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-6000 | Orbit of Element under Conjugacy Action is Conjugacy Class | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*$ be the conjugacy group action on $G$:
: $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
Let $x \in G$.
Then the orbit of $x$ under this group action is:
:$\Orb x = \conjclass x$
where $\conjclass x$ is the conjugacy class of $x$. | Follows from the definition of the conjugacy class.
{{explain|Needs expansion.}}
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $*$ be the [[Definition:Conjugacy Action|conjugacy group action]] on $G$:
: $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
Let $x \in G$.
Then the [[Definition:Orbit (Group Theory)|orbit]] of $x... | Follows from the definition of the [[Definition:Conjugacy Class|conjugacy class]].
{{explain|Needs expansion.}}
{{qed}} | Orbit of Element under Conjugacy Action is Conjugacy Class | https://proofwiki.org/wiki/Orbit_of_Element_under_Conjugacy_Action_is_Conjugacy_Class | https://proofwiki.org/wiki/Orbit_of_Element_under_Conjugacy_Action_is_Conjugacy_Class | [
"Conjugacy Action"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Conjugacy Action",
"Definition:Orbit (Group Theory)",
"Definition:Group Action",
"Definition:Conjugacy Class"
] | [
"Definition:Conjugacy Class"
] |
proofwiki-6001 | Axiom of Foundation (Strong Form) | Let $B$ be a class.
Suppose $B$ is non-empty.
Then $B$ has a strictly minimal element under $\in$. | {{NotZFC}}
By Epsilon Relation is Strictly Well-Founded, $\Epsilon$, the epsilon relation, is a strictly well-founded relation on $B$.
By Epsilon Relation is Proper, $\struct {\mathbb U, \Epsilon}$ is a proper relational structure, where $\mathbb U$ is the universal class.
By Well-Founded Proper Relational Structure De... | Let $B$ be a [[Definition:Class (Class Theory)|class]].
Suppose $B$ is [[Definition:Non-Empty Set|non-empty]].
Then $B$ has a [[Definition:Strictly Minimal Element|strictly minimal element]] under $\in$. | {{NotZFC}}
By [[Epsilon Relation is Strictly Well-Founded]], $\Epsilon$, the [[Definition:Epsilon Relation|epsilon relation]], is a [[Definition:Strictly Well-Founded Relation|strictly well-founded relation]] on $B$.
By [[Epsilon Relation is Proper]], $\struct {\mathbb U, \Epsilon}$ is a [[Definition:Proper Relationa... | Axiom of Foundation (Strong Form)/Proof 1 | https://proofwiki.org/wiki/Axiom_of_Foundation_(Strong_Form) | https://proofwiki.org/wiki/Axiom_of_Foundation_(Strong_Form)/Proof_1 | [
"Class Theory",
"Axiom of Foundation (Strong Form)"
] | [
"Definition:Class (Class Theory)",
"Definition:Non-Empty Set",
"Definition:Strictly Minimal Element"
] | [
"Epsilon Relation is Strictly Well-Founded",
"Definition:Epsilon Relation",
"Definition:Strictly Well-Founded Relation",
"Epsilon Relation is Proper",
"Definition:Proper Relational Structure",
"Definition:Universal Class",
"Well-Founded Proper Relational Structure Determines Minimal Elements",
"Defini... |
proofwiki-6002 | Axiom of Foundation (Strong Form) | Let $B$ be a class.
Suppose $B$ is non-empty.
Then $B$ has a strictly minimal element under $\in$. | {{NotZFC}}
Let $x \in B$.
Let $x'$ be the transitive closure of $x$.
Let $L = x' \cap B$.
Then $x \in L$, so $L$ is not empty.
Since $x'$ is a set, so is $L$, by the axiom of subset.
Thus by the Axiom of Foundation, $L$ has an $\in$-minimal element $m$.
By the definition of intersection, $m \in B$.
{{AimForCont}} that ... | Let $B$ be a [[Definition:Class (Class Theory)|class]].
Suppose $B$ is [[Definition:Non-Empty Set|non-empty]].
Then $B$ has a [[Definition:Strictly Minimal Element|strictly minimal element]] under $\in$. | {{NotZFC}}
Let $x \in B$.
Let $x'$ be the [[Definition:Transitive Closure of Set/Definition 2|transitive closure of $x$]].
Let $L = x' \cap B$.
Then $x \in L$, so $L$ is not empty.
Since $x'$ is a set, so is $L$, by the axiom of subset.
Thus by the [[Axiom:Axiom of Foundation|Axiom of Foundation]], $L$ has an $\i... | Axiom of Foundation (Strong Form)/Proof 2 | https://proofwiki.org/wiki/Axiom_of_Foundation_(Strong_Form) | https://proofwiki.org/wiki/Axiom_of_Foundation_(Strong_Form)/Proof_2 | [
"Class Theory",
"Axiom of Foundation (Strong Form)"
] | [
"Definition:Class (Class Theory)",
"Definition:Non-Empty Set",
"Definition:Strictly Minimal Element"
] | [
"Definition:Transitive Closure of Set/Definition 2",
"Axiom:Axiom of Foundation",
"Definition:Set Intersection"
] |
proofwiki-6003 | Trivial Group is Initial Object | Let $\mathbf{Grp}$ be the category of groups.
Let $1 = \set e$ be the trivial group.
Then $1$ is an initial object of $\mathbf{Grp}$. | Let $\struct {G, \circ}$ be a group with identity $e_G$.
By Group Homomorphism Preserves Identity, any hypothetical group homomorphism $\phi: 1 \to G$ must satisfy:
:$\map \phi e = e_G$
Let us define the mapping $\phi$ in this way.
By Equality of Mappings, only one such mapping $1 \to G$ can exist, establishing uniquen... | Let $\mathbf{Grp}$ be the [[Definition:Category of Groups|category of groups]].
Let $1 = \set e$ be the [[Definition:Trivial Group|trivial group]].
Then $1$ is an [[Definition:Initial Object|initial object]] of $\mathbf{Grp}$. | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] with [[Definition:Identity Element|identity]] $e_G$.
By [[Group Homomorphism Preserves Identity]], any hypothetical [[Definition:Group Homomorphism|group homomorphism]] $\phi: 1 \to G$ must satisfy:
:$\map \phi e = e_G$
Let us define the [[Definition:Mapping|m... | Trivial Group is Initial Object | https://proofwiki.org/wiki/Trivial_Group_is_Initial_Object | https://proofwiki.org/wiki/Trivial_Group_is_Initial_Object | [
"Category of Groups",
"Trivial Group"
] | [
"Definition:Category of Groups",
"Definition:Trivial Group",
"Definition:Initial Object"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Group Homomorphism Preserves Identity",
"Definition:Group Homomorphism",
"Definition:Mapping",
"Equality of Mappings",
"Definition:Mapping",
"Definition:Unique",
"Definition:Group Homomorphism",
"Definition:Group Pro... |
proofwiki-6004 | Trivial Group is Terminal Object of Category of Groups | Let $\mathbf {Grp}$ be the category of groups.
Let $\set e$ be the trivial group.
Then $\set e$ is a terminal object of $\mathbf {Grp}$. | Let $\struct {G, \circ}$ be any group.
By Singleton is Terminal Object of Category of Sets, there is precisely one mapping:
:$!: G \to \set e$
defined by:
:$\forall g \in G: ! (g) = e$
By definition, any group homomorphism is also a mapping.
Hence, there is at most one morphism $\struct {G, \circ} \to \set e$ in $\math... | Let $\mathbf {Grp}$ be the [[Definition:Category of Groups|category of groups]].
Let $\set e$ be the [[Definition:Trivial Group|trivial group]].
Then $\set e$ is a [[Definition:Terminal Object|terminal object]] of $\mathbf {Grp}$. | Let $\struct {G, \circ}$ be any [[Definition:Group|group]].
By [[Singleton is Terminal Object of Category of Sets]], there is precisely one [[Definition:Mapping|mapping]]:
:$!: G \to \set e$
defined by:
:$\forall g \in G: ! (g) = e$
By definition, any [[Definition:Group Homomorphism|group homomorphism]] is also a ... | Trivial Group is Terminal Object of Category of Groups | https://proofwiki.org/wiki/Trivial_Group_is_Terminal_Object_of_Category_of_Groups | https://proofwiki.org/wiki/Trivial_Group_is_Terminal_Object_of_Category_of_Groups | [
"Category of Groups",
"Trivial Group"
] | [
"Definition:Category of Groups",
"Definition:Trivial Group",
"Definition:Terminal Object"
] | [
"Definition:Group",
"Singleton is Terminal Object of Category of Sets",
"Definition:Mapping",
"Definition:Group Homomorphism",
"Definition:Mapping",
"Definition:Morphism",
"Definition:Mapping",
"Definition:Group Homomorphism",
"Definition:Group Product/Group Law",
"Definition:Group Homomorphism",
... |
proofwiki-6005 | Orbit of Subgroup under Coset Action is Coset Space | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\powerset G$ be the power set of $G$.
Let $H \in \powerset G$ be a subgroup of $G$.
Let $*$ be the group action on $H$ defined as:
:$\forall g \in G: g * H = g \circ H$
where $g \circ H$ is the (left) coset of $g$ by $H$.
Then the orbit of $H$ in $\powerse... | From the definition of orbit:
:$\Orb H = \set {y \in G: \exists g \in G: y = g * H}$
By the definition of $*$:
:$\Orb H = \set {y \in G: \exists g \in G: y = g \circ H}$
The result follows from the definition of (left) coset space.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\powerset G$ be the [[Definition:Power Set|power set]] of $G$.
Let $H \in \powerset G$ be a [[Definition:Subgroup|subgroup]] of $G$.
Let $*$ be the [[Definition:Group Action|group action]] on $H$ def... | From the definition of [[Definition:Orbit (Group Theory)|orbit]]:
:$\Orb H = \set {y \in G: \exists g \in G: y = g * H}$
By the definition of $*$:
:$\Orb H = \set {y \in G: \exists g \in G: y = g \circ H}$
The result follows from the definition of [[Definition:Left Coset Space|(left) coset space]].
{{qed}} | Orbit of Subgroup under Coset Action is Coset Space | https://proofwiki.org/wiki/Orbit_of_Subgroup_under_Coset_Action_is_Coset_Space | https://proofwiki.org/wiki/Orbit_of_Subgroup_under_Coset_Action_is_Coset_Space | [
"Subset Product Action"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Power Set",
"Definition:Subgroup",
"Definition:Group Action",
"Definition:Coset/Left Coset",
"Definition:Orbit (Group Theory)",
"Definition:Coset Space/Left Coset Space"
] | [
"Definition:Orbit (Group Theory)",
"Definition:Coset Space/Left Coset Space"
] |
proofwiki-6006 | Orbit of Conjugacy Action on Subgroup is Set of Conjugate Subgroups | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $X$ be the set of all subgroups of $G$.
Let $*$ be the conjugacy action on $H$ defined as:
: $\forall g \in G, H \in X: g * H = g \circ H \circ g^{-1}$
Then the orbit $\Orb H$ of $H$ in $\powerset G$ is the set of subgroups of $G$ conjugate to $H$. | We have that:
:$\Orb H = \set {g \circ H \circ g^{-1}: g \in G}$
from the definition.
The result follows by definition of conjugate subgroup.
{{Qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $X$ be the set of all [[Definition:Subgroup|subgroups]] of $G$.
Let $*$ be the [[Definition:Conjugacy Action|conjugacy action]] on $H$ defined as:
: $\forall g \in G, H \in X: g * H = g \circ H \circ g... | We have that:
:$\Orb H = \set {g \circ H \circ g^{-1}: g \in G}$
from the definition.
The result follows by definition of [[Definition:Conjugate of Group Subset|conjugate subgroup]].
{{Qed}} | Orbit of Conjugacy Action on Subgroup is Set of Conjugate Subgroups | https://proofwiki.org/wiki/Orbit_of_Conjugacy_Action_on_Subgroup_is_Set_of_Conjugate_Subgroups | https://proofwiki.org/wiki/Orbit_of_Conjugacy_Action_on_Subgroup_is_Set_of_Conjugate_Subgroups | [
"Conjugacy Action"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Conjugacy Action",
"Definition:Orbit (Group Theory)",
"Definition:Set",
"Definition:Subgroup",
"Definition:Conjugate (Group Theory)/Subset"
] | [
"Definition:Conjugate (Group Theory)/Subset"
] |
proofwiki-6007 | Closure for Finite Collection of Relations and Operations | Let $\RR_1, \RR_2, \ldots \RR_n$ be relations.
Let $\SS_1, \SS_2, \ldots \SS_m$ be operations.
Let $T$ be a small class.
Let the image of $\RR_i$ over any small class $x$ be small classes for $1 \le i \le n$.
Let the image of $\SS_i$ over any Cartesian product $x \times x$ be small classes for $1 \le i \le m$.
Then the... | Let $R \sqbrk x$ denote the image of $x$ under $R$.
Set:
:$\ds \map G x = x \cup \bigcup_{i \mathop = 1}^n \RR_i \sqbrk x \cup \bigcup_{i \mathop = 1}^m \SS_i \sqbrk x$
Using the Principle of Recursive Definition, construct the function $F$ as follows:
:$\map F 0 = T$
:$\map F {n + 1} = \map G {\map F n}$
Define $X$ as... | Let $\RR_1, \RR_2, \ldots \RR_n$ be [[Definition:Relation|relations]].
Let $\SS_1, \SS_2, \ldots \SS_m$ be [[Definition:Operation|operations]].
Let $T$ be a [[Definition:Small Class|small class]].
Let the [[Definition:Image of Relation|image]] of $\RR_i$ over any [[Definition:Small Class|small class]] $x$ be [[Defi... | Let $R \sqbrk x$ denote the [[Definition:Image of Subset under Relation|image]] of $x$ under $R$.
Set:
:$\ds \map G x = x \cup \bigcup_{i \mathop = 1}^n \RR_i \sqbrk x \cup \bigcup_{i \mathop = 1}^m \SS_i \sqbrk x$
Using the [[Principle of Recursive Definition]], construct the function $F$ as follows:
:$\map F 0 = T... | Closure for Finite Collection of Relations and Operations | https://proofwiki.org/wiki/Closure_for_Finite_Collection_of_Relations_and_Operations | https://proofwiki.org/wiki/Closure_for_Finite_Collection_of_Relations_and_Operations | [
"Relational Closures"
] | [
"Definition:Relation",
"Definition:Operation",
"Definition:Small Class",
"Definition:Image (Set Theory)/Relation/Relation",
"Definition:Small Class",
"Definition:Small Class",
"Definition:Cartesian Product",
"Definition:Small Class",
"Definition:Small Class",
"Definition:Closed Relation",
"Defin... | [
"Definition:Image (Set Theory)/Relation/Subset",
"Principle of Recursive Definition",
"Definition:Image (Set Theory)/Relation/Subset"
] |
proofwiki-6008 | Von Neumann Hierarchy is Supertransitive | Let $V$ denote the Von Neumann Hierarchy.
Let $x$ be an ordinal.
Then $\map V x$ is supertransitive. | {{NotZFC}}
The proof shall proceed by Transfinite Induction on $x$. | Let $V$ denote the [[Definition:Von Neumann Hierarchy|Von Neumann Hierarchy]].
Let $x$ be an [[Definition:Ordinal|ordinal]].
Then $\map V x$ is [[Definition:Supertransitive Class|supertransitive]]. | {{NotZFC}}
The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $x$. | Von Neumann Hierarchy is Supertransitive | https://proofwiki.org/wiki/Von_Neumann_Hierarchy_is_Supertransitive | https://proofwiki.org/wiki/Von_Neumann_Hierarchy_is_Supertransitive | [
"Von Neumann Hierarchy"
] | [
"Definition:Von Neumann Hierarchy",
"Definition:Ordinal",
"Definition:Supertransitive Class"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-6009 | Von Neumann Hierarchy Comparison | Let $x$ and $y$ be ordinals such that $x < y$.
Then:
:$\map V x \in \map V y$
:$\map V x \subset \map V y$
{{explain|$\map V x$ etc.}} | {{NotZFC}}
The proof shall proceed by Transfinite Induction on $y$. | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]] such that $x < y$.
Then:
:$\map V x \in \map V y$
:$\map V x \subset \map V y$
{{explain|$\map V x$ etc.}} | {{NotZFC}}
The proof shall proceed by [[Transfinite Induction/Schema 2|Transfinite Induction]] on $y$. | Von Neumann Hierarchy Comparison | https://proofwiki.org/wiki/Von_Neumann_Hierarchy_Comparison | https://proofwiki.org/wiki/Von_Neumann_Hierarchy_Comparison | [
"Von Neumann Hierarchy"
] | [
"Definition:Ordinal"
] | [
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2",
"Transfinite Induction/Schema 2"
] |
proofwiki-6010 | Rank is Ordinal | Let $S$ be a small class
The rank of $S$ is an ordinal. | {{NotZFC}}
The rank of $S$ is an intersection of a set of ordinals $B$.
$B$ is non-empty by the fact that Set has Rank.
Thus, $B$ has a minimal element, which is the rank of $S$ plus $1$.
Therefore, the rank is itself an ordinal.
{{qed}} | Let $S$ be a [[Definition:Small Class|small class]]
The [[Definition:Rank (Set Theory)|rank]] of $S$ is an [[Definition:Ordinal|ordinal]]. | {{NotZFC}}
The [[Definition:Rank (Set Theory)|rank]] of $S$ is an [[Definition:Set Intersection|intersection]] of a [[Definition:Set|set]] of [[Definition:Ordinal|ordinals]] $B$.
$B$ is [[Definition:Non-Empty|non-empty]] by the fact that [[Set has Rank]].
Thus, $B$ has a [[Definition:Minimal Element|minimal element]... | Rank is Ordinal | https://proofwiki.org/wiki/Rank_is_Ordinal | https://proofwiki.org/wiki/Rank_is_Ordinal | [
"Von Neumann Hierarchy"
] | [
"Definition:Small Class",
"Definition:Rank (Set Theory)",
"Definition:Ordinal"
] | [
"Definition:Rank (Set Theory)",
"Definition:Set Intersection",
"Definition:Set",
"Definition:Ordinal",
"Definition:Non-Empty",
"Set has Rank",
"Definition:Minimal/Element",
"Definition:Rank (Set Theory)",
"Definition:Rank (Set Theory)",
"Definition:Ordinal"
] |
proofwiki-6011 | Ordinal Equal to Rank | Let $x$ be an ordinal.
Let $S$ be a small class.
Let $\map V x$ denote the von Neumann hierarchy on the ordinal $x$.
Then $x$ equals the rank of $S$ {{iff}} $S \in \map V {x + 1} \land S \notin \map V x$. | {{NotZFC}} | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $S$ be a [[Definition:Small Class|small class]].
Let $\map V x$ denote the [[Definition:Von Neumann Hierarchy|von Neumann hierarchy]] on the [[Definition:Ordinal|ordinal]] $x$.
Then $x$ equals the [[Definition:Rank (Set Theory)|rank]] of $S$ {{iff}} $S \in \map V {x... | {{NotZFC}} | Ordinal Equal to Rank | https://proofwiki.org/wiki/Ordinal_Equal_to_Rank | https://proofwiki.org/wiki/Ordinal_Equal_to_Rank | [
"Von Neumann Hierarchy"
] | [
"Definition:Ordinal",
"Definition:Small Class",
"Definition:Von Neumann Hierarchy",
"Definition:Ordinal",
"Definition:Rank (Set Theory)"
] | [] |
proofwiki-6012 | Group Direct Product of Cyclic Groups/Corollary | Let $n_1, n_2, \ldots, n_s$ be a finite sequence of integers, all greater than $1$, such that for any pair of them $n_i$ and $n_j$, $n_i \perp n_j$.
Let $G_i$ be a cyclic group of order $n_i$ for each $i: 1 \le i \le s$.
Then $G_1 \times G_2 \times \cdots \times G_s$ is cyclic of order $n_1 n_2 \ldots n_s$. | When $s = 1$ the result is trivial.
Assume the result holds for $s = k$.
Then $H = G_1 \times G_2 \times \ldots \times G_k$ is cyclic of order $n_1 n_2 \ldots n_k$.
Applying the main result to $H \times G_{k + 1}$ gives us the result for $s = k + 1$.
The result follows by induction.
{{qed}} | Let $n_1, n_2, \ldots, n_s$ be a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Integer|integers]], all greater than $1$, such that for any pair of them $n_i$ and $n_j$, $n_i \perp n_j$.
Let $G_i$ be a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $n_i$ for each $i: 1 ... | When $s = 1$ the result is trivial.
Assume the result holds for $s = k$.
Then $H = G_1 \times G_2 \times \ldots \times G_k$ is [[Definition:Cyclic Group|cyclic]] of [[Definition:Order of Group|order]] $n_1 n_2 \ldots n_k$.
Applying the main result to $H \times G_{k + 1}$ gives us the result for $s = k + 1$.
