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-5000 | Existence of Hartogs Number | Let $S$ be a set.
Then $S$ has a Hartogs number. | Follows immediately from Cardinal Equal to Collection of All Dominated Ordinals.
The collection of all dominated ordinals:
:$\set {y \in \On: y \preccurlyeq S}$
is the Hartogs number of $S$.
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Then $S$ has a [[Definition:Hartogs Number|Hartogs number]]. | Follows immediately from [[Cardinal Equal to Collection of All Dominated Ordinals]].
The collection of all [[Definition:Dominate (Set Theory)|dominated]] [[Definition:Ordinal|ordinals]]:
:$\set {y \in \On: y \preccurlyeq S}$
is the [[Definition:Hartogs Number|Hartogs number]] of $S$.
{{qed}} | Existence of Hartogs Number/Proof 2 | https://proofwiki.org/wiki/Existence_of_Hartogs_Number | https://proofwiki.org/wiki/Existence_of_Hartogs_Number/Proof_2 | [
"Ordinals",
"Existence of Hartogs Number"
] | [
"Definition:Set",
"Definition:Hartogs Number"
] | [
"Cardinal Equal to Collection of All Dominated Ordinals",
"Definition:Dominate (Set Theory)",
"Definition:Ordinal",
"Definition:Hartogs Number"
] |
proofwiki-5001 | Left and Right Inverse Mappings Implies Bijection | Let $f: S \to T$ be a mapping.
Let $f$ be such that:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
where both $g_1$ and $g_2$ are mappings.
Then $f$ is a bijection. | Suppose:
:$\exists g_1: T \to S: g_1 \circ f = I_S$
:$\exists g_2: T \to S: f \circ g_2 = I_T$
We have that the Identity Mapping is Bijection, thus $I_S$ and $I_T$ are both bijections.
From Injection if Composite is Injection, if $I_S = g_1 \circ f$ is an injection, then so is $f$.
From Surjection if Composite is Surje... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $f$ be such that:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
where both $g_1$ and $g_2$ are [[Definition:Mapping|mappings]].
Then $f$ is a [[Definition:Bijection|bijection]]. | Suppose:
:$\exists g_1: T \to S: g_1 \circ f = I_S$
:$\exists g_2: T \to S: f \circ g_2 = I_T$
We have that the [[Identity Mapping is Bijection]], thus $I_S$ and $I_T$ are both [[Definition:Bijection|bijections]].
From [[Injection if Composite is Injection]], if $I_S = g_1 \circ f$ is an [[Definition:Injection|inject... | Left and Right Inverse Mappings Implies Bijection/Proof 1 | https://proofwiki.org/wiki/Left_and_Right_Inverse_Mappings_Implies_Bijection | https://proofwiki.org/wiki/Left_and_Right_Inverse_Mappings_Implies_Bijection/Proof_1 | [
"Bijections",
"Inverse Mappings",
"Left and Right Inverse Mappings Implies Bijection"
] | [
"Definition:Mapping",
"Definition:Mapping",
"Definition:Bijection"
] | [
"Identity Mapping is Bijection",
"Definition:Bijection",
"Injection if Composite is Injection",
"Definition:Injection",
"Surjection if Composite is Surjection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection"
] |
proofwiki-5002 | Left and Right Inverse Mappings Implies Bijection | Let $f: S \to T$ be a mapping.
Let $f$ be such that:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
where both $g_1$ and $g_2$ are mappings.
Then $f$ is a bijection. | Suppose:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
From Injection iff Left Inverse, it follows that $f$ is an injection.
From Surjection iff Right Inverse, it follows that $f$ is a surjection.
So $f$ is both an injection and a surjection and, by definition, therefore also a... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $f$ be such that:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
where both $g_1$ and $g_2$ are [[Definition:Mapping|mappings]].
Then $f$ is a [[Definition:Bijection|bijection]]. | Suppose:
: $\exists g_1: T \to S: g_1 \circ f = I_S$
: $\exists g_2: T \to S: f \circ g_2 = I_T$
From [[Injection iff Left Inverse]], it follows that $f$ is an [[Definition:Injection|injection]].
From [[Surjection iff Right Inverse]], it follows that $f$ is a [[Definition:Surjection|surjection]].
So $f$ is both an [... | Left and Right Inverse Mappings Implies Bijection/Proof 2 | https://proofwiki.org/wiki/Left_and_Right_Inverse_Mappings_Implies_Bijection | https://proofwiki.org/wiki/Left_and_Right_Inverse_Mappings_Implies_Bijection/Proof_2 | [
"Bijections",
"Inverse Mappings",
"Left and Right Inverse Mappings Implies Bijection"
] | [
"Definition:Mapping",
"Definition:Mapping",
"Definition:Bijection"
] | [
"Injection iff Left Inverse",
"Definition:Injection",
"Surjection iff Right Inverse",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection"
] |
proofwiki-5003 | Left and Right Inverses of Mapping are Inverse Mapping | Let $f: S \to T$ be a mapping such that:
:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
Then:
: $g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the inverse of $f$. | {{begin-eqn}}
{{eqn | l = g_2
| r = I_S \circ g_2
| c = {{Defof|Identity Mapping}}
}}
{{eqn | r = \paren {g_1 \circ f} \circ g_2
| c = by hypothesis
}}
{{eqn | r = g_1 \circ \paren {f \circ g_2}
| c = Composition of Mappings is Associative
}}
{{eqn | r = g_1 \circ I_T
| c = by hypothesis
}... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]] such that:
:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
Then:
: $g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the [[Definition:Inverse Mapping|inverse]] of $f$. | {{begin-eqn}}
{{eqn | l = g_2
| r = I_S \circ g_2
| c = {{Defof|Identity Mapping}}
}}
{{eqn | r = \paren {g_1 \circ f} \circ g_2
| c = [[Definition:By Hypothesis|by hypothesis]]
}}
{{eqn | r = g_1 \circ \paren {f \circ g_2}
| c = [[Composition of Mappings is Associative]]
}}
{{eqn | r = g_1 \cir... | Left and Right Inverses of Mapping are Inverse Mapping/Proof 1 | https://proofwiki.org/wiki/Left_and_Right_Inverses_of_Mapping_are_Inverse_Mapping | https://proofwiki.org/wiki/Left_and_Right_Inverses_of_Mapping_are_Inverse_Mapping/Proof_1 | [
"Bijections",
"Inverse Mappings",
"Left and Right Inverses of Mapping are Inverse Mapping"
] | [
"Definition:Mapping",
"Definition:Inverse Mapping"
] | [
"Definition:By Hypothesis",
"Composition of Mappings is Associative",
"Definition:By Hypothesis",
"Left and Right Inverse Mappings Implies Bijection",
"Definition:Bijection",
"Composite of Bijection with Inverse is Identity Mapping"
] |
proofwiki-5004 | Left and Right Inverses of Mapping are Inverse Mapping | Let $f: S \to T$ be a mapping such that:
:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
Then:
: $g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the inverse of $f$. | From Left and Right Inverse Mappings Implies Bijection, we have that $f$ is a bijection.
Let $y \in T$ and let $x = \map {f^{-1} } y$.
Such an $x$ exists because $f$ is surjective, and unique within $S$ as $f$ is injective.
Then $y = \map f x$ and so:
{{begin-eqn}}
{{eqn | l = \map {f^{-1} } y
| r = x
| c =... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]] such that:
:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
Then:
: $g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the [[Definition:Inverse Mapping|inverse]] of $f$. | From [[Left and Right Inverse Mappings Implies Bijection]], we have that $f$ is a [[Definition:Bijection|bijection]].
Let $y \in T$ and let $x = \map {f^{-1} } y$.
Such an $x$ exists because $f$ is [[Definition:Surjection|surjective]], and [[Definition:Unique|unique]] within $S$ as $f$ is [[Definition:Injection|injec... | Left and Right Inverses of Mapping are Inverse Mapping/Proof 2 | https://proofwiki.org/wiki/Left_and_Right_Inverses_of_Mapping_are_Inverse_Mapping | https://proofwiki.org/wiki/Left_and_Right_Inverses_of_Mapping_are_Inverse_Mapping/Proof_2 | [
"Bijections",
"Inverse Mappings",
"Left and Right Inverses of Mapping are Inverse Mapping"
] | [
"Definition:Mapping",
"Definition:Inverse Mapping"
] | [
"Left and Right Inverse Mappings Implies Bijection",
"Definition:Bijection",
"Definition:Surjection",
"Definition:Unique",
"Definition:Injection",
"Definition:Identity Mapping",
"Definition:By Hypothesis",
"Definition:Identity Mapping",
"Definition:By Hypothesis",
"Definition:Injection"
] |
proofwiki-5005 | Left and Right Inverses of Mapping are Inverse Mapping | Let $f: S \to T$ be a mapping such that:
:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
Then:
: $g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the inverse of $f$. | Because Composition of Mappings is Associative, brackets do not need to be used.
{{begin-eqn}}
{{eqn | n = 1
| l = g_1 \circ f
| r = I_S
| c =
}}
{{eqn | ll= \leadsto
| l = g_1 \circ f \circ g_2
| r = I_S \circ g_2
| c =
}}
{{eqn | r = g_2
| c = {{Defof|Identity Mapping}}
}}
... | Let $f: S \to T$ be a [[Definition:Mapping|mapping]] such that:
:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
Then:
: $g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the [[Definition:Inverse Mapping|inverse]] of $f$. | Because [[Composition of Mappings is Associative]], brackets do not need to be used.
{{begin-eqn}}
{{eqn | n = 1
| l = g_1 \circ f
| r = I_S
| c =
}}
{{eqn | ll= \leadsto
| l = g_1 \circ f \circ g_2
| r = I_S \circ g_2
| c =
}}
{{eqn | r = g_2
| c = {{Defof|Identity Mapping}... | Left and Right Inverses of Mapping are Inverse Mapping/Proof 3 | https://proofwiki.org/wiki/Left_and_Right_Inverses_of_Mapping_are_Inverse_Mapping | https://proofwiki.org/wiki/Left_and_Right_Inverses_of_Mapping_are_Inverse_Mapping/Proof_3 | [
"Bijections",
"Inverse Mappings",
"Left and Right Inverses of Mapping are Inverse Mapping"
] | [
"Definition:Mapping",
"Definition:Inverse Mapping"
] | [
"Composition of Mappings is Associative",
"Definition:Left Inverse Mapping",
"Definition:Right Inverse Mapping",
"Composite of Bijection with Inverse is Identity Mapping",
"Definition:Inverse Mapping"
] |
proofwiki-5006 | Composition of Mappings is Composition of Relations | Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that the domain of $f_2$ is the same as the codomain of $f_1$.
Let $f_2 \circ f_1$ be the composite of $f_1$ and $f_2$.
Let $f_1$ and $f_2$ be considered as relations on $S_1 \times S_2$ and $S_2 \times S_3$ respectively.
Then $f_2 \circ f_1$ defined as the... | By definition of composition of mappings:
:$f_2 \circ f_1 = \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f_1 \land \tuple {y, z} \in f_2}$
Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be relations.
The composite of $\RR_1$ and $\RR_2$ is defined as:
:$\RR_2 \... | Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be [[Definition:Mapping|mappings]] such that the [[Definition:Domain of Mapping|domain]] of $f_2$ is the same as the [[Definition:Codomain of Mapping|codomain]] of $f_1$.
Let $f_2 \circ f_1$ be the [[Definition:Composition of Mappings|composite of $f_1$ and $f_2$]].
