X}^2}$
Let $\mu = \expect X$.
Let $\map f X = \paren {X - \mu}^2$ be considered as a function of $X$.
Then by applying Expectation of Function of Discrete Random Variable:
:$\ds \expect {\map f X} = \sum... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]].
Let $\mu = \expect X$ be the [[Definition:Expectation|expectation of $X$]].
{{TFAE|def = Variance of Discrete Random Variable}} | === Definition 1 equivalent to Definition 2 ===
Let $\var X$ be defined as:
:$\var X := \expect {\paren {X - \expect X}^2}$
Let $\mu = \expect X$.
Let $\map f X = \paren {X - \mu}^2$ be considered as a [[Definition:Function|function]] of $X$.
Then by applying [[Expectation of Function of Discrete Random Variable]]... | Equivalence of Definitions of Variance of Discrete Random Variable | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Variance_of_Discrete_Random_Variable | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Variance_of_Discrete_Random_Variable | [
"Variance"
] | [
"Definition:Random Variable/Discrete",
"Definition:Expectation"
] | [
"Definition:Function",
"Expectation of Function of Discrete Random Variable",
"Definition:Variance/Discrete/Definition 1",
"Definition:Variance/Discrete/Definition 2",
"Definition:Variance/Discrete/Definition 2"
] |
proofwiki-7673 | Derivative of Geometric Sequence/Corollary | :$\ds \sum_{n \mathop \ge 1} n \paren {n + 1} x^{n - 1} = \frac 2 {\paren {1 - x}^3}$ | We have from Power Rule for Derivatives that:
:$\ds \frac {\d} {\d x} \sum_{n \mathop \ge 1} \paren {n + 1} x^n = \sum_{n \mathop \ge 1} n \paren {n + 1} x^{n - 1}$
But from Sum of Infinite Geometric Sequence:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop \ge 1} \paren {n + 1} x^n
| r = \sum_{m \mathop \ge 2} m x^{m ... | :$\ds \sum_{n \mathop \ge 1} n \paren {n + 1} x^{n - 1} = \frac 2 {\paren {1 - x}^3}$ | We have from [[Power Rule for Derivatives]] that:
:$\ds \frac {\d} {\d x} \sum_{n \mathop \ge 1} \paren {n + 1} x^n = \sum_{n \mathop \ge 1} n \paren {n + 1} x^{n - 1}$
But from [[Sum of Infinite Geometric Sequence]]:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop \ge 1} \paren {n + 1} x^n
| r = \sum_{m \mathop \ge 2... | Derivative of Geometric Sequence/Corollary | https://proofwiki.org/wiki/Derivative_of_Geometric_Sequence/Corollary | https://proofwiki.org/wiki/Derivative_of_Geometric_Sequence/Corollary | [
"Analysis",
"Differential Calculus",
"Geometric Sequences"
] | [] | [
"Power Rule for Derivatives",
"Sum of Infinite Geometric Sequence",
"Power Rule for Derivatives",
"Derivative of Composite Function",
"Category:Analysis",
"Category:Differential Calculus",
"Category:Geometric Sequences"
] |
proofwiki-7674 | Equivalence of Definitions of Countable Set | Let $S$ be a set.
{{TFAE|def = Countable Set}} | === Definition 1 implies Definition 2 ===
Let $S$ be a countable set by Definition 1.
Then there is an injection $f: S \to \N$.
By Law of Excluded Middle, $S$ is finite or infinite.
If $S$ is finite, then it trivially satisfies Definition 2.
{{explain|Implicitly we take $S \cong f[S]$ here}}
If $S$ is infinite, then it... | Let $S$ be a [[Definition:Set|set]].
{{TFAE|def = Countable Set}} | === Definition 1 implies Definition 2 ===
Let $S$ be a countable set by [[Definition:Countable Set/Definition 1|Definition 1]].
Then there is an [[Definition:Injection|injection]] $f: S \to \N$.
By [[Law of Excluded Middle]], $S$ is [[Definition:Finite Set|finite]] or [[Definition:Infinite Set|infinite]].
If $S$ is... | Equivalence of Definitions of Countable Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Countable_Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Countable_Set | [
"Countable Sets"
] | [
"Definition:Set"
] | [
"Definition:Countable Set/Definition 1",
"Definition:Injection",
"Law of Excluded Middle",
"Definition:Finite Set",
"Definition:Infinite Set",
"Definition:Finite Set",
"Definition:Countable Set/Definition 2",
"Definition:Infinite Set",
"Definition:Countably Infinite/Set",
"Infinite Set of Natural ... |
proofwiki-7675 | Infinite Set of Natural Numbers is Countably Infinite | Let $\N$ be the set of natural numbers.
Let $S$ be an infinite subset of $\N$.
Then $S$ is countably infinite.
That is, there is a bijection $f: \N \to S$. | By Infinite Set has Countably Infinite Subset, we have an injection $g: \N \to S$
But by Cantor-Bernstein-Schröder Theorem/Lemma this produces a bijection $f: \N \to S$
{{qed}} | Let $\N$ be the [[Definition:Natural Numbers|set of natural numbers]].
Let $S$ be an [[Definition:Infinite Set|infinite]] [[Definition:subset|subset]] of $\N$.
Then $S$ is [[Definition:Countably Infinite Set|countably infinite]].