The re... | Group Direct Product of Cyclic Groups/Corollary | https://proofwiki.org/wiki/Group_Direct_Product_of_Cyclic_Groups/Corollary | https://proofwiki.org/wiki/Group_Direct_Product_of_Cyclic_Groups/Corollary | [
"Group Direct Products",
"Cyclic Groups"
] | [
"Definition:Finite Sequence",
"Definition:Integer",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Cyclic Group",
"Definition:Order of Structure"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Principle of Mathematical Induction"
] |
proofwiki-6013 | Divisor Count Function of Prime Number | Let $p \in \Z_{> 0}$.
Then $p$ is a prime number {{iff}}:
:$\map {\sigma_0} p = 2$
where $\map {\sigma_0} p$ denotes the divisor count function of $p$. | === Necessary Condition ===
Let $p$ be a prime number.
Then, by definition, the only positive divisors of $p$ are $1$ and $p$.
Hence by definition of the divisor count function:
:$\map {\sigma_0} p = 2$
{{qed|lemma}}
=== Sufficient Condition ===
Suppose $\map {\sigma_0} p = 2$.
Then by One Divides all Integers we have:... | Let $p \in \Z_{> 0}$.
Then $p$ is a [[Definition:Prime Number|prime number]] {{iff}}:
:$\map {\sigma_0} p = 2$
where $\map {\sigma_0} p$ denotes the [[Definition:Divisor Count Function|divisor count function]] of $p$. | === Necessary Condition ===
Let $p$ be a [[Definition:Prime Number|prime number]].
Then, by [[Definition:Prime Number/Definition 1|definition]], the only [[Definition:Positive Integer|positive]] [[Definition:Divisor of Integer|divisors]] of $p$ are $1$ and $p$.
Hence by definition of the [[Definition:Divisor Count F... | Divisor Count Function of Prime Number | https://proofwiki.org/wiki/Divisor_Count_Function_of_Prime_Number | https://proofwiki.org/wiki/Divisor_Count_Function_of_Prime_Number | [
"Prime Numbers",
"Divisor Count Function"
] | [
"Definition:Prime Number",
"Definition:Divisor Count Function"
] | [
"Definition:Prime Number",
"Definition:Prime Number/Definition 1",
"Definition:Positive/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor Count Function",
"Integer Divisor Results/One Divides all Integers",
"Integer Divisor Results/Integer Divides Itself",
"Definition:Divisor (Alge... |
proofwiki-6014 | Join is Commutative | Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\vee$ is commutative. | Let $a, b \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \vee b
| r = \sup \set {a, b}
| c = {{Defof|Join (Order Theory)}}
}}
{{eqn | r = \sup \set {b, a}
| c = {{Defof|Set Equality}}
}}
{{eqn | r = b \vee a
| c = {{Defof|Join (Order Theory)}}
}}
{{end-eqn}}
Hence the result.
{{qed}} | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
Then $\vee$ is [[Definition:Commutative Operation|commutative]]. | Let $a, b \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \vee b
| r = \sup \set {a, b}
| c = {{Defof|Join (Order Theory)}}
}}
{{eqn | r = \sup \set {b, a}
| c = {{Defof|Set Equality}}
}}
{{eqn | r = b \vee a
| c = {{Defof|Join (Order Theory)}}
}}
{{end-eqn}}
Hence the result.
{{qed}} | Join is Commutative | https://proofwiki.org/wiki/Join_is_Commutative | https://proofwiki.org/wiki/Join_is_Commutative | [
"Join Operation",
"Examples of Commutative Operations"
] | [
"Definition:Join Semilattice",
"Definition:Commutative/Operation"
] | [] |
proofwiki-6015 | Meet is Commutative | Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Then $\wedge$ is commutative. | Let $a, b \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \wedge b
| r = \inf \set {a, b}
| c = {{Defof|Meet (Order Theory)|Meet}}
}}
{{eqn | r = \inf \set {b, a}
| c = {{Defof|Set Equality}}
}}
{{eqn | r = b \wedge a
| c = {{Defof|Meet (Order Theory)|Meet}}
}}
{{end-eqn}}
Hence the resul... | Let $\struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
Then $\wedge$ is [[Definition:Commutative Operation|commutative]]. | Let $a, b \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \wedge b
| r = \inf \set {a, b}
| c = {{Defof|Meet (Order Theory)|Meet}}
}}
{{eqn | r = \inf \set {b, a}
| c = {{Defof|Set Equality}}
}}
{{eqn | r = b \wedge a
| c = {{Defof|Meet (Order Theory)|Meet}}
}}
{{end-eqn}}
Hence the re... | Meet is Commutative | https://proofwiki.org/wiki/Meet_is_Commutative | https://proofwiki.org/wiki/Meet_is_Commutative | [
"Meet Operation",
"Examples of Commutative Operations"
] | [
"Definition:Meet Semilattice",
"Definition:Commutative/Operation"
] | [] |
proofwiki-6016 | Join Succeeds Operands | Let $\struct {S, \preceq}$ be an ordered set.
Let $a, b \in S$ admit a join $a \vee b \in S$.
Then:
:$a \preceq a \vee b$
:$b \preceq a \vee b$
That is, $a \vee b$ succeeds its operands $a$ and $b$. | By definition of join:
:$a \vee b = \sup \set {a, b}$
where $\sup$ denotes supremum.
Since a supremum is {{afortiori}} an upper bound:
:$a \preceq \sup \set {a, b}$
:$b \preceq \sup \set {a, b}$
as desired.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $a, b \in S$ admit a [[Definition:Join (Order Theory)|join]] $a \vee b \in S$.
Then:
:$a \preceq a \vee b$
:$b \preceq a \vee b$
That is, $a \vee b$ [[Definition:Succeed|succeeds]] its [[Definition:Operand|operands]] $a$ and $b$. | By definition of [[Definition:Join (Order Theory)|join]]:
:$a \vee b = \sup \set {a, b}$
where $\sup$ denotes [[Definition:Supremum of Set|supremum]].
Since a [[Definition:Supremum of Set|supremum]] is {{afortiori}} an [[Definition:Upper Bound of Set|upper bound]]:
:$a \preceq \sup \set {a, b}$
:$b \preceq \sup \s... | Join Succeeds Operands | https://proofwiki.org/wiki/Join_Succeeds_Operands | https://proofwiki.org/wiki/Join_Succeeds_Operands | [
"Join Operation"
] | [
"Definition:Ordered Set",
"Definition:Join (Order Theory)",
"Definition:Succeed",
"Definition:Operation/Operand"
] | [
"Definition:Join (Order Theory)",
"Definition:Supremum of Set",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set"
] |
proofwiki-6017 | Meet Precedes Operands | Let $\struct {S, \preceq}$ be an ordered set.
Let $a, b \in S$ admit a meet $a \wedge b \in S$.
Then:
:$a \wedge b \preceq a$
:$a \wedge b \preceq b$
That is, $a \wedge b$ precedes its operands $a$ and $b$. | By definition of meet:
:$a \wedge b = \inf \set {a, b}$
where $\inf$ denotes infimum.
Since an infimum is {{afortiori}} a lower bound:
:$\inf \set {a, b} \preceq a$
:$\inf \set {a, b} \preceq b$
as desired.
{{qed}}
Category:Meet Operation
qzj9efw5t71pagp8isyhwlqpnr9eccw | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $a, b \in S$ admit a [[Definition:Meet (Order Theory)|meet]] $a \wedge b \in S$.
Then:
:$a \wedge b \preceq a$
:$a \wedge b \preceq b$
That is, $a \wedge b$ [[Definition:Precede|precedes]] its [[Definition:Operand|operands]] $a$ and $b$. | By definition of [[Definition:Meet (Order Theory)|meet]]:
:$a \wedge b = \inf \set {a, b}$
where $\inf$ denotes [[Definition:Infimum of Set|infimum]].
Since an [[Definition:Infimum of Set|infimum]] is {{afortiori}} a [[Definition:Lower Bound of Set|lower bound]]:
:$\inf \set {a, b} \preceq a$
:$\inf \set {a, b} \p... | Meet Precedes Operands | https://proofwiki.org/wiki/Meet_Precedes_Operands | https://proofwiki.org/wiki/Meet_Precedes_Operands | [
"Meet Operation"
] | [
"Definition:Ordered Set",
"Definition:Meet (Order Theory)",
"Definition:Precede",
"Definition:Operation/Operand"
] | [
"Definition:Meet (Order Theory)",
"Definition:Infimum of Set",
"Definition:Infimum of Set",
"Definition:Lower Bound of Set",
"Category:Meet Operation"
] |
proofwiki-6018 | Meet is Associative | Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Then $\wedge$ is associative. | Let $a, b, c \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \wedge \paren {b \wedge c}
| r = \inf \set {a, b \wedge c}
| c = {{Defof|Meet (Order Theory)|Meet}}
}}
{{eqn | r = \inf \set {\inf \set a, \inf \set {b, c} }
| c = Infimum of Singleton
}}
{{eqn | r = \inf \set {a, b, c}
| c = In... | Let $\struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
Then $\wedge$ is [[Definition:Associative Operation|associative]]. | Let $a, b, c \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \wedge \paren {b \wedge c}
| r = \inf \set {a, b \wedge c}
| c = {{Defof|Meet (Order Theory)|Meet}}
}}
{{eqn | r = \inf \set {\inf \set a, \inf \set {b, c} }
| c = [[Infimum of Singleton]]
}}
{{eqn | r = \inf \set {a, b, c}
| ... | Meet is Associative | https://proofwiki.org/wiki/Meet_is_Associative | https://proofwiki.org/wiki/Meet_is_Associative | [
"Meet Operation",
"Examples of Associative Operations"
] | [
"Definition:Meet Semilattice",
"Definition:Associative Operation"
] | [
"Infimum of Singleton",
"Infimum of Infima",
"Infimum of Infima",
"Infimum of Singleton"
] |
proofwiki-6019 | Poset Elements Equal iff Equal Weak Lower Closure | Let $\left({S, \preccurlyeq}\right)$ be an ordered set.
Let $s, t \in S$.
Then $s = t$ {{iff}}:
:$s^\preccurlyeq = t^\preccurlyeq$
where $s^\preccurlyeq$ denotes weak lower closure of $s$.
That is, {{iff}}, for all $r \in S$:
:$r \preccurlyeq s \iff r \preccurlyeq t$ | === Necessary Condition ===
If $s = t$, then trivially also:
:$s^\preccurlyeq = t^\preccurlyeq$
{{qed|lemma}} | Let $\left({S, \preccurlyeq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $s, t \in S$.
Then $s = t$ {{iff}}:
:$s^\preccurlyeq = t^\preccurlyeq$
where $s^\preccurlyeq$ denotes [[Definition:Weak Lower Closure of Element|weak lower closure of $s$]].
That is, {{iff}}, for all $r \in S$:
:$r \preccurlye... | === Necessary Condition ===
If $s = t$, then trivially also:
:$s^\preccurlyeq = t^\preccurlyeq$
{{qed|lemma}} | Poset Elements Equal iff Equal Weak Lower Closure | https://proofwiki.org/wiki/Poset_Elements_Equal_iff_Equal_Weak_Lower_Closure | https://proofwiki.org/wiki/Poset_Elements_Equal_iff_Equal_Weak_Lower_Closure | [
"Lower Closures"
] | [
"Definition:Ordered Set",
"Definition:Lower Closure/Element"
] | [] |
proofwiki-6020 | Poset Elements Equal iff Equal Weak Upper Closure | Let $\left({S, \preccurlyeq}\right)$ be an ordered set.
Let $s, t \in S$.
Then $s = t$ {{iff}}:
:$s^\succcurlyeq = t^\succcurlyeq$
where $s^\succcurlyeq$ denotes weak upper closure of $s$.
That is, {{iff}}, for all $r \in S$:
:$s \preccurlyeq r \iff t \preccurlyeq r$ | === Necessary Condition ===
If $s = t$, then trivially also:
:$s^\succcurlyeq = t^\succcurlyeq$
{{qed|lemma}} | Let $\left({S, \preccurlyeq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $s, t \in S$.
Then $s = t$ {{iff}}:
:$s^\succcurlyeq = t^\succcurlyeq$
where $s^\succcurlyeq$ denotes [[Definition:Weak Upper Closure of Element|weak upper closure of $s$]].
That is, {{iff}}, for all $r \in S$:
:$s \preccurlye... | === Necessary Condition ===
If $s = t$, then trivially also:
:$s^\succcurlyeq = t^\succcurlyeq$
{{qed|lemma}} | Poset Elements Equal iff Equal Weak Upper Closure | https://proofwiki.org/wiki/Poset_Elements_Equal_iff_Equal_Weak_Upper_Closure | https://proofwiki.org/wiki/Poset_Elements_Equal_iff_Equal_Weak_Upper_Closure | [
"Upper Closures"
] | [
"Definition:Ordered Set",
"Definition:Upper Closure/Element"
] | [] |
proofwiki-6021 | Join is Associative | Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Then $\vee$ is associative. | Let $a, b, c \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \vee \paren {b \vee c}
| r = a \vee \sup \set {b, c}
| c = {{Defof|Join (Order Theory)}}
}}
{{eqn | r = \sup \set {\sup \set a, \sup \set {b, c} }
| c = Supremum of Singleton
}}
{{eqn | r = \sup \set {a, b, c}
| c = Supremum of ... | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
Then $\vee$ is [[Definition:Associative Operation|associative]]. | Let $a, b, c \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \vee \paren {b \vee c}
| r = a \vee \sup \set {b, c}
| c = {{Defof|Join (Order Theory)}}
}}
{{eqn | r = \sup \set {\sup \set a, \sup \set {b, c} }
| c = [[Supremum of Singleton]]
}}
{{eqn | r = \sup \set {a, b, c}
| c = [[Supr... | Join is Associative | https://proofwiki.org/wiki/Join_is_Associative | https://proofwiki.org/wiki/Join_is_Associative | [
"Join Operation",
"Examples of Associative Operations"
] | [
"Definition:Join Semilattice",
"Definition:Associative Operation"
] | [
"Supremum of Singleton",
"Supremum of Suprema",
"Supremum of Suprema",
"Supremum of Singleton"
] |
proofwiki-6022 | Prime iff Equal to Product | Let $p \in \Z$ be an integer such that $p \ne 0$ and $p \ne \pm 1$.
Then $p$ is prime {{iff}}:
:$\forall a, b \in \Z: p = ab \implies p = \pm a \lor p = \pm b$ | === Necessary Condition ===
Let $p$ be a prime number.
Then by definition, the only divisors of $p$ are $\pm 1$ and $\pm p$.
Thus, if $p = a b$ then either $a = \pm 1$ and $b = \pm p$ or $a = \pm p$ and $b = \pm 1$.
{{qed|lemma}} | Let $p \in \Z$ be an [[Definition:Integer|integer]] such that $p \ne 0$ and $p \ne \pm 1$.
Then $p$ is [[Definition:Prime Number|prime]] {{iff}}:
:$\forall a, b \in \Z: p = ab \implies p = \pm a \lor p = \pm b$ | === Necessary Condition ===
Let $p$ be a [[Definition:Prime Number|prime number]].
Then by definition, the only [[Definition:Divisor of Integer|divisors]] of $p$ are $\pm 1$ and $\pm p$.
Thus, if $p = a b$ then either $a = \pm 1$ and $b = \pm p$ or $a = \pm p$ and $b = \pm 1$.
{{qed|lemma}} | Prime iff Equal to Product | https://proofwiki.org/wiki/Prime_iff_Equal_to_Product | https://proofwiki.org/wiki/Prime_iff_Equal_to_Product | [
"Prime Numbers"
] | [
"Definition:Integer",
"Definition:Prime Number"
] | [
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Number"
] |
proofwiki-6023 | Multiple of Divisor Divides Multiple | Let $a, b, c \in \Z$.
Let:
:$a \divides b$
where $\divides$ denotes divisibility.
Then:
:$a c \divides b c$ | We have that Integers form Integral Domain.
The result then follows from Multiple of Divisor in Integral Domain Divides Multiple.
{{qed}} | Let $a, b, c \in \Z$.
Let:
:$a \divides b$
where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Then:
:$a c \divides b c$ | We have that [[Integers form Integral Domain]].
The result then follows from [[Multiple of Divisor in Integral Domain Divides Multiple]].
{{qed}} | Multiple of Divisor Divides Multiple/Proof 1 | https://proofwiki.org/wiki/Multiple_of_Divisor_Divides_Multiple | https://proofwiki.org/wiki/Multiple_of_Divisor_Divides_Multiple/Proof_1 | [
"Divisors",
"Multiple of Divisor Divides Multiple"
] | [
"Definition:Divisor (Algebra)/Integer"
] | [
"Integers form Integral Domain",
"Multiple of Divisor in Integral Domain Divides Multiple"
] |
proofwiki-6024 | Multiple of Divisor Divides Multiple | Let $a, b, c \in \Z$.
Let:
:$a \divides b$
where $\divides$ denotes divisibility.
Then:
:$a c \divides b c$ | By definition, if $a \divides b$ then $\exists d \in \Z: a d = b$.
Then:
:$\paren {a d} c = b c$
that is:
:$\paren {a c} d = b c$
which follows because Integer Multiplication is Commutative and Integer Multiplication is Associative.
Hence the result.
{{qed}} | Let $a, b, c \in \Z$.
Let:
:$a \divides b$
where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Then:
:$a c \divides b c$ | By definition, if $a \divides b$ then $\exists d \in \Z: a d = b$.
Then:
:$\paren {a d} c = b c$
that is:
:$\paren {a c} d = b c$
which follows because [[Integer Multiplication is Commutative]] and [[Integer Multiplication is Associative]].
Hence the result.
{{qed}} | Multiple of Divisor Divides Multiple/Proof 2 | https://proofwiki.org/wiki/Multiple_of_Divisor_Divides_Multiple | https://proofwiki.org/wiki/Multiple_of_Divisor_Divides_Multiple/Proof_2 | [
"Divisors",
"Multiple of Divisor Divides Multiple"
] | [
"Definition:Divisor (Algebra)/Integer"
] | [
"Integer Multiplication is Commutative",
"Integer Multiplication is Associative"
] |
proofwiki-6025 | Divisor Relation is Transitive | The divisibility relation is a transitive relation on $\Z$, the set of integers.
That is:
:$\forall x, y, z \in \Z: x \divides y \land y \divides z \implies x \divides z$ | We have that Integers form Integral Domain.
The result then follows directly from Divisor Relation in Integral Domain is Transitive.
{{qed}} | The [[Definition:Divisor of Integer|divisibility]] relation is a [[Definition:Transitive Relation|transitive relation]] on $\Z$, the set of [[Definition:Integer|integers]].
That is:
:$\forall x, y, z \in \Z: x \divides y \land y \divides z \implies x \divides z$ | We have that [[Integers form Integral Domain]].
The result then follows directly from [[Divisor Relation in Integral Domain is Transitive]].
{{qed}} | Divisor Relation is Transitive/Proof 1 | https://proofwiki.org/wiki/Divisor_Relation_is_Transitive | https://proofwiki.org/wiki/Divisor_Relation_is_Transitive/Proof_1 | [
"Divisors",
"Divisor Relation is Transitive",
"Examples of Transitive Relations"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Transitive Relation",
"Definition:Integer"
] | [
"Integers form Integral Domain",
"Divisor Relation in Integral Domain is Transitive"
] |
proofwiki-6026 | Divisor Relation is Transitive | The divisibility relation is a transitive relation on $\Z$, the set of integers.
That is:
:$\forall x, y, z \in \Z: x \divides y \land y \divides z \implies x \divides z$ | {{begin-eqn}}
{{eqn | l = x
| o = \divides
| r = y
| c =
}}
{{eqn | ll= \leadsto
| q = \exists q_1 \in \Z
| l = q_1 x
| r = y
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | l = y
| o = \divides
| r = z
| c =
}}
{{eqn | ll= \leadsto
| q = \exists q_2 \in... | The [[Definition:Divisor of Integer|divisibility]] relation is a [[Definition:Transitive Relation|transitive relation]] on $\Z$, the set of [[Definition:Integer|integers]].
That is:
:$\forall x, y, z \in \Z: x \divides y \land y \divides z \implies x \divides z$ | {{begin-eqn}}
{{eqn | l = x
| o = \divides
| r = y
| c =
}}
{{eqn | ll= \leadsto
| q = \exists q_1 \in \Z
| l = q_1 x
| r = y
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | l = y
| o = \divides
| r = z
| c =
}}
{{eqn | ll= \leadsto
| q = \exists q_2 \in... | Divisor Relation is Transitive/Proof 2 | https://proofwiki.org/wiki/Divisor_Relation_is_Transitive | https://proofwiki.org/wiki/Divisor_Relation_is_Transitive/Proof_2 | [
"Divisors",
"Divisor Relation is Transitive",
"Examples of Transitive Relations"
] | [
"Definition:Divisor (Algebra)/Integer",
"Definition:Transitive Relation",
"Definition:Integer"
] | [
"Integer Multiplication is Associative"
] |
proofwiki-6027 | Polynomial Forms over Field form Integral Domain/Formulation 1 | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental in $F$.
Let $F \sqbrk X$ be the ring of polynomial forms in $X$ over $F$.
Then $F \sqbrk X$ is an integral domain. | We already have from Ring of Polynomial Forms is Commutative Ring with Unity that $F \sqbrk X$ is a ring.
Suppose $f$ and $g$ are polynomials in $F \sqbrk X$ such that $f \ne 0_F, g \ne 0_F$.