Let ... | By definition of [[Definition:Composition of Mappings|composition of mappings]]:
:$f_2 \circ f_1 = \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f_1 \land \tuple {y, z} \in f_2}$
Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be [[Definition:Relation|relations... | Composition of Mappings is Composition of Relations | https://proofwiki.org/wiki/Composition_of_Mappings_is_Composition_of_Relations | https://proofwiki.org/wiki/Composition_of_Mappings_is_Composition_of_Relations | [
"Composite Mappings",
"Composite Relations"
] | [
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Composition of Mappings",
"Definition:Relation",
"Definition:Composition of Relations",
"Definition:Composition of Mappings"
] | [
"Definition:Composition of Mappings",
"Definition:Relation",
"Definition:Composition of Relations"
] |
proofwiki-5007 | Null Sequence in Exponential Sequence | Let $\sequence {a_n}_{n \mathop \in \N} \in \C$ be a sequence of complex numbers such that:
:$\ds \lim_{n \mathop \to +\infty}a_n = 0$
Then:
:$\ds \lim_{n \mathop \to +\infty} \paren {1 + \dfrac {a_n} n}^n = 1$ | {{begin-eqn}}
{{eqn | l = \paren {1 + \frac {a_n} n}^n
| r = \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac {a_n} n}^k
| c = Binomial Theorem
}}
{{eqn | r = {n \choose 0} \paren {\frac {a_n} n}^0 + \sum_{k \mathop = 1}^n {n \choose k} \paren {\frac {a_n} n}^k
}}
{{eqn | r = 1 + a_n \sum_{k \mathop = 1}^... | Let $\sequence {a_n}_{n \mathop \in \N} \in \C$ be a [[Definition:Complex Sequence|sequence of complex numbers]] such that:
:$\ds \lim_{n \mathop \to +\infty}a_n = 0$
Then:
:$\ds \lim_{n \mathop \to +\infty} \paren {1 + \dfrac {a_n} n}^n = 1$ | {{begin-eqn}}
{{eqn | l = \paren {1 + \frac {a_n} n}^n
| r = \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac {a_n} n}^k
| c = [[Binomial Theorem]]
}}
{{eqn | r = {n \choose 0} \paren {\frac {a_n} n}^0 + \sum_{k \mathop = 1}^n {n \choose k} \paren {\frac {a_n} n}^k
}}
{{eqn | r = 1 + a_n \sum_{k \mathop =... | Null Sequence in Exponential Sequence/Proof 1 | https://proofwiki.org/wiki/Null_Sequence_in_Exponential_Sequence | https://proofwiki.org/wiki/Null_Sequence_in_Exponential_Sequence/Proof_1 | [
"Exponential Function",
"Sequences",
"Null Sequence in Exponential Sequence"
] | [
"Definition:Complex Sequence"
] | [
"Binomial Theorem",
"Combination Theorem for Sequences"
] |
proofwiki-5008 | Null Sequence in Exponential Sequence | Let $\sequence {a_n}_{n \mathop \in \N} \in \C$ be a sequence of complex numbers such that:
:$\ds \lim_{n \mathop \to +\infty}a_n = 0$
Then:
:$\ds \lim_{n \mathop \to +\infty} \paren {1 + \dfrac {a_n} n}^n = 1$ | Let $\sequence {E_n}$ be the sequence of complex functions $E_n: \C \to \C$ defined by:
:$\map {E_n} z = \paren {1 + \dfrac z n}^n$
We have that:
:$\ds \lim_{n \mathop \to \infty} \map {E_n} z = \map \exp z$
where $\map \exp z$ is the complex exponential.
We also have that:
:$E_n \paren {a_n} = \paren {1 + \dfrac {a_n}... | Let $\sequence {a_n}_{n \mathop \in \N} \in \C$ be a [[Definition:Complex Sequence|sequence of complex numbers]] such that:
:$\ds \lim_{n \mathop \to +\infty}a_n = 0$
Then:
:$\ds \lim_{n \mathop \to +\infty} \paren {1 + \dfrac {a_n} n}^n = 1$ | Let $\sequence {E_n}$ be the [[Definition:Sequence|sequence]] of [[Definition:Complex Function|complex functions]] $E_n: \C \to \C$ defined by:
:$\map {E_n} z = \paren {1 + \dfrac z n}^n$
We have that:
:$\ds \lim_{n \mathop \to \infty} \map {E_n} z = \map \exp z$
where $\map \exp z$ is the [[Definition:Exponential Fun... | Null Sequence in Exponential Sequence/Proof 2 | https://proofwiki.org/wiki/Null_Sequence_in_Exponential_Sequence | https://proofwiki.org/wiki/Null_Sequence_in_Exponential_Sequence/Proof_2 | [
"Exponential Function",
"Sequences",
"Null Sequence in Exponential Sequence"
] | [
"Definition:Complex Sequence"
] | [
"Definition:Sequence",
"Definition:Complex Function",
"Definition:Exponential Function/Complex/Limit of Sequence",
"Convergent Sequence in Metric Space is Bounded",
"Definition:Bounded Sequence/Complex",
"Definition:Complex Disk/Closed",
"Definition:Disk/Radius",
"Closed Disk is Compact",
"Definitio... |
proofwiki-5009 | Identity Mapping is Right Identity | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$f \circ I_S = f$
where $I_S$ is the identity mapping on $S$, and $\circ$ signifies composition of mappings. | === Equality of Codomains ===
The codomains of $f$ and $f \circ I_S$ are both equal to $T$ from Codomain of Composite Relation.
{{qed|lemma}}
=== Equality of Domains ===
From Domain of Composite Relation:
:$\Dom {f \circ I_S} = \Dom {I_S}$
But from the definition of the identity mapping:
:$\Dom {I_S} = \Img {I_S} = S$
... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$f \circ I_S = f$
where $I_S$ is the [[Definition:Identity Mapping|identity mapping]] on $S$, and $\circ$ signifies [[Definition:Composition of Mappings|composition of mappings]]. | === Equality of Codomains ===
The [[Definition:Codomain of Mapping|codomains]] of $f$ and $f \circ I_S$ are both equal to $T$ from [[Codomain of Composite Relation]].
{{qed|lemma}}
=== Equality of Domains ===
From [[Domain of Composite Relation]]:
:$\Dom {f \circ I_S} = \Dom {I_S}$
But from the definition of the [... | Identity Mapping is Right Identity/Proof 1 | https://proofwiki.org/wiki/Identity_Mapping_is_Right_Identity | https://proofwiki.org/wiki/Identity_Mapping_is_Right_Identity/Proof_1 | [
"Identity Mappings",
"Identity Mapping is Right Identity"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Identity Mapping",
"Definition:Composition of Mappings"
] | [
"Definition:Codomain (Set Theory)/Mapping",
"Codomain of Composite Relation",
"Domain of Composite Relation",
"Definition:Identity Mapping",
"Definition:Composition of Mappings",
"Definition:Identity Mapping",
"Equality of Mappings"
] |
proofwiki-5010 | Identity Mapping is Right Identity | Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$f \circ I_S = f$
where $I_S$ is the identity mapping on $S$, and $\circ$ signifies composition of mappings. | By definition, a mapping is also a relation.
Also by definition, the identity mapping is the same as the diagonal relation.
Thus Diagonal Relation is Right Identity can be applied directly.
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Then:
:$f \circ I_S = f$
where $I_S$ is the [[Definition:Identity Mapping|identity mapping]] on $S$, and $\circ$ signifies [[Definition:Composition of Mappings|composition of mappings]]. | By definition, a [[Definition:Mapping|mapping]] is also a [[Definition:Relation|relation]].
Also by definition, the [[Definition:Identity Mapping|identity mapping]] is the same as the [[Definition:Diagonal Relation|diagonal relation]].
Thus [[Diagonal Relation is Right Identity]] can be applied directly.
{{qed}} | Identity Mapping is Right Identity/Proof 2 | https://proofwiki.org/wiki/Identity_Mapping_is_Right_Identity | https://proofwiki.org/wiki/Identity_Mapping_is_Right_Identity/Proof_2 | [
"Identity Mappings",
"Identity Mapping is Right Identity"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Identity Mapping",
"Definition:Composition of Mappings"
] | [
"Definition:Mapping",
"Definition:Relation",
"Definition:Identity Mapping",
"Definition:Diagonal Relation",
"Diagonal Relation is Right Identity"
] |
proofwiki-5011 | Diagonal Relation is Left Identity | Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Then:
:$\Delta_T \circ \RR = \RR$
where $\Delta_T$ is the diagonal relation on $T$, and $\circ$ signifies composition of relations. | We use the definition of relation equality, as follows: | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$.
Then:
:$\Delta_T \circ \RR = \RR$
where $\Delta_T$ is the [[Definition:Diagonal Relation|diagonal relation]] on $T$, and $\circ$ signifies [[Definition:Composition of Relations|composition of relations]]. | We use the [[Equality of Relations|definition of relation equality]], as follows: | Diagonal Relation is Left Identity | https://proofwiki.org/wiki/Diagonal_Relation_is_Left_Identity | https://proofwiki.org/wiki/Diagonal_Relation_is_Left_Identity | [
"Diagonal Relation",
"Identity Mappings"
] | [
"Definition:Relation",
"Definition:Diagonal Relation",
"Definition:Composition of Relations"
] | [
"Equality of Relations"
] |
proofwiki-5012 | Isomorphism Preserves Associativity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is associative {{iff}} $*$ is associative. | Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
As an isomorphism is surjective, it follows that:
:$\forall u, v, w \in T: \exists x, y, z \in S: \map \phi x = u, \map \phi y = v, \map \phi z = w$
So:
{{begin-eqn}}
... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then $\circ$ is [[Definition:Associative Operation|associative]] {{iff}} $*$ ... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $\circ$ is [[Definition:Associative Operation|associative]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism|isomorphism]].
As an [[Definition:Isomorphism (Abstract Algebr... | Isomorphism Preserves Associativity/Proof 1 | https://proofwiki.org/wiki/Isomorphism_Preserves_Associativity | https://proofwiki.org/wiki/Isomorphism_Preserves_Associativity/Proof_1 | [
"Isomorphism Preserves Associativity",
"Isomorphisms (Abstract Algebra)",
"Associativity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Associative Operation",
"Definition:Associative Operation"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Associative Operation",
"Definition:Isomorphism",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Associative Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Inverse of Al... |
proofwiki-5013 | Isomorphism Preserves Associativity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is associative {{iff}} $*$ is associative. | An isomorphism is {{afortiori}} an epimorphism.
The result follows from Epimorphism Preserves Associativity.
{{qed}} | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then $\circ$ is [[Definition:Associative Operation|associative]] {{iff}} $*$ ... | An [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is {{afortiori}} an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
The result follows from [[Epimorphism Preserves Associativity]].
{{qed}} | Isomorphism Preserves Associativity/Proof 2 | https://proofwiki.org/wiki/Isomorphism_Preserves_Associativity | https://proofwiki.org/wiki/Isomorphism_Preserves_Associativity/Proof_2 | [
"Isomorphism Preserves Associativity",
"Isomorphisms (Abstract Algebra)",
"Associativity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Associative Operation",
"Definition:Associative Operation"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Epimorphism (Abstract Algebra)",
"Epimorphism Preserves Associativity"
] |
proofwiki-5014 | Isomorphism Preserves Commutativity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is commutative {{iff}} $*$ is commutative. | Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
As an isomorphism is surjective, it follows that:
:$\forall u, v \in T: \exists x, y \in S: \map \phi x = u, \map \phi y = v$
So:
{{begin-eqn}}
{{eqn | l = u * v
... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then $\circ$ is [[Definition:Commutative Operation|commutative]] {{iff}} $*$ ... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $\circ$ is [[Definition:Commutative Operation|commutative]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
As an [[Definition:Isomorphi... | Isomorphism Preserves Commutativity/Proof 1 | https://proofwiki.org/wiki/Isomorphism_Preserves_Commutativity | https://proofwiki.org/wiki/Isomorphism_Preserves_Commutativity/Proof_1 | [
"Isomorphism Preserves Commutativity",
"Isomorphisms (Abstract Algebra)",
"Commutativity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Commutative/Operation",
"Definition:Commutative/Operation"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Commutative/Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Surjection",
"Definition:Surjection",
"Definition:Morphism Property",
"Definition:Commutative/Operation",
"Definit... |
proofwiki-5015 | Isomorphism Preserves Commutativity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is commutative {{iff}} $*$ is commutative. | We have that an isomorphism is a homomorphism which is also a bijection.
By definition, an epimorphism is a homomorphism which is also a surjection.
That is, an isomorphism is an epimorphism which is also an injection.
Thus Epimorphism Preserves Commutativity can be applied.
{{qed}} | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then $\circ$ is [[Definition:Commutative Operation|commutative]] {{iff}} $*$ ... | We have that an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also a [[Definition:Bijection|bijection]].
By definition, an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|ho... | Isomorphism Preserves Commutativity/Proof 2 | https://proofwiki.org/wiki/Isomorphism_Preserves_Commutativity | https://proofwiki.org/wiki/Isomorphism_Preserves_Commutativity/Proof_2 | [
"Isomorphism Preserves Commutativity",
"Isomorphisms (Abstract Algebra)",
"Commutativity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Commutative/Operation",
"Definition:Commutative/Operation"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Bijection",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Surjection",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Epimorphism (A... |
proofwiki-5016 | Isomorphism Preserves Inverses | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Then $x^{-1}$ is an inverse of $x$ for $\circ$ {{iff}} $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
That is, {{iff... | An group isomorphism is by definition a group epimorphism.
The result follows from Epimorphism Preserves Inverses.
{{Qed}} | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity]] $e_S... | An [[Definition:Group Isomorphism|group isomorphism]] is by definition a [[Definition:Group Epimorphism|group epimorphism]].
The result follows from [[Epimorphism Preserves Inverses]].
{{Qed}} | Group Isomorphism Preserves Inverses/Proof 1 | https://proofwiki.org/wiki/Isomorphism_Preserves_Inverses | https://proofwiki.org/wiki/Group_Isomorphism_Preserves_Inverses/Proof_1 | [
"Isomorphism Preserves Inverses",
"Isomorphisms (Abstract Algebra)",
"Inverse Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Epimorphism",
"Epimorphism Preserves Inverses"
] |
proofwiki-5017 | Isomorphism Preserves Inverses | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Then $x^{-1}$ is an inverse of $x$ for $\circ$ {{iff}} $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
That is, {{iff... | Let $g \in G$.