That is, there is a [[Definition:bijection|bijection]] $f: \N \to S$. | By [[Infinite Set has Countably Infinite Subset]], we have an [[Definition:injection|injection]] $g: \N \to S$
But by [[Cantor-Bernstein-Schröder Theorem/Lemma]] this produces a [[Definition:bijection|bijection]] $f: \N \to S$
{{qed}} | Infinite Set of Natural Numbers is Countably Infinite | https://proofwiki.org/wiki/Infinite_Set_of_Natural_Numbers_is_Countably_Infinite | https://proofwiki.org/wiki/Infinite_Set_of_Natural_Numbers_is_Countably_Infinite | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Infinite Set",
"Definition:subset",
"Definition:Countably Infinite/Set",
"Definition:bijection"
] | [
"Infinite Set has Countably Infinite Subset",
"Definition:injection",
"Cantor-Bernstein-Schröder Theorem/Lemma",
"Definition:bijection"
] |
proofwiki-7676 | Generating Function for Powers of Two | Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N: a_n = 2^n$
That is:
:$\sequence {a_n} = 1, 2, 4, 8, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac 1 {1 - 2 z}$ for $\size z < \dfrac 1 2$ | {{begin-eqn}}
{{eqn | o =
| r = 1 + 2 z + 4 z^2 + \cdots
| c =
}}
{{eqn | r = \sum_{n \mathop \ge 0} \paren {2 z}^n
| c =
}}
{{eqn | r = \frac 1 {1 - 2 z}
| c = Sum of Infinite Geometric Sequence
}}
{{end-eqn}}
This is valid for:
:$\size {2 z} < 1$
from which:
:$\size z < \dfrac 1 2$
follows ... | Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
:$\forall n \in \N: a_n = 2^n$
That is:
:$\sequence {a_n} = 1, 2, 4, 8, \ldots$
Then the [[Definition:Generating Function|generating function]] for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac 1 {1 - 2 z}$ for $\size z < \dfrac 1 2$ | {{begin-eqn}}
{{eqn | o =
| r = 1 + 2 z + 4 z^2 + \cdots
| c =
}}
{{eqn | r = \sum_{n \mathop \ge 0} \paren {2 z}^n
| c =
}}
{{eqn | r = \frac 1 {1 - 2 z}
| c = [[Sum of Infinite Geometric Sequence]]
}}
{{end-eqn}}
This is valid for:
:$\size {2 z} < 1$
from which:
:$\size z < \dfrac 1 2$
fo... | Generating Function for Powers of Two | https://proofwiki.org/wiki/Generating_Function_for_Powers_of_Two | https://proofwiki.org/wiki/Generating_Function_for_Powers_of_Two | [
"Examples of Generating Functions"
] | [
"Definition:Sequence",
"Definition:Generating Function"
] | [
"Sum of Infinite Geometric Sequence",
"Definition:Generating Function"
] |
proofwiki-7677 | Generating Function for Natural Numbers | Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{> 0}: a_n = n - 1$
That is:
:$\sequence {a_n} = 0, 1, 2, 3, 4, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$G \paren z = \dfrac 1 {\paren {1 - z}^2}$ | Take the sequence:
:$S_n = 1, 1, 1, \ldots$
From Generating Function for Constant Sequence, this has the generating function:
:$\ds G \paren z = \sum_{n \mathop = 1}^\infty z^n = \frac 1 {1 - z}$
By Derivative of Generating Function:
:$\ds \dfrac \d {\d z} G \paren z = 0 + 1 + 2 z + 3 z^2 \cdots = \sum_{n \mathop \ge 0... | Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
:$\forall n \in \N_{> 0}: a_n = n - 1$
That is:
:$\sequence {a_n} = 0, 1, 2, 3, 4, \ldots$
Then the [[Definition:Generating Function|generating function]] for $\sequence {a_n}$ is given as:
:$G \paren z = \dfrac 1 {\paren {1 - z}^2}$ | Take the [[Definition:Sequence|sequence]]:
:$S_n = 1, 1, 1, \ldots$
From [[Generating Function for Constant Sequence]], this has the [[Definition:Generating Function|generating function]]:
:$\ds G \paren z = \sum_{n \mathop = 1}^\infty z^n = \frac 1 {1 - z}$
By [[Derivative of Generating Function]]:
:$\ds \dfrac ... | Generating Function for Natural Numbers | https://proofwiki.org/wiki/Generating_Function_for_Natural_Numbers | https://proofwiki.org/wiki/Generating_Function_for_Natural_Numbers | [
"Examples of Generating Functions",
"Natural Numbers"
] | [
"Definition:Sequence",
"Definition:Generating Function"
] | [
"Definition:Sequence",
"Generating Function for Constant Sequence",
"Definition:Generating Function",
"Derivative of Generating Function",
"Definition:Generating Function",
"Definition:Sequence",
"Power Rule for Derivatives",
"Derivative of Composite Function",
"Definition:Generating Function"
] |
proofwiki-7678 | Injection has Surjective Left Inverse Mapping | Let $S$ and $T$ be sets such that $S \ne \O$.
Let $f: S \to T$ be a injection.
Then there exists a surjection $g: T \to S$ such that:
:$g \circ f = I_S$ | Since $S$ is non-empty, we can choose an element $x \in S$.
Since $f$ is an injection, for each $t \in \Img f$ there exists a unique $s \in S$ such that:
:$\map f s = t$
Thus by Law of Excluded Middle there exists a well-defined mapping $T \to S$ given by:
:$\map g t = \begin {cases} s & : \paren {t \in \Img f} \land \... | Let $S$ and $T$ be [[Definition:Set|sets]] such that $S \ne \O$.
Let $f: S \to T$ be a [[Definition:Injection|injection]].
Then there exists a [[Definition:Surjection|surjection]] $g: T \to S$ such that:
:$g \circ f = I_S$ | Since $S$ is [[Definition:Non-Empty Set|non-empty]], we can choose an [[Definition:Element|element]] $x \in S$.
Since $f$ is an [[Definition:injection|injection]], for each $t \in \Img f$ there exists a [[Definition:Unique|unique]] $s \in S$ such that:
:$\map f s = t$
Thus by [[Law of Excluded Middle]] there exists a... | Injection has Surjective Left Inverse Mapping/Proof 1 | https://proofwiki.org/wiki/Injection_has_Surjective_Left_Inverse_Mapping | https://proofwiki.org/wiki/Injection_has_Surjective_Left_Inverse_Mapping/Proof_1 | [
"Injections",
"Surjections",
"Left Inverse Mappings",
"Injection has Surjective Left Inverse Mapping"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Surjection"
] | [
"Definition:Non-Empty Set",
"Definition:Element",
"Definition:injection",
"Definition:Unique",
"Law of Excluded Middle",
"Definition:Well-Defined/Mapping",
"Definition:Element",
"Definition:Surjection"
] |
proofwiki-7679 | Injection has Surjective Left Inverse Mapping | Let $S$ and $T$ be sets such that $S \ne \O$.