If $\map \deg f = \map \deg g = 0$ then $f$ and $g$ are elements of $F$.
As $F$ is a field and a field is an integral domain, $f... | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $X$ be [[Definition:Transcendental over Field|transcendental]] in $F$.
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|... | We already have from [[Ring of Polynomial Forms is Commutative Ring with Unity]] that $F \sqbrk X$ is a [[Definition:Ring (Abstract Algebra)|ring]].
Suppose $f$ and $g$ are [[Definition:Polynomial|polynomials]] in $F \sqbrk X$ such that $f \ne 0_F, g \ne 0_F$.
If $\map \deg f = \map \deg g = 0$ then $f$ and $g$ are e... | Polynomial Forms over Field form Integral Domain/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_form_Integral_Domain/Formulation_1 | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_form_Integral_Domain/Formulation_1/Proof_1 | [
"Field Theory",
"Polynomial Theory",
"Ideal Theory",
"Polynomial Forms over Field form Integral Domain"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Transcendental (Abstract Algebra)/Field Extension/Element",
"Definition:Ring of Polynomial Forms",
"Definition:Integral Domain"
] | [
"Ring of Polynomial Forms is Commutative Ring with Unity",
"Definition:Ring (Abstract Algebra)",
"Definition:Polynomial",
"Definition:Field (Abstract Algebra)",
"Field is Integral Domain",
"Degree of Product of Polynomials over Ring/Corollary 2",
"Definition:Integral Domain"
] |
proofwiki-6028 | Polynomial Forms over Field form Integral Domain/Formulation 1 | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental in $F$.
Let $F \sqbrk X$ be the ring of polynomial forms in $X$ over $F$.
Then $F \sqbrk X$ is an integral domain. | We have from Ring of Polynomial Forms is Commutative Ring with Unity that $F \sqbrk X$ is a commutative ring with unity.
The result follows from Ring of Polynomial Forms is Integral Domain.
{{qed}} | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Let $X$ be [[Definition:Transcendental over Field|transcendental]] in $F$.
Let $F \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|... | We have from [[Ring of Polynomial Forms is Commutative Ring with Unity]] that $F \sqbrk X$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
The result follows from [[Ring of Polynomial Forms is Integral Domain]].
{{qed}} | Polynomial Forms over Field form Integral Domain/Formulation 1/Proof 2 | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_form_Integral_Domain/Formulation_1 | https://proofwiki.org/wiki/Polynomial_Forms_over_Field_form_Integral_Domain/Formulation_1/Proof_2 | [
"Field Theory",
"Polynomial Theory",
"Ideal Theory",
"Polynomial Forms over Field form Integral Domain"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Transcendental (Abstract Algebra)/Field Extension/Element",
"Definition:Ring of Polynomial Forms",
"Definition:Integral Domain"
] | [
"Ring of Polynomial Forms is Commutative Ring with Unity",
"Definition:Commutative and Unitary Ring",
"Ring of Polynomial Forms is Integral Domain"
] |
proofwiki-6029 | Sum of Indices of Real Number/Positive Integers | Let $n, m \in \Z_{\ge 0}$ be positive integers.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
: $r^{n + m} = r^n \times r^m$ | Proof by induction on $m$:
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\forall n \in \Z_{\ge 0}: r^{n + m} = r^n \times r^m$
$\map P 0$ is true, as this just says:
:$r^{n + 0} = r^n = r^n \times 1 = r^n \times r^0$ | Let $n, m \in \Z_{\ge 0}$ be [[Definition:Positive Integer|positive integers]].
Let $r^n$ be defined as [[Definition:Integer Power|$r$ to the power of $n$]].
Then:
: $r^{n + m} = r^n \times r^m$ | Proof by [[Principle of Mathematical Induction|induction]] on $m$:
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the [[Definition:Proposition|proposition]]:
:$\forall n \in \Z_{\ge 0}: r^{n + m} = r^n \times r^m$
$\map P 0$ is true, as this just says:
:$r^{n + 0} = r^n = r^n \times 1 = r^n \times r^0$ | Sum of Indices of Real Number/Positive Integers | https://proofwiki.org/wiki/Sum_of_Indices_of_Real_Number/Positive_Integers | https://proofwiki.org/wiki/Sum_of_Indices_of_Real_Number/Positive_Integers | [
"Sum of Indices of Real Number"
] | [
"Definition:Positive/Integer",
"Definition:Power (Algebra)/Integer"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-6030 | Product of Indices of Real Number/Positive Integers | Let $n, m \in \Z_{\ge 0}$ be positive integers.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
:$\paren {r^n}^m = r^{n m}$ | Proof by induction on $m$:
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\forall n \in \Z_{\ge 0}: \paren {r^n}^m = r^{n m}$
$\map P 0$ is true, as this just says:
:$\paren {r^n}^0 = 1 = r^0 = r^{n \times 0}$ | Let $n, m \in \Z_{\ge 0}$ be [[Definition:Positive Integer|positive integers]].
Let $r^n$ be defined as [[Definition:Integer Power|$r$ to the power of $n$]].
Then:
:$\paren {r^n}^m = r^{n m}$ | Proof by [[Principle of Mathematical Induction|induction]] on $m$:
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the [[Definition:Proposition|proposition]]:
:$\forall n \in \Z_{\ge 0}: \paren {r^n}^m = r^{n m}$
$\map P 0$ is true, as this just says:
:$\paren {r^n}^0 = 1 = r^0 = r^{n \times 0}$ | Product of Indices of Real Number/Positive Integers | https://proofwiki.org/wiki/Product_of_Indices_of_Real_Number/Positive_Integers | https://proofwiki.org/wiki/Product_of_Indices_of_Real_Number/Positive_Integers | [
"Product of Indices of Real Number"
] | [
"Definition:Positive/Integer",
"Definition:Power (Algebra)/Integer"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-6031 | Equivalence of Formulations of Axiom of Empty Set | The following formulations of the '''{{axiom-link|the Empty Set|Set Theory}}''' are equivalent: | === Formulation 1 implies Formulation 2 ===
{{begin-eqn}}
{{eqn | n = 1
| o =
| r = \forall y: y = y
| c = Equality is Reflexive
}}
{{eqn | n = 2
| o =
| r = \neg \exists y: y \ne y
| c = From $(1)$: Assertion of Universality
}}
{{eqn | n = 3
| o =
| r = x := \set { {y... | The following formulations of the '''{{axiom-link|the Empty Set|Set Theory}}''' are [[Definition:Logical Equivalence|equivalent]]: | === Formulation 1 implies Formulation 2 ===
{{begin-eqn}}
{{eqn | n = 1
| o =
| r = \forall y: y = y
| c = [[Equality is Reflexive]]
}}
{{eqn | n = 2
| o =
| r = \neg \exists y: y \ne y
| c = From $(1)$: [[Assertion of Universality]]
}}
{{eqn | n = 3
| o =
| r = x := ... | Equivalence of Formulations of Axiom of Empty Set | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Empty_Set | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Empty_Set | [
"Empty Set",
"Definition Equivalences"
] | [
"Definition:Logical Equivalence"
] | [
"Equality is Reflexive",
"De Morgan's Laws (Predicate Logic)/Assertion of Universality"
] |
proofwiki-6032 | Smallest Element is Initial Object | Let $\mathbf P$ be an order category.
Suppose the objects $\mathbf P_0$ of $\mathbf P$, considered as an ordered set, have a smallest element $p$.
Then $p$ is an initial object of $\mathbf P$. | Since $p$ is the smallest element of $\mathbf P_0$, we have:
:$\forall q \in \mathbf P_0: p \le q$
that is, for every object $q$ of $\mathbf P$ there is a unique morphism $p \to q$.
That is, $p$ is initial.
{{qed}} | Let $\mathbf P$ be an [[Definition:Order Category|order category]].
Suppose the [[Definition:Object (Category Theory)|objects]] $\mathbf P_0$ of $\mathbf P$, considered as an [[Definition:Ordered Set|ordered set]], have a [[Definition:Smallest Element|smallest element]] $p$.
Then $p$ is an [[Definition:Initial Objec... | Since $p$ is the [[Definition:Smallest Element|smallest element]] of $\mathbf P_0$, we have:
:$\forall q \in \mathbf P_0: p \le q$
that is, for every [[Definition:Object (Category Theory)|object]] $q$ of $\mathbf P$ there is a unique [[Definition:Morphism (Category Theory)|morphism]] $p \to q$.
That is, $p$ is [[De... | Smallest Element is Initial Object | https://proofwiki.org/wiki/Smallest_Element_is_Initial_Object | https://proofwiki.org/wiki/Smallest_Element_is_Initial_Object | [
"Order Categories"
] | [
"Definition:Order Category",
"Definition:Object (Category Theory)",
"Definition:Ordered Set",
"Definition:Smallest Element",
"Definition:Initial Object"
] | [
"Definition:Smallest Element",
"Definition:Object (Category Theory)",
"Definition:Morphism",
"Definition:Initial Object"
] |
proofwiki-6033 | Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy | Let $x$ be an ordinal.
Let $S$ be a small class.
Let $\map V x$ denote the von Neumann hierarchy on the ordinal $x$.
Then $x$ is a subset of the rank of $S$ {{iff}} $S \notin \map V x$. | {{NotZFC}} | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $S$ be a [[Definition:Small Class|small class]].
Let $\map V x$ denote the [[Definition:Von Neumann Hierarchy|von Neumann hierarchy]] on the [[Definition:Ordinal|ordinal]] $x$.
Then $x$ is a [[Definition:Subset|subset]] of the [[Definition:Rank (Set Theory)|rank]] o... | {{NotZFC}} | Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy | https://proofwiki.org/wiki/Ordinal_is_Subset_of_Rank_of_Small_Class_iff_Not_in_Von_Neumann_Hierarchy | https://proofwiki.org/wiki/Ordinal_is_Subset_of_Rank_of_Small_Class_iff_Not_in_Von_Neumann_Hierarchy | [
"Von Neumann Hierarchy"
] | [
"Definition:Ordinal",
"Definition:Small Class",
"Definition:Von Neumann Hierarchy",
"Definition:Ordinal",
"Definition:Subset",
"Definition:Rank (Set Theory)"
] | [] |
proofwiki-6034 | Membership Rank Inequality | Let $S$ and $T$ be sets.
Let $\map {\operatorname{rank} } S$ denote the rank of $S$.
Then:
:$S \in T \implies \map {\operatorname{rank} } S < \map {\operatorname{rank} } T$ | {{NotZFC}}
By Ordinal Equal to Rank:
:$T \in \map V {\map {\operatorname{rank} } T + 1}$
By the definition of rank:
:$T \subseteq \map V {\map {\operatorname{rank} } T}$
Since $S \in T$:
:$S \in \map V {\map {\operatorname{rank} } T}$
By Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy:
:$\map ... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\map {\operatorname{rank} } S$ denote the [[Definition:Rank (Set Theory)|rank]] of $S$.
Then:
:$S \in T \implies \map {\operatorname{rank} } S < \map {\operatorname{rank} } T$ | {{NotZFC}}
By [[Ordinal Equal to Rank]]:
:$T \in \map V {\map {\operatorname{rank} } T + 1}$
By the definition of [[Definition:Rank (Set Theory)|rank]]:
:$T \subseteq \map V {\map {\operatorname{rank} } T}$
Since $S \in T$:
:$S \in \map V {\map {\operatorname{rank} } T}$
By [[Ordinal is Subset of Rank of Small Cla... | Membership Rank Inequality | https://proofwiki.org/wiki/Membership_Rank_Inequality | https://proofwiki.org/wiki/Membership_Rank_Inequality | [
"Von Neumann Hierarchy"
] | [
"Definition:Set",
"Definition:Rank (Set Theory)"
] | [
"Ordinal Equal to Rank",
"Definition:Rank (Set Theory)",
"Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy",
"Ordinal Membership is Trichotomy",
"Transitive Set is Proper Subset of Ordinal iff Element of Ordinal"
] |
proofwiki-6035 | Greatest Element is Terminal Object | Let $\mathbf P$ be an order category.
Let $p$ be the greatest element of the objects $\mathbf P_0$ of $\mathbf P$, considered as a ordered set.
Then $p$ is a terminal object of $\mathbf P$. | Since $p$ is the greatest element of $\mathbf P_0$, we have:
:$\forall q \in \mathbf P_0: q \le p$
that is, for every object $q$ of $\mathbf P$ there is a unique morphism $q \to p$.
That is, $p$ is terminal.
{{qed}} | Let $\mathbf P$ be an [[Definition:Order Category|order category]].
Let $p$ be the [[Definition:Greatest Element|greatest element]] of the [[Definition:Object (Category Theory)|objects]] $\mathbf P_0$ of $\mathbf P$, considered as a [[Definition:Ordered Set|ordered set]].
Then $p$ is a [[Definition:Terminal Object|t... | Since $p$ is the [[Definition:Greatest Element|greatest element]] of $\mathbf P_0$, we have:
:$\forall q \in \mathbf P_0: q \le p$
that is, for every [[Definition:Object (Category Theory)|object]] $q$ of $\mathbf P$ there is a unique [[Definition:Morphism (Category Theory)|morphism]] $q \to p$.
That is, $p$ is [[De... | Greatest Element is Terminal Object | https://proofwiki.org/wiki/Greatest_Element_is_Terminal_Object | https://proofwiki.org/wiki/Greatest_Element_is_Terminal_Object | [
"Order Categories"
] | [
"Definition:Order Category",
"Definition:Greatest Element",
"Definition:Object (Category Theory)",
"Definition:Ordered Set",
"Definition:Terminal Object"
] | [
"Definition:Greatest Element",
"Definition:Object (Category Theory)",
"Definition:Morphism",
"Definition:Terminal Object"
] |
proofwiki-6036 | Identity Morphism is Terminal Object in Slice Category | Let $\mathbf C$ be a metacategory, and let $C \in \mathbf C_0$ be an object of $\mathbf C$.
Let $\operatorname{id}_C: C \to C$ be the identity morphism for $C$.
Then $\operatorname{id}_C$ is a terminal object in the slice category $\mathbf C \mathop / C$. | Let $f: B \to C$ be an object of $\mathbf C \mathop / C$.
Then there is a morphism $a: f \to \operatorname{id}_C$ {{iff}}:
:$f = \operatorname{id}_C \circ a = a$
Thus, $f$ itself defines the unique morphism $f \to \operatorname{id}_C$ in $\mathbf C \mathop / C$.
We therefore have the following commutative diagram in $\... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]], and let $C \in \mathbf C_0$ be an [[Definition:Object|object]] of $\mathbf C$.
Let $\operatorname{id}_C: C \to C$ be the [[Definition:Identity Morphism|identity morphism]] for $C$.
Then $\operatorname{id}_C$ is a [[Definition:Terminal Object|terminal obj... | Let $f: B \to C$ be an [[Definition:Object|object]] of $\mathbf C \mathop / C$.
Then there is a [[Definition:Morphism (Category Theory)|morphism]] $a: f \to \operatorname{id}_C$ {{iff}}:
:$f = \operatorname{id}_C \circ a = a$
Thus, $f$ itself defines the unique [[Definition:Morphism (Category Theory)|morphism]] $f \... | Identity Morphism is Terminal Object in Slice Category | https://proofwiki.org/wiki/Identity_Morphism_is_Terminal_Object_in_Slice_Category | https://proofwiki.org/wiki/Identity_Morphism_is_Terminal_Object_in_Slice_Category | [
"Slice Categories"
] | [
"Definition:Metacategory",
"Definition:Object",
"Definition:Identity Morphism",
"Definition:Terminal Object",
"Definition:Slice Category"
] | [
"Definition:Object",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Commutative Diagram",
"Definition:Terminal Object"
] |
proofwiki-6037 | Identity Morphism is Initial Object in Coslice Category | Let $\mathbf C$ be a metacategory, and let $C \in \mathbf C_0$ be an object of $\mathbf C$.
Let $\operatorname{id}_C: C \to C$ be the identity morphism for $C$.
Then $\operatorname{id}_C$ is an initial object in the coslice category $C / \mathbf C$. | Let $f: C \to D$ be an object of $C / \mathbf C$.
Then there is a morphism $a: \operatorname{id}_C \to f$ {{iff}}:
:$f = a \circ \operatorname{id}_C = a$
Thus, $f$ itself defines the unique morphism $\operatorname{id}_C \to f$ in $C \mathop / \mathbf C$.
We therefore have the following commutative diagram in $\mathbf C... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]], and let $C \in \mathbf C_0$ be an [[Definition:Object|object]] of $\mathbf C$.
Let $\operatorname{id}_C: C \to C$ be the [[Definition:Identity Morphism|identity morphism]] for $C$.
Then $\operatorname{id}_C$ is an [[Definition:Initial Object|initial obje... | Let $f: C \to D$ be an [[Definition:Object|object]] of $C / \mathbf C$.
Then there is a [[Definition:Morphism (Category Theory)|morphism]] $a: \operatorname{id}_C \to f$ {{iff}}:
:$f = a \circ \operatorname{id}_C = a$
Thus, $f$ itself defines the unique [[Definition:Morphism (Category Theory)|morphism]] $\operatorna... | Identity Morphism is Initial Object in Coslice Category | https://proofwiki.org/wiki/Identity_Morphism_is_Initial_Object_in_Coslice_Category | https://proofwiki.org/wiki/Identity_Morphism_is_Initial_Object_in_Coslice_Category | [
"Slice Categories"
] | [
"Definition:Metacategory",
"Definition:Object",
"Definition:Identity Morphism",
"Definition:Initial Object",
"Definition:Coslice Category"
] | [
"Definition:Object",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Commutative Diagram",
"Definition:Initial Object"
] |
proofwiki-6038 | Rank of Set Determined by Members | Let $S$ be a set.
Let $\map {\operatorname{rank} } S$ denote the rank of $S$.
Then:
:$\map {\operatorname{rank} } S = \bigcap \set {x \in \On: \forall y \in S: \map {\operatorname{rank} } y < x}$ | {{NotZFC}}
Let:
:$T = \bigcap \set {x \in \On: \forall y \in S: \map {\operatorname{rank} } y < x}$
Let $y \in S$.
Then by Membership Rank Inequality:
:$\map {\operatorname{rank} } x < \map {\operatorname{rank} } S$
{{explain|$x$ or $y$?}}
Therefore:
:$T \subseteq \map {\operatorname{rank} } S$
Conversely, take any $x ... | Let $S$ be a [[Definition:Set|set]].
Let $\map {\operatorname{rank} } S$ denote the [[Definition:Rank (Set Theory)|rank]] of $S$.
Then:
:$\map {\operatorname{rank} } S = \bigcap \set {x \in \On: \forall y \in S: \map {\operatorname{rank} } y < x}$ | {{NotZFC}}
Let:
:$T = \bigcap \set {x \in \On: \forall y \in S: \map {\operatorname{rank} } y < x}$
Let $y \in S$.
Then by [[Membership Rank Inequality]]:
:$\map {\operatorname{rank} } x < \map {\operatorname{rank} } S$
{{explain|$x$ or $y$?}}
Therefore:
:$T \subseteq \map {\operatorname{rank} } S$
Conversely, ... | Rank of Set Determined by Members | https://proofwiki.org/wiki/Rank_of_Set_Determined_by_Members | https://proofwiki.org/wiki/Rank_of_Set_Determined_by_Members | [
"Von Neumann Hierarchy"
] | [
"Definition:Set",
"Definition:Rank (Set Theory)"
] | [
"Membership Rank Inequality",
"Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy",
"Definition:Von Neumann Hierarchy",
"Definition:Subset",
"Definition:Rank (Set Theory)"
] |
proofwiki-6039 | Rank of Ordinal | Let $x$ be an ordinal.
Let $\map {\operatorname {rank} } x$ denote the rank of $x$.
Then:
:$\map {\operatorname {rank} } x = x$ | {{NotZFC}}
The proof shall proceed by Transfinite Induction (Strong Induction) on $x$.
Suppose $\forall y \in x: \map {\operatorname {rank} } y = y$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\operatorname {rank} } x
| r = \bigcap \set {z \in \On: \forall y \in x: y < z}
| c = Rank of Set Determined by Members
... | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $\map {\operatorname {rank} } x$ denote the [[Definition:Rank (Set Theory)|rank]] of $x$.
Then:
:$\map {\operatorname {rank} } x = x$ | {{NotZFC}}
The proof shall proceed by [[Transfinite Induction/Schema 1|Transfinite Induction (Strong Induction)]] on $x$.
Suppose $\forall y \in x: \map {\operatorname {rank} } y = y$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\operatorname {rank} } x
| r = \bigcap \set {z \in \On: \forall y \in x: y < z}
... | Rank of Ordinal | https://proofwiki.org/wiki/Rank_of_Ordinal | https://proofwiki.org/wiki/Rank_of_Ordinal | [
"Von Neumann Hierarchy"
] | [
"Definition:Ordinal",
"Definition:Rank (Set Theory)"
] | [
"Transfinite Induction/Schema 1",
"Rank of Set Determined by Members"
] |
proofwiki-6040 | Bounded Rank implies Small Class | Let $S$ be a class.
Suppose the rank, denoted $\map {\operatorname{rank} } x$, of each $x \in S$ is bounded above by some ordinal $y$.