{{begin-eqn}}
{{eqn | l = \map \phi g * \map \phi {g^{-1} }
| r = \map \phi {g \circ g^{-1} }
| c = {{Defof|Group Isomorphism}}
}}
{{eqn | r = \map \phi {e_G}
| c = {{Defof|Inverse Element}}
}}
{{eqn | r = e_H
| c = Group Isomorphism Preserves Identity
}}
{{end-eqn}}
It follows fr... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity]] $e_S... | Let $g \in G$.
{{begin-eqn}}
{{eqn | l = \map \phi g * \map \phi {g^{-1} }
| r = \map \phi {g \circ g^{-1} }
| c = {{Defof|Group Isomorphism}}
}}
{{eqn | r = \map \phi {e_G}
| c = {{Defof|Inverse Element}}
}}
{{eqn | r = e_H
| c = [[Group Isomorphism Preserves Identity]]
}}
{{end-eqn}}
It foll... | Group Isomorphism Preserves Inverses/Proof 2 | https://proofwiki.org/wiki/Isomorphism_Preserves_Inverses | https://proofwiki.org/wiki/Group_Isomorphism_Preserves_Inverses/Proof_2 | [
"Isomorphism Preserves Inverses",
"Isomorphisms (Abstract Algebra)",
"Inverse Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Group Isomorphism Preserves Identity",
"Inverse in Group is Unique",
"Definition:Unique",
"Definition:Inverse (Abstract Algebra)/Inverse"
] |
proofwiki-5018 | Isomorphism Preserves Inverses | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Then $x^{-1}$ is an inverse of $x$ for $\circ$ {{iff}} $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
That is, {{iff... | Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
From Epimorphism Preserves Identity, it follows that $\struct {T, *}$ also has an identity, which is $\map \phi {e_S}$.
Let $y$ be an inverse of $x$ in $\struct {S, \circ}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi x * \map \ph... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity]] $e_S... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity]] $e_S$.
From [[Epimorphism Preserves Identity]], it follows that $\struct {T, *}$ also has an identity, which is $\map \phi {e_S}$.
Let $y$ be an [[D... | Isomorphism Preserves Inverses/Proof 1 | https://proofwiki.org/wiki/Isomorphism_Preserves_Inverses | https://proofwiki.org/wiki/Isomorphism_Preserves_Inverses/Proof_1 | [
"Isomorphism Preserves Inverses",
"Isomorphisms (Abstract Algebra)",
"Inverse Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Epimorphism Preserves Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Isomorphism (Abstract Algebra)",
"Inverse of Algebr... |
proofwiki-5019 | Isomorphism Preserves Inverses | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Then $x^{-1}$ is an inverse of $x$ for $\circ$ {{iff}} $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
That is, {{iff... | We have that an isomorphism is {{afortiori}} an epimorphism.
The result follows from Epimorphism Preserves Inverses.
{{qed}} | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity]] $e_S... | We have that an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is {{afortiori}} an [[Definition:Epimorphism|epimorphism]].
The result follows from [[Epimorphism Preserves Inverses]].
{{qed}} | Isomorphism Preserves Inverses/Proof 2 | https://proofwiki.org/wiki/Isomorphism_Preserves_Inverses | https://proofwiki.org/wiki/Isomorphism_Preserves_Inverses/Proof_2 | [
"Isomorphism Preserves Inverses",
"Isomorphisms (Abstract Algebra)",
"Inverse Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Epimorphism",
"Epimorphism Preserves Inverses"
] |
proofwiki-5020 | Isomorphism Preserves Identity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ has an identity $e_S$ {{iff}} $\map \phi {e_S}$ is the identity for $*$. | An group isomorphism is by definition a group epimorphism.
The result follows from Epimorphism Preserves Identity.
{{Qed}} | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then $\circ$ has an [[Definition:Identity Element|identity]] $e_S$ {{iff}} $\... | An [[Definition:Group Isomorphism|group isomorphism]] is by definition a [[Definition:Group Epimorphism|group epimorphism]].
The result follows from [[Epimorphism Preserves Identity]].
{{Qed}} | Group Isomorphism Preserves Identity/Proof 1 | https://proofwiki.org/wiki/Isomorphism_Preserves_Identity | https://proofwiki.org/wiki/Group_Isomorphism_Preserves_Identity/Proof_1 | [
"Isomorphism Preserves Identity",
"Isomorphisms (Abstract Algebra)",
"Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Epimorphism",
"Epimorphism Preserves Identity"
] |
proofwiki-5021 | Isomorphism Preserves Identity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ has an identity $e_S$ {{iff}} $\map \phi {e_S}$ is the identity for $*$. | {{begin-eqn}}
{{eqn | l = \map \phi {e_G}
| r = \map \phi {e_G \circ e_G}
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \map \phi {e_G} * \map \phi {e_G}
| c = {{Defof|Group Isomorphism}}
}}
{{eqn | r = \map \phi {e_G} * \map \phi {e_G}
| c = {{Defof|Group Isomorphism}}
}}
{{end-eqn}}
It follo... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then $\circ$ has an [[Definition:Identity Element|identity]] $e_S$ {{iff}} $\... | {{begin-eqn}}
{{eqn | l = \map \phi {e_G}
| r = \map \phi {e_G \circ e_G}
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \map \phi {e_G} * \map \phi {e_G}
| c = {{Defof|Group Isomorphism}}
}}
{{eqn | r = \map \phi {e_G} * \map \phi {e_G}
| c = {{Defof|Group Isomorphism}}
}}
{{end-eqn}}
It foll... | Group Isomorphism Preserves Identity/Proof 2 | https://proofwiki.org/wiki/Isomorphism_Preserves_Identity | https://proofwiki.org/wiki/Group_Isomorphism_Preserves_Identity/Proof_2 | [
"Isomorphism Preserves Identity",
"Isomorphisms (Abstract Algebra)",
"Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Identity is only Idempotent Element in Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-5022 | Isomorphism Preserves Identity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ has an identity $e_S$ {{iff}} $\map \phi {e_S}$ is the identity for $*$. | Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Then $\forall x \in S: x \circ e_S = x = e_S \circ x$.
The result follows directly from the morphism property of $\circ$ under $\phi$:
{{begin-eqn}}
{{eqn | l = \map \phi {x \circ e_S}
| r = \map \phi x
| c =
}}
{{eqn... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then $\circ$ has an [[Definition:Identity Element|identity]] $e_S$ {{iff}} $\... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity]] $e_S$.
Then $\forall x \in S: x \circ e_S = x = e_S \circ x$.
The result follows directly from the [[Definition:Morphism Property|morphism property]... | Isomorphism Preserves Identity/Proof 1 | https://proofwiki.org/wiki/Isomorphism_Preserves_Identity | https://proofwiki.org/wiki/Isomorphism_Preserves_Identity/Proof_1 | [
"Isomorphism Preserves Identity",
"Isomorphisms (Abstract Algebra)",
"Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Morphism Property",
"Definition:Isomorphism (Abstract Algebra)",
"Inverse of Algebraic Structure Isomorphism is Isomorphism",
"Definition:Isomorphism (Abstract Algebra)"
] |
proofwiki-5023 | Isomorphism Preserves Identity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ has an identity $e_S$ {{iff}} $\map \phi {e_S}$ is the identity for $*$. | We have that an isomorphism is a homomorphism which is also a bijection.
By definition, an epimorphism is a homomorphism which is also a surjection.
That is, an isomorphism is an epimorphism which is also an injection.
Thus Epimorphism Preserves Identity can be applied.
{{qed}} | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then $\circ$ has an [[Definition:Identity Element|identity]] $e_S$ {{iff}} $\... | We have that an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is also a [[Definition:Bijection|bijection]].
By definition, an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] is a [[Definition:Homomorphism (Abstract Algebra)|ho... | Isomorphism Preserves Identity/Proof 2 | https://proofwiki.org/wiki/Isomorphism_Preserves_Identity | https://proofwiki.org/wiki/Isomorphism_Preserves_Identity/Proof_2 | [
"Isomorphism Preserves Identity",
"Isomorphisms (Abstract Algebra)",
"Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Bijection",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Surjection",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Epimorphism (A... |
proofwiki-5024 | Non-Zero Natural Numbers under Multiplication form Commutative Semigroup | Let $\N_{>0}$ be the set of natural numbers without zero, that is, $\N_{>0} = \N \setminus \set 0$.
The structure $\struct {\N_{>0}, \times}$ forms an infinite commutative semigroup. | === {{Semigroup-axiom|0|nolink}} ===
From Natural Numbers have No Proper Zero Divisors:
:$\forall m, n \in \N: m \times n = 0 \iff m = 0 \lor n = 0$
It follows that:
:$\forall m, n \in \N_{>0}: m \times n \ne 0$
and so:
:$\forall m, n \in \N_{>0}: m \times n \in \N_{>0}$
That is, $\struct {\N_{>0}, \times}$ is closed.
... | Let $\N_{>0}$ be the set of [[Definition:Natural Numbers|natural numbers]] without [[Definition:Zero (Number)|zero]], that is, $\N_{>0} = \N \setminus \set 0$.
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\N_{>0}, \times}$ forms an [[Definition:Infinite Set|infinite]] [[Definition:Comm... | === {{Semigroup-axiom|0|nolink}} ===
From [[Natural Numbers have No Proper Zero Divisors]]:
:$\forall m, n \in \N: m \times n = 0 \iff m = 0 \lor n = 0$
It follows that:
:$\forall m, n \in \N_{>0}: m \times n \ne 0$
and so:
:$\forall m, n \in \N_{>0}: m \times n \in \N_{>0}$
That is, $\struct {\N_{>0}, \times}$ is ... | Non-Zero Natural Numbers under Multiplication form Commutative Semigroup | https://proofwiki.org/wiki/Non-Zero_Natural_Numbers_under_Multiplication_form_Commutative_Semigroup | https://proofwiki.org/wiki/Non-Zero_Natural_Numbers_under_Multiplication_form_Commutative_Semigroup | [
"Natural Number Multiplication",
"Examples of Commutative Semigroups"
] | [
"Definition:Natural Numbers",
"Definition:Zero (Number)",
"Definition:Algebraic Structure/One Operation",
"Definition:Infinite Set",
"Definition:Commutative Semigroup"
] | [
"Natural Numbers have No Proper Zero Divisors",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-5025 | Isomorphism Preserves Groups | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
If $\struct {S, \circ}$ is a group, then so is $\struct {T, *}$. | From Isomorphism Preserves Semigroups, if $\struct {S, \circ}$ is a semigroup then so is $\struct {T, *}$.
From Isomorphism Preserves Identity, if $\struct {S, \circ}$ has an identity $e_S$, then $\map \phi {e_S}$ is the identity for $*$.
From Isomorphism Preserves Inverses, if $x^{-1}$ is an inverse of $x$ for $\circ$... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
If $\struct {S, \circ}$ is a [[Definition:Group|group]], then so is $\struct ... | From [[Isomorphism Preserves Semigroups]], if $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]] then so is $\struct {T, *}$.
From [[Isomorphism Preserves Identity]], if $\struct {S, \circ}$ has an [[Definition:Identity Element|identity]] $e_S$, then $\map \phi {e_S}$ is the [[Definition:Identity Element|ide... | Isomorphism Preserves Groups/Proof 1 | https://proofwiki.org/wiki/Isomorphism_Preserves_Groups | https://proofwiki.org/wiki/Isomorphism_Preserves_Groups/Proof_1 | [
"Isomorphisms (Abstract Algebra)",
"Group Theory",
"Isomorphism Preserves Groups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Group"
] | [
"Isomorphism Preserves Semigroups",
"Definition:Semigroup",
"Isomorphism Preserves Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Isomorphism Preserves Inverses",
"Definition:Inverse (Abstract Algebra)/Inverse",
"De... |
proofwiki-5026 | Isomorphism Preserves Groups | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
If $\struct {S, \circ}$ is a group, then so is $\struct {T, *}$. | An isomorphism is an epimorphism.
The result follows as a direct corollary of Epimorphism Preserves Groups.
{{qed}} | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
If $\struct {S, \circ}$ is a [[Definition:Group|group]], then so is $\struct ... | An [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
The result follows as a direct [[Definition:Corollary|corollary]] of [[Epimorphism Preserves Groups]].
{{qed}} | Isomorphism Preserves Groups/Proof 2 | https://proofwiki.org/wiki/Isomorphism_Preserves_Groups | https://proofwiki.org/wiki/Isomorphism_Preserves_Groups/Proof_2 | [
"Isomorphisms (Abstract Algebra)",
"Group Theory",
"Isomorphism Preserves Groups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Group"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Corollary",
"Epimorphism Preserves Groups"
] |
proofwiki-5027 | Isomorphism Preserves Semigroups | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: S \to T$ be an isomorphism.