Let $f: S \to T$ be a injection.
Then there exists a surjection $g: T \to S$ such that:
:$g \circ f = I_S$ | By Injection iff Left Inverse, $f$ has a left inverse $g: T \to S$.
By Left Inverse Mapping is Surjection, $g$ is a surjection.
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]] such that $S \ne \O$.
Let $f: S \to T$ be a [[Definition:Injection|injection]].
Then there exists a [[Definition:Surjection|surjection]] $g: T \to S$ such that:
:$g \circ f = I_S$ | By [[Injection iff Left Inverse]], $f$ has a [[Definition:Left Inverse Mapping|left inverse]] $g: T \to S$.
By [[Left Inverse Mapping is Surjection]], $g$ is a [[Definition:Surjection|surjection]].
{{qed}} | Injection has Surjective Left Inverse Mapping/Proof 2 | https://proofwiki.org/wiki/Injection_has_Surjective_Left_Inverse_Mapping | https://proofwiki.org/wiki/Injection_has_Surjective_Left_Inverse_Mapping/Proof_2 | [
"Injections",
"Surjections",
"Left Inverse Mappings",
"Injection has Surjective Left Inverse Mapping"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Surjection"
] | [
"Injection iff Left Inverse",
"Definition:Left Inverse Mapping",
"Left Inverse Mapping is Surjection",
"Definition:Surjection"
] |
proofwiki-7680 | Many-to-One Relation Extends to Mapping | Let $S$ and $T$ be sets.
Let $T$ be non-empty.
Let $\RR \subset S \times T$ be a many-to-one relation.
Then there exists a mapping $f: S \to T$ such that $\RR \subseteq f$. | Since $T$ is not empty, it has an element $t_0$.
Define a mapping $g: S \setminus \Preimg \RR$ by letting $\map g x = t_0$ for all $x \in S$.
Let $f = \RR \cup g$ be a relation on $S \times T$.
By Union of Many-to-One Relations with Disjoint Domains is Many-to-One, $f$ is a many-to-one relation.
By Union with Relative ... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $T$ be [[Definition:Non-Empty Set|non-empty]].
Let $\RR \subset S \times T$ be a [[Definition:Many-to-One Relation|many-to-one relation]].
Then there exists a [[Definition:Mapping|mapping]] $f: S \to T$ such that $\RR \subseteq f$. | Since $T$ is not [[Definition:Empty Set|empty]], it has an [[Definition:Element|element]] $t_0$.
Define a [[Definition:Mapping|mapping]] $g: S \setminus \Preimg \RR$ by letting $\map g x = t_0$ for all $x \in S$.
Let $f = \RR \cup g$ be a [[Definition:Relation|relation]] on $S \times T$.
By [[Union of Many-to-One Re... | Many-to-One Relation Extends to Mapping | https://proofwiki.org/wiki/Many-to-One_Relation_Extends_to_Mapping | https://proofwiki.org/wiki/Many-to-One_Relation_Extends_to_Mapping | [
"Relation Theory",
"Mapping Theory"
] | [
"Definition:Set",
"Definition:Non-Empty Set",
"Definition:Many-to-One Relation",
"Definition:Mapping"
] | [
"Definition:Empty Set",
"Definition:Element",
"Definition:Mapping",
"Definition:Relation",
"Union of Many-to-One Relations with Disjoint Domains is Many-to-One",
"Definition:Many-to-One Relation",
"Union with Relative Complement",
"Definition:Total Relation",
"Definition:Mapping",
"Category:Relati... |
proofwiki-7681 | Generating Function for Binomial Coefficients | Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N: a_n = \begin{cases}
\dbinom m n & : n = 0, 1, 2, \ldots, m \\
0 & : \text{otherwise}\end{cases}$
where $\dbinom m n$ denotes a binomial coefficient.
Then the generating function for $\sequence {a_n}$ is given as:
:$\ds \map G z = \sum_{n \mathop = 0}... | {{begin-eqn}}
{{eqn | l = \paren {1 + z}^m
| r = \sum_{n \mathop = 0}^m \binom m n z^n
| c = Binomial Theorem
}}
{{end-eqn}}
The result follows from the definition of a generating function.
{{qed}} | Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
:$\forall n \in \N: a_n = \begin{cases}
\dbinom m n & : n = 0, 1, 2, \ldots, m \\
0 & : \text{otherwise}\end{cases}$
where $\dbinom m n$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
Then the [[Definition:Generating Funct... | {{begin-eqn}}
{{eqn | l = \paren {1 + z}^m
| r = \sum_{n \mathop = 0}^m \binom m n z^n
| c = [[Binomial Theorem]]
}}
{{end-eqn}}
The result follows from the definition of a [[Definition:Generating Function|generating function]].