{{MissingLinks|rank}}
Then $S$ is a small class. | {{NotZFC}}
Let $V$ denote the von Neumann hierarchy.
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\operatorname{rank} } x
| o = \le
| r = y
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \map V {y + 1}
| c = Ordinal is Subset of Rank o... | Let $S$ be a [[Definition:Class (Class Theory)|class]].
Suppose the rank, denoted $\map {\operatorname{rank} } x$, of each $x \in S$ is bounded above by some [[Definition:Ordinal|ordinal]] $y$.
{{MissingLinks|rank}}
Then $S$ is a [[Definition:Small Class|small class]]. | {{NotZFC}}
Let $V$ denote the [[Definition:Von Neumann Hierarchy|von Neumann hierarchy]].
Then:
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\operatorname{rank} } x
| o = \le
| r = y
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = \map V {y + 1}... | Bounded Rank implies Small Class | https://proofwiki.org/wiki/Bounded_Rank_implies_Small_Class | https://proofwiki.org/wiki/Bounded_Rank_implies_Small_Class | [
"Von Neumann Hierarchy"
] | [
"Definition:Class (Class Theory)",
"Definition:Ordinal",
"Definition:Small Class"
] | [
"Definition:Von Neumann Hierarchy",
"Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy",
"Ordinal Equal to Rank",
"Axiom of Subsets Equivalents",
"Definition:Small Class"
] |
proofwiki-6041 | Count of Subsets with Even Cardinality | Let $S$ be a set whose cardinality is $n$.
Then the number of subsets of $S$ whose cardinality is even is $2^{n-1}$. | Let $E$ be the total number of subsets of $S$ whose cardinality is even.
From Cardinality of Set of Subsets, the number of subsets of $S$ with $m$ elements is $\dbinom n m$:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m} }$
where $\dbinom n m$ is a binomial coefficient.
Thus the total number of subsets of $S$ whose car... | Let $S$ be a [[Definition:Set|set]] whose [[Definition:Cardinality|cardinality]] is $n$.
Then the number of [[Definition:Subset|subsets]] of $S$ whose [[Definition:Cardinality|cardinality]] is [[Definition:Even Integer|even]] is $2^{n-1}$. | Let $E$ be the total number of [[Definition:Subset|subsets]] of $S$ whose [[Definition:Cardinality|cardinality]] is [[Definition:Even Integer|even]].
From [[Cardinality of Set of Subsets]], the number of [[Definition:Subset|subsets]] of $S$ with $m$ elements is $\dbinom n m$:
:$\dbinom n m = \dfrac {n!} {m! \paren {n ... | Count of Subsets with Even Cardinality/Proof 1 | https://proofwiki.org/wiki/Count_of_Subsets_with_Even_Cardinality | https://proofwiki.org/wiki/Count_of_Subsets_with_Even_Cardinality/Proof_1 | [
"Combinatorics",
"Count of Subsets with Even Cardinality"
] | [
"Definition:Set",
"Definition:Cardinality",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Even Integer"
] | [
"Definition:Subset",
"Definition:Cardinality",
"Definition:Even Integer",
"Cardinality of Set of Subsets",
"Definition:Subset",
"Definition:Binomial Coefficient",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Even Integer",
"Definition:Binomial Coefficient",
"Sum of Even Index Binom... |
proofwiki-6042 | Count of Subsets with Even Cardinality | Let $S$ be a set whose cardinality is $n$.
Then the number of subsets of $S$ whose cardinality is even is $2^{n-1}$. | Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:The number of subsets of $S$ whose cardinality is even is $2^{n - 1}$, where $\card S = n$.
=== Basis for the Induction ===
When $n = 1$ we have from Cardinality of Power Set of Finite Set that $S$ has $2^1 = 1$ subsets: $\O$ and $S$ itself.
We... | Let $S$ be a [[Definition:Set|set]] whose [[Definition:Cardinality|cardinality]] is $n$.
Then the number of [[Definition:Subset|subsets]] of $S$ whose [[Definition:Cardinality|cardinality]] is [[Definition:Even Integer|even]] is $2^{n-1}$. | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:The number of [[Definition:Subset|subsets]] of $S$ whose [[Definition:Cardinality|cardinality]] is [[Definition:Even Integer|even]] is $2^{n - 1}$, where $\card S = n$.
=== B... | Count of Subsets with Even Cardinality/Proof 2 | https://proofwiki.org/wiki/Count_of_Subsets_with_Even_Cardinality | https://proofwiki.org/wiki/Count_of_Subsets_with_Even_Cardinality/Proof_2 | [
"Combinatorics",
"Count of Subsets with Even Cardinality"
] | [
"Definition:Set",
"Definition:Cardinality",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Even Integer"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Even Integer",
"Cardinality of Power Set of Finite Set",
"Definition:Subset",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Even Integer",
"Principle of Mathem... |
proofwiki-6043 | Count of Subsets with Odd Cardinality | Let $S$ be a set whose cardinality is $n$.
Then the number of subsets of $S$ whose cardinality is odd is $2^{n-1}$. | Let $F$ be the total number of subsets of $S$ whose cardinality is odd.
From Cardinality of Set of Subsets, the number of subsets of $S$ with $m$ elements is $\dbinom n m$:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - m} }$
where $\dbinom n m$ is a binomial coefficient.
Thus the total number of subsets of $S$ whose card... | Let $S$ be a [[Definition:Set|set]] whose [[Definition:Cardinality|cardinality]] is $n$.
Then the number of [[Definition:Subset|subsets]] of $S$ whose [[Definition:Cardinality|cardinality]] is [[Definition:Odd Integer|odd]] is $2^{n-1}$. | Let $F$ be the total number of [[Definition:Subset|subsets]] of $S$ whose [[Definition:Cardinality|cardinality]] is [[Definition:Odd Integer|odd]].
From [[Cardinality of Set of Subsets]], the number of [[Definition:Subset|subsets]] of $S$ with $m$ elements is $\dbinom n m$:
:$\dbinom n m = \dfrac {n!} {m! \paren {n - ... | Count of Subsets with Odd Cardinality/Proof 1 | https://proofwiki.org/wiki/Count_of_Subsets_with_Odd_Cardinality | https://proofwiki.org/wiki/Count_of_Subsets_with_Odd_Cardinality/Proof_1 | [
"Combinatorics",
"Count of Subsets with Odd Cardinality"
] | [
"Definition:Set",
"Definition:Cardinality",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Odd Integer"
] | [
"Definition:Subset",
"Definition:Cardinality",
"Definition:Odd Integer",
"Cardinality of Set of Subsets",
"Definition:Subset",
"Definition:Binomial Coefficient",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Odd Integer",
"Definition:Binomial Coefficient",
"Sum of Odd Index Binomial... |
proofwiki-6044 | Count of Subsets with Odd Cardinality | Let $S$ be a set whose cardinality is $n$.
Then the number of subsets of $S$ whose cardinality is odd is $2^{n-1}$. | Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:The number of subsets of $S$ whose cardinality is odd is $2^{n - 1}$, where $\card S = n$.
=== Basis for the Induction ===
When $n = 1$ we have from Cardinality of Power Set of Finite Set that $S$ has $2^1 = 2$ subsets: $\O$ and $S$ itself.
We ... | Let $S$ be a [[Definition:Set|set]] whose [[Definition:Cardinality|cardinality]] is $n$.
Then the number of [[Definition:Subset|subsets]] of $S$ whose [[Definition:Cardinality|cardinality]] is [[Definition:Odd Integer|odd]] is $2^{n-1}$. | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:The number of [[Definition:Subset|subsets]] of $S$ whose [[Definition:Cardinality|cardinality]] is [[Definition:Odd Integer|odd]] is $2^{n - 1}$, where $\card S = n$.
=== Bas... | Count of Subsets with Odd Cardinality/Proof 2 | https://proofwiki.org/wiki/Count_of_Subsets_with_Odd_Cardinality | https://proofwiki.org/wiki/Count_of_Subsets_with_Odd_Cardinality/Proof_2 | [
"Combinatorics",
"Count of Subsets with Odd Cardinality"
] | [
"Definition:Set",
"Definition:Cardinality",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Odd Integer"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Odd Integer",
"Cardinality of Power Set of Finite Set",
"Definition:Subset",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Odd Integer",
"Definition:Basis for ... |
proofwiki-6045 | Strictly Well-Founded Relation determines Strictly Minimal Elements | Let $A$ be a class.
Let $\RR$ be a strictly well-founded relation on $A$.
Let $B$ be a nonempty class such that $B \subseteq A$.
Then $B$ has a strictly minimal element under $\RR$. | {{NotZFC}}
First a lemma: | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\RR$ be a [[Definition:Strictly Well-Founded Relation|strictly well-founded relation]] on $A$.
Let $B$ be a [[Definition:Non-Empty Set|nonempty]] [[Definition:Class (Class Theory)|class]] such that $B \subseteq A$.
Then $B$ has a [[Definition:Strictly Min... | {{NotZFC}}
First a [[Definition:Lemma|lemma]]: | Strictly Well-Founded Relation determines Strictly Minimal Elements | https://proofwiki.org/wiki/Strictly_Well-Founded_Relation_determines_Strictly_Minimal_Elements | https://proofwiki.org/wiki/Strictly_Well-Founded_Relation_determines_Strictly_Minimal_Elements | [
"Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Strictly Well-Founded Relation",
"Definition:Non-Empty Set",
"Definition:Class (Class Theory)",
"Definition:Strictly Minimal Element"
] | [
"Definition:Lemma"
] |
proofwiki-6046 | Category of Ordered Sets has Enough Constants | Let $\mathbf{OrdSet}$ be the category of ordered sets.
Then $\mathbf{OrdSet}$ has enough constants. | By Singleton Ordered Set is Terminal Object, we have that any ordered set with a singleton as underlying set is terminal in $\mathbf{OrdSet}$.
Let $1$ be such a singleton ordered set.
To show that $\mathbf{OrdSet}$ has enough constants, it is to be shown that if:
:$f: P \to Q \ne g: P \to Q$
then there exists an $x: 1 ... | Let $\mathbf{OrdSet}$ be the [[Definition:Category of Ordered Sets|category of ordered sets]].
Then $\mathbf{OrdSet}$ has [[Definition:Enough Constants|enough constants]]. | By [[Singleton Ordered Set is Terminal Object]], we have that any [[Definition:Ordered Set|ordered set]] with a [[Definition:Singleton|singleton]] as [[Definition:Underlying Set of Relational Structure|underlying set]] is [[Definition:Terminal Object|terminal]] in $\mathbf{OrdSet}$.
Let $1$ be such a [[Definition:Sing... | Category of Ordered Sets has Enough Constants | https://proofwiki.org/wiki/Category_of_Ordered_Sets_has_Enough_Constants | https://proofwiki.org/wiki/Category_of_Ordered_Sets_has_Enough_Constants | [
"Category of Ordered Sets"
] | [
"Definition:Category of Ordered Sets",
"Definition:Enough Constants"
] | [
"Singleton Ordered Set is Terminal Object",
"Definition:Ordered Set",
"Definition:Singleton",
"Definition:Underlying Set/Relational Structure",
"Definition:Terminal Object",
"Definition:Singleton",
"Definition:Ordered Set",
"Definition:Enough Constants",
"Definition:Category of Ordered Sets",
"Def... |
proofwiki-6047 | Unique Constant in Category of Monoids | Let $\mathbf{Mon}$ be the category of monoids.
Then every object $M$ of $\mathbf{Mon}$ has precisely one constant. | From Trivial Monoid is Terminal Element, we obtain that a constant of $M$ is a morphism $f: \left\{{e}\right\} \to M$.
By Trivial Monoid is Initial Element, there is precisely one such morphism.
Hence $M$ has one constant.
{{qed}} | Let $\mathbf{Mon}$ be the [[Definition:Category of Monoids|category of monoids]].
Then every [[Definition:Object|object]] $M$ of $\mathbf{Mon}$ has precisely one [[Definition:Constant (Category Theory)|constant]]. | From [[Trivial Monoid is Terminal Element]], we obtain that a [[Definition:Constant (Category Theory)|constant]] of $M$ is a [[Definition:Morphism (Category Theory)|morphism]] $f: \left\{{e}\right\} \to M$.
By [[Trivial Monoid is Initial Element]], there is precisely one such [[Definition:Morphism (Category Theory)|mo... | Unique Constant in Category of Monoids | https://proofwiki.org/wiki/Unique_Constant_in_Category_of_Monoids | https://proofwiki.org/wiki/Unique_Constant_in_Category_of_Monoids | [
"Category of Monoids"
] | [
"Definition:Category of Monoids",
"Definition:Object",
"Definition:Constant (Category Theory)",
"Definition:Category of Monoids"
] | [
"Trivial Monoid is Terminal Element",
"Definition:Constant (Category Theory)",
"Definition:Morphism",
"Trivial Monoid is Initial Element",
"Definition:Morphism",
"Definition:Constant (Category Theory)"
] |
proofwiki-6048 | Well-Founded Induction | Let $\struct {A, \RR}$ be a strictly well-founded relation.
Let $\RR^{-1} \sqbrk x$ denote the preimage of $x$ for each $x \in A$.
Let $B$ be a class such that $B \subseteq A$.
Suppose that:
:$(1): \quad \forall x \in A: \paren {\RR^{-1} \sqbrk x \subseteq B \implies x \in B}$
Then:
:$A = B$
That is, if a property pass... | {{AimForCont}} $A \nsubseteq B$.
Then:
:$A \setminus B \ne 0$
By Strictly Well-Founded Relation determines Strictly Minimal Elements, $A \setminus B$ must have some strictly minimal element under $\RR$.
Let $\map \complement B$ be the set complement of $B$.
Then:
{{begin-eqn}}
{{eqn | q = \exists x \in A \setminus B
... | Let $\struct {A, \RR}$ be a [[Definition:Strictly Well-Founded Relation|strictly well-founded relation]].
Let $\RR^{-1} \sqbrk x$ denote the [[Definition:Preimage of Element under Relation|preimage]] of $x$ for each $x \in A$.
Let $B$ be a [[Definition:Class (Class Theory)|class]] such that $B \subseteq A$.
Suppose ... | {{AimForCont}} $A \nsubseteq B$.
Then:
:$A \setminus B \ne 0$
By [[Strictly Well-Founded Relation determines Strictly Minimal Elements]], $A \setminus B$ must have some [[Definition:Strictly Minimal Element|strictly minimal element]] under $\RR$.
Let $\map \complement B$ be the [[Definition:Set Complement|set comple... | Well-Founded Induction | https://proofwiki.org/wiki/Well-Founded_Induction | https://proofwiki.org/wiki/Well-Founded_Induction | [
"Axiom of Foundation",
"Set Theory"
] | [
"Definition:Strictly Well-Founded Relation",
"Definition:Preimage/Relation/Element",
"Definition:Class (Class Theory)",
"Definition:Preimage/Relation/Element"
] | [
"Strictly Well-Founded Relation determines Strictly Minimal Elements",
"Definition:Strictly Minimal Element",
"Definition:Set Complement",
"Intersection is Associative",
"Intersection is Commutative",
"Intersection is Associative",
"Intersection with Complement is Empty iff Subset",
"Definition:Subset... |
proofwiki-6049 | Cardinal Number is Ordinal | Let $S$ be a set such that $S \sim x$ for some ordinal $x$.
Let $\card S$ denote the cardinality of $S$.
Then:
:$\card S \in \On$
where $\On$ denotes the class of all ordinals. | If $S \sim x$, then $\set {x \in \On: S \sim x}$ is a non-empty set of ordinals.
It follows that this set has a minimal element, its intersection.
{{explain|Needs a link to the fact that $\set {x \in \On: S \sim x}$ is the subset of a well-ordered set, and also that the intersection is that minimal element.}}
This mini... | Let $S$ be a [[Definition:Set|set]] such that $S \sim x$ for some [[Definition:Ordinal|ordinal]] $x$.
Let $\card S$ denote the [[Definition:Cardinality|cardinality]] of $S$.
Then:
:$\card S \in \On$
where $\On$ denotes the [[Definition:Class of All Ordinals|class of all ordinals]]. | If $S \sim x$, then $\set {x \in \On: S \sim x}$ is a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Ordinal|ordinals]].
It follows that this set has a [[Definition:Minimal Element|minimal element]], its intersection.
{{explain|Needs a link to the fact that $\set {x \in \On: S \sim x}$ is the subset of a ... | Cardinal Number is Ordinal | https://proofwiki.org/wiki/Cardinal_Number_is_Ordinal | https://proofwiki.org/wiki/Cardinal_Number_is_Ordinal | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Ordinal",
"Definition:Cardinality",
"Definition:Class of All Ordinals"
] | [
"Definition:Non-Empty Set",
"Definition:Ordinal",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Cardinal Number",
"Definition:Cardinal Number",
"Definition:Ordinal"
] |
proofwiki-6050 | Condition for Set Equivalent to Cardinal Number | Let $S$ be a set.
Let $\card S$ denote the cardinality of $S$.
That is, let $\card S$ be the intersection of all ordinals equivalent to $S$.
{{ExtractTheorem|This following statement needs to go somewhere else as it's a non-sequitur in this context.}}
Note that in the absence of the Axiom of Choice, $\card S$ may be th... | === $(2) \implies (1)$ ===
If $\exists x \in \On: S \sim x$, then by Class of All Ordinals is Well-Ordered by Subset Relation there is a smallest ordinal $x$ such that $S \sim x$.
This smallest ordinal $x$ is the cardinal number of $S$, by definition.
{{qed|lemma}} | Let $S$ be a [[Definition:Set|set]].
Let $\card S$ denote the [[Definition:Cardinality|cardinality]] of $S$.
That is, let $\card S$ be the [[Definition:Intersection of Set of Sets|intersection]] of all [[Definition:Ordinal|ordinals]] [[Definition:Set Equivalence|equivalent]] to $S$.
{{ExtractTheorem|This following s... | === $(2) \implies (1)$ ===
If $\exists x \in \On: S \sim x$, then by [[Class of All Ordinals is Well-Ordered by Subset Relation]] there is a smallest [[Definition:Ordinal|ordinal]] $x$ such that $S \sim x$.
This smallest [[Definition:Ordinal|ordinal]] $x$ is the [[Definition:Cardinal Number|cardinal number]] of $S$, ... | Condition for Set Equivalent to Cardinal Number | https://proofwiki.org/wiki/Condition_for_Set_Equivalent_to_Cardinal_Number | https://proofwiki.org/wiki/Condition_for_Set_Equivalent_to_Cardinal_Number | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Cardinality",
"Definition:Set Intersection/Set of Sets",
"Definition:Ordinal",
"Definition:Set Equivalence",
"Axiom:Axiom of Choice",
"Definition:Universal Class",
"Definition:Bijection"
] | [
"Class of All Ordinals is Well-Ordered by Subset Relation",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Cardinal Number",
"Definition:Ordinal",
"Definition:Ordinal",
"Definition:Cardinal Number"
] |
proofwiki-6051 | Cardinal Number Equivalence or Equal to Universe | Let $S$ be a set.
Let $\card S$ denote the cardinal number of $S$.
Let $\mathbb U$ denote the universal class.
Then:
:$S \sim \card S \lor \card S = \mathbb U$ | By Condition for Set Equivalent to Cardinal Number:
If $\exists x \in \On: S \sim x$, then:
:$S \sim \card S$
If $\neg \exists x \in \On: S \sim x$, then:
{{begin-eqn}}
{{eqn | l = \bigcap \set {x \in \On : S \sim x}
| r = \bigcap \O
| c =
}}
{{eqn | r = \mathbb U
| c = Intersection of Empty Set
}}
{... | Let $S$ be a [[Definition:Set|set]].
Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Let $\mathbb U$ denote the [[Definition:Universal Class|universal class]].
Then:
:$S \sim \card S \lor \card S = \mathbb U$ | By [[Condition for Set Equivalent to Cardinal Number]]:
If $\exists x \in \On: S \sim x$, then:
:$S \sim \card S$
If $\neg \exists x \in \On: S \sim x$, then:
{{begin-eqn}}
{{eqn | l = \bigcap \set {x \in \On : S \sim x}
| r = \bigcap \O
| c =
}}
{{eqn | r = \mathbb U
| c = [[Intersection of Emp... | Cardinal Number Equivalence or Equal to Universe | https://proofwiki.org/wiki/Cardinal_Number_Equivalence_or_Equal_to_Universe | https://proofwiki.org/wiki/Cardinal_Number_Equivalence_or_Equal_to_Universe | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Cardinal Number",
"Definition:Universal Class"
] | [
"Condition for Set Equivalent to Cardinal Number",
"Intersection of Empty Set",
"Definition:Cardinal Number"
] |
proofwiki-6052 | Ordinal Number Equivalent to Cardinal Number | Let $x$ be an ordinal.
Let $\card x$ denote the cardinal number of $x$.
Then:
:$x \sim \card x$
where $\sim$ denotes set equivalence. | From Set is Equivalent to Itself:
:$x \sim x$
Therefore, $x$ is equivalent to some ordinal.
By Condition for Set Equivalent to Cardinal Number:
:$x \sim \card x$
{{qed}} | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $\card x$ denote the [[Definition:Cardinal Number|cardinal number]] of $x$.