If $\struct {S, \circ}$ is a semigroup, then so is $\struct {T, *}$. | If $\struct {S, \circ}$ is a semigroup, then by definition it is closed.
From Morphism Property Preserves Closure, $\struct {T, *}$ is therefore also closed.
If $\struct {S, \circ}$ is a semigroup, then by definition $\circ$ is associative.
From Isomorphism Preserves Associativity, $*$ is therefore also associative.
So... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: S \to T$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
If $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]], then so is $\struct {T, *}$. | If $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]], then by definition it is [[Definition:Closed Algebraic Structure|closed]].
From [[Morphism Property Preserves Closure]], $\struct {T, *}$ is therefore also [[Definition:Closed Algebraic Structure|closed]].
If $\struct {S, \circ}$ is a [[Definition:Semi... | Isomorphism Preserves Semigroups/Proof 1 | https://proofwiki.org/wiki/Isomorphism_Preserves_Semigroups | https://proofwiki.org/wiki/Isomorphism_Preserves_Semigroups/Proof_1 | [
"Isomorphism Preserves Semigroups",
"Isomorphisms (Abstract Algebra)",
"Semigroups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Semigroup"
] | [
"Definition:Semigroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Morphism Property Preserves Closure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Semigroup",
"Definition:Associative Operation",
"Isomorphism Preserves Associativity",
"Definition:Associ... |
proofwiki-5028 | Isomorphism Preserves Semigroups | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: S \to T$ be an isomorphism.
If $\struct {S, \circ}$ is a semigroup, then so is $\struct {T, *}$. | An isomorphism is an epimorphism.
The result follows as a direct corollary of Epimorphism Preserves Semigroups.
{{qed}} | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: S \to T$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
If $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]], then so is $\struct {T, *}$. | An [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
The result follows as a direct [[Definition:Corollary|corollary]] of [[Epimorphism Preserves Semigroups]].
{{qed}} | Isomorphism Preserves Semigroups/Proof 2 | https://proofwiki.org/wiki/Isomorphism_Preserves_Semigroups | https://proofwiki.org/wiki/Isomorphism_Preserves_Semigroups/Proof_2 | [
"Isomorphism Preserves Semigroups",
"Isomorphisms (Abstract Algebra)",
"Semigroups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Semigroup"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Corollary",
"Epimorphism Preserves Semigroups"
] |
proofwiki-5029 | Left Cancellable iff Left Regular Representation Injective | Let $\struct {S, \circ}$ be an algebraic structure.
Then $a \in S$ is left cancellable {{iff}} the left regular representation $\map {\lambda_a} x$ is injective. | Suppose $a \in S$ is left cancellable.
Then:
:$\forall x, y \in S: a \circ x = a \circ y \implies x = y$
From the definition of the left regular representation:
:$\map {\lambda_a} x = a \circ x$
Thus:
:$\forall x, y \in S: \map {\lambda_a} x = \map {\lambda_a} y \implies x = y$
and so the left regular representation is... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Then $a \in S$ is [[Definition:Left Cancellable Element|left cancellable]] {{iff}} the [[Definition:Left Regular Representation|left regular representation]] $\map {\lambda_a} x$ is [[Definition:Injection|injectiv... | Suppose $a \in S$ is [[Definition:Left Cancellable Element|left cancellable]].
Then:
:$\forall x, y \in S: a \circ x = a \circ y \implies x = y$
From the definition of the [[Definition:Left Regular Representation|left regular representation]]:
:$\map {\lambda_a} x = a \circ x$
Thus:
:$\forall x, y \in S: \map {\lamb... | Left Cancellable iff Left Regular Representation Injective | https://proofwiki.org/wiki/Left_Cancellable_iff_Left_Regular_Representation_Injective | https://proofwiki.org/wiki/Left_Cancellable_iff_Left_Regular_Representation_Injective | [
"Regular Representations",
"Cancellability"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Regular Representations/Left Regular Representation",
"Definition:Injection"
] | [
"Definition:Cancellable Element/Left Cancellable",
"Definition:Regular Representations/Left Regular Representation",
"Definition:Regular Representations/Left Regular Representation",
"Definition:Injection",
"Definition:Injection",
"Definition:Regular Representations/Left Regular Representation",
"Defini... |
proofwiki-5030 | Right Cancellable iff Right Regular Representation Injective | Let $\struct {S, \circ}$ be an algebraic structure.
Then $a \in S$ is right cancellable {{iff}} the right regular representation $\map {\rho_a} x$ is injective. | Suppose $a \in S$ is right cancellable.
Then:
:$\forall x, y \in S: x \circ a = y \circ a \implies x = y$
From the definition of the right regular representation:
:$\map {\rho_a} x = x \circ a$
Thus:
:$\forall x, y \in S: \map {\rho_a} x = \map {\rho_a} y \implies x = y$
and so the right regular representation is injec... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Then $a \in S$ is [[Definition:Right Cancellable Element|right cancellable]] {{iff}} the [[Definition:Right Regular Representation|right regular representation]] $\map {\rho_a} x$ is [[Definition:Injection|injecti... | Suppose $a \in S$ is [[Definition:Right Cancellable Element|right cancellable]].
Then:
:$\forall x, y \in S: x \circ a = y \circ a \implies x = y$
From the definition of the [[Definition:Right Regular Representation|right regular representation]]:
:$\map {\rho_a} x = x \circ a$
Thus:
:$\forall x, y \in S: \map {\rho... | Right Cancellable iff Right Regular Representation Injective | https://proofwiki.org/wiki/Right_Cancellable_iff_Right_Regular_Representation_Injective | https://proofwiki.org/wiki/Right_Cancellable_iff_Right_Regular_Representation_Injective | [
"Regular Representations",
"Cancellability"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Cancellable Element/Right Cancellable",
"Definition:Regular Representations/Right Regular Representation",
"Definition:Injection"
] | [
"Definition:Cancellable Element/Right Cancellable",
"Definition:Regular Representations/Right Regular Representation",
"Definition:Regular Representations/Right Regular Representation",
"Definition:Injection",
"Definition:Injection",
"Definition:Regular Representations/Right Regular Representation",
"De... |
proofwiki-5031 | Isomorphism Preserves Cancellability | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then:
:$a \in S$ is cancellable in $\struct {S, \circ}$ {{iff}} $\map \phi a \in T$ is cancellable in $\struct {T, *}$. | Let $\struct {S, \circ}$ be an algebraic structure in which $a$ is cancellable.
From Isomorphism Preserves Left Cancellability and Isomorphism Preserves Right Cancellability:
:$a \in S$ is left cancellable in $\struct {S, \circ}$ {{iff}} $\map \phi a \in T$ is left cancellable in $\struct {T, *}$
and
:$a \in S$ is righ... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then:
:$a \in S$ is [[Definition:Cancellable Element|cancellable]] in $\struct {S, \circ}$ {{iff... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]] in which $a$ is [[Definition:Cancellable Element|cancellable]].
From [[Isomorphism Preserves Left Cancellability]] and [[Isomorphism Preserves Right Cancellability]]:
:$a \in S$ is [[Definition:Left Cancellable Element|left cancellab... | Isomorphism Preserves Cancellability | https://proofwiki.org/wiki/Isomorphism_Preserves_Cancellability | https://proofwiki.org/wiki/Isomorphism_Preserves_Cancellability | [
"Isomorphism Preserves Cancellability",
"Isomorphisms (Abstract Algebra)",
"Cancellability"
] | [
"Definition:Algebraic Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Cancellable Element",
"Definition:Cancellable Element"
] | [
"Definition:Algebraic Structure",
"Definition:Cancellable Element",
"Isomorphism Preserves Left Cancellability",
"Isomorphism Preserves Right Cancellability",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Cancellable Element/Right Cance... |
proofwiki-5032 | Isomorphism Preserves Left Cancellability | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then:
:$a \in S$ is left cancellable in $\struct {S, \circ}$
{{iff}}:
:$\map \phi a \in T$ is left cancellable in $\struct {T, *}$. | Let $a$ be left cancellable in $\struct {S, \circ}$.
Let $x \in S$ and $y \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi a * \map \phi x
| r = \map \phi a * \map \phi y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi {a \circ x}
| r = \map \phi {a \circ y}
| c = Morphism Pro... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then:
:$a \in S$ is [[Definition:Left Cancellable Element|left cancellable]] in $\struct {S, \ci... | Let $a$ be [[Definition:Left Cancellable Element|left cancellable]] in $\struct {S, \circ}$.
Let $x \in S$ and $y \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi a * \map \phi x
| r = \map \phi a * \map \phi y
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi {a \circ x}
| r = \map... | Isomorphism Preserves Left Cancellability | https://proofwiki.org/wiki/Isomorphism_Preserves_Left_Cancellability | https://proofwiki.org/wiki/Isomorphism_Preserves_Left_Cancellability | [
"Isomorphism Preserves Cancellability"
] | [
"Definition:Algebraic Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Cancellable Element/Left Cancellable"
] | [
"Definition:Cancellable Element/Left Cancellable",
"Definition:Morphism Property",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Isomorphism (Abstract Algebra)"
] |
proofwiki-5033 | Isomorphism Preserves Right Cancellability | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then:
:$a \in S$ is right cancellable in $\struct {S, \circ}$ {{iff}} $\map \phi a \in T$ is right cancellable in $\struct {T, *}$. | Let $a$ be right cancellable in $\struct {S, \circ}$.
Let $x \in S$ and $y \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi x * \map \phi a
| r = \map \phi y * \map \phi a
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi {x \circ a}
| r = \map \phi {y \circ a}
| c = Morphism pr... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then:
:$a \in S$ is [[Definition:Right Cancellable Element|right cancellable]] in $\struct {S, \... | Let $a$ be [[Definition:Right Cancellable Element|right cancellable]] in $\struct {S, \circ}$.
Let $x \in S$ and $y \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi x * \map \phi a
| r = \map \phi y * \map \phi a
| c =
}}
{{eqn | ll= \leadsto
| l = \map \phi {x \circ a}
| r = \m... | Isomorphism Preserves Right Cancellability | https://proofwiki.org/wiki/Isomorphism_Preserves_Right_Cancellability | https://proofwiki.org/wiki/Isomorphism_Preserves_Right_Cancellability | [
"Isomorphism Preserves Cancellability"
] | [
"Definition:Algebraic Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Cancellable Element/Right Cancellable",
"Definition:Cancellable Element/Right Cancellable"
] | [
"Definition:Cancellable Element/Right Cancellable",
"Definition:Morphism Property",
"Definition:Cancellable Element/Right Cancellable",
"Definition:Cancellable Element/Right Cancellable",
"Definition:Isomorphism (Abstract Algebra)"
] |
proofwiki-5034 | Differentiation of Real Power Series | Let $\xi \in \R$ be a real number.
Let $\sequence {a_n}$ be a sequence in $\R$.
Let $\ds \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m$ be the power series in $x$ about the point $\xi$.
Then within the interval of convergence:
:$\ds \frac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m = \sum_{m \mathop \... | First we can make the substitution $z = x - \xi$ and convert the expression into:
:$\ds \dfrac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m z^m$
We then use $n$th Derivative of $m$th Power:
:$\dfrac {\d^n} {\d z^n} z^m = \begin{cases} m^{\underline n} z^{m - n} & : n \le m \\ 0 & : n > m \end{cases}$
$x$ is {{hypothesis}... | Let $\xi \in \R$ be a [[Definition:Real Number|real number]].
Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence in $\R$]].
Let $\ds \sum_{m \mathop \ge 0} a_m \paren {x - \xi}^m$ be the [[Definition:Power Series|power series]] in $x$ about the point $\xi$.
Then within the [[Definition:Interval of Convergen... | First we can make the substitution $z = x - \xi$ and convert the expression into:
:$\ds \dfrac {\d^n} {\d x^n} \sum_{m \mathop \ge 0} a_m z^m$
We then use [[Nth Derivative of Mth Power|$n$th Derivative of $m$th Power]]:
:$\dfrac {\d^n} {\d z^n} z^m = \begin{cases} m^{\underline n} z^{m - n} & : n \le m \\ 0 & : n > m... | Differentiation of Real Power Series | https://proofwiki.org/wiki/Differentiation_of_Real_Power_Series | https://proofwiki.org/wiki/Differentiation_of_Real_Power_Series | [
"Differentiation of Real Power Series",
"Differential Calculus",
"Real Power Series"
] | [
"Definition:Real Number",
"Definition:Sequence",
"Definition:Power Series",
"Definition:Interval of Convergence",
"Definition:Falling Factorial"
] | [
"Nth Derivative of Mth Power",
"Definition:Interval of Convergence",
"Abel's Theorem",
"Derivative of Identity Function",
"Derivative of Composite Function",
"Abel's Theorem",
"Category:Differentiation of Real Power Series",
"Category:Differential Calculus",
"Category:Real Power Series"
] |
proofwiki-5035 | General Associativity Theorem/Formulation 1 | Let $\struct {S, \circ}$ be a semigroup.