{{qed}} | Generating Function for Binomial Coefficients | https://proofwiki.org/wiki/Generating_Function_for_Binomial_Coefficients | https://proofwiki.org/wiki/Generating_Function_for_Binomial_Coefficients | [
"Examples of Generating Functions"
] | [
"Definition:Sequence",
"Definition:Binomial Coefficient",
"Definition:Generating Function"
] | [
"Binomial Theorem",
"Definition:Generating Function"
] |
proofwiki-7682 | Probability Generating Function of Zero | :$\map {\Pi_X} 0 = \map {p_X} 0$ | {{begin-eqn}}
{{eqn | l = \map {\Pi_X} 0
| r = \map {p_X} 0 + 0^1 \cdot \map {p_X} 1 + 0^2 \cdot \map {p_X} 2 + \cdots
| c = {{Defof|Probability Generating Function}}
}}
{{eqn | r = \map {p_X} 0 + 0 + 0 + \cdots
| c = as $\forall n \in \N_{>0}: s^n = 0$
}}
{{eqn | r = \map {p_X} 0
| c =
}}
{{en... | :$\map {\Pi_X} 0 = \map {p_X} 0$ | {{begin-eqn}}
{{eqn | l = \map {\Pi_X} 0
| r = \map {p_X} 0 + 0^1 \cdot \map {p_X} 1 + 0^2 \cdot \map {p_X} 2 + \cdots
| c = {{Defof|Probability Generating Function}}
}}
{{eqn | r = \map {p_X} 0 + 0 + 0 + \cdots
| c = as $\forall n \in \N_{>0}: s^n = 0$
}}
{{eqn | r = \map {p_X} 0
| c =
}}
{{en... | Probability Generating Function of Zero | https://proofwiki.org/wiki/Probability_Generating_Function_of_Zero | https://proofwiki.org/wiki/Probability_Generating_Function_of_Zero | [
"Probability Generating Functions"
] | [] | [] |
proofwiki-7683 | Probability Generating Function of One | :$\map {\Pi_X} 1 = 1$ | {{begin-eqn}}
{{eqn | l=\map {\Pi_X} 1
| r=\map {p_X} 0 + 1^1 \cdot \map {p_X} 1 + 1^2 \cdot \map {p_X} 2 + \cdots
| c={{Defof|Probability Generating Function}}
}}
{{eqn | r=\map {p_X} 0 + \map {p_X} 1 + \map {p_X} 2 + \cdots
| c=as $\forall n \in \N_{>0}: s^n = 1$
}}
{{eqn | r=\sum_{n \mathop = 0}^\i... | :$\map {\Pi_X} 1 = 1$ | {{begin-eqn}}
{{eqn | l=\map {\Pi_X} 1
| r=\map {p_X} 0 + 1^1 \cdot \map {p_X} 1 + 1^2 \cdot \map {p_X} 2 + \cdots
| c={{Defof|Probability Generating Function}}
}}
{{eqn | r=\map {p_X} 0 + \map {p_X} 1 + \map {p_X} 2 + \cdots
| c=as $\forall n \in \N_{>0}: s^n = 1$
}}
{{eqn | r=\sum_{n \mathop = 0}^\i... | Probability Generating Function of One | https://proofwiki.org/wiki/Probability_Generating_Function_of_One | https://proofwiki.org/wiki/Probability_Generating_Function_of_One | [
"Probability Generating Functions"
] | [] | [] |
proofwiki-7684 | Probability Generating Function as Expectation | Let $X$ be a discrete random variable whose codomain, $\Omega_X$, is a subset of the natural numbers $\N$.
Let $p_X$ be the probability mass function for $X$.
Let $\map {\Pi_X} s$ be the probability generating function for $X$.
Then:
:$\map {\Pi_X} s = \expect {s^X}$
where $\expect {s^X}$ denotes the expectation of $s^... | By definition of probability generating function:
:$\ds \map {\Pi_X} s := \sum_{n \mathop = 0}^\infty \map {p_X} n s^n = \map {p_X} 0 + \map {p_X} 1 s + \map {p_X} 2 s^2 + \cdots$
where $p_X$ is the probability mass function for $X$.
For any real function $g: \R \to \R$, by Expectation of Function of Discrete Random Va... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] whose [[Definition:Codomain of Mapping|codomain]], $\Omega_X$, is a [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]] $\N$.
Let $p_X$ be the [[Definition:Probability Mass Function|probability mass function]] ... | By definition of [[Definition:Probability Generating Function|probability generating function]]:
:$\ds \map {\Pi_X} s := \sum_{n \mathop = 0}^\infty \map {p_X} n s^n = \map {p_X} 0 + \map {p_X} 1 s + \map {p_X} 2 s^2 + \cdots$
where $p_X$ is the [[Definition:Probability Mass Function|probability mass function]] for $X... | Probability Generating Function as Expectation | https://proofwiki.org/wiki/Probability_Generating_Function_as_Expectation | https://proofwiki.org/wiki/Probability_Generating_Function_as_Expectation | [
"Probability Generating Functions",
"Expectation"
] | [
"Definition:Random Variable/Discrete",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Subset",
"Definition:Natural Numbers",
"Definition:Probability Mass Function",
"Definition:Probability Generating Function",
"Definition:Expectation"
] | [
"Definition:Probability Generating Function",
"Definition:Probability Mass Function",
"Definition:Real Function",
"Expectation of Function of Discrete Random Variable",
"Definition:Absolutely Convergent Series",
"Definition:Absolutely Convergent Series",
"Definition:Probability Mass Function",
"Defini... |
proofwiki-7685 | Probability Generating Function defines Probability Distribution | Let $X$ and $Y$ be discrete random variables whose codomain, $\Omega_X$, is a subset of the natural numbers $\N$.
Let the probability generating functions of $X$ and $Y$ be $\map {\Pi_X} s$ and $\map {\Pi_Y} s$ respectively.
Then:
:$\forall s \in \closedint {-1} 1: \map {\Pi_X} s = \map {\Pi_Y} s$
{{iff}}:
:$\forall k... | By the definition of PGF, it follows that if:
:$\forall k \in \N: \map \Pr {X = k} = \map \Pr {Y = k}$
then:
:$\forall s \in \closedint {-1} 1: \map {\Pi_X} s = \map {\Pi_Y} s$
Suppose that $\map {\Pi_X} s = \map {\Pi_Y} s$ for all $s \in \closedint {-1} 1$.
From Probability Generating Function as Expectation the radiu... | Let $X$ and $Y$ be [[Definition:Discrete Random Variable|discrete random variables]] whose [[Definition:Codomain of Mapping|codomain]], $\Omega_X$, is a [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]] $\N$.
Let the [[Definition:Probability Generating Function|probability generating ... | By the definition of [[Definition:Probability Generating Function|PGF]], it follows that if:
:$\forall k \in \N: \map \Pr {X = k} = \map \Pr {Y = k}$
then:
:$\forall s \in \closedint {-1} 1: \map {\Pi_X} s = \map {\Pi_Y} s$
Suppose that $\map {\Pi_X} s = \map {\Pi_Y} s$ for all $s \in \closedint {-1} 1$.