Then:
:$x \sim \card x$
where $\sim$ denotes [[Definition:Set Equivalence|set equivalence]]. | From [[Set is Equivalent to Itself]]:
:$x \sim x$
Therefore, $x$ is [[Definition:Set Equivalence|equivalent]] to some [[Definition:Ordinal|ordinal]].
By [[Condition for Set Equivalent to Cardinal Number]]:
:$x \sim \card x$
{{qed}} | Ordinal Number Equivalent to Cardinal Number | https://proofwiki.org/wiki/Ordinal_Number_Equivalent_to_Cardinal_Number | https://proofwiki.org/wiki/Ordinal_Number_Equivalent_to_Cardinal_Number | [
"Cardinals"
] | [
"Definition:Ordinal",
"Definition:Cardinal Number",
"Definition:Set Equivalence"
] | [
"Set is Equivalent to Itself",
"Definition:Set Equivalence",
"Definition:Ordinal",
"Condition for Set Equivalent to Cardinal Number"
] |
proofwiki-6053 | Cardinal Number Less than Ordinal | Let $S$ be a set.
Let $\card S$ denote the cardinal number of $S$.
Let $x$ be an ordinal such that $S \sim x$.
Then:
:$\card S \le x$ | Since $S \sim x$, it follows that:
: $x \in \set {y \in \On : S \sim y}$
By Intersection is Subset: General Result, it follows that:
: $\ds \bigcap \set {y \in \On: S \sim y} \subseteq x$
{{explain|Link to the fact that $\le$ is isomorphic to $\subseteq$ on ordinals.}}
Therefore $\card S \le x$ by the definition of car... | Let $S$ be a [[Definition:Set|set]].
Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Let $x$ be an [[Definition:Ordinal|ordinal]] such that $S \sim x$.
Then:
:$\card S \le x$ | Since $S \sim x$, it follows that:
: $x \in \set {y \in \On : S \sim y}$
By [[Intersection is Subset/General Result|Intersection is Subset: General Result]], it follows that:
: $\ds \bigcap \set {y \in \On: S \sim y} \subseteq x$
{{explain|Link to the fact that $\le$ is isomorphic to $\subseteq$ on ordinals.}}
Ther... | Cardinal Number Less than Ordinal | https://proofwiki.org/wiki/Cardinal_Number_Less_than_Ordinal | https://proofwiki.org/wiki/Cardinal_Number_Less_than_Ordinal | [
"Cardinals",
"Ordinals"
] | [
"Definition:Set",
"Definition:Cardinal Number",
"Definition:Ordinal"
] | [
"Intersection is Subset/General Result",
"Definition:Cardinal Number"
] |
proofwiki-6054 | Equivalent Sets have Equal Cardinal Numbers | Let $S$ and $T$ be sets.
Let $\card S$ denote the cardinal number of $S$.
Then:
:$S \sim T \implies \card S = \card T$ | Let $x$ be an arbitrary set that is an ordinal:
{{begin-eqn}}
{{eqn | l = S
| o = \sim
| r = T
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = S \sim x
| o = \iff
| r = T \sim x
| c = Set Equivalence behaves like Equivalence Relation
}}
{{eqn | ll= \leadsto
| l = \set ... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Then:
:$S \sim T \implies \card S = \card T$ | Let $x$ be an arbitrary [[Definition:Set|set]] that is an [[Definition:Ordinal|ordinal]]:
{{begin-eqn}}
{{eqn | l = S
| o = \sim
| r = T
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = S \sim x
| o = \iff
| r = T \sim x
| c = [[Set Equivalence behaves like Equivalence Relat... | Equivalent Sets have Equal Cardinal Numbers | https://proofwiki.org/wiki/Equivalent_Sets_have_Equal_Cardinal_Numbers | https://proofwiki.org/wiki/Equivalent_Sets_have_Equal_Cardinal_Numbers | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Cardinal Number"
] | [
"Definition:Set",
"Definition:Ordinal",
"Set Equivalence behaves like Equivalence Relation",
"Substitutivity of Class Equality"
] |
proofwiki-6055 | Condition for Set Union Equivalent to Associated Cardinal Number | Let $S$ and $T$ be sets.
Let $\card S$ denote the cardinal number of $S$.
Let $\sim$ denote set equivalence.
Then:
:$S \cup T \sim \card {S \cup T} \iff S \sim \card S \land T \sim \card T$ | === Necessary Condition ===
Let $S \cup T \sim \card {S \cup T}$.
By definition of set equivalence, there exists a bijection:
:$f: S \cup T \to \card {S \cup T}$
Since $f$ is a bijection, it follows that:
:$S$ is equivalent to the image of $S$ under $f$.
{{explain|Link to a result proving the above.}}
This, in turn, is... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Let $\sim$ denote [[Definition:Set Equivalence|set equivalence]].
Then:
:$S \cup T \sim \card {S \cup T} \iff S \sim \card S \land T \sim \card T$ | === Necessary Condition ===
Let $S \cup T \sim \card {S \cup T}$.
By definition of [[Definition:Set Equivalence|set equivalence]], there exists a [[Definition:Bijection|bijection]]:
:$f: S \cup T \to \card {S \cup T}$
Since $f$ is a [[Definition:Bijection|bijection]], it follows that:
:$S$ is equivalent to the ima... | Condition for Set Union Equivalent to Associated Cardinal Number | https://proofwiki.org/wiki/Condition_for_Set_Union_Equivalent_to_Associated_Cardinal_Number | https://proofwiki.org/wiki/Condition_for_Set_Union_Equivalent_to_Associated_Cardinal_Number | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Cardinal Number",
"Definition:Set Equivalence"
] | [
"Definition:Set Equivalence",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Subset",
"Definition:Ordinal",
"Definition:Ordinal",
"Cardinal Number is Ordinal",
"Condition for Set Equivalent to Cardinal Number",
"Definition:Bijection",
"Definition:Set Equivalence",
"Definition:Biject... |
proofwiki-6056 | Condition for Cartesian Product Equivalent to Associated Cardinal Number | Let $S$ and $T$ be nonempty sets.
Let $\card S$ denote the cardinal number of $S$.
Then:
:$S \times T \sim \card {S \times T} \iff S \sim \card S \land T \sim \card T$
where $S \times T$ denotes the cartesian product of $S$ and $T$. | === Necessary Condition ===
If $S \times T \sim \card {S \times T}$, then there is a mapping $f$ such that:
:$f : S \times T \to \card {S \times T}$ is a bijection.
Since $f$ is a bijection, it follows that:
:$S$ is equivalent to the image of $S \times \set x$ under $f$ where $x \in T$.
This, in turn, is a subset of th... | Let $S$ and $T$ be [[Definition:Empty Set|nonempty]] [[Definition:Set|sets]].
Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Then:
:$S \times T \sim \card {S \times T} \iff S \sim \card S \land T \sim \card T$
where $S \times T$ denotes the [[Definition:Cartesian Product|cartesian p... | === Necessary Condition ===
If $S \times T \sim \card {S \times T}$, then there is a [[Definition:Mapping|mapping]] $f$ such that:
:$f : S \times T \to \card {S \times T}$ is a [[Definition:Bijection|bijection]].
Since $f$ is a [[Definition:Bijection|bijection]], it follows that:
:$S$ is equivalent to the image of... | Condition for Cartesian Product Equivalent to Associated Cardinal Number | https://proofwiki.org/wiki/Condition_for_Cartesian_Product_Equivalent_to_Associated_Cardinal_Number | https://proofwiki.org/wiki/Condition_for_Cartesian_Product_Equivalent_to_Associated_Cardinal_Number | [
"Cardinals"
] | [
"Definition:Empty Set",
"Definition:Set",
"Definition:Cardinal Number",
"Definition:Cartesian Product"
] | [
"Definition:Mapping",
"Definition:Bijection",
"Definition:Bijection",
"Definition:Subset",
"Definition:Ordinal",
"Definition:Ordinal",
"Cardinal Number is Ordinal",
"Condition for Set Equivalent to Cardinal Number",
"Definition:Bijection",
"Definition:Bijection",
"Condition for Set Equivalent to... |
proofwiki-6057 | Cardinal of Cardinal Equal to Cardinal | Let $S$ be a set such that $S$ is equivalent to its cardinal.
If the axiom of choice holds, then this condition holds for any set.
Then:
:$\card {\paren {\card S} } = \card S$
where $\card S$ denotes the cardinal number of $S$. | By Condition for Set Equivalent to Cardinal Number:
:$S \sim \card S$
Therefore, by Equivalent Sets have Equal Cardinal Numbers:
:$\card S = \card {\paren {\card S} }$
{{qed}} | Let $S$ be a [[Definition:Set|set]] such that $S$ is [[Definition:Set Equivalence|equivalent]] to its [[Definition:Cardinal Number|cardinal]].
If the [[Axiom:Axiom of Choice|axiom of choice]] holds, then this condition holds for any set.
Then:
:$\card {\paren {\card S} } = \card S$
where $\card S$ denotes the [[Defi... | By [[Condition for Set Equivalent to Cardinal Number]]:
:$S \sim \card S$
Therefore, by [[Equivalent Sets have Equal Cardinal Numbers]]:
:$\card S = \card {\paren {\card S} }$
{{qed}} | Cardinal of Cardinal Equal to Cardinal | https://proofwiki.org/wiki/Cardinal_of_Cardinal_Equal_to_Cardinal | https://proofwiki.org/wiki/Cardinal_of_Cardinal_Equal_to_Cardinal | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Cardinal Number",
"Axiom:Axiom of Choice",
"Definition:Cardinal Number"
] | [
"Condition for Set Equivalent to Cardinal Number",
"Equivalent Sets have Equal Cardinal Numbers"
] |
proofwiki-6058 | Cardinal of Finite Ordinal | Let $n$ be a finite ordinal.
Let $\card n$ denote the cardinal number of $n$.
Then:
:$\card n = n$ | Since $n$ is an ordinal, it follows that $\card n \le n$ by {{Corollary|Cardinal Number Less than Ordinal}}.
{{MissingLinks|Predecessor of Finite Ordinal is Finite Ordinal/Element of Finite Ordinal is Finite Ordinal}}
Hence, $\card n$ is also a finite ordinal.
Since $n$ is an ordinal, it also follows that $n \sim \card... | Let $n$ be a [[Definition:Finite Ordinal|finite ordinal]].
Let $\card n$ denote the [[Definition:Cardinal Number|cardinal number]] of $n$.
Then:
:$\card n = n$ | Since $n$ is an [[Definition:Ordinal|ordinal]], it follows that $\card n \le n$ by {{Corollary|Cardinal Number Less than Ordinal}}.
{{MissingLinks|[[Predecessor of Finite Ordinal is Finite Ordinal]]/[[Element of Finite Ordinal is Finite Ordinal]]}}
Hence, $\card n$ is also a [[Definition:Finite Ordinal|finite ordinal]... | Cardinal of Finite Ordinal | https://proofwiki.org/wiki/Cardinal_of_Finite_Ordinal | https://proofwiki.org/wiki/Cardinal_of_Finite_Ordinal | [
"Cardinals",
"Finite Ordinals"
] | [
"Definition:Finite Ordinal",
"Definition:Cardinal Number"
] | [
"Definition:Ordinal",
"Predecessor of Finite Ordinal is Finite Ordinal",
"Element of Finite Ordinal is Finite Ordinal",
"Definition:Finite Ordinal",
"Definition:Ordinal",
"Ordinal Number Equivalent to Cardinal Number",
"Equality of Finite Ordinals",
"Equality of Natural Numbers"
] |
proofwiki-6059 | Finite Ordinal is equal to Natural Number | Let $n$ be an element of the minimally inductive set.
Let $x$ be an ordinal.
Then:
:$n \sim x \implies n = x$ | Let $n \ne x$.
Then either $n < x$ or $x < n$ by Ordinal Membership is Trichotomy.
{{explain|Link to the result which states that $\le$ is isomorphic to $\subseteq$ on ordinals.}}
If $x < n$, then by Subset of Finite Set is Finite both $x$ and $n$ are finite.
Therefore by No Bijection between Finite Set and Proper Subs... | Let $n$ be an [[Definition:Element|element]] of the [[Definition:Minimally Inductive Set|minimally inductive set]].
Let $x$ be an [[Definition:Ordinal|ordinal]].
Then:
:$n \sim x \implies n = x$ | Let $n \ne x$.
Then either $n < x$ or $x < n$ by [[Ordinal Membership is Trichotomy]].
{{explain|Link to the result which states that $\le$ is isomorphic to $\subseteq$ on ordinals.}}
If $x < n$, then by [[Subset of Finite Set is Finite]] both $x$ and $n$ are [[Definition:Finite Set|finite]].
Therefore by [[No Bij... | Finite Ordinal is equal to Natural Number | https://proofwiki.org/wiki/Finite_Ordinal_is_equal_to_Natural_Number | https://proofwiki.org/wiki/Finite_Ordinal_is_equal_to_Natural_Number | [
"Cardinals",
"Minimally Inductive Set"
] | [
"Definition:Element",
"Definition:Minimally Inductive Set",
"Definition:Ordinal"
] | [
"Ordinal Membership is Trichotomy",
"Subset of Finite Set is Finite",
"Definition:Finite Set",
"No Bijection between Finite Set and Proper Subset",
"Definition:Finite Set",
"No Bijection between Finite Set and Proper Subset",
"Rule of Transposition"
] |
proofwiki-6060 | Product (Category Theory) is Unique | Let $\mathbf C$ be a metacategory.
Let $A$ and $B$ be objects of $\mathbf C$.
Let $A \times B$ and $A \times' B$ both be products of $A$ and $B$.
Then there is a unique isomorphism $u: A \times B \to A \times' B$.
That is, products are unique up to unique isomorphism.
{{expand|cover extended theorem (for general def)}} | Denote the implicit projections of the two products by:
<nowiki>
$\begin{xy}\xymatrix@C=2em@L+4mu{
&
A \times B
\ar@/_/[ddl]_*{\textrm{pr}_1}
\ar@/^/[ddr]^*{\textrm{pr}_2}
\\ &
A \times' B
\ar[dl]^*{\textrm{pr}'_1}
\ar[dr]_*{\textrm{pr}'_2}
\\
A
& &
B
}\end{xy}$
</nowiki>
By the universal mapping property o... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $A$ and $B$ be [[Definition:Object (Category Theory)|objects]] of $\mathbf C$.
Let $A \times B$ and $A \times' B$ both be [[Definition:Product (Category Theory)|products]] of $A$ and $B$.
Then there is a unique [[Definition:Isomorphism (Category The... | Denote the implicit [[Definition:Projection (Category Theory)|projections]] of the two [[Definition:Product (Category Theory)|products]] by:
<nowiki>
$\begin{xy}\xymatrix@C=2em@L+4mu{
&
A \times B
\ar@/_/[ddl]_*{\textrm{pr}_1}
\ar@/^/[ddr]^*{\textrm{pr}_2}
\\ &
A \times' B
\ar[dl]^*{\textrm{pr}'_1}
\ar[dr]... | Product (Category Theory) is Unique | https://proofwiki.org/wiki/Product_(Category_Theory)_is_Unique | https://proofwiki.org/wiki/Product_(Category_Theory)_is_Unique | [
"Products (Category Theory)"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Product (Category Theory)",
"Definition:Isomorphism (Category Theory)",
"Definition:Product (Category Theory)",
"Definition:Isomorphism (Category Theory)"
] | [
"Definition:Product (Category Theory)/Projection",
"Definition:Product (Category Theory)",
"Definition:Product UMP (Category Theory)",
"Definition:Unique",
"Definition:Product UMP (Category Theory)",
"Definition:Unique",
"Definition:Identity Morphism",
"Definition:Product UMP (Category Theory)",
"De... |
proofwiki-6061 | Ordinal Subset is Well-Ordered | Let $S$ be a class.
{{explain|Is the $S$ really a class? All links below are for a set.}}
Let every element of $S$ be an ordinal.
Then $\struct {S, \in}$ is a strict well-ordering. | {{improve|wrong use of links}}
Let $\On$ denote the class of all ordinals.
By definition of subset, $S \subseteq \On$.
But by Class of All Ordinals is Ordinal $\On$ is an ordinal.
Therefore $\On$ is well-ordered by $\in$.
This means that $S$ is also well-ordered by $\in$.
{{qed}}
Category:Ordinals
t5z8hn3ab8s3knfjl0g0y... | Let $S$ be a [[Definition:Class (Class Theory)|class]].
{{explain|Is the $S$ really a class? All links below are for a set.}}
Let every [[Definition:Element|element]] of $S$ be an [[Definition:Ordinal|ordinal]].
Then $\struct {S, \in}$ is a [[Definition:Strict Well-Ordering|strict well-ordering]]. | {{improve|wrong use of links}}
Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
By definition of [[Definition:Subset|subset]], $S \subseteq \On$.
But by [[Class of All Ordinals is Ordinal]] $\On$ is an [[Definition:Ordinal|ordinal]].
Therefore $\On$ is [[Alternative Definition of Or... | Ordinal Subset is Well-Ordered | https://proofwiki.org/wiki/Ordinal_Subset_is_Well-Ordered | https://proofwiki.org/wiki/Ordinal_Subset_is_Well-Ordered | [
"Ordinals"
] | [
"Definition:Class (Class Theory)",
"Definition:Element",
"Definition:Ordinal",
"Definition:Strict Well-Ordering"
] | [
"Definition:Class of All Ordinals",
"Definition:Subset",
"Class of All Ordinals is Ordinal",
"Definition:Ordinal",
"Alternative Definition of Ordinal",
"Subset of Well-Ordered Set is Well-Ordered",
"Category:Ordinals"
] |
proofwiki-6062 | Subset implies Cardinal Inequality | Let $S$ and $T$ be sets such that $S \subseteq T$.
Furthermore, let:
:$T \sim \card T$
where $\card T$ denotes the cardinality of $T$.
Then:
:$\card S \le \card T$ | For the proof:
:the ordering relation $\le$ for ordinals
and
:the subset relation $\subseteq$
shall be used interchangeably.
Let $f: T \to \card T$ be a bijection.
It follows that $f \restriction_S : S \to \card T$ is an injection.
The image of $S$ under $f$ is a subset of $\card T$ and thus is a subset of an ordinal.
... | Let $S$ and $T$ be [[Definition:Set|sets]] such that $S \subseteq T$.
Furthermore, let:
:$T \sim \card T$
where $\card T$ denotes the [[Definition:Cardinality|cardinality]] of $T$.
Then:
:$\card S \le \card T$ | For the proof:
:the ordering relation $\le$ for [[Definition:Ordinal|ordinals]]
and
:the [[Definition:Subset|subset relation]] $\subseteq$
shall be used interchangeably.
Let $f: T \to \card T$ be a [[Definition:Bijection|bijection]].
It follows that $f \restriction_S : S \to \card T$ is an [[Definition:Injection|in... | Subset implies Cardinal Inequality | https://proofwiki.org/wiki/Subset_implies_Cardinal_Inequality | https://proofwiki.org/wiki/Subset_implies_Cardinal_Inequality | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Cardinality"
] | [
"Definition:Ordinal",
"Definition:Subset",
"Definition:Bijection",
"Definition:Injection",
"Definition:Image (Set Theory)/Mapping/Subset",
"Image of Subset under Relation is Subset of Image",
"Definition:Subset",
"Definition:Ordinal",
"Unique Isomorphism between Ordinal Subset and Unique Ordinal",
... |
proofwiki-6063 | Subset of Ordinal implies Cardinal Inequality | Let $S$ be a set.
Let $x$ be an ordinal such that $S \subseteq x$.
Then:
:$\card S \le \card x$
where $\card S$ denotes the cardinality of $S$. | Since $x$ is an ordinal, it follows that $x \sim \card x$ by Ordinal Number Equivalent to Cardinal Number.
This satisfies the hypothesis for Subset implies Cardinal Inequality.
Therefore:
:$\card S \le \card x$
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let $x$ be an [[Definition:Ordinal|ordinal]] such that $S \subseteq x$.
Then:
:$\card S \le \card x$
where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$. | Since $x$ is an [[Definition:Ordinal|ordinal]], it follows that $x \sim \card x$ by [[Ordinal Number Equivalent to Cardinal Number]].
This satisfies the hypothesis for [[Subset implies Cardinal Inequality]].
Therefore:
:$\card S \le \card x$
{{qed}} | Subset of Ordinal implies Cardinal Inequality | https://proofwiki.org/wiki/Subset_of_Ordinal_implies_Cardinal_Inequality | https://proofwiki.org/wiki/Subset_of_Ordinal_implies_Cardinal_Inequality | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Ordinal",
"Definition:Cardinality"
] | [
"Definition:Ordinal",
"Ordinal Number Equivalent to Cardinal Number",
"Subset implies Cardinal Inequality"
] |
proofwiki-6064 | Set Less than Cardinal Product | Let $S$ and $T$ be sets.
Let $T$ be nonempty.
Suppose that $S \times T \sim \card {S \times T}$.
Then:
:$\card S \le \card {S \times T}$ | Let $y \in T$.
Define the mapping $f : S \to S \times T$ as follows:
:$\map f x = \tuple {x, y}$
If $\map f {x_1} = \map f {x_2}$, then $\tuple {x_1, y} = \tuple {x_2, y}$ by the definition of $f$.