Let $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + n}$ be a sequence of elements of $S$.
Let $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ be a strictly increasing sequence of natural numbers such that $r_0 = p$ and $r_s = p+n$.
Suppose:
:$\ds \forall k \in \closedint... | The proof will proceed by the Principle of Mathematical Induction on $\N$.
Let $T$ be the set of all $n \in \N_{>0}$ such that:
:$(1): \quad$ for every sequence $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + n}$ of elements of $S$
and:
:$(2): \quad$ for every strictly increasing sequence $\sequence {r_k}_{0 \mat... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Element|elements]] of $S$.
Let $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ be a [[Definition:Strictly Increasing Sequence|strictly... | The proof will proceed by the [[Principle of Mathematical Induction]] on $\N$.
Let $T$ be the [[Definition:Set|set]] of all $n \in \N_{>0}$ such that:
:$(1): \quad$ for every [[Definition:Sequence|sequence]] $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + n}$ of [[Definition:Element|elements]] of $S$
and:
:$(2)... | General Associativity Theorem/Formulation 1 | https://proofwiki.org/wiki/General_Associativity_Theorem/Formulation_1 | https://proofwiki.org/wiki/General_Associativity_Theorem/Formulation_1 | [
"General Associativity Theorem"
] | [
"Definition:Semigroup",
"Definition:Sequence",
"Definition:Element",
"Definition:Strictly Increasing/Sequence",
"Definition:Natural Numbers"
] | [
"Principle of Mathematical Induction",
"Definition:Set",
"Definition:Sequence",
"Definition:Element",
"Definition:Strictly Increasing/Sequence",
"Definition:Natural Numbers",
"Definition:Sequence",
"Definition:Element",
"Definition:Strictly Increasing/Sequence",
"Definition:Natural Numbers",
"De... |
proofwiki-5036 | General Associativity Theorem/Formulation 2 | Let $n \in \N_{>0}$ and let $a_1, \ldots, a_n$ be elements of a set $S$.
Let $\circ$ be an associative operation on $S$.
Let the set $\map {P_n} {a_1, a_2, \ldots, a_n}$ be defined inductively by:
:$\map {P_1} {a_1} = \set {a_1}$
:$\map {P_2} {a_1, a_2} = \set {a_1 \circ a_2}$
:$\map {P_n} {a_1, a_2, \ldots, a_n} = \se... | The cases where $n = 1$ and $n = 2$ are clear.
Let $a = x \circ y \in P_n: x \in P_r, y \in P_s$.
If $r > 1$ then we write $x = a_1 \circ z$ where $z = a_2 \circ a_3 \circ \ldots \circ a_r$ by induction.
Then $x \circ y = \paren {a_1 \circ z} \circ y = a_1 \circ \paren {z \circ y} = a_1 \circ \paren {a_2 \circ a_3 \cir... | Let $n \in \N_{>0}$ and let $a_1, \ldots, a_n$ be elements of a set $S$.
Let $\circ$ be an [[Definition:Associative Operation|associative operation]] on $S$.
Let the [[Definition:Set|set]] $\map {P_n} {a_1, a_2, \ldots, a_n}$ be defined inductively by:
:$\map {P_1} {a_1} = \set {a_1}$
:$\map {P_2} {a_1, a_2} = \set... | The cases where $n = 1$ and $n = 2$ are clear.
Let $a = x \circ y \in P_n: x \in P_r, y \in P_s$.
If $r > 1$ then we write $x = a_1 \circ z$ where $z = a_2 \circ a_3 \circ \ldots \circ a_r$ by [[Principle of Mathematical Induction|induction]].
Then $x \circ y = \paren {a_1 \circ z} \circ y = a_1 \circ \paren {z \cir... | General Associativity Theorem/Formulation 2/Proof 1 | https://proofwiki.org/wiki/General_Associativity_Theorem/Formulation_2 | https://proofwiki.org/wiki/General_Associativity_Theorem/Formulation_2/Proof_1 | [
"General Associativity Theorem"
] | [
"Definition:Associative Operation",
"Definition:Set"
] | [
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-5037 | General Associativity Theorem/Formulation 2 | Let $n \in \N_{>0}$ and let $a_1, \ldots, a_n$ be elements of a set $S$.
Let $\circ$ be an associative operation on $S$.
Let the set $\map {P_n} {a_1, a_2, \ldots, a_n}$ be defined inductively by:
:$\map {P_1} {a_1} = \set {a_1}$
:$\map {P_2} {a_1, a_2} = \set {a_1 \circ a_2}$
:$\map {P_n} {a_1, a_2, \ldots, a_n} = \se... | Proof by strong induction:
For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
:The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le n$.
$P \left({1}\right)$ is trivially true, as this just says $a_1 = a_1$.
$P \left({2}\right)$ is the cas... | Let $n \in \N_{>0}$ and let $a_1, \ldots, a_n$ be elements of a set $S$.
Let $\circ$ be an [[Definition:Associative Operation|associative operation]] on $S$.
Let the [[Definition:Set|set]] $\map {P_n} {a_1, a_2, \ldots, a_n}$ be defined inductively by:
:$\map {P_1} {a_1} = \set {a_1}$
:$\map {P_2} {a_1, a_2} = \set... | Proof by [[Second Principle of Mathematical Induction|strong induction]]:
For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
:The [[General Associativity Theorem]] holds for all [[Definition:Composite (Abstract Algebra)|composites]] $a_1 \circ a_2 \circ \cdots \circ a_r$ ... | General Associativity Theorem/Formulation 2/Proof 2 | https://proofwiki.org/wiki/General_Associativity_Theorem/Formulation_2 | https://proofwiki.org/wiki/General_Associativity_Theorem/Formulation_2/Proof_2 | [
"General Associativity Theorem"
] | [
"Definition:Associative Operation",
"Definition:Set"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"General Associativity Theorem",
"Definition:Iterated Binary Operation",
"Definition:Associative Operation",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"General Associativity Theorem",
"Definit... |
proofwiki-5038 | Monoid of Self-Inverse Elements is Abelian Group | Let $\struct {S, \circ}$ be a monoid such that:
:$\forall x \in S: x \circ x = e$
where $e$ is the identity element of $\struct {S, \circ}$.
Then $\struct {S, \circ}$ is an abelian group. | From Equivalence of Definitions of Self-Inverse, $x \circ x = e \implies x = x^{-1}$.
From Inverse in Monoid is Unique, it follows that every element of $\struct {S, \circ}$ has a unique inverse.
So by definition, $\struct {S, \circ}$ is a group.
From All Elements Self-Inverse then Abelian, it follows that $\struct {S,... | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] such that:
:$\forall x \in S: x \circ x = e$
where $e$ is the [[Definition:Identity Element|identity element]] of $\struct {S, \circ}$.
Then $\struct {S, \circ}$ is an [[Definition:Abelian Group|abelian group]]. | From [[Equivalence of Definitions of Self-Inverse]], $x \circ x = e \implies x = x^{-1}$.
From [[Inverse in Monoid is Unique]], it follows that every element of $\struct {S, \circ}$ has a unique [[Definition:Inverse Element|inverse]].
So by definition, $\struct {S, \circ}$ is a [[Definition:Group|group]].
From [[Al... | Monoid of Self-Inverse Elements is Abelian Group | https://proofwiki.org/wiki/Monoid_of_Self-Inverse_Elements_is_Abelian_Group | https://proofwiki.org/wiki/Monoid_of_Self-Inverse_Elements_is_Abelian_Group | [
"Monoids",
"Abelian Groups"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Abelian Group"
] | [
"Equivalence of Definitions of Self-Inverse",
"Inverse in Monoid is Unique",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Group",
"All Elements Self-Inverse then Abelian",
"Definition:Abelian Group"
] |
proofwiki-5039 | Product is Left Identity Therefore Left Cancellable | Let $\struct {S, \circ}$ be a semigroup.
Let $e_L \in S$ be a left identity of $S$.
Let $a \in S$ such that:
:$\exists b \in S: b \circ a = e_L$
Then $a$ is left cancellable in $\struct {S, \circ}$. | Let $x, y \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \circ x
| r = a \circ y
| c = {{Semigroup-axiom|0}}
}}
{{eqn | ll= \leadsto
| l = b \circ \paren {a \circ x}
| r = b \circ \paren {a \circ y}
| c = {{Semigroup-axiom|0}}
}}
{{eqn | ll= \leadsto
| l = \paren {b \circ a} ... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $e_L \in S$ be a [[Definition:Left Identity|left identity]] of $S$.
Let $a \in S$ such that:
:$\exists b \in S: b \circ a = e_L$
Then $a$ is [[Definition:Left Cancellable Element|left cancellable]] in $\struct {S, \circ}$. | Let $x, y \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = a \circ x
| r = a \circ y
| c = {{Semigroup-axiom|0}}
}}
{{eqn | ll= \leadsto
| l = b \circ \paren {a \circ x}
| r = b \circ \paren {a \circ y}
| c = {{Semigroup-axiom|0}}
}}
{{eqn | ll= \leadsto
| l = \paren {b \circ ... | Product is Left Identity Therefore Left Cancellable | https://proofwiki.org/wiki/Product_is_Left_Identity_Therefore_Left_Cancellable | https://proofwiki.org/wiki/Product_is_Left_Identity_Therefore_Left_Cancellable | [
"Semigroups",
"Cancellability",
"Identity Elements"
] | [
"Definition:Semigroup",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Cancellable Element/Left Cancellable"
] | [
"Definition:Cancellable Element/Left Cancellable"
] |
proofwiki-5040 | Product is Right Identity Therefore Right Cancellable | Let $\struct {S, \circ}$ be a semigroup.
Let $e_R \in S$ be a right identity of $S$.
Let $a \in S$ such that:
:$\exists b \in S: a \circ b = e_R$
Then $a$ is right cancellable in $\struct {S, \circ}$. | Let $x, y \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = x \circ a
| r = y \circ a
| c = {{Semigroup-axiom|0}}
}}
{{eqn | ll= \leadsto
| l = \paren {x \circ a} \circ b
| r = \paren {y \circ a} \circ b
| c = {{Semigroup-axiom|0}}
}}
{{eqn | ll= \leadsto
| l = x \circ \paren {a \... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $e_R \in S$ be a [[Definition:Right Identity|right identity]] of $S$.
Let $a \in S$ such that:
:$\exists b \in S: a \circ b = e_R$
Then $a$ is [[Definition:Right Cancellable Element|right cancellable]] in $\struct {S, \circ}$. | Let $x, y \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = x \circ a
| r = y \circ a
| c = {{Semigroup-axiom|0}}
}}
{{eqn | ll= \leadsto
| l = \paren {x \circ a} \circ b
| r = \paren {y \circ a} \circ b
| c = {{Semigroup-axiom|0}}
}}
{{eqn | ll= \leadsto
| l = x \circ \paren {... | Product is Right Identity Therefore Right Cancellable | https://proofwiki.org/wiki/Product_is_Right_Identity_Therefore_Right_Cancellable | https://proofwiki.org/wiki/Product_is_Right_Identity_Therefore_Right_Cancellable | [
"Semigroups",
"Cancellability",
"Identity Elements"
] | [
"Definition:Semigroup",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Cancellable Element/Right Cancellable"
] | [
"Definition:Cancellable Element/Right Cancellable"
] |
proofwiki-5041 | Integer Multiples Greater than Positive Integer Closed under Addition | Let $n \Z$ be the set of integer multiples of $n$.
Let $p \in \Z: p \ge 0$ be a positive integer.
Let $S \subseteq n \Z$ be defined as:
:$S := \set {x \in n \Z: x > p}$
that is, the set of integer multiples of $n$ greater than $p$.
Then the algebraic structure $\struct {S, +}$ is closed under addition. | Let $x, y \in S$.
From Integer Multiples Closed under Addition, $x + y \in n \Z$.
As $x, y > p$ we have that:
:$\exists r \in \Z_{>0}: x = p + r$
:$\exists s \in \Z_{>0}: y = p + s$
Thus it follows that:
{{begin-eqn}}
{{eqn | l = x + y
| r = \paren {p + r} + \paren {p + s}
| c =
}}
{{eqn | r = 2 p + r + s
... | Let $n \Z$ be the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
Let $p \in \Z: p \ge 0$ be a [[Definition:Positive Integer|positive integer]].
Let $S \subseteq n \Z$ be defined as:
:$S := \set {x \in n \Z: x > p}$
that is, the [[Definition:Set|set]] of [[Definition:Set of Integer Multiples... | Let $x, y \in S$.
From [[Integer Multiples Closed under Addition]], $x + y \in n \Z$.