From [[Prob... | Probability Generating Function defines Probability Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_defines_Probability_Distribution | https://proofwiki.org/wiki/Probability_Generating_Function_defines_Probability_Distribution | [
"Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Subset",
"Definition:Natural Numbers",
"Definition:Probability Generating Function",
"Definition:Random Variable/Discrete",
"Definition:Integer",
"Definition:Probability Generating Function",
"Definition:P... | [
"Definition:Probability Generating Function",
"Probability Generating Function as Expectation",
"Definition:Radius of Convergence",
"Definition:Unique",
"Definition:Power Series Expansion"
] |
proofwiki-7686 | Probability Generating Function of Negative Binomial Distribution/Type 2 | Let $X$ be a discrete random variable with the type $2$ negative binomial distribution with parameters $n$ and $p$.
Then the p.g.f. of $X$ is:
:$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^n$
where $q = 1 - p$. | From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map {p_X} k s^k$
From the definition of the type $2$ negative binomial distribution:
:$\map {p_X} k = \dbinom {n + k - 1} {n - 1} p^k q^n$
where $q = 1 - p$.
So:
{{begin-eqn}}
{{eqn | l = \map {\Pi_X} s
| r = \sum_{k \mathop \ge 0} \bino... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 2)|type $2$ negative binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Probability Generating Function|p.g.f.]] of $X$ is:
:$\map {\Pi_X} s = \paren {\dfrac... | From the definition of [[Definition:Probability Generating Function|p.g.f]]:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map {p_X} k s^k$
From the definition of the [[Definition:Negative Binomial Distribution (Type 2)|type $2$ negative binomial distribution]]:
:$\map {p_X} k = \dbinom {n + k - 1} {n - 1} p^k q^n$... | Probability Generating Function of Negative Binomial Distribution/Type 2 | https://proofwiki.org/wiki/Probability_Generating_Function_of_Negative_Binomial_Distribution/Type_2 | https://proofwiki.org/wiki/Probability_Generating_Function_of_Negative_Binomial_Distribution/Type_2 | [
"Probability Generating Function of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 2)",
"Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 2",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Definition:Negative Binomial Distribution/Type 2"
] |
proofwiki-7687 | Probability Generating Function of Negative Binomial Distribution/Type 1 | Let $X$ be a discrete random variable with the type 1 negative binomial distribution with parameters $n$ and $p$.
Then the p.g.f. of $X$ is:
:$\ds \map {\Pi_X} s = \paren {\frac {p s} {1 - q s} }^n$
where $q = 1 - p$. | From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map {p_X} k s^k$
From the definition of the type $1$ negative binomial distribution:
:$\map {p_X} k = \dbinom {k - 1} {n - 1} p^n q^{k - n}$
where $q = 1 - p$.
So:
{{begin-eqn}}
{{eqn | l = \map {\Pi_X} s
| r = \sum_{k \mathop \ge n} \bi... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Negative Binomial Distribution (Type 1)|type 1 negative binomial distribution with parameters $n$ and $p$]].
Then the [[Definition:Probability Generating Function|p.g.f.]] of $X$ is:
:$\ds \map {\Pi_X} s = \paren {\fra... | From the definition of [[Definition:Probability Generating Function|p.g.f]]:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map {p_X} k s^k$
From the definition of the [[Definition:Negative Binomial Distribution (Type 1)|type $1$ negative binomial distribution]]:
:$\map {p_X} k = \dbinom {k - 1} {n - 1} p^n q^{k - n... | Probability Generating Function of Negative Binomial Distribution/Type 1 | https://proofwiki.org/wiki/Probability_Generating_Function_of_Negative_Binomial_Distribution/Type_1 | https://proofwiki.org/wiki/Probability_Generating_Function_of_Negative_Binomial_Distribution/Type_1 | [
"Probability Generating Function of Negative Binomial Distribution",
"Negative Binomial Distribution (Type 1)",
"Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Negative Binomial Distribution/Type 1",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Definition:Negative Binomial Distribution/Type 1"
] |
proofwiki-7688 | Fundamental Theorem of Calculus for Contour Integrals/Corollary | Let $D \subseteq \C$ be an open set.
Let $f: D \to \C$ be a continuous function.
Suppose that $F: D \to \C$ is an antiderivative of $f$.
Let $\gamma: \closedint a b \to D$ be a contour that consists of one directed smooth curve.
Then the contour integral:
:$\ds \int_\gamma \map f z \rd z = \map F {\map \gamma b} - \ma... | By the chain rule:
:$\dfrac \d {\d t} \map F {\map \gamma t} = \map {F'} {\map \gamma t} \map {\gamma'} t = \map f {\map \gamma t} \map {\gamma'} t$
Thus:
{{begin-eqn}}
{{eqn | l = \int_\gamma \map f z \rd z
| r = \int_a^b \map f {\map \gamma t} \map {\gamma'} t \rd t
| c = {{Defof|Complex Primitive|Antider... | Let $D \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open set]].
Let $f: D \to \C$ be a [[Definition:Continuous Complex Function|continuous function]].
Suppose that $F: D \to \C$ is an [[Definition:Complex Primitive|antiderivative]] of $f$.
Let $\gamma: \closedint a b \to D$ be a [[Definition:Contour ... | By the [[Derivative of Complex Composite Function|chain rule]]:
:$\dfrac \d {\d t} \map F {\map \gamma t} = \map {F'} {\map \gamma t} \map {\gamma'} t = \map f {\map \gamma t} \map {\gamma'} t$
Thus:
{{begin-eqn}}
{{eqn | l = \int_\gamma \map f z \rd z
| r = \int_a^b \map f {\map \gamma t} \map {\gamma'} t \rd t... | Fundamental Theorem of Calculus for Contour Integrals/Corollary | https://proofwiki.org/wiki/Fundamental_Theorem_of_Calculus_for_Contour_Integrals/Corollary | https://proofwiki.org/wiki/Fundamental_Theorem_of_Calculus_for_Contour_Integrals/Corollary | [
"Fundamental Theorem of Calculus for Contour Integrals"
] | [
"Definition:Open Set/Complex Analysis",
"Definition:Continuous Complex Function",
"Definition:Primitive (Calculus)/Complex",
"Definition:Contour/Complex Plane",
"Definition:Directed Smooth Curve/Complex Plane",
"Definition:Contour Integral/Complex"
] | [
"Derivative of Complex Composite Function",
"Fundamental Theorem of Calculus",
"Category:Fundamental Theorem of Calculus for Contour Integrals"
] |
proofwiki-7689 | Probability Generating Function of Scalar Multiple of Random Variable | Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Let $k \in \Z_{\ge 0}$ be a positive integer.