It follows that $x_1 = x_2$ by Equality of Ordered Pairs.
Thus, $f: S \to S \times T$ is an injection.
By Injection implie... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $T$ be [[Definition:Empty Set|nonempty]].
Suppose that $S \times T \sim \card {S \times T}$.
Then:
:$\card S \le \card {S \times T}$ | Let $y \in T$.
Define the [[Definition:Mapping|mapping]] $f : S \to S \times T$ as follows:
:$\map f x = \tuple {x, y}$
If $\map f {x_1} = \map f {x_2}$, then $\tuple {x_1, y} = \tuple {x_2, y}$ by the definition of $f$.
It follows that $x_1 = x_2$ by [[Equality of Ordered Pairs]].
Thus, $f: S \to S \times T$ is ... | Set Less than Cardinal Product | https://proofwiki.org/wiki/Set_Less_than_Cardinal_Product | https://proofwiki.org/wiki/Set_Less_than_Cardinal_Product | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Empty Set"
] | [
"Definition:Mapping",
"Equality of Ordered Pairs",
"Definition:Injection",
"Injection implies Cardinal Inequality"
] |
proofwiki-6065 | Injection implies Cardinal Inequality | Let $S$ and $T$ be sets.
Let $f: S \to T$ be an injection.
Let $\card T$ denote the cardinal number of $T$.
Let:
:$T \sim \card T$
where $\sim$ denotes set equivalence
Then:
:$\card S \le \card T$ | Let $f \sqbrk S$ denote the image of $S$ under $f$.
{{begin-eqn}}
{{eqn | l = S
| o = \sim
| r = f \sqbrk S
| c = Set is Equivalent to Image under Injection
}}
{{eqn | o = \subseteq
| r = T
| c = Image Preserves Subsets
}}
{{eqn | ll= \leadsto
| l = \card S
| r = \card {f \sqbr... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be an [[Definition:Injection|injection]].
Let $\card T$ denote the [[Definition:Cardinal Number|cardinal number]] of $T$.
Let:
:$T \sim \card T$
where $\sim$ denotes [[Definition:Set Equivalence|set equivalence]]
Then:
:$\card S \le \card T$ | Let $f \sqbrk S$ denote the [[Definition:Image of Subset under Mapping|image]] of $S$ under $f$.
{{begin-eqn}}
{{eqn | l = S
| o = \sim
| r = f \sqbrk S
| c = [[Set is Equivalent to Image under Injection]]
}}
{{eqn | o = \subseteq
| r = T
| c = [[Image Preserves Subsets]]
}}
{{eqn | ll= ... | Injection implies Cardinal Inequality | https://proofwiki.org/wiki/Injection_implies_Cardinal_Inequality | https://proofwiki.org/wiki/Injection_implies_Cardinal_Inequality | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Cardinal Number",
"Definition:Set Equivalence"
] | [
"Definition:Image (Set Theory)/Mapping/Subset",
"Set is Equivalent to Image under Injection",
"Image of Subset under Relation is Subset of Image",
"Equivalent Sets have Equal Cardinal Numbers",
"Subset implies Cardinal Inequality",
"Category:Cardinals"
] |
proofwiki-6066 | Set is Equivalent to Image under Injection | Let $S$ and $T$ be sets.
Let $f: S \to T$ be an injection.
Then the image of $S$ under $f$ is equivalent to $S$. | {{ProofWanted}}
Category:Set Theory
Category:Injections
n03hbds288gm7nqe1bo4sqrh78r8y1g | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be an [[Definition:Injection|injection]].
Then the [[Definition:Image of Subset under Mapping|image]] of $S$ under $f$ is [[Definition:Set Equivalence|equivalent]] to $S$. | {{ProofWanted}}
[[Category:Set Theory]]
[[Category:Injections]]
n03hbds288gm7nqe1bo4sqrh78r8y1g | Set is Equivalent to Image under Injection | https://proofwiki.org/wiki/Set_is_Equivalent_to_Image_under_Injection | https://proofwiki.org/wiki/Set_is_Equivalent_to_Image_under_Injection | [
"Set Theory",
"Injections"
] | [
"Definition:Set",
"Definition:Injection",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Set Equivalence"
] | [
"Category:Set Theory",
"Category:Injections"
] |
proofwiki-6067 | Cardinality of Image of Mapping not greater than Cardinality of Domain | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $\card S$ denote the cardinal number of $S$.
Then:
:$\card {\Img f} \le \card S$ | By Restriction of Mapping to Image is Surjection, the mapping:
:$f: S \to \Img f$
is a surjection.
Let $h$ be a mapping such that:
:$h: \card S \to S$
is a bijection.
By Composite of Surjections is Surjection:
:$f \circ h: \card S \to \Img f$
is a surjection.
Construct a set $R$ such that:
:$R = \set {x \in \card S: \f... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Then:
:$\card {\Img f} \le \card S$ | By [[Restriction of Mapping to Image is Surjection]], the [[Definition:Mapping|mapping]]:
:$f: S \to \Img f$
is a [[Definition:Surjection|surjection]].
Let $h$ be a [[Definition:Mapping|mapping]] such that:
:$h: \card S \to S$
is a [[Definition:Bijection|bijection]].
By [[Composite of Surjections is Surjection]]:
:... | Cardinality of Image of Mapping not greater than Cardinality of Domain | https://proofwiki.org/wiki/Cardinality_of_Image_of_Mapping_not_greater_than_Cardinality_of_Domain | https://proofwiki.org/wiki/Cardinality_of_Image_of_Mapping_not_greater_than_Cardinality_of_Domain | [
"Images",
"Cardinality"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Cardinal Number"
] | [
"Restriction of Mapping to Image is Surjection",
"Definition:Mapping",
"Definition:Surjection",
"Definition:Mapping",
"Definition:Bijection",
"Composite of Surjections is Surjection",
"Definition:Surjection",
"Definition:Set",
"Subset of Ordinal implies Cardinal Inequality",
"Definition:Ordinal",
... |
proofwiki-6068 | Injection iff Cardinal Inequality | Let $\card T$ denote the cardinal number of $T$.
Let $S$ and $T$ be sets such that $S \sim \card S$ and $T \sim \card T$.
Then:
:$\card S \le \card T $ {{iff}} there exists an injection $f: S \to T$. | === Necessary Condition ===
Suppose that $\card S \le \card T$.
Let $g : S \to \card S$ be a bijection and $h: \card T \to T$ be a bijection.
It follows that $g: S \to \card T$ is an injection by the fact that $\card T \le \card S$.
Then from Composite of Injections is Injection, $h \circ g: S \to T$ is an injection.
{... | Let $\card T$ denote the [[Definition:Cardinal Number|cardinal number]] of $T$.
Let $S$ and $T$ be [[Definition:Set|sets]] such that $S \sim \card S$ and $T \sim \card T$.
Then:
:$\card S \le \card T $ {{iff}} there exists an [[Definition:Injection|injection]] $f: S \to T$. | === Necessary Condition ===
Suppose that $\card S \le \card T$.
Let $g : S \to \card S$ be a [[Definition:Bijection|bijection]] and $h: \card T \to T$ be a [[Definition:Bijection|bijection]].
It follows that $g: S \to \card T$ is an [[Definition:Injection|injection]] by the fact that $\card T \le \card S$.
Then f... | Injection iff Cardinal Inequality | https://proofwiki.org/wiki/Injection_iff_Cardinal_Inequality | https://proofwiki.org/wiki/Injection_iff_Cardinal_Inequality | [
"Cardinals"
] | [
"Definition:Cardinal Number",
"Definition:Set",
"Definition:Injection"
] | [
"Definition:Bijection",
"Definition:Bijection",
"Definition:Injection",
"Composite of Injections is Injection",
"Definition:Injection"
] |
proofwiki-6069 | Surjection iff Cardinal Inequality | Let $S$ and $T$ be sets.
Let $S$ be non-empty.
Then:
:$0 < \card T \le \card S$
{{iff}}:
:there exists a surjection $f: S \to T$
where $\card S$ denotes the cardinality of $S$. | === Necessary Condition ===
Suppose $f: S \to T$ is a surjection.
Then $\Img f = T$ by definition.
From Cardinality of Image of Mapping not greater than Cardinality of Domain:
:$\card T \le \card S$
Furthermore, if $S$ is non-empty, then $\map f x \in T$ for some $x \in S$.
Thus, $T$ is non-empty and $0 < \card T$ by C... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $S$ be [[Definition:Non-Empty Set|non-empty]].
Then:
:$0 < \card T \le \card S$
{{iff}}:
:there exists a [[Definition:Surjection|surjection]] $f: S \to T$
where $\card S$ denotes the [[Definition:Cardinality of Set|cardinality]] of $S$. | === Necessary Condition ===
Suppose $f: S \to T$ is a [[Definition:Surjection|surjection]].
Then $\Img f = T$ by definition.
From [[Cardinality of Image of Mapping not greater than Cardinality of Domain]]:
:$\card T \le \card S$
Furthermore, if $S$ is [[Definition:Non-Empty Set|non-empty]], then $\map f x \in T$ f... | Surjection iff Cardinal Inequality | https://proofwiki.org/wiki/Surjection_iff_Cardinal_Inequality | https://proofwiki.org/wiki/Surjection_iff_Cardinal_Inequality | [
"Surjections",
"Cardinals"
] | [
"Definition:Set",
"Definition:Non-Empty Set",
"Definition:Surjection",
"Definition:Cardinality"
] | [
"Definition:Surjection",
"Cardinality of Image of Mapping not greater than Cardinality of Domain",
"Definition:Non-Empty Set",
"Definition:Non-Empty Set",
"Cardinality of Empty Set",
"Definition:Surjection"
] |
proofwiki-6070 | Right Identity while exists Right Inverse for All is Identity | Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$ such that:
:$\forall x \in S: \exists x_R: x \circ x_R = e_R$
That is, every element of $S$ has a right inverse with respect to the right identity.
Then $e_R$ is also a left identity, that is, is an identity. | Let $x \in S$ be any element of $S$.
From Right Inverse for All is Left Inverse we have that $x_R \circ x = e_R$.
Then:
{{begin-eqn}}
{{eqn | l = e_R \circ x
| r = \paren {x \circ x_R} \circ x
| c = {{Defof|Right Inverse Element}}
}}
{{eqn | r = x \circ \paren {x_R \circ x}
| c = {{Semigroup-axiom|1}}... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] with a [[Definition:Right Identity|right identity]] $e_R$ such that:
:$\forall x \in S: \exists x_R: x \circ x_R = e_R$
That is, every [[Definition:Element|element]] of $S$ has a [[Definition:Right Inverse Element|right inverse]] with respect to the [[... | Let $x \in S$ be any element of $S$.
From [[Right Inverse for All is Left Inverse]] we have that $x_R \circ x = e_R$.
Then:
{{begin-eqn}}
{{eqn | l = e_R \circ x
| r = \paren {x \circ x_R} \circ x
| c = {{Defof|Right Inverse Element}}
}}
{{eqn | r = x \circ \paren {x_R \circ x}
| c = {{Semigroup-axi... | Right Identity while exists Right Inverse for All is Identity | https://proofwiki.org/wiki/Right_Identity_while_exists_Right_Inverse_for_All_is_Identity | https://proofwiki.org/wiki/Right_Identity_while_exists_Right_Inverse_for_All_is_Identity | [
"Semigroups"
] | [
"Definition:Semigroup",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Identity (Abstract Al... | [
"Right Inverse for All is Left Inverse",
"Right Inverse for All is Left Inverse",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-6071 | Left Identity while exists Left Inverse for All is Identity | Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$ such that:
:$\forall x \in S: \exists x_L: x_L \circ x = e_L$
That is, every element of $S$ has a left inverse with respect to the left identity.
Then $e_L$ is also a right identity, that is, is an identity. | From Left Inverse for All is Right Inverse we have that:
:$x \circ x_L = e_L$
Then:
{{begin-eqn}}
{{eqn | l = x \circ e_L
| r = x \circ \paren {x_L \circ x}
| c = {{Defof|Left Inverse Element}}
}}
{{eqn | r = \paren {x \circ x_L} \circ x
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = e_L \circ x
| c... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] with a [[Definition:Left Identity|left identity]] $e_L$ such that:
:$\forall x \in S: \exists x_L: x_L \circ x = e_L$
That is, every [[Definition:Element|element]] of $S$ has a [[Definition:Left Inverse Element|left inverse]] with respect to the [[Defin... | From [[Left Inverse for All is Right Inverse]] we have that:
:$x \circ x_L = e_L$
Then:
{{begin-eqn}}
{{eqn | l = x \circ e_L
| r = x \circ \paren {x_L \circ x}
| c = {{Defof|Left Inverse Element}}
}}
{{eqn | r = \paren {x \circ x_L} \circ x
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = e_L \circ x
... | Left Identity while exists Left Inverse for All is Identity | https://proofwiki.org/wiki/Left_Identity_while_exists_Left_Inverse_for_All_is_Identity | https://proofwiki.org/wiki/Left_Identity_while_exists_Left_Inverse_for_All_is_Identity | [
"Semigroups"
] | [
"Definition:Semigroup",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Identity (Abstract Alge... | [
"Left Inverse for All is Right Inverse",
"Left Inverse for All is Right Inverse",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-6072 | Right Inverse for All is Left Inverse | Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$ such that:
:$\forall x \in S: \exists x_R: x \circ x_R = e_R$
That is, every element of $S$ has a right inverse with respect to the right identity.
Then $x_R \circ x = e_R$, that is, $x_R$ is also a left inverse with respect to the right identity. | Let $y = x_R \circ x$. Then:
{{begin-eqn}}
{{eqn | l = y \circ e_R
| r = y \circ \paren {y \circ y_R}
| c = {{Defof|Right Inverse Element}}
}}
{{eqn | r = \paren {y \circ y} \circ y_R
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = y \circ y_R
| c = Product of Semigroup Element with Right Inverse is ... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] with a [[Definition:Right Identity|right identity]] $e_R$ such that:
:$\forall x \in S: \exists x_R: x \circ x_R = e_R$
That is, every [[Definition:Element|element]] of $S$ has a [[Definition:Right Inverse Element|right inverse]] with respect to the [[D... | Let $y = x_R \circ x$. Then:
{{begin-eqn}}
{{eqn | l = y \circ e_R
| r = y \circ \paren {y \circ y_R}
| c = {{Defof|Right Inverse Element}}
}}
{{eqn | r = \paren {y \circ y} \circ y_R
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = y \circ y_R
| c = [[Product of Semigroup Element with Right Inverse ... | Right Inverse for All is Left Inverse | https://proofwiki.org/wiki/Right_Inverse_for_All_is_Left_Inverse | https://proofwiki.org/wiki/Right_Inverse_for_All_is_Left_Inverse | [
"Semigroups",
"Inverse Elements"
] | [
"Definition:Semigroup",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Identity (Abstract Alge... | [
"Product of Semigroup Element with Right Inverse is Idempotent",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Definition:Identity (Abstract Algebra)/Right Identity"
] |
proofwiki-6073 | Equal Powers of Group Element implies Finite Order | Let $\struct {G, \circ}$ be a group.
Let $g \in G$ such that $g^r = g^s$ where $r, s \in \Z: r \ne s$.
Then there exists $m \in \Z_{>0}$ such that:
:$(1): \quad g^m = e$
:$(2): \quad 0 \le i < j < m \implies g^i \ne g^j$ | {{WLOG}}, suppose that $r > s$.
From $g^r = g^s$ it follows from Powers of Group Elements that:
: $g^{r - s} = g^r g^{-s} = g^s g^{-s} = e$
Thus there exists $t \in \Z_{>0}$ such that $g^t = e$.
Let $m \in \Z_{>0}$ be the smallest such that $g^m = e$.
Suppose $0 \le i < j < m$ such that $g^i = g^j$.
Then $g^{j - i} = e... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $g \in G$ such that $g^r = g^s$ where $r, s \in \Z: r \ne s$.
Then there exists $m \in \Z_{>0}$ such that:
:$(1): \quad g^m = e$
:$(2): \quad 0 \le i < j < m \implies g^i \ne g^j$ | {{WLOG}}, suppose that $r > s$.
From $g^r = g^s$ it follows from [[Powers of Group Elements]] that:
: $g^{r - s} = g^r g^{-s} = g^s g^{-s} = e$
Thus there exists $t \in \Z_{>0}$ such that $g^t = e$.
Let $m \in \Z_{>0}$ be the smallest such that $g^m = e$.
Suppose $0 \le i < j < m$ such that $g^i = g^j$.
Then $g^{j... | Equal Powers of Group Element implies Finite Order | https://proofwiki.org/wiki/Equal_Powers_of_Group_Element_implies_Finite_Order | https://proofwiki.org/wiki/Equal_Powers_of_Group_Element_implies_Finite_Order | [
"Order of Group Elements"
] | [
"Definition:Group"
] | [
"Powers of Group Elements"
] |
proofwiki-6074 | Exponential on Complex Plane is Group Homomorphism | Let $\struct {\C, +}$ be the additive group of complex numbers.
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $\exp: \struct {\C, +} \to \struct {\C_{\ne 0}, \times}$ be the mapping:
:$x \mapsto \map \exp x$
where $\exp$ is the complex exponential function.
Then $\exp$ is a grou... | If $z \in \C$, then by the definition of the complex exponential function, $\exp$ is a mapping $\C \to \C_{\ne 0}$.
Let $z_1, z_2 \in \C$.
By Exponential of Sum:
:$\map \exp {z_1 + z_2} = \map \exp {z_1} \, \map \exp {z_2}$
Therefore $\exp: \struct {\C, +} \to \struct {\C_{\ne 0}, \times}$ is a group homomorphism.
{{qe... | Let $\struct {\C, +}$ be the [[Definition:Additive Group of Complex Numbers|additive group of complex numbers]].
Let $\struct {\C_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Complex Numbers|multiplicative group of complex numbers]].
Let $\exp: \struct {\C, +} \to \struct {\C_{\ne 0}, \times}$ be th... | If $z \in \C$, then by the definition of the [[Definition:Complex Exponential Function|complex exponential function]], $\exp$ is a [[Definition:Mapping|mapping]] $\C \to \C_{\ne 0}$.
Let $z_1, z_2 \in \C$.
By [[Exponential of Sum/Complex Numbers|Exponential of Sum]]:
:$\map \exp {z_1 + z_2} = \map \exp {z_1} \, \map ... | Exponential on Complex Plane is Group Homomorphism | https://proofwiki.org/wiki/Exponential_on_Complex_Plane_is_Group_Homomorphism | https://proofwiki.org/wiki/Exponential_on_Complex_Plane_is_Group_Homomorphism | [
"Examples of Group Homomorphisms",
"Exponential Function"
] | [
"Definition:Additive Group of Complex Numbers",
"Definition:Multiplicative Group of Complex Numbers",
"Definition:Mapping",
"Definition:Exponential Function/Complex",
"Definition:Group Homomorphism"
] | [
"Definition:Exponential Function/Complex",
"Definition:Mapping",
"Exponential of Sum/Complex Numbers",
"Definition:Group Homomorphism"
] |
proofwiki-6075 | Group Direct Product is Product in Category of Groups | Let $\mathbf{Grp}$ be the category of groups.
Let $G$ and $H$ be groups, and let $G \times H$ be their direct product.
Then $G \times H$ is a binary product of $G$ and $H$ in $\mathbf{Grp}$. | Let $F$ be a group.
By Direct Product of Group Homomorphisms is Homomorphism, given group homomorphisms:
:$g: F \to G, h: F \to H$
their direct product $g \times h: F \to G \times H$ is a group homomorphism.
From Projections on Direct Product of Group Homomorphisms, the following diagram is commutative:
:<nowiki>$\begi... | Let $\mathbf{Grp}$ be the [[Definition:Category of Groups|category of groups]].
Let $G$ and $H$ be [[Definition:Group|groups]], and let $G \times H$ be their [[Definition:Group Direct Product|direct product]].
Then $G \times H$ is a [[Definition:Binary Product (Category Theory)|binary product]] of $G$ and $H$ in $\m... | Let $F$ be a [[Definition:Group|group]].
By [[Direct Product of Group Homomorphisms is Homomorphism]], given [[Definition:Group Homomorphism|group homomorphisms]]:
:$g: F \to G, h: F \to H$
their [[Definition:Direct Product of Group Homomorphisms|direct product]] $g \times h: F \to G \times H$ is a [[Definition:Grou... | Group Direct Product is Product in Category of Groups | https://proofwiki.org/wiki/Group_Direct_Product_is_Product_in_Category_of_Groups | https://proofwiki.org/wiki/Group_Direct_Product_is_Product_in_Category_of_Groups | [
"Category of Groups",
"Group Theory"
] | [
"Definition:Category of Groups",
"Definition:Group",
"Definition:Group Direct Product",
"Definition:Product (Category Theory)/Binary Product"
] | [
"Definition:Group",
"Direct Product of Group Homomorphisms is Homomorphism",
"Definition:Group Homomorphism",
"Definition:Direct Product of Group Homomorphisms",
"Definition:Group Homomorphism",
"Projections on Direct Product of Group Homomorphisms",
"Cartesian Product is Set Product",
"Definition:Map... |
proofwiki-6076 | Product Category is Product in Category of Categories | Let $\mathbf{Cat}$ be the category of categories.