As $x, y > p$ we have that:
:$\exists r \in \Z_{>0}: x = p + r$
:$\exists s \in \Z_{>0}: y = p + s$
Thus it follows that:
{{begin-eqn}}
{{eqn | l = x + y
| r = \paren {p + r} + \paren {p + s}
| c =
}}
{{eqn | r = 2 p ... | Integer Multiples Greater than Positive Integer Closed under Addition | https://proofwiki.org/wiki/Integer_Multiples_Greater_than_Positive_Integer_Closed_under_Addition | https://proofwiki.org/wiki/Integer_Multiples_Greater_than_Positive_Integer_Closed_under_Addition | [
"Integer Addition",
"Algebraic Closure"
] | [
"Definition:Set of Integer Multiples",
"Definition:Positive/Integer",
"Definition:Set",
"Definition:Set of Integer Multiples",
"Definition:Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Addition/Integers"
] | [
"Integer Multiples Closed under Addition",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Addition/Integers"
] |
proofwiki-5042 | Integer Multiples Greater than Positive Integer Closed under Multiplication | Let $n \Z$ be the set of integer multiples of $n$.
Let $p \in \Z: p \ge 0$ be a positive integer.
Let $S \subseteq n \Z$ be defined as:
:$S := \set {x \in n \Z: x > p}$
that is, the set of integer multiples of $n$ greater than $p$.
Then the algebraic structure $\struct {S, \times}$ is closed under multiplication. | Let $x, y \in S$.
From Integer Multiples Closed under Multiplication, $x y \in n \Z$.
As $x, y > p$ we have that:
:$\exists r \in \Z_{>0}: x = p + r$
:$\exists s \in \Z_{>0}: y = p + s$
Thus it follows that:
{{begin-eqn}}
{{eqn | l = x + y
| r = \paren {p + r} \paren {p + s}
| c =
}}
{{eqn | r = p^2 + p \p... | Let $n \Z$ be the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
Let $p \in \Z: p \ge 0$ be a [[Definition:Positive Integer|positive integer]].
Let $S \subseteq n \Z$ be defined as:
:$S := \set {x \in n \Z: x > p}$
that is, the [[Definition:Set|set]] of [[Definition:Set of Integer Multiples... | Let $x, y \in S$.
From [[Integer Multiples Closed under Multiplication]], $x y \in n \Z$.
As $x, y > p$ we have that:
:$\exists r \in \Z_{>0}: x = p + r$
:$\exists s \in \Z_{>0}: y = p + s$
Thus it follows that:
{{begin-eqn}}
{{eqn | l = x + y
| r = \paren {p + r} \paren {p + s}
| c =
}}
{{eqn | r = p^... | Integer Multiples Greater than Positive Integer Closed under Multiplication | https://proofwiki.org/wiki/Integer_Multiples_Greater_than_Positive_Integer_Closed_under_Multiplication | https://proofwiki.org/wiki/Integer_Multiples_Greater_than_Positive_Integer_Closed_under_Multiplication | [
"Integer Multiplication",
"Algebraic Closure"
] | [
"Definition:Set of Integer Multiples",
"Definition:Positive/Integer",
"Definition:Set",
"Definition:Set of Integer Multiples",
"Definition:Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Integers"
] | [
"Integer Multiples Closed under Multiplication",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Integers"
] |
proofwiki-5043 | Non-Zero Rational Numbers Closed under Multiplication | The set of non-zero rational numbers is closed under multiplication. | Recall that Rational Numbers form Field under the operations of addition and multiplication.
By definition of a field, the algebraic structure $\struct {\Q_{\ne 0}, \times}$ is a group.
Thus, by definition, $\times$ is closed in $\struct {\Q_{\ne 0}, \times}$.
{{qed}} | The [[Definition:Set|set]] of [[Definition:Zero (Algebra)|non-zero]] [[Definition:Rational Number|rational numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Rational Multiplication|multiplication]]. | Recall that [[Rational Numbers form Field]] under the [[Definition:Binary Operation|operations]] of [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]].
By definition of a [[Definition:Field (Abstract Algebra)|field]], the [[Definition:Algebraic Structure|algebraic struc... | Non-Zero Rational Numbers Closed under Multiplication | https://proofwiki.org/wiki/Non-Zero_Rational_Numbers_Closed_under_Multiplication | https://proofwiki.org/wiki/Non-Zero_Rational_Numbers_Closed_under_Multiplication | [
"Rational Multiplication",
"Algebraic Closure"
] | [
"Definition:Set",
"Definition:Zero (Number)",
"Definition:Rational Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Rational Numbers"
] | [
"Rational Numbers form Field",
"Definition:Operation/Binary Operation",
"Definition:Addition/Rational Numbers",
"Definition:Multiplication/Rational Numbers",
"Definition:Field (Abstract Algebra)",
"Definition:Algebraic Structure",
"Definition:Group",
"Definition:Closure (Abstract Algebra)/Algebraic St... |
proofwiki-5044 | Non-Zero Real Numbers Closed under Multiplication | The set of non-zero real numbers is closed under multiplication:
:$\forall x, y \in \R_{\ne 0}: x \times y \in \R_{\ne 0}$ | Recall that Real Numbers form Field under the operations of addition and multiplication.
By definition of a field, the algebraic structure $\struct {\R_{\ne 0}, \times}$ is a group.
Thus, by definition, $\times$ is closed in $\struct {\R_{\ne 0}, \times}$.
{{qed}} | The [[Definition:Set|set]] of [[Definition:Zero (Number)|non-zero]] [[Definition:Real Number|real numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Multiplication|multiplication]]:
:$\forall x, y \in \R_{\ne 0}: x \times y \in \R_{\ne 0}$ | Recall that [[Real Numbers form Field]] under the [[Definition:Binary Operation|operations]] of [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]].
By definition of a [[Definition:Field (Abstract Algebra)|field]], the [[Definition:Algebraic Structure with One Operation|algebrai... | Non-Zero Real Numbers Closed under Multiplication/Proof 1 | https://proofwiki.org/wiki/Non-Zero_Real_Numbers_Closed_under_Multiplication | https://proofwiki.org/wiki/Non-Zero_Real_Numbers_Closed_under_Multiplication/Proof_1 | [
"Real Multiplication",
"Algebraic Closure",
"Non-Zero Real Numbers Closed under Multiplication"
] | [
"Definition:Set",
"Definition:Zero (Number)",
"Definition:Real Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Real Numbers"
] | [
"Real Numbers form Field",
"Definition:Operation/Binary Operation",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Field (Abstract Algebra)",
"Definition:Algebraic Structure/One Operation",
"Definition:Group",
"Definition:Closure (Abstract Algebra)/Algebraic ... |
proofwiki-5045 | Non-Zero Real Numbers Closed under Multiplication | The set of non-zero real numbers is closed under multiplication:
:$\forall x, y \in \R_{\ne 0}: x \times y \in \R_{\ne 0}$ | Let $x \times y = 0$.
{{WLOG}}, suppose that $x \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = x \times y
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = \frac 1 x \times \paren {x \times y}
| r = \frac 1 x \times 0
| c = as $x \ne 0$
}}
{{eqn | ll= \leadsto
| l = \paren {\frac 1 x \times x} \... | The [[Definition:Set|set]] of [[Definition:Zero (Number)|non-zero]] [[Definition:Real Number|real numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Multiplication|multiplication]]:
:$\forall x, y \in \R_{\ne 0}: x \times y \in \R_{\ne 0}$ | Let $x \times y = 0$.
{{WLOG}}, suppose that $x \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = x \times y
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = \frac 1 x \times \paren {x \times y}
| r = \frac 1 x \times 0
| c = as $x \ne 0$
}}
{{eqn | ll= \leadsto
| l = \paren {\frac 1 x \times x}... | Non-Zero Real Numbers Closed under Multiplication/Proof 2 | https://proofwiki.org/wiki/Non-Zero_Real_Numbers_Closed_under_Multiplication | https://proofwiki.org/wiki/Non-Zero_Real_Numbers_Closed_under_Multiplication/Proof_2 | [
"Real Multiplication",
"Algebraic Closure",
"Non-Zero Real Numbers Closed under Multiplication"
] | [
"Definition:Set",
"Definition:Zero (Number)",
"Definition:Real Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Real Numbers"
] | [
"Real Zero is Zero Element",
"De Morgan's Laws (Logic)/Disjunction",
"Rule of Transposition"
] |
proofwiki-5046 | Non-Zero Complex Numbers are Closed under Multiplication | The set of non-zero complex numbers is closed under multiplication. | Recall that Complex Numbers form Field under the operations of addition and multiplication.
By definition of a field, the algebraic structure $\struct {\C_{\ne 0}, \times}$ is a group.
Thus, by definition, $\times$ is closed in $\struct {\C_{\ne 0}, \times}$.
{{qed}} | The [[Definition:Set|set]] of non-[[Definition:Zero (Number)|zero]] [[Definition:Complex Number|complex numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Multiplication|multiplication]]. | Recall that [[Complex Numbers form Field]] under the [[Definition:Binary Operation|operations]] of [[Definition:Complex Addition|addition]] and [[Definition:Complex Multiplication|multiplication]].
By definition of a [[Definition:Field (Abstract Algebra)|field]], the [[Definition:Algebraic Structure with One Operation... | Non-Zero Complex Numbers are Closed under Multiplication/Proof 1 | https://proofwiki.org/wiki/Non-Zero_Complex_Numbers_are_Closed_under_Multiplication | https://proofwiki.org/wiki/Non-Zero_Complex_Numbers_are_Closed_under_Multiplication/Proof_1 | [
"Non-Zero Complex Numbers are Closed under Multiplication",
"Complex Multiplication",
"Algebraic Closure"
] | [
"Definition:Set",
"Definition:Zero (Number)",
"Definition:Complex Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Real Numbers"
] | [
"Complex Numbers form Field",
"Definition:Operation/Binary Operation",
"Definition:Addition/Complex Numbers",
"Definition:Multiplication/Complex Numbers",
"Definition:Field (Abstract Algebra)",
"Definition:Algebraic Structure/One Operation",
"Definition:Group",
"Definition:Closure (Abstract Algebra)/A... |
proofwiki-5047 | Non-Zero Complex Numbers are Closed under Multiplication | The set of non-zero complex numbers is closed under multiplication. | Let $z_1, z_2 \in \C_{\ne 0}$.
Then by definition of complex number:
:$z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$
for some $x_1, y_1, x_2, y_2 \in \R$ such that:
:$x_1 \ne 0$ or $y_1 \ne 0$
:$x_2 \ne 0$ or $y_2 \ne 0$
Expressing $z_1$ and $z_2$ in exponential form (although polar form is equally adequate):
:$z_1 = r_1 e^{i ... | The [[Definition:Set|set]] of non-[[Definition:Zero (Number)|zero]] [[Definition:Complex Number|complex numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Multiplication|multiplication]]. | Let $z_1, z_2 \in \C_{\ne 0}$.
Then by definition of [[Definition:Complex Number/Definition 1|complex number]]:
:$z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$
for some $x_1, y_1, x_2, y_2 \in \R$ such that:
:$x_1 \ne 0$ or $y_1 \ne 0$
:$x_2 \ne 0$ or $y_2 \ne 0$
Expressing $z_1$ and $z_2$ in [[Definition:Exponential Form of... | Non-Zero Complex Numbers are Closed under Multiplication/Proof 2 | https://proofwiki.org/wiki/Non-Zero_Complex_Numbers_are_Closed_under_Multiplication | https://proofwiki.org/wiki/Non-Zero_Complex_Numbers_are_Closed_under_Multiplication/Proof_2 | [
"Non-Zero Complex Numbers are Closed under Multiplication",
"Complex Multiplication",
"Algebraic Closure"
] | [
"Definition:Set",
"Definition:Zero (Number)",
"Definition:Complex Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Real Numbers"
] | [
"Definition:Complex Number/Definition 1",
"Definition:Complex Number/Polar Form/Exponential Form",
"Definition:Complex Number/Polar Form",
"Product of Complex Numbers in Polar Form",
"Definition:Complex Number/Polar Form/Exponential Form"
] |
proofwiki-5048 | Non-Zero Complex Numbers are Closed under Multiplication | The set of non-zero complex numbers is closed under multiplication. | Equivalently this is to say:
:$z_1 z_2 = 0 \implies z_1 = 0 \lor z_2 = 0$
Let $z_1 z_2 = 0$.
{{begin-eqn}}
{{eqn | l = z_1
| r = \tuple {x_1, y_1}
| c = {{Defof|Complex Number|index = 2}}: for some $x_1, y_1 \in \R$
}}
{{eqn | l = z_2
| r = \tuple {x_2, y_2}
| c = {{Defof|Complex Number|index = ... | The [[Definition:Set|set]] of non-[[Definition:Zero (Number)|zero]] [[Definition:Complex Number|complex numbers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Multiplication|multiplication]]. | Equivalently this is to say:
:$z_1 z_2 = 0 \implies z_1 = 0 \lor z_2 = 0$
Let $z_1 z_2 = 0$.