Let $Y$ be a discrete random variable such that $Y = m X$.
Then
:$\map {\Pi_Y} s = \map {\Pi_X} {s^m}$
where $\map {\Pi_Y} s$ is the probability generating function of $Y$. | From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map \Pr {X = k} s^k$
We have {{hypothesis}}:
:$\map \Pr {Y = m k} = \map \Pr {X = k}$
Thus:
:$\ds \map {\Pi_Y} s = \sum_{m k \mathop \ge 0} \map \Pr {X = k} s^{m k}$
From the definition of a probability generating function:
:$\map {\Pi_Y} s = ... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] whose [[Definition:Probability Generating Function|probability generating function]] is $\map {\Pi_X} s$.
Let $k \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Let $Y$ be a [[Definition:Discrete Random Variable|discr... | From the definition of [[Definition:Probability Generating Function|p.g.f]]:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map \Pr {X = k} s^k$
We have {{hypothesis}}:
:$\map \Pr {Y = m k} = \map \Pr {X = k}$
Thus:
:$\ds \map {\Pi_Y} s = \sum_{m k \mathop \ge 0} \map \Pr {X = k} s^{m k}$
From the definition of a... | Probability Generating Function of Scalar Multiple of Random Variable | https://proofwiki.org/wiki/Probability_Generating_Function_of_Scalar_Multiple_of_Random_Variable | https://proofwiki.org/wiki/Probability_Generating_Function_of_Scalar_Multiple_of_Random_Variable | [
"Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Generating Function",
"Definition:Positive/Integer",
"Definition:Random Variable/Discrete",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Definition:Probability Generating Function"
] |
proofwiki-7690 | Probability Generating Function of Shifted Random Variable | Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Let $k \in \Z_{\ge 0}$ be a positive integer.
Let $Y$ be a discrete random variable such that $Y = X + m$.
Then
:$\map {\Pi_Y} s = s^m \map {\Pi_X} s$
where $\map {\Pi_Y} s$ is the probability generating function of $Y$. | From the definition of p.g.f:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map \Pr {X = k} s^k$
By hypothesis:
:$\map \Pr {Y = k + m} = \map \Pr {X = k}$
Thus:
{{begin-eqn}}
{{eqn | l = \map {\Pi_Y} s
| r = \sum_{k + m \mathop \ge 0} \map \Pr {X = k} s^{k + m}
| c =
}}
{{eqn | r = \sum_{k + m \mathop \ge... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] whose [[Definition:Probability Generating Function|probability generating function]] is $\map {\Pi_X} s$.
Let $k \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Let $Y$ be a [[Definition:Discrete Random Variable|discr... | From the definition of [[Definition:Probability Generating Function|p.g.f]]:
:$\ds \map {\Pi_X} s = \sum_{k \mathop \ge 0} \map \Pr {X = k} s^k$
[[Definition:By Hypothesis|By hypothesis]]:
:$\map \Pr {Y = k + m} = \map \Pr {X = k}$
Thus:
{{begin-eqn}}
{{eqn | l = \map {\Pi_Y} s
| r = \sum_{k + m \mathop \ge 0... | Probability Generating Function of Shifted Random Variable | https://proofwiki.org/wiki/Probability_Generating_Function_of_Shifted_Random_Variable | https://proofwiki.org/wiki/Probability_Generating_Function_of_Shifted_Random_Variable | [
"Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Generating Function",
"Definition:Positive/Integer",
"Definition:Random Variable/Discrete",
"Definition:Probability Generating Function"
] | [
"Definition:Probability Generating Function",
"Definition:By Hypothesis",
"Translation of Index Variable of Summation",
"Definition:Probability Generating Function"
] |
proofwiki-7691 | Derivatives of Probability Generating Function at One | Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Then the $n$th derivative of $\map {\Pi_X} s$ at $s = 1$ is given by:
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$
for $n = 1, 2, \ldots$ | Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$ | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] whose [[Definition:Probability Generating Function|probability generating function]] is $\map {\Pi_X} s$.
Then the $n$th [[Definition:Higher Derivative|derivative]] of $\map {\Pi_X} s$ at $s = 1$ is given by:
:$\dfrac {\d^n} {\d s^n} \map {... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$ | Derivatives of Probability Generating Function at One | https://proofwiki.org/wiki/Derivatives_of_Probability_Generating_Function_at_One | https://proofwiki.org/wiki/Derivatives_of_Probability_Generating_Function_at_One | [
"Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Generating Function",
"Definition:Derivative/Higher Derivatives/Higher Order"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-7692 | Expectation of Square of Discrete Random Variable | Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.
Then the square of the expectation of $X$ is given by the expression:
:$\expect {X^2} = \map { {\Pi_X}''} 1 + \map { {\Pi_X}'} 1$
where $\map { {\Pi_X}''} 1$ and $\map { {\Pi_X}'} 1$ denote the second and first derivative r... | From Derivatives of Probability Generating Function at One:
:$\map { {\Pi_X}''} 1 = \expect {X \paren {X - 1} }$
and from Expectation of Discrete Random Variable from PGF:
:$\map { {\Pi_X}'} 1 = \expect X$
So:
{{begin-eqn}}
{{eqn | l = \expect {X^2}
| r = \expect {X \paren {X - 1} + X}
| c = Algebra: $X \pa... | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] whose [[Definition:Probability Generating Function|probability generating function]] is $\map {\Pi_X} s$.