Let $\mathbf C$ and $\mathbf D$ be small categories, and let $\mathbf C \times \mathbf D$ be their product category.
Then $\mathbf C \times \mathbf D$ is a binary product of $\mathbf C$ and $\mathbf D$ in $\mathbf{Cat}$. | Let $F: \mathbf A \to \mathbf C$ and $G: \mathbf A \to \mathbf D$ be morphisms in $\mathbf{Cat}$, i.e. functors.
Suppose $X: \mathbf A \to \mathbf C \times \mathbf D$ is a functor such that $\pr_1 \circ X = F$ and $\pr_2 \circ X = G$:
::<nowiki>$\begin{xy}\xymatrix@C=3em@R=3em@L+3mu{
&
\mathbf A
\ar[dl]_*{F}
\ar[d... | Let $\mathbf{Cat}$ be the [[Definition:Category of Categories|category of categories]].
Let $\mathbf C$ and $\mathbf D$ be [[Definition:Small Category|small categories]], and let $\mathbf C \times \mathbf D$ be their [[Definition:Product Category|product category]].
Then $\mathbf C \times \mathbf D$ is a [[Definitio... | Let $F: \mathbf A \to \mathbf C$ and $G: \mathbf A \to \mathbf D$ be [[Definition:Morphism (Category Theory)|morphisms]] in $\mathbf{Cat}$, i.e. [[Definition:Covariant Functor|functors]].
Suppose $X: \mathbf A \to \mathbf C \times \mathbf D$ is a functor such that $\pr_1 \circ X = F$ and $\pr_2 \circ X = G$:
::<nowik... | Product Category is Product in Category of Categories | https://proofwiki.org/wiki/Product_Category_is_Product_in_Category_of_Categories | https://proofwiki.org/wiki/Product_Category_is_Product_in_Category_of_Categories | [
"Category of Categories"
] | [
"Definition:Category of Categories",
"Definition:Small Category",
"Definition:Product Category",
"Definition:Product (Category Theory)/Binary Product"
] | [
"Definition:Morphism",
"Definition:Functor/Covariant",
"Definition:Projection Functor",
"Definition:Object (Category Theory)",
"Definition:Product Category",
"Definition:Morphism",
"Definition:Product Category",
"Definition:Functor/Covariant",
"Functor to Product Category",
"Definition:Product (Ca... |
proofwiki-6077 | Infimum is Product in Order Category | Let $\mathbf P$ be an order category whose ordering is $\preceq$.
Let $p, q \in \mathbf P_0$, and suppose that they have some infimum $r = \inf \set {p, q}$.
Then $r$ is a binary product of $p$ and $q$ in $\mathbf P$. | Suppose that there are morphisms $l \to p$ and $l \to q$ in $\mathbf P$.
That is to say, suppose $l \preceq p$ and $l \preceq q$.
Then $l$ is a lower bound for $\set {p, q}$.
By definition of infimum, we then have:
:$l \preceq r = \inf \set {p, q}$
By definition of $\mathbf P$ as an order category, this means there is ... | Let $\mathbf P$ be an [[Definition:Order Category|order category]] whose [[Definition:Ordering|ordering]] is $\preceq$.
Let $p, q \in \mathbf P_0$, and suppose that they have some [[Definition:Infimum of Set|infimum]] $r = \inf \set {p, q}$.
Then $r$ is a [[Definition:Binary Product (Category Theory)|binary product]... | Suppose that there are [[Definition:Morphism (Category Theory)|morphisms]] $l \to p$ and $l \to q$ in $\mathbf P$.
That is to say, suppose $l \preceq p$ and $l \preceq q$.
Then $l$ is a [[Definition:Lower Bound of Set|lower bound]] for $\set {p, q}$.
By definition of [[Definition:Infimum of Set|infimum]], we then ha... | Infimum is Product in Order Category | https://proofwiki.org/wiki/Infimum_is_Product_in_Order_Category | https://proofwiki.org/wiki/Infimum_is_Product_in_Order_Category | [
"Order Categories"
] | [
"Definition:Order Category",
"Definition:Ordering",
"Definition:Infimum of Set",
"Definition:Product (Category Theory)/Binary Product"
] | [
"Definition:Morphism",
"Definition:Lower Bound of Set",
"Definition:Infimum of Set",
"Definition:Order Category",
"Definition:Morphism",
"Definition:Morphism",
"Definition:Product UMP (Category Theory)",
"Definition:Product (Category Theory)/Binary Product"
] |
proofwiki-6078 | Cardinal of Union Less than Cardinal of Cartesian Product | Let $S$ and $T$ be sets that are equivalent to their cardinal numbers.
Let $\card S$ denote the cardinal number of $S$.
Let $\card S > 1$ and $\card T > 1$.
Then:
:$\card {S \cup T} \le \card {S \times T}$ | Let $x_1$ and $x_2$ be distinct elements of $S$.
Let $y_1$ and $y_2$ be distinct elements of $T$.
Define the mapping $f : S \times T \to S \cup T$ as follows:
$\quad\map f {x, y} = \begin {cases}
y &: x = x_1 \\
x_1 &: x = x_2 \land y = y_1 \\
x &: \text {otherwise}
\end {cases}$
If $x \in S$, then we have that either ... | Let $S$ and $T$ be [[Definition:Set|sets]] that are [[Definition:Set Equivalence|equivalent]] to their [[Definition:Cardinal Number|cardinal numbers]].
Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Let $\card S > 1$ and $\card T > 1$.
Then:
:$\card {S \cup T} \le \card {S \times T... | Let $x_1$ and $x_2$ be [[Definition:Distinct Elements|distinct elements]] of $S$.
Let $y_1$ and $y_2$ be [[Definition:Distinct Elements|distinct elements]] of $T$.
Define the [[Definition:Mapping|mapping]] $f : S \times T \to S \cup T$ as follows:
$\quad\map f {x, y} = \begin {cases}
y &: x = x_1 \\
x_1 &: x = x_2... | Cardinal of Union Less than Cardinal of Cartesian Product | https://proofwiki.org/wiki/Cardinal_of_Union_Less_than_Cardinal_of_Cartesian_Product | https://proofwiki.org/wiki/Cardinal_of_Union_Less_than_Cardinal_of_Cartesian_Product | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Cardinal Number",
"Definition:Cardinal Number"
] | [
"Definition:Distinct/Plural",
"Definition:Distinct/Plural",
"Definition:Mapping",
"Definition:Surjection",
"Surjection iff Cardinal Inequality"
] |
proofwiki-6079 | Cardinal Product Equinumerous to Ordinal Product | Let $S$ and $T$ be sets that are equivalent to their cardinal numbers.
Let $\card S$ denote the cardinal number of $S$.
Let $\cdot$ denote ordinal multiplication and let $\times$ denote the Cartesian product.
Then:
:$S \times T \sim \card S \cdot \card T$ | Let $f: S \to \card S$ and $g: T \to \card T$ be bijections.
Let $\cdot$ denote ordinal multiplication, while $\times$ shall denote the Cartesian product.
Define the function $F$ to be:
:$\forall x \in S, y \in T: \map F {x, y} = \card S \cdot \map g y + \map f x$
Suppose $\map F {x_1, y_1} = \map F {x_2, y_2}$.
{{begi... | Let $S$ and $T$ be [[Definition:Set|sets]] that are [[Definition:Set Equivalence|equivalent]] to their [[Definition:Cardinal Number|cardinal numbers]].
Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Let $\cdot$ denote [[Definition:Ordinal Multiplication|ordinal multiplication]] and le... | Let $f: S \to \card S$ and $g: T \to \card T$ be [[Definition:Bijection|bijections]].
Let $\cdot$ denote [[Definition:Ordinal Multiplication|ordinal multiplication]], while $\times$ shall denote the [[Definition:Cartesian Product|Cartesian product]].
Define the function $F$ to be:
:$\forall x \in S, y \in T: \map F... | Cardinal Product Equinumerous to Ordinal Product | https://proofwiki.org/wiki/Cardinal_Product_Equinumerous_to_Ordinal_Product | https://proofwiki.org/wiki/Cardinal_Product_Equinumerous_to_Ordinal_Product | [
"Cardinals",
"Ordinal Arithmetic"
] | [
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Cardinal Number",
"Definition:Cardinal Number",
"Definition:Ordinal Multiplication",
"Definition:Cartesian Product"
] | [
"Definition:Bijection",
"Definition:Ordinal Multiplication",
"Definition:Cartesian Product",
"Division Theorem for Ordinals",
"Definition:Injection",
"Division Theorem for Ordinals",
"Definition:Surjection",
"Definition:Bijection",
"Condition for Set Equivalent to Cardinal Number",
"Category:Cardi... |
proofwiki-6080 | Product of Finite Sets is Finite | Let $S$ and $T$ be finite sets.
Then $S \times T$ is a finite set. | By the definition of Cartesian product:
:$S \times T = \set {\tuple {s, t}: s \in S, t \in T}$
Then by definition of set union:
:$S \times T = \ds \bigcup_{s \mathop \in S} \set s \times T$
Also, the mapping $g_s: \set s \times T \to T$ defined by:
:$\map {g_s} {s, t} = t$
is a bijection.
Therefore, since $T$ is finite... | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
Then $S \times T$ is a [[Definition:Finite Set|finite set]]. | By the definition of [[Definition:Cartesian Product|Cartesian product]]:
:$S \times T = \set {\tuple {s, t}: s \in S, t \in T}$
Then by definition of [[Definition:Set Union|set union]]:
:$S \times T = \ds \bigcup_{s \mathop \in S} \set s \times T$
Also, the [[Definition:Mapping|mapping]] $g_s: \set s \times T \to T... | Product of Finite Sets is Finite/Proof 1 | https://proofwiki.org/wiki/Product_of_Finite_Sets_is_Finite | https://proofwiki.org/wiki/Product_of_Finite_Sets_is_Finite/Proof_1 | [
"Cartesian Product",
"Product of Finite Sets is Finite"
] | [
"Definition:Finite Set",
"Definition:Finite Set"
] | [
"Definition:Cartesian Product",
"Definition:Set Union",
"Definition:Mapping",
"Definition:Bijection",
"Definition:Finite Set",
"Definition:Finite Set",
"Finite Union of Finite Sets is Finite"
] |
proofwiki-6081 | Product of Finite Sets is Finite | Let $S$ and $T$ be finite sets.
Then $S \times T$ is a finite set. | Let $\card S$ denote the cardinal number of $S$.
Let $\cdot$ denote ordinal multiplication.
By Cardinal Product Equinumerous to Ordinal Product, it follows that $S \times T \sim \card S \cdot \card T$.
But then $\card S$ and $\card T$ are members of the minimally inductive set.
Therefore, $\card S \cdot \card T \in \om... | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
Then $S \times T$ is a [[Definition:Finite Set|finite set]]. | Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Let $\cdot$ denote [[Definition:Ordinal Multiplication|ordinal multiplication]].
By [[Cardinal Product Equinumerous to Ordinal Product]], it follows that $S \times T \sim \card S \cdot \card T$.
But then $\card S$ and $\card T$ are memb... | Product of Finite Sets is Finite/Proof 2 | https://proofwiki.org/wiki/Product_of_Finite_Sets_is_Finite | https://proofwiki.org/wiki/Product_of_Finite_Sets_is_Finite/Proof_2 | [
"Cartesian Product",
"Product of Finite Sets is Finite"
] | [
"Definition:Finite Set",
"Definition:Finite Set"
] | [
"Definition:Cardinal Number",
"Definition:Ordinal Multiplication",
"Cardinal Product Equinumerous to Ordinal Product",
"Definition:Minimally Inductive Set",
"Natural Number Multiplication is Closed",
"Definition:Set Equivalence",
"Definition:Minimally Inductive Set",
"Definition:Finite Set"
] |
proofwiki-6082 | Union of Finite Sets is Finite | Let $S$ and $T$ be finite sets.
Then $S \cup T$ is a finite set. | The proof proceeds by induction.
Let $S$ be a finite set with cardinality $n$.
If $n = 0$ then $S = \O$, so $\bigcup S = \O$, which is finite.
Suppose that an arbitrary finite set with cardinality $n$ of finite sets has a finite union.
Let $S$ have cardinality $n^+$.
Then there is a bijection $f: n^+ \to S$.
Then:
:$\d... | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
Then $S \cup T$ is a [[Definition:Finite Set|finite set]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
Let $S$ be a [[Definition:Finite Set|finite set]] with [[Definition:Cardinality|cardinality]] $n$.
If $n = 0$ then $S = \O$, so $\bigcup S = \O$, which is [[Definition:Finite Set|finite]].
Suppose that an arbitrary [[Definition:Finite Set|finit... | Finite Union of Finite Sets is Finite/Proof 1 | https://proofwiki.org/wiki/Union_of_Finite_Sets_is_Finite | https://proofwiki.org/wiki/Finite_Union_of_Finite_Sets_is_Finite/Proof_1 | [
"Set Union",
"Finite Sets",
"Union of Finite Sets is Finite"
] | [
"Definition:Finite Set",
"Definition:Finite Set"
] | [
"Principle of Mathematical Induction",
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Finite Set",
"Definition:Set Union/Finite Union",
"Definition:Cardinality",
"Definition:Bijection",
"Union of Finite S... |
proofwiki-6083 | Union of Finite Sets is Finite | Let $S$ and $T$ be finite sets.
Then $S \cup T$ is a finite set. | Let $S = \set {A_1, \ldots, A_n}$ such that $A_k$ is finite $\forall k = 1, \ldots, n$.
Set:
: $m = \max \set {\card {A_1}, \ldots, \card {A_n} }$
Then:
: $\ds \card {\bigcup_{k \mathop = 1}^n A_k} \le \sum_{k \mathop = 1}^n \card {A_k} \le \sum_{k \mathop = 1}^n m = n m$
{{explain|needs to invoke a link to result that... | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
Then $S \cup T$ is a [[Definition:Finite Set|finite set]]. | Let $S = \set {A_1, \ldots, A_n}$ such that $A_k$ is [[Definition:Finite Set|finite]] $\forall k = 1, \ldots, n$.
Set:
: $m = \max \set {\card {A_1}, \ldots, \card {A_n} }$
Then:
: $\ds \card {\bigcup_{k \mathop = 1}^n A_k} \le \sum_{k \mathop = 1}^n \card {A_k} \le \sum_{k \mathop = 1}^n m = n m$
{{explain|needs to... | Finite Union of Finite Sets is Finite/Proof 2 | https://proofwiki.org/wiki/Union_of_Finite_Sets_is_Finite | https://proofwiki.org/wiki/Finite_Union_of_Finite_Sets_is_Finite/Proof_2 | [
"Set Union",
"Finite Sets",
"Union of Finite Sets is Finite"
] | [
"Definition:Finite Set",
"Definition:Finite Set"
] | [
"Definition:Finite Set"
] |
proofwiki-6084 | Union of Finite Sets is Finite | Let $S$ and $T$ be finite sets.
Then $S \cup T$ is a finite set. | If $S$ or $T$ is empty, the result is trivial.
Otherwise, let $f: \N_{<n} \to S$ and $g: \N_{<m} \to T$ be bijections, where $\N_{<n}$ is an initial segment of $\N$.
Now define $h: \N_{< n + m} \to S \cup T$ by:
:$\map h i = \begin{cases} \map f i : & \text {if $i < n$} \\ \map g {i - n} : & \text{if $i \ge n$} \end{ca... | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
Then $S \cup T$ is a [[Definition:Finite Set|finite set]]. | If $S$ or $T$ is [[Definition:Empty Set|empty]], the result is trivial.
Otherwise, let $f: \N_{<n} \to S$ and $g: \N_{<m} \to T$ be [[Definition:Bijection|bijections]], where $\N_{<n}$ is an [[Definition:Initial Segment of Natural Numbers|initial segment of $\N$]].
Now define $h: \N_{< n + m} \to S \cup T$ by:
:$\ma... | Union of Finite Sets is Finite/Proof 1 | https://proofwiki.org/wiki/Union_of_Finite_Sets_is_Finite | https://proofwiki.org/wiki/Union_of_Finite_Sets_is_Finite/Proof_1 | [
"Set Union",
"Finite Sets",
"Union of Finite Sets is Finite"
] | [
"Definition:Finite Set",
"Definition:Finite Set"
] | [
"Definition:Empty Set",
"Definition:Bijection",
"Definition:Initial Segment of Natural Numbers",
"Set Finite iff Surjection from Initial Segment of Natural Numbers",
"Definition:Surjection",
"Definition:Element",
"Definition:Mapping",
"Definition:Set Union",
"Definition:Surjection",
"Definition:Fi... |
proofwiki-6085 | Union of Finite Sets is Finite | Let $S$ and $T$ be finite sets.
Then $S \cup T$ is a finite set. | Note that $\card {S \cup T} \le \card {S \times T}$ by Cardinal of Union Less than Cardinal of Cartesian Product.
The theorem follows from the fact that $S \times T$ is finite by Product of Finite Sets is Finite.
{{qed}} | Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
Then $S \cup T$ is a [[Definition:Finite Set|finite set]]. | Note that $\card {S \cup T} \le \card {S \times T}$ by [[Cardinal of Union Less than Cardinal of Cartesian Product]].
The theorem follows from the fact that $S \times T$ is [[Definition:Finite Set|finite]] by [[Product of Finite Sets is Finite]].
{{qed}} | Union of Finite Sets is Finite/Proof 2 | https://proofwiki.org/wiki/Union_of_Finite_Sets_is_Finite | https://proofwiki.org/wiki/Union_of_Finite_Sets_is_Finite/Proof_2 | [
"Set Union",
"Finite Sets",
"Union of Finite Sets is Finite"
] | [
"Definition:Finite Set",
"Definition:Finite Set"
] | [
"Cardinal of Union Less than Cardinal of Cartesian Product",
"Definition:Finite Set",
"Product of Finite Sets is Finite"
] |
proofwiki-6086 | Ordinal is Finite iff Natural Number | Let $x$ be an ordinal.
Then $x$ is a finite set {{iff}} $x$ is an element of the minimally inductive set. | $x$ is finite {{iff}} $x \sim \N_n$ for some $n \in \N$, by definition.
But $x$ is an ordinal, and by definition, it is equal to its initial segment.
By definition of the von Neumann construction of natural numbers, it follows that $x \sim n$ for some $n$.
By Finite Ordinal is equal to Natural Number, it follows that $... | Let $x$ be an [[Definition:Ordinal|ordinal]].
Then $x$ is a [[Definition:Finite Set|finite set]] {{iff}} $x$ is an [[Definition:Element|element]] of the [[Definition:Minimally Inductive Set|minimally inductive set]]. | $x$ is [[Definition:Finite Set|finite]] {{iff}} $x \sim \N_n$ for some $n \in \N$, by definition.
But $x$ is an [[Definition:Ordinal|ordinal]], and by definition, it is equal to its [[Definition:Initial Segment|initial segment]].
By definition of the [[Definition:Von Neumann Construction of Natural Numbers|von Neuma... | Ordinal is Finite iff Natural Number | https://proofwiki.org/wiki/Ordinal_is_Finite_iff_Natural_Number | https://proofwiki.org/wiki/Ordinal_is_Finite_iff_Natural_Number | [
"Ordinals",
"Minimally Inductive Set"
] | [
"Definition:Ordinal",
"Definition:Finite Set",
"Definition:Element",
"Definition:Minimally Inductive Set"
] | [
"Definition:Finite Set",
"Definition:Ordinal",
"Definition:Initial Segment",
"Definition:Natural Numbers/Von Neumann Construction",
"Finite Ordinal is equal to Natural Number",
"Definition:Element",
"Definition:Minimally Inductive Set"
] |
proofwiki-6087 | Cardinal Inequality implies Ordinal Inequality | Let $T$ be a set.
Let $\card T$ denote the cardinal number of $T$.
Let $x$ be an ordinal.
Then:
:$x < \card T \iff \card x < \card T$ | === Sufficient Condition ===
By Cardinal Number Less than Ordinal, it follows that $\card x \le x$.
So if $x < \card T$, then $\card x < \card T$.
{{qed|lemma}} | Let $T$ be a [[Definition:Set|set]].
Let $\card T$ denote the [[Definition:Cardinal Number|cardinal number]] of $T$.
Let $x$ be an [[Definition:Ordinal|ordinal]].
Then:
:$x < \card T \iff \card x < \card T$ | === Sufficient Condition ===
By [[Cardinal Number Less than Ordinal]], it follows that $\card x \le x$.
So if $x < \card T$, then $\card x < \card T$.
{{qed|lemma}} | Cardinal Inequality implies Ordinal Inequality | https://proofwiki.org/wiki/Cardinal_Inequality_implies_Ordinal_Inequality | https://proofwiki.org/wiki/Cardinal_Inequality_implies_Ordinal_Inequality | [
"Cardinals",
"Ordinals"
] | [
"Definition:Set",
"Definition:Cardinal Number",
"Definition:Ordinal"
] | [
"Cardinal Number Less than Ordinal"
] |
proofwiki-6088 | Cardinal Number Plus One Less than Cardinal Product | Let $x$ be an ordinal such that $x > 1$.