{{begin-eqn}}
{{eqn | l = z_1
| r = \tuple {x_1, y_1}
| c = {{Defof|Complex Number|index = 2}}: for some $x_1, y_1 \in \R$
}}
{{eqn | l = z_2
| r = \tuple {x_2, y_2}
| c = {{Defof|Complex Number|index... | Non-Zero Complex Numbers are Closed under Multiplication/Proof 3 | https://proofwiki.org/wiki/Non-Zero_Complex_Numbers_are_Closed_under_Multiplication | https://proofwiki.org/wiki/Non-Zero_Complex_Numbers_are_Closed_under_Multiplication/Proof_3 | [
"Non-Zero Complex Numbers are Closed under Multiplication",
"Complex Multiplication",
"Algebraic Closure"
] | [
"Definition:Set",
"Definition:Zero (Number)",
"Definition:Complex Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Multiplication/Real Numbers"
] | [
"Definition:Multiplication",
"Definition:Multiplication",
"Equality is Transitive",
"Real Multiplication Distributes over Addition",
"Non-Zero Real Numbers Closed under Multiplication",
"Non-Zero Real Numbers Closed under Multiplication",
"Definition:Zero (Number)"
] |
proofwiki-5049 | Non-Zero Modulo Numbers Closed under Multiplication then Modulo is Prime | Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$ for $m > 1$.
Let $\Z'_m$ be the set of non-zero integers modulo $m$.
Let $\struct {\Z_m, \times_m}$ be closed under modulo multiplication.
Then $m$ is prime. | Suppose $m$ is not prime.
Then $m = r s$ for some $r, s \in \Z: 1 < r < m, 1 < s < m$.
So $r, s \in \Z'_m$.
But:
:$r \times_m s \equiv 0 \pmod m$
and so $r \times_m s \notin \Z'_m$.
So if $m$ is not prime, $\struct {\Z_m, \times_m}$ is not closed.
The result follows from the Rule of Transposition.
{{qed}} | Let $\struct {\Z_m, +_m, \times_m}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]] for $m > 1$.
Let $\Z'_m$ be the [[Definition:Set|set]] of [[Definition:Zero (Algebra)|non-zero]] [[Definition:Integers Modulo m|integers modulo $m$]].
Let $\struct {\Z_m, \times_m}$ be [[Definition:Closed ... | Suppose $m$ is not [[Definition:Prime Number|prime]].
Then $m = r s$ for some $r, s \in \Z: 1 < r < m, 1 < s < m$.
So $r, s \in \Z'_m$.
But:
:$r \times_m s \equiv 0 \pmod m$
and so $r \times_m s \notin \Z'_m$.
So if $m$ is not [[Definition:Prime Number|prime]], $\struct {\Z_m, \times_m}$ is not [[Definition:Closed... | Non-Zero Modulo Numbers Closed under Multiplication then Modulo is Prime | https://proofwiki.org/wiki/Non-Zero_Modulo_Numbers_Closed_under_Multiplication_then_Modulo_is_Prime | https://proofwiki.org/wiki/Non-Zero_Modulo_Numbers_Closed_under_Multiplication_then_Modulo_is_Prime | [
"Modulo Arithmetic",
"Ring of Integers Modulo m",
"Prime Numbers"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Set",
"Definition:Zero (Number)",
"Definition:Integers Modulo m",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Modulo Multiplication",
"Definition:Prime Number"
] | [
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Rule of Transposition"
] |
proofwiki-5050 | Magma is Submagma of Itself | Let $\struct {S, \circ}$ be a magma.
Then $\struct {S, \circ}$ is a submagma of itself. | By definition, a magma is an algebraic structure $\struct {S, \circ}$ where $\circ$ is closed.
That is:
:$\forall x, y \in S: x \circ y \in S$
From Set is Subset of Itself, $S \subseteq S$.
By definition, $\struct {T, \circ}$ is a submagma of $S$ if:
:$\forall x, y \in T: x \circ y \in T$
It follows from the above that... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Then $\struct {S, \circ}$ is a [[Definition:Submagma|submagma]] of itself. | By definition, a [[Definition:Magma|magma]] is an [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, \circ}$ where $\circ$ is [[Definition:Closed Algebraic Structure|closed]].
That is:
:$\forall x, y \in S: x \circ y \in S$
From [[Set is Subset of Itself]], $S \subseteq S$.
By definition, $\struct ... | Magma is Submagma of Itself | https://proofwiki.org/wiki/Magma_is_Submagma_of_Itself | https://proofwiki.org/wiki/Magma_is_Submagma_of_Itself | [
"Magmas",
"Submagmas"
] | [
"Definition:Magma",
"Definition:Submagma"
] | [
"Definition:Magma",
"Definition:Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Set is Subset of Itself",
"Definition:Submagma",
"Definition:Submagma"
] |
proofwiki-5051 | Empty Set is Submagma of Magma | Let $\struct {S, \circ}$ be a magma.
Then:
:$\struct {\O, \circ}$ is a submagma of $\struct {S, \circ}$
where $\O$ is the empty set. | By definition, a magma is an algebraic structure $\struct {S, \circ}$ where $\circ$ is closed.
That is:
:$\forall x, y \in S: x \circ y \in S$
By definition, $\struct {T, \circ}$ is a submagma of $S$ if:
:$\forall x, y \in T: x \circ y \in T$
But:
:$\not \exists x, y \in \O: x \circ y \notin \O$
it follows vacuously th... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Then:
:$\struct {\O, \circ}$ is a [[Definition:Submagma|submagma]] of $\struct {S, \circ}$
where $\O$ is the [[Definition:Empty Set|empty set]]. | By definition, a [[Definition:Magma|magma]] is an [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, \circ}$ where $\circ$ is [[Definition:Closed Algebraic Structure|closed]].
That is:
:$\forall x, y \in S: x \circ y \in S$
By definition, $\struct {T, \circ}$ is a [[Definition:Submagma|submagma]] of... | Empty Set is Submagma of Magma | https://proofwiki.org/wiki/Empty_Set_is_Submagma_of_Magma | https://proofwiki.org/wiki/Empty_Set_is_Submagma_of_Magma | [
"Magmas",
"Submagmas",
"Empty Set"
] | [
"Definition:Magma",
"Definition:Submagma",
"Definition:Empty Set"
] | [
"Definition:Magma",
"Definition:Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Submagma",
"Definition:Vacuous Truth"
] |
proofwiki-5052 | Max Operation on Toset forms Semigroup | Let $\struct{S, \preceq}$ be a totally ordered set.
Let $\map \max {x, y}$ denote the max operation on $x, y \in S$.
Then $\struct{S, \max}$ is a semigroup. | By the definition of the max operation, either:
:$\map \max {x, y}= x$
or
:$\map \max {x, y}= y$
So $\max$ is closed on $S$.
From Max Operation is Associative:
:$\forall x, y, z \in S: \map \max {x, \map \max {y, z}} = \map \max {\map \max {x, y}, z}$
Hence the result, by definition of semigroup.
{{qed}} | Let $\struct{S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\map \max {x, y}$ denote the [[Definition:Max Operation|max operation]] on $x, y \in S$.
Then $\struct{S, \max}$ is a [[Definition:Semigroup|semigroup]]. | By the definition of the [[Definition:Max Operation|max operation]], either:
:$\map \max {x, y}= x$
or
:$\map \max {x, y}= y$
So $\max$ is [[Definition:Closed Algebraic Structure|closed on $S$]].
From [[Max Operation is Associative]]:
:$\forall x, y, z \in S: \map \max {x, \map \max {y, z}} = \map \max {\map \max {... | Max Operation on Toset forms Semigroup | https://proofwiki.org/wiki/Max_Operation_on_Toset_forms_Semigroup | https://proofwiki.org/wiki/Max_Operation_on_Toset_forms_Semigroup | [
"Max Operation"
] | [
"Definition:Totally Ordered Set",
"Definition:Max Operation",
"Definition:Semigroup"
] | [
"Definition:Max Operation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Max Operation is Associative",
"Definition:Semigroup"
] |
proofwiki-5053 | Min Semigroup on Toset forms Semilattice | Let $\struct {S, \preceq}$ be a totally ordered set.
Then the min semigroup $\struct {S, \min}$ is a semilattice. | We have:
:Min Semigroup is Commutative
:Min Semigroup is Idempotent.
Hence the result, by definition of a semilattice.
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then the [[Min Operation on Toset forms Semigroup|min semigroup]] $\struct {S, \min}$ is a [[Definition:Semilattice|semilattice]]. | We have:
:[[Min Semigroup is Commutative]]
:[[Min Semigroup is Idempotent]].
Hence the result, by definition of a [[Definition:Semilattice|semilattice]].
{{qed}} | Min Semigroup on Toset forms Semilattice | https://proofwiki.org/wiki/Min_Semigroup_on_Toset_forms_Semilattice | https://proofwiki.org/wiki/Min_Semigroup_on_Toset_forms_Semilattice | [
"Min Operation",
"Semilattices"
] | [
"Definition:Totally Ordered Set",
"Min Operation on Toset forms Semigroup",
"Definition:Semilattice"
] | [
"Min Semigroup is Commutative",
"Min Semigroup is Idempotent",
"Definition:Semilattice"
] |
proofwiki-5054 | Min Semigroup is Commutative | Let $\struct {S, \preceq}$ be a totally ordered set.
Then the semigroup $\struct {S, \min}$ is commutative. | Let $x, y \in S$.
From Min Operation is Commutative:
:$\map \min {x, y} = \map \min {y, x}$
Hence the result, by definition of commutative semigroup.
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then the [[Min Operation on Toset forms Semigroup|semigroup $\struct {S, \min}$]] is [[Definition:Commutative Semigroup|commutative]]. | Let $x, y \in S$.
From [[Min Operation is Commutative]]:
:$\map \min {x, y} = \map \min {y, x}$
Hence the result, by definition of [[Definition:Commutative Semigroup|commutative semigroup]].
{{qed}} | Min Semigroup is Commutative | https://proofwiki.org/wiki/Min_Semigroup_is_Commutative | https://proofwiki.org/wiki/Min_Semigroup_is_Commutative | [
"Min Operation",
"Examples of Commutative Semigroups"
] | [
"Definition:Totally Ordered Set",
"Min Operation on Toset forms Semigroup",
"Definition:Commutative Semigroup"
] | [
"Min Operation is Commutative",
"Definition:Commutative Semigroup"
] |
proofwiki-5055 | Min Semigroup is Idempotent | Let $\struct {S, \preceq}$ be a totally ordered set.
Then the semigroup $\struct {S, \min}$ is an idempotent semigroup. | The fact that $\struct {S, \min}$ is a semigroup is demonstrated in Min Operation on Toset forms Semigroup.
Then the min operation is idempotent:
:$\forall x \in S: \min \set {x, x} = x$
The result follows by the definition of idempotent semigroup.
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then the [[Definition:Semigroup|semigroup]] $\struct {S, \min}$ is an [[Definition:Idempotent Semigroup|idempotent semigroup]]. | The fact that $\struct {S, \min}$ is a [[Definition:Semigroup|semigroup]] is demonstrated in [[Min Operation on Toset forms Semigroup]].
Then the [[Max and Min are Idempotent|min operation is idempotent]]:
:$\forall x \in S: \min \set {x, x} = x$
The result follows by the definition of [[Definition:Idempotent Semigro... | Min Semigroup is Idempotent | https://proofwiki.org/wiki/Min_Semigroup_is_Idempotent | https://proofwiki.org/wiki/Min_Semigroup_is_Idempotent | [
"Min Operation",
"Examples of Idempotent Semigroups"
] | [
"Definition:Totally Ordered Set",
"Definition:Semigroup",
"Definition:Idempotent Semigroup"
] | [
"Definition:Semigroup",
"Min Operation on Toset forms Semigroup",
"Max and Min are Idempotent",
"Definition:Idempotent Semigroup"
] |
proofwiki-5056 | Max Operation on Woset is Monoid | Let $\struct {S, \preceq}$ be a well-ordered set.
Let $\max \set {x, y}$ denote the max operation on $x, y \in S$.
Then $\struct {S, \max}$ is a monoid.
Its identity element is the smallest element of $S$. | From Well-Ordering is Total Ordering, we have that $\struct {S, \preceq}$ is a totally ordered set.
Thus, by Max Operation on Toset forms Semigroup, $\struct {S, \max}$ is a semigroup.
By definition, a well-ordered set has a smallest element.
So:
:$\exists m \in S: \forall x \in S: m \preceq x$
where $m$ is that smalle... | Let $\struct {S, \preceq}$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Let $\max \set {x, y}$ denote the [[Definition:Max Operation|max operation]] on $x, y \in S$.
Then $\struct {S, \max}$ is a [[Definition:Monoid|monoid]].
Its [[Definition:Identity Element|identity element]] is the [[Definition:Smalle... | From [[Well-Ordering is Total Ordering]], we have that $\struct {S, \preceq}$ is a [[Definition:Totally Ordered Set|totally ordered set]].
Thus, by [[Max Operation on Toset forms Semigroup]], $\struct {S, \max}$ is a [[Definition:Semigroup|semigroup]].