Then the [[Definition:Square (Algebra)|square]] of the [[Definition:Expectation|expectation]] of $X$ is given by the expression:
:$\ex... | From [[Derivatives of Probability Generating Function at One]]:
:$\map { {\Pi_X}''} 1 = \expect {X \paren {X - 1} }$
and from [[Expectation of Discrete Random Variable from PGF]]:
:$\map { {\Pi_X}'} 1 = \expect X$
So:
{{begin-eqn}}
{{eqn | l = \expect {X^2}
| r = \expect {X \paren {X - 1} + X}
| c = Algebr... | Expectation of Square of Discrete Random Variable | https://proofwiki.org/wiki/Expectation_of_Square_of_Discrete_Random_Variable | https://proofwiki.org/wiki/Expectation_of_Square_of_Discrete_Random_Variable | [
"Expectation",
"Probability Generating Functions"
] | [
"Definition:Random Variable/Discrete",
"Definition:Probability Generating Function",
"Definition:Square/Function",
"Definition:Expectation",
"Definition:Derivative/Higher Derivatives/Second Derivative",
"Definition:Derivative",
"Definition:Probability Generating Function"
] | [
"Derivatives of Probability Generating Function at One",
"Expectation of Discrete Random Variable from PGF",
"Expectation is Linear"
] |
proofwiki-7693 | Non-Equivalence of Proposition and Negation/Formulation 1 | :$p \implies \neg p, \neg p \implies p \vdash \bot$ | {{BeginTableau|p \implies \neg p, \neg p \implies p \vdash \bot}}
{{Premise|1|p \implies \neg p}}
{{Premise|2|\neg p \implies p}}
{{SequentIntro|3|1|\neg p|1|Proof by Contradiction: Variant 3}}
{{ModusPonens|4|1, 2|p|2|3}}
{{NonContradiction|5|1, 2|4|3}}
{{EndTableau}}
{{qed}}
Category:Contradiction
Category:Conditiona... | :$p \implies \neg p, \neg p \implies p \vdash \bot$ | {{BeginTableau|p \implies \neg p, \neg p \implies p \vdash \bot}}
{{Premise|1|p \implies \neg p}}
{{Premise|2|\neg p \implies p}}
{{SequentIntro|3|1|\neg p|1|[[Proof by Contradiction/Variant 3|Proof by Contradiction: Variant 3]]}}
{{ModusPonens|4|1, 2|p|2|3}}
{{NonContradiction|5|1, 2|4|3}}
{{EndTableau}}
{{qed}}
[[Ca... | Non-Equivalence of Proposition and Negation/Formulation 1 | https://proofwiki.org/wiki/Non-Equivalence_of_Proposition_and_Negation/Formulation_1 | https://proofwiki.org/wiki/Non-Equivalence_of_Proposition_and_Negation/Formulation_1 | [
"Contradiction",
"Conditional"
] | [] | [
"Proof by Contradiction/Variant 3",
"Category:Contradiction",
"Category:Conditional"
] |
proofwiki-7694 | Non-Equivalence of Proposition and Negation/Formulation 2 | :$\vdash \neg \left({p \iff \neg p}\right)$ | {{BeginTableau|$\vdash \neg \left({p \iff \neg p}\right)$}}
{{Assumption|1|p \iff \neg p}}
{{BiconditionalElimination|2|1|p \implies \neg p|1|1}}
{{BiconditionalElimination|3|1|\neg p \implies p|1|2}}
{{SequentIntro|4|1|\bot|2,3|Non-Equivalence of Proposition and Negation: Formulation 1}}
{{Contradiction|5||\neg \left(... | :$\vdash \neg \left({p \iff \neg p}\right)$ | {{BeginTableau|$\vdash \neg \left({p \iff \neg p}\right)$}}
{{Assumption|1|p \iff \neg p}}
{{BiconditionalElimination|2|1|p \implies \neg p|1|1}}
{{BiconditionalElimination|3|1|\neg p \implies p|1|2}}
{{SequentIntro|4|1|\bot|2,3|[[Non-Equivalence of Proposition and Negation/Formulation 1|Non-Equivalence of Proposition ... | Non-Equivalence of Proposition and Negation/Formulation 2 | https://proofwiki.org/wiki/Non-Equivalence_of_Proposition_and_Negation/Formulation_2 | https://proofwiki.org/wiki/Non-Equivalence_of_Proposition_and_Negation/Formulation_2 | [
"Biconditional",
"Contradiction"
] | [] | [
"Non-Equivalence of Proposition and Negation/Formulation 1",
"Category:Biconditional",
"Category:Contradiction"
] |
proofwiki-7695 | No Injection from Power Set to Set/Lemma | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then there does not exist a set $B$ such that there is an injection from $B$ into $S$ and a surjection from $B$ onto $\powerset S$. | {{AimForCont}} there exists such a $B$.
Let $i: B \to S$ be an injection.
Let $f: B \to \powerset S$ be a surjection.
Let $i^\gets: \powerset S \to \powerset B$ be the inverse image mapping of $i$.
By Mapping Induced by Inverse of Injection is Surjection, $i^\gets$ is a surjection.
Let $f^\to: \powerset B \to \powerset... | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Then there does not exist a [[Definition:Set|set]] $B$ such that there is an [[Definition:Injection|injection]] from $B$ into $S$ and a [[Definition:Surjection|surjection]] from $B$ onto $\powerset S$. | {{AimForCont}} there exists such a $B$.
Let $i: B \to S$ be an [[Definition:Injection|injection]].
Let $f: B \to \powerset S$ be a [[Definition:Surjection|surjection]].
Let $i^\gets: \powerset S \to \powerset B$ be the [[Definition:Inverse Image Mapping of Mapping|inverse image mapping]] of $i$.
By [[Mapping Induce... | No Injection from Power Set to Set/Lemma | https://proofwiki.org/wiki/No_Injection_from_Power_Set_to_Set/Lemma | https://proofwiki.org/wiki/No_Injection_from_Power_Set_to_Set/Lemma | [
"No Injection from Power Set to Set"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Set",
"Definition:Injection",
"Definition:Surjection"
] | [
"Definition:Injection",
"Definition:Surjection",
"Definition:Inverse Image Mapping/Mapping",
"Inverse Image Mapping of Injection is Surjection",
"Definition:Surjection",
"Definition:Direct Image Mapping/Mapping",
"Direct Image Mapping of Surjection is Surjection",
"Definition:Surjection",
"Definitio... |
proofwiki-7696 | No Injection from Power Set to Set | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then there is no injection from $\powerset S$ into $S$. | {{AimForCont}} $f: \powerset S \to S$ is an injection.