Then:
:$\card {x + 1} \le \card {x \times x}$
Where $\times$ denotes the Cartesian product. | Since $x > 1$, then $0 < x$ and $1 < x$.
Define the function $f: x + 1 \to x \times x$ as follows:
:$\map f y = \begin{cases}
\tuple {y, 0} &: y < x \\
\tuple {0, 1} &: y = x
\end{cases}$
If $\map f y = \map f z$, then $y = z$ by cases. | Let $x$ be an [[Definition:Ordinal|ordinal]] such that $x > 1$.
Then:
:$\card {x + 1} \le \card {x \times x}$
Where $\times$ denotes the [[Definition:Cartesian Product|Cartesian product]]. | Since $x > 1$, then $0 < x$ and $1 < x$.
Define the function $f: x + 1 \to x \times x$ as follows:
:$\map f y = \begin{cases}
\tuple {y, 0} &: y < x \\
\tuple {0, 1} &: y = x
\end{cases}$
If $\map f y = \map f z$, then $y = z$ by [[Proof by Cases|cases]]. | Cardinal Number Plus One Less than Cardinal Product | https://proofwiki.org/wiki/Cardinal_Number_Plus_One_Less_than_Cardinal_Product | https://proofwiki.org/wiki/Cardinal_Number_Plus_One_Less_than_Cardinal_Product | [
"Ordinals",
"Cardinals"
] | [
"Definition:Ordinal",
"Definition:Cartesian Product"
] | [
"Proof by Cases"
] |
proofwiki-6089 | Non-Finite Cardinal is equal to Cardinal Product | Let $\omega$ denote the minimally inductive set.
Let $x$ be an ordinal such that $x \ge \omega$.
Then:
:$\card x = \card {x \times x}$
where $\times$ denotes the Cartesian product. | The proof shall proceed by Transfinite Induction on $x$.
Let:
:$\forall y \in x: y < \omega \lor \card y = \card {y \times y}$
There are two cases: | Let $\omega$ denote the [[Definition:Minimally Inductive Set|minimally inductive set]].
Let $x$ be an [[Definition:Ordinal|ordinal]] such that $x \ge \omega$.
Then:
:$\card x = \card {x \times x}$
where $\times$ denotes the [[Definition:Cartesian Product|Cartesian product]]. | The proof shall proceed by [[Transfinite Induction/Schema 1|Transfinite Induction]] on $x$.
Let:
:$\forall y \in x: y < \omega \lor \card y = \card {y \times y}$
There are two cases: | Non-Finite Cardinal is equal to Cardinal Product | https://proofwiki.org/wiki/Non-Finite_Cardinal_is_equal_to_Cardinal_Product | https://proofwiki.org/wiki/Non-Finite_Cardinal_is_equal_to_Cardinal_Product | [
"Ordinals",
"Cardinals"
] | [
"Definition:Minimally Inductive Set",
"Definition:Ordinal",
"Definition:Cartesian Product"
] | [
"Transfinite Induction/Schema 1"
] |
proofwiki-6090 | Permutation is Cyclic iff At Most One Non-Trivial Orbit | Let $S$ be a set.
Let $\rho: S \to S$ be a permutation on $S$.
Then:
:$\rho$ is a cyclic permutation
{{iff}}:
:$S$ has no more than one orbit under $\rho$ with more than one element. | === Necessary Condition ===
{{ProofWanted}} | Let $S$ be a [[Definition:Set|set]].
Let $\rho: S \to S$ be a [[Definition:Permutation|permutation]] on $S$.
Then:
:$\rho$ is a [[Definition:Cyclic Permutation|cyclic permutation]]
{{iff}}:
:$S$ has no more than one [[Definition:Orbit (Group Theory)|orbit]] under $\rho$ with more than one [[Definition:Element|elemen... | === Necessary Condition ===
{{ProofWanted}} | Permutation is Cyclic iff At Most One Non-Trivial Orbit | https://proofwiki.org/wiki/Permutation_is_Cyclic_iff_At_Most_One_Non-Trivial_Orbit | https://proofwiki.org/wiki/Permutation_is_Cyclic_iff_At_Most_One_Non-Trivial_Orbit | [
"Cyclic Permutations",
"Permutations"
] | [
"Definition:Set",
"Definition:Permutation",
"Definition:Cyclic Permutation",
"Definition:Orbit (Group Theory)",
"Definition:Element"
] | [] |
proofwiki-6091 | Infinite Ramsey's Theorem implies Finite Ramsey's Theorem | :$\forall l, n, r \in \N: \exists m \in \N: m \to \left({l}\right)_r^n$
where $\alpha \to \left({\beta}\right)^n_r$ means that:
:for any assignment of $r$-colors to the $n$-subsets of $\alpha$
::there is a particular color $\gamma$ and a subset $X$ of $\alpha$ of size $\beta$ such that all $n$-subsets of $X$ are $\gamm... | {{AimForCont}} there is a $l$ such that:
:$\forall m \in \N: m \nrightarrow \left({l}\right)_r^n$
Let $\hat{K_i}$ denote a hypergraph on $i$ vertices where all possible $n$-subsets of the vertices are the hyperedges.
Let $G$ be a hypergraph with vertices $V = \left\{ {v_i: i \in \N}\right\}$.
Let the hyperedges of $G <... | :$\forall l, n, r \in \N: \exists m \in \N: m \to \left({l}\right)_r^n$
where $\alpha \to \left({\beta}\right)^n_r$ means that:
:for any assignment of [[Definition:Coloring|$r$-colors]] to the [[Definition:N-Subset|$n$-subsets]] of $\alpha$
::there is a particular [[Definition:Color|color]] $\gamma$ and a [[Definition... | {{AimForCont}} there is a $l$ such that:
:$\forall m \in \N: m \nrightarrow \left({l}\right)_r^n$
Let $\hat{K_i}$ denote a [[Definition:Hypergraph|hypergraph]] on $i$ [[Definition:Vertex of Graph|vertices]] where all possible [[Definition:N-Subset|$n$-subsets]] of the [[Definition:Vertex of Graph|vertices]] are the [[... | Infinite Ramsey's Theorem implies Finite Ramsey's Theorem | https://proofwiki.org/wiki/Infinite_Ramsey's_Theorem_implies_Finite_Ramsey's_Theorem | https://proofwiki.org/wiki/Infinite_Ramsey's_Theorem_implies_Finite_Ramsey's_Theorem | [
"Ramsey Theory"
] | [
"Definition:Coloring",
"Definition:N-Subset",
"Definition:Color",
"Definition:Subset",
"Definition:Cardinality",
"Definition:N-Subset"
] | [
"Definition:Hypergraph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:N-Subset",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Hyperedge",
"Definition:Hypergraph",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Hyperedge",
"Definition:Rooted Tree",
"Definition:Rooted Tree/R... |
proofwiki-6092 | Identity Morphism of Product | Let $\mathbf C$ be a metacategory.
Let $C$ and $D$ be objects of $\mathbf C$, and let $C \times D$ be a binary product for $C$ and $D$.
Then:
:$\operatorname{id}_{\paren {C \mathop \times D} } = \operatorname{id}_C \times \operatorname{id}_D$
where $\operatorname{id}$ denotes an identity morphism, and $\times$ signifie... | By definition of the product morphism $\operatorname{id}_C \times \operatorname{id}_D$, it is the unique morphism making:
$\quad\quad \begin{xy}\xymatrix@+1em@L+5px{
C
\ar[d]_*+{\operatorname{id}_C}
&
C \times D
\ar[l]_*+{\pr_1}
\ar[r]^*+{\pr_2}
\ar@{-->}[d]^*+{\operatorname{id}_C \times \operatorname{id}_D}
&
D
\ar[d]... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $C$ and $D$ be [[Definition:Object (Category Theory)|objects]] of $\mathbf C$, and let $C \times D$ be a [[Definition:Binary Product (Category Theory)|binary product]] for $C$ and $D$.
Then:
:$\operatorname{id}_{\paren {C \mathop \times D} } = \oper... | By definition of the [[Definition:Product of Morphisms|product morphism]] $\operatorname{id}_C \times \operatorname{id}_D$, it is the [[Definition:Unique|unique]] [[Definition:Morphism (Category Theory)|morphism]] making:
$\quad\quad \begin{xy}\xymatrix@+1em@L+5px{
C
\ar[d]_*+{\operatorname{id}_C}
&
C \times D
\ar[l]_... | Identity Morphism of Product | https://proofwiki.org/wiki/Identity_Morphism_of_Product | https://proofwiki.org/wiki/Identity_Morphism_of_Product | [
"Morphisms"
] | [
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Product (Category Theory)/Binary Product",
"Definition:Identity Morphism",
"Definition:Product of Morphisms"
] | [
"Definition:Product of Morphisms",
"Definition:Unique",
"Definition:Morphism",
"Definition:Commutative Diagram",
"Definition:Identity Morphism",
"Definition:Unique",
"Definition:Morphism",
"Category:Morphisms"
] |
proofwiki-6093 | Cartesian Product Preserves Cardinality | Let $R$, $S$, and $T$ be sets.
Suppose that $S$ is equivalent to $T$.
Then:
:$R \times S \sim R \times T$
:$S \times R \sim T \times R$ | Since $S$ and $T$ are equivalent, there exists a bijection $f: S \to T$.
Let $g: T \to S$ be the inverse of $f$; its existence is assured by Bijection iff Left and Right Inverse.
Define $\hat f: R \times S \to R \times T$ by:
:$\map {\hat f} {r, s} := \tuple {r, \map f s}$
Next, define $\hat g: R \times T \to R \times ... | Let $R$, $S$, and $T$ be [[Definition:Set|sets]].
Suppose that $S$ is [[Definition:Set Equivalence|equivalent]] to $T$.
Then:
:$R \times S \sim R \times T$
:$S \times R \sim T \times R$ | Since $S$ and $T$ are [[Definition:Set Equivalence|equivalent]], there exists a [[Definition:Bijection|bijection]] $f: S \to T$.
Let $g: T \to S$ be the [[Definition:Inverse Mapping|inverse]] of $f$; its existence is assured by [[Bijection iff Left and Right Inverse]].
Define $\hat f: R \times S \to R \times T$ by:
... | Cartesian Product Preserves Cardinality | https://proofwiki.org/wiki/Cartesian_Product_Preserves_Cardinality | https://proofwiki.org/wiki/Cartesian_Product_Preserves_Cardinality | [
"Cartesian Product"
] | [
"Definition:Set",
"Definition:Set Equivalence"
] | [
"Definition:Set Equivalence",
"Definition:Bijection",
"Definition:Inverse Mapping",
"Bijection iff Left and Right Inverse",
"Definition:Inverse Mapping",
"Definition:Inverse Mapping",
"Definition:Inverse Mapping",
"Bijection iff Left and Right Inverse",
"Definition:Bijection",
"Definition:Set Equi... |
proofwiki-6094 | Product of Composite Morphisms | Let $\mathbf C$ be a metacategory.
Let $f \times f': A \times A' \to B \times B'$ and $g \times g': B \times B' \to C \times C'$ be two composable products of morphisms in $\mathbf C$.
Then:
:$\paren {g \circ f} \times \paren {g' \circ f'} = \paren {g \times g'} \circ \paren {f \times f'}$
where $\times$ signifies prod... | The situation is efficiently captured in the following commutative diagram:
$\quad\quad \begin{xy}
<-5em,0em>*+{A} = "A",
<0em,0em>*+{A \times A'} = "P",
<5em,0em>*+{A'} = "A2",
<-5em,-5em>*+{B} = "B",
<0em,-5em>*+{B \times B'} = "Q",
<5em,-5em>*+{B'} = "B2",
<-5em,-10em>*+{... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $f \times f': A \times A' \to B \times B'$ and $g \times g': B \times B' \to C \times C'$ be two [[Definition:Composable Morphisms|composable]] [[Definition:Product of Morphisms|products of morphisms]] in $\mathbf C$.
Then:
:$\paren {g \circ f} \tim... | The situation is efficiently captured in the following [[Definition:Commutative Diagram|commutative diagram]]:
$\quad\quad \begin{xy}
<-5em,0em>*+{A} = "A",
<0em,0em>*+{A \times A'} = "P",
<5em,0em>*+{A'} = "A2",
<-5em,-5em>*+{B} = "B",
<0em,-5em>*+{B \times B'} = "Q",
<5em,-5em>*+{B... | Product of Composite Morphisms | https://proofwiki.org/wiki/Product_of_Composite_Morphisms | https://proofwiki.org/wiki/Product_of_Composite_Morphisms | [
"Product Categories",
"Morphisms"
] | [
"Definition:Metacategory",
"Definition:Composable Morphisms",
"Definition:Product of Morphisms",
"Definition:Product of Morphisms"
] | [
"Definition:Commutative Diagram",
"Category:Product Categories",
"Category:Morphisms"
] |
proofwiki-6095 | Cardinal Product Equal to Maximum | Let $S$ and $T$ be sets that are equinumerous to their cardinal number.
Let $\card S$ denote the cardinal number of $S$.
Suppose $S$ is infinite.
Suppose $T > 0$.
Then:
:$\card {S \times T} = \map \max {\card S, \card T}$ | Let $x$ denote $\map \max {\card S, \card T}$.
Then by Cartesian Product Preserves Cardinality:
:$S \times T \sim \card S \times \card T$
Let $f: S \times T \to \card S \times \card T$ be a bijection.
It follows that $f: S \times T \to x \times x$ is an injection.
Hence:
{{begin-eqn}}
{{eqn | l = \card {S \times T}
... | Let $S$ and $T$ be [[Definition:Set|sets]] that are [[Definition:Set Equivalence|equinumerous]] to their [[Definition:Cardinal Number|cardinal number]].
Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Suppose $S$ is [[Definition:Infinite Set|infinite]].
Suppose $T > 0$.
Then:
:$\ca... | Let $x$ denote $\map \max {\card S, \card T}$.
Then by [[Cartesian Product Preserves Cardinality]]:
:$S \times T \sim \card S \times \card T$
Let $f: S \times T \to \card S \times \card T$ be a [[Definition:Bijection|bijection]].
It follows that $f: S \times T \to x \times x$ is an [[Definition:Injection|injection]]... | Cardinal Product Equal to Maximum | https://proofwiki.org/wiki/Cardinal_Product_Equal_to_Maximum | https://proofwiki.org/wiki/Cardinal_Product_Equal_to_Maximum | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Cardinal Number",
"Definition:Cardinal Number",
"Definition:Infinite Set"
] | [
"Cartesian Product Preserves Cardinality",
"Definition:Bijection",
"Definition:Injection",
"Injection iff Cardinal Inequality",
"Non-Finite Cardinal is equal to Cardinal Product",
"Cardinal Number Less than Ordinal/Corollary",
"Relation between Two Ordinals",
"Set Less than Cardinal Product"
] |
proofwiki-6096 | Cardinal of Union Equal to Maximum | Let $S$ and $T$ be sets that are equinumerous to their cardinal number.
Let $\card S$ denote the cardinal number of $S$.
Suppose $S$ is infinite.
Then:
:$\card {S \cup T} = \map \max {\card S, \card T}$ | Let $x$ denote $\map \max {\card S, \card T}$.
$x = \card S$ if $\card T \le \card S$
$x = \card T$ if $\card S \le \card T$
By Relation between Two Ordinals:
:$x = \card S$ or $x = \card T$
In either case, it follows by Subset of Union that:
:$x \le \card {S \cup T}$
{{qed|lemma}}
If $\card T = 1$ or $\card T = 0$, i... | Let $S$ and $T$ be [[Definition:Set|sets]] that are [[Definition:Set Equivalence|equinumerous]] to their [[Definition:Cardinal Number|cardinal number]].
Let $\card S$ denote the [[Definition:Cardinal Number|cardinal number]] of $S$.
Suppose $S$ is [[Definition:Infinite Set|infinite]].
Then:
:$\card {S \cup T} = \m... | Let $x$ denote $\map \max {\card S, \card T}$.
$x = \card S$ if $\card T \le \card S$
$x = \card T$ if $\card S \le \card T$
By [[Relation between Two Ordinals]]:
:$x = \card S$ or $x = \card T$
In either case, it follows by [[Subset of Union]] that:
:$x \le \card {S \cup T}$
{{qed|lemma}}
If $\card T = 1$ or $... | Cardinal of Union Equal to Maximum | https://proofwiki.org/wiki/Cardinal_of_Union_Equal_to_Maximum | https://proofwiki.org/wiki/Cardinal_of_Union_Equal_to_Maximum | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Set Equivalence",
"Definition:Cardinal Number",
"Definition:Cardinal Number",
"Definition:Infinite Set"
] | [
"Relation between Two Ordinals",
"Set is Subset of Union",
"Definition:Infinite Set",
"Cardinal of Union Less than Cardinal of Cartesian Product",
"Cardinal Product Equal to Maximum"
] |
proofwiki-6097 | Class of All Cardinals is Subclass of Class of All Ordinals | Let $\NN$ denote the class of all cardinals.
Let $\On$ denote the class of all ordinals.
Then:
:$\NN \subseteq \On$ | By definition of the class of all cardinals:
:$\NN = \set {x \in \On: \exists y: x = \card y}$
Every element of $\NN$ is thus an element of $\On$.
{{qed}} | Let $\NN$ denote the [[Definition:Class of All Cardinals|class of all cardinals]].
Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Then:
:$\NN \subseteq \On$ | By definition of the [[Definition:Class of All Cardinals|class of all cardinals]]:
:$\NN = \set {x \in \On: \exists y: x = \card y}$
Every [[Definition:Element of Class|element]] of $\NN$ is thus an [[Definition:Element of Class|element]] of $\On$.
{{qed}} | Class of All Cardinals is Subclass of Class of All Ordinals | https://proofwiki.org/wiki/Class_of_All_Cardinals_is_Subclass_of_Class_of_All_Ordinals | https://proofwiki.org/wiki/Class_of_All_Cardinals_is_Subclass_of_Class_of_All_Ordinals | [
"Class of All Cardinals",
"Class of All Ordinals"
] | [
"Definition:Class of All Cardinals",
"Definition:Class of All Ordinals"
] | [
"Definition:Class of All Cardinals",
"Definition:Element/Class",
"Definition:Element/Class"
] |
proofwiki-6098 | Cardinal of Cardinal Equal to Cardinal/Corollary | Let $\NN$ denote the class of all cardinal numbers.
Let $x$ be an ordinal.
Then:
:$x \in \NN \iff x = \card x$ | === Necessary Condition ===
Suppose $x = \card x$.
Then $x = \card y$ for some $y$ by Existential Generalisation.
By definition of class of all cardinals:
:$\NN = \set {x \in \On: \exists y: x = \card y}$
It follows that $x \in \NN$.
{{qed|lemma}} | Let $\NN$ denote the [[Definition:Class of All Cardinals|class of all cardinal numbers]].
Let $x$ be an [[Definition:Ordinal|ordinal]].
Then:
:$x \in \NN \iff x = \card x$ | === Necessary Condition ===
Suppose $x = \card x$.
Then $x = \card y$ for some $y$ by [[Existential Generalisation]].
By definition of [[Definition:Class of All Cardinals|class of all cardinals]]:
:$\NN = \set {x \in \On: \exists y: x = \card y}$
It follows that $x \in \NN$.
{{qed|lemma}} | Cardinal of Cardinal Equal to Cardinal/Corollary | https://proofwiki.org/wiki/Cardinal_of_Cardinal_Equal_to_Cardinal/Corollary | https://proofwiki.org/wiki/Cardinal_of_Cardinal_Equal_to_Cardinal/Corollary | [
"Cardinals"
] | [
"Definition:Class of All Cardinals",
"Definition:Ordinal"
] | [
"Existential Generalisation",
"Definition:Class of All Cardinals"
] |
proofwiki-6099 | Class of All Cardinals Contains Minimally Inductive Set | Let $\NN$ denote the class of all cardinal numbers.
Then:
:$\omega \subseteq \NN$
where $\omega$ denotes the minimally inductive set. | Suppose $n \in \omega$.
By Cardinal of Finite Ordinal, $n = \card n$.
By Cardinal of Cardinal Equal to Cardinal/Corollary, $n \in \NN$.
{{qed}} | Let $\NN$ denote the [[Definition:Class of All Cardinals|class of all cardinal numbers]].
Then:
:$\omega \subseteq \NN$
where $\omega$ denotes the [[Definition:Minimally Inductive Set|minimally inductive set]]. | Suppose $n \in \omega$.
By [[Cardinal of Finite Ordinal]], $n = \card n$.
By [[Cardinal of Cardinal Equal to Cardinal/Corollary]], $n \in \NN$.
{{qed}} | Class of All Cardinals Contains Minimally Inductive Set | https://proofwiki.org/wiki/Class_of_All_Cardinals_Contains_Minimally_Inductive_Set | https://proofwiki.org/wiki/Class_of_All_Cardinals_Contains_Minimally_Inductive_Set | [
"Class of All Cardinals",
"Minimally Inductive Set"
] | [
"Definition:Class of All Cardinals",
"Definition:Minimally Inductive Set"
] | [
"Cardinal of Finite Ordinal",
"Cardinal of Cardinal Equal to Cardinal/Corollary"
] |
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