By Injection has Surjective Left Inverse Mapping, there is a surjection $g: S \to \powerset S$.
But this contradicts Cantor's Theorem.
Thus there can be no such injection.
{{qed}}
{{LEM|Injection has Surjective Left Inverse Mapping|3}} | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Then there is no [[Definition:Injection|injection]] from $\powerset S$ into $S$. | {{AimForCont}} $f: \powerset S \to S$ is an [[Definition:Injection|injection]].
By [[Injection has Surjective Left Inverse Mapping]], there is a [[Definition:Surjection|surjection]] $g: S \to \powerset S$.
But this [[Definition:Contradiction|contradicts]] [[Cantor's Theorem]].
Thus there can be no such [[Definition:... | No Injection from Power Set to Set/Proof 1 | https://proofwiki.org/wiki/No_Injection_from_Power_Set_to_Set | https://proofwiki.org/wiki/No_Injection_from_Power_Set_to_Set/Proof_1 | [
"Power Set",
"Injections",
"No Injection from Power Set to Set"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Injection"
] | [
"Definition:Injection",
"Injection has Surjective Left Inverse Mapping",
"Definition:Surjection",
"Definition:Contradiction",
"Cantor's Theorem",
"Definition:Injection"
] |
proofwiki-7697 | No Injection from Power Set to Set | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then there is no injection from $\powerset S$ into $S$. | === Lemma ===
{{:No Injection from Power Set to Set/Lemma}}{{qed|lemma}}
The identity mapping $I_{\powerset S}: \powerset S \to \powerset S$ is a surjection by Identity Mapping is Surjection.
Thus, by the lemma, there can be no injection from $\powerset S$ into $S$.
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Then there is no [[Definition:Injection|injection]] from $\powerset S$ into $S$. | === [[No Injection from Power Set to Set/Lemma|Lemma]] ===
{{:No Injection from Power Set to Set/Lemma}}{{qed|lemma}}
The [[Definition:Identity Mapping|identity mapping]] $I_{\powerset S}: \powerset S \to \powerset S$ is a [[Definition:Surjection|surjection]] by [[Identity Mapping is Surjection]].
Thus, by the [[No ... | No Injection from Power Set to Set/Proof 2 | https://proofwiki.org/wiki/No_Injection_from_Power_Set_to_Set | https://proofwiki.org/wiki/No_Injection_from_Power_Set_to_Set/Proof_2 | [
"Power Set",
"Injections",
"No Injection from Power Set to Set"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Injection"
] | [
"No Injection from Power Set to Set/Lemma",
"Definition:Identity Mapping",
"Definition:Surjection",
"Identity Mapping is Surjection",
"No Injection from Power Set to Set/Lemma",
"Definition:injection"
] |
proofwiki-7698 | Inverse Image Mapping of Injection is Surjection | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a injection.
Let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$.
Then $f^\gets$ is a surjection. | Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping by $f$.
Let $X \in \powerset S$.
Let $Y = \map {f^\to} X$.
By Subset equals Preimage of Image iff Mapping is Injection:
:$\map {f^\gets} Y = X$
As such a $Y$ exists for each $X \in \powerset S$, $f^\gets$ is surjective.
{{qed}}
Category:Injections
Cat... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Injection|injection]].
Let $f^\gets: \powerset T \to \powerset S$ be the [[Definition:Inverse Image Mapping of Mapping|inverse image mapping]] of $f$.
Then $f^\gets$ is a [[Definition:Surjection|surjection]]. | Let $f^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] by $f$.
Let $X \in \powerset S$.
Let $Y = \map {f^\to} X$.
By [[Subset equals Preimage of Image iff Mapping is Injection]]:
:$\map {f^\gets} Y = X$
As such a $Y$ exists for each $X \in \powerset S$, $... | Inverse Image Mapping of Injection is Surjection | https://proofwiki.org/wiki/Inverse_Image_Mapping_of_Injection_is_Surjection | https://proofwiki.org/wiki/Inverse_Image_Mapping_of_Injection_is_Surjection | [
"Injections",
"Surjections",
"Inverse Image Mappings"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Inverse Image Mapping/Mapping",
"Definition:Surjection"
] | [
"Definition:Direct Image Mapping/Mapping",
"Subset equals Preimage of Image iff Mapping is Injection",
"Definition:Surjection",
"Category:Injections",
"Category:Surjections",
"Category:Inverse Image Mappings"
] |
proofwiki-7699 | Expectation and Variance of Poisson Distribution equal its Parameter | Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.
Then the expectation of $X$ equals the variance of $X$, that is, $\lambda$ itself. | From Expectation of Poisson Distribution:
:$\expect X = \lambda$
From Variance of Poisson Distribution:
:$\var X = \lambda$
{{qed}} | Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with the [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
Then the [[Definition:Expectation|expectation]] of $X$ equals the [[Definition:Variance of Discrete Random Variable|variance]] of $X$, that is, $\la... | From [[Expectation of Poisson Distribution]]:
:$\expect X = \lambda$
From [[Variance of Poisson Distribution]]:
:$\var X = \lambda$
{{qed}} | Expectation and Variance of Poisson Distribution equal its Parameter | https://proofwiki.org/wiki/Expectation_and_Variance_of_Poisson_Distribution_equal_its_Parameter | https://proofwiki.org/wiki/Expectation_and_Variance_of_Poisson_Distribution_equal_its_Parameter | [
"Poisson Distribution"
] | [
"Definition:Random Variable/Discrete",
"Definition:Poisson Distribution",
"Definition:Expectation",
"Definition:Variance/Discrete"
] | [
"Expectation of Poisson Distribution",
"Variance of Poisson Distribution"
] |
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