By definition, a [[Definition:Well-Ordered Set|well-ordered set]... | Max Operation on Woset is Monoid | https://proofwiki.org/wiki/Max_Operation_on_Woset_is_Monoid | https://proofwiki.org/wiki/Max_Operation_on_Woset_is_Monoid | [
"Max Operation",
"Well-Orderings",
"Examples of Monoids"
] | [
"Definition:Well-Ordered Set",
"Definition:Max Operation",
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Smallest Element"
] | [
"Equivalence of Definitions of Well-Ordering/Definition 1 implies Definition 2",
"Definition:Totally Ordered Set",
"Max Operation on Toset forms Semigroup",
"Definition:Semigroup",
"Definition:Well-Ordered Set",
"Definition:Smallest Element",
"Definition:Smallest Element",
"Definition:Max Operation",
... |
proofwiki-5057 | Max Operation on Natural Numbers forms Monoid | Let $\struct {\N, \max}$ denote the algebraic structure formed from the natural numbers $\N$ and the max operation.
Then $\struct {\N, \max}$ is a monoid.
Its identity element is zero. | By the Well-Ordering Principle, $\N$ is a well-ordered set.
The result follows from Max Operation on Woset is Monoid.
{{qed}} | Let $\struct {\N, \max}$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed from the [[Definition:Natural Numbers|natural numbers]] $\N$ and the [[Definition:Max Operation|max operation]].
Then $\struct {\N, \max}$ is a [[Definition:Monoid|monoid]].
Its [[Definition:Identity Element|identity e... | By the [[Well-Ordering Principle]], $\N$ is a [[Definition:Well-Ordered Set|well-ordered set]].
The result follows from [[Max Operation on Woset is Monoid]].
{{qed}} | Max Operation on Natural Numbers forms Monoid | https://proofwiki.org/wiki/Max_Operation_on_Natural_Numbers_forms_Monoid | https://proofwiki.org/wiki/Max_Operation_on_Natural_Numbers_forms_Monoid | [
"Natural Numbers",
"Examples of Monoids",
"Max Operation"
] | [
"Definition:Algebraic Structure",
"Definition:Natural Numbers",
"Definition:Max Operation",
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Zero (Number)"
] | [
"Well-Ordering Principle",
"Definition:Well-Ordered Set",
"Max Operation on Woset is Monoid"
] |
proofwiki-5058 | Complex Modulus Function is Continuous | Let $z_0 \in \C$ be a complex number.
Then the complex modulus function is continuous at $z_0$. | Let $\epsilon>0$.
Let $z \in \C$ be a complex number satisfying $\left\vert{z - z_0}\right\vert < \epsilon$.
By the Reverse Triangle Inequality:
: $\left\vert{ \left\vert{z}\right\vert - \left\vert{z_0}\right\vert }\right\vert \le \left\vert{z - z_0}\right\vert < \epsilon$
Hence the result, by the $\epsilon$-$\delta$ d... | Let $z_0 \in \C$ be a [[Definition:Complex Number|complex number]].
Then the [[Definition:Modulus of Complex Number|complex modulus function]] is [[Definition:Continuous Complex Function|continuous]] at $z_0$. | Let $\epsilon>0$.
Let $z \in \C$ be a [[Definition:Complex Number|complex number]] satisfying $\left\vert{z - z_0}\right\vert < \epsilon$.
By the [[Reverse Triangle Inequality/Real and Complex Fields|Reverse Triangle Inequality]]:
: $\left\vert{ \left\vert{z}\right\vert - \left\vert{z_0}\right\vert }\right\vert \le \... | Complex Modulus Function is Continuous | https://proofwiki.org/wiki/Complex_Modulus_Function_is_Continuous | https://proofwiki.org/wiki/Complex_Modulus_Function_is_Continuous | [
"Complex Modulus",
"Continuous Functions"
] | [
"Definition:Complex Number",
"Definition:Complex Modulus",
"Definition:Continuous Complex Function"
] | [
"Definition:Complex Number",
"Reverse Triangle Inequality/Real and Complex Fields",
"Definition:Continuous Complex Function",
"Category:Complex Modulus",
"Category:Continuous Functions"
] |
proofwiki-5059 | Subset of Natural Numbers under Max Operation is Monoid | Let $S \subseteq \N$ be a subset of the natural numbers $\N$.
Let $\struct {S, \max}$ denote the algebraic structure formed from $S$ and the max operation.
Then $\struct {S, \max}$ is a monoid.
Its identity element is the smallest element of $S$. | By the Well-Ordering Principle, $\N$ is a well-ordered set.
By definition, every subset of a well-ordered set is also well-ordered.
Thus $S$ is a well-ordered set.
The result follows from Max Operation on Woset is Monoid.
{{qed}} | Let $S \subseteq \N$ be a [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]] $\N$.
Let $\struct {S, \max}$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed from $S$ and the [[Definition:Max Operation|max operation]].
Then $\struct {S, \max}$ is a [[Definition:M... | By the [[Well-Ordering Principle]], $\N$ is a [[Definition:Well-Ordered Set|well-ordered set]].
By definition, every [[Definition:Subset|subset]] of a [[Definition:Well-Ordered Set|well-ordered set]] is also [[Definition:Well-Ordered Set|well-ordered]].
Thus $S$ is a [[Definition:Well-Ordered Set|well-ordered set]].
... | Subset of Natural Numbers under Max Operation is Monoid | https://proofwiki.org/wiki/Subset_of_Natural_Numbers_under_Max_Operation_is_Monoid | https://proofwiki.org/wiki/Subset_of_Natural_Numbers_under_Max_Operation_is_Monoid | [
"Natural Numbers",
"Examples of Monoids",
"Max Operation"
] | [
"Definition:Subset",
"Definition:Natural Numbers",
"Definition:Algebraic Structure",
"Definition:Max Operation",
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Smallest Element"
] | [
"Well-Ordering Principle",
"Definition:Well-Ordered Set",
"Definition:Subset",
"Definition:Well-Ordered Set",
"Definition:Well-Ordered Set",
"Definition:Well-Ordered Set",
"Max Operation on Woset is Monoid"
] |
proofwiki-5060 | Euclidean Space is Banach Space | Let $m$ be a positive integer.
Then the Euclidean space $\R^m$, along with the Euclidean norm, forms a Banach space over $\R$. | The Euclidean space $\R^m$ is a vector space over $\R$.
That the norm axioms are satisfied is proven in Euclidean Space is Normed Vector Space.
Then we have Euclidean Space is Complete Metric Space.
The result follows by the definition of a Banach space.
{{qed}} | Let $m$ be a [[Definition:Positive Integer|positive integer]].
Then the [[Definition:Euclidean Space|Euclidean space]] $\R^m$, along with the [[Definition:Euclidean Norm|Euclidean norm]], forms a [[Definition:Banach Space|Banach space]] over $\R$. | The [[Definition:Euclidean Space|Euclidean space]] $\R^m$ is a [[Definition:Vector Space|vector space]] over $\R$.
That the [[Axiom:Vector Space Norm Axioms|norm axioms]] are satisfied is proven in [[Euclidean Space is Normed Vector Space]].
Then we have [[Euclidean Space is Complete Metric Space]].
The result follo... | Euclidean Space is Banach Space/Proof 1 | https://proofwiki.org/wiki/Euclidean_Space_is_Banach_Space | https://proofwiki.org/wiki/Euclidean_Space_is_Banach_Space/Proof_1 | [
"Euclidean Space is Banach Space",
"Euclidean Spaces",
"Banach Spaces"
] | [
"Definition:Positive/Integer",
"Definition:Euclidean Space",
"Definition:Euclidean Norm",
"Definition:Banach Space"
] | [
"Definition:Euclidean Space",
"Definition:Vector Space",
"Axiom:Vector Space Norm Axioms",
"Euclidean Space is Normed Vector Space",
"Euclidean Space is Complete Metric Space",
"Definition:Banach Space"
] |
proofwiki-5061 | Euclidean Space is Banach Space | Let $m$ be a positive integer.
Then the Euclidean space $\R^m$, along with the Euclidean norm, forms a Banach space over $\R$. | By definition, Euclidean norm is the same as $p$-norm with $p = 2$.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N} = \sequence {\tuple {x_n^{\paren 1}, x_n^{\paren 2}, \ldots, x_n^{\paren m} } }_{n \mathop \in \N} $ be a Cauchy sequence in $\R^m$.
Let $k \in \N_{> 0} : k \le m$.
Then:
{{begin-eqn}}
{{eqn | l = \size {... | Let $m$ be a [[Definition:Positive Integer|positive integer]].
Then the [[Definition:Euclidean Space|Euclidean space]] $\R^m$, along with the [[Definition:Euclidean Norm|Euclidean norm]], forms a [[Definition:Banach Space|Banach space]] over $\R$. | By definition, [[Definition:Euclidean Norm|Euclidean norm]] is the same as [[Definition:P-Norm|$p$-norm]] with $p = 2$.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N} = \sequence {\tuple {x_n^{\paren 1}, x_n^{\paren 2}, \ldots, x_n^{\paren m} } }_{n \mathop \in \N} $ be a [[Definition:Cauchy Sequence|Cauchy sequence]... | Euclidean Space is Banach Space/Proof 2 | https://proofwiki.org/wiki/Euclidean_Space_is_Banach_Space | https://proofwiki.org/wiki/Euclidean_Space_is_Banach_Space/Proof_2 | [
"Euclidean Space is Banach Space",
"Euclidean Spaces",
"Banach Spaces"
] | [
"Definition:Positive/Integer",
"Definition:Euclidean Space",
"Definition:Euclidean Norm",
"Definition:Banach Space"
] | [
"Definition:Euclidean Norm",
"Definition:P-Norm",
"Definition:Cauchy Sequence",
"Definition:Cauchy Sequence/Real Numbers"
] |
proofwiki-5062 | Complex Plane is Complete Metric Space | The complex plane, along with the metric induced by the norm given by the complex modulus, forms a complete metric space. | Let $z = x + iy$ be a complex number, where $x, y \in \R$.
Now, we can identify the complex number $z$ with the ordered pair $\left( x, y \right) \in \R^2$.
The norm on $\C$ given by the complex modulus is then identical to the Euclidean norm on $\R^2$.
Therefore, metric on $\C$ induced by the norm given by the complex... | The [[Definition:Complex Number|complex plane]], along with the [[Definition:Metric Induced by Norm|metric induced]] by the norm given by the [[Definition:Modulus of Complex Number|complex modulus]], forms a [[Definition:Complete Metric Space|complete metric space]]. | Let $z = x + iy$ be a [[Definition:Complex Number|complex number]], where $x, y \in \R$.
Now, we can identify the complex number $z$ with the [[Definition:Ordered Pair|ordered pair]] $\left( x, y \right) \in \R^2$.
The norm on $\C$ given by the [[Definition:Modulus of Complex Number|complex modulus]] is then identica... | Complex Plane is Complete Metric Space | https://proofwiki.org/wiki/Complex_Plane_is_Complete_Metric_Space | https://proofwiki.org/wiki/Complex_Plane_is_Complete_Metric_Space | [
"Complete Metric Spaces"
] | [
"Definition:Complex Number",
"Definition:Metric Induced by Norm",
"Definition:Complex Modulus",
"Definition:Complete Metric Space"
] | [
"Definition:Complex Number",
"Definition:Ordered Pair",
"Definition:Complex Modulus",
"Definition:Euclidean Norm",
"Definition:Metric Induced by Norm",
"Definition:Complex Modulus",
"Definition:Euclidean Metric/Real Number Plane",
"Definition:Euclidean Norm",
"Euclidean Space is Complete Metric Spac... |
proofwiki-5063 | Euclidean Space is Complete Metric Space | Let $m$ be a positive integer.
Then the Euclidean space $\R^m$, along with the Euclidean metric, forms a complete metric space. | Let $\sequence {\tuple {x_{n, 1}, x_{n, 2}, \ldots, x_{n, m} } }_{n \mathop \in \N} $ be a Cauchy sequence in $\R^m$.
Let $\epsilon > 0$.
Then $\dfrac \epsilon m > 0$.
We have that Real Number Line is Complete Metric Space.
Therefore:
:$\exists y_1, y_2, \ldots, y_m \in \R$ and $N_1, N_2, \ldots, N_m \in \N$ (depending... | Let $m$ be a [[Definition:Positive Integer|positive integer]].
Then the [[Definition:Euclidean Space|Euclidean space]] $\R^m$, along with the [[Definition:Euclidean Metric on Real Vector Space|Euclidean metric]], forms a [[Definition:Complete Metric Space|complete metric space]]. | Let $\sequence {\tuple {x_{n, 1}, x_{n, 2}, \ldots, x_{n, m} } }_{n \mathop \in \N} $ be a [[Definition:Cauchy Sequence|Cauchy sequence]] in $\R^m$.
Let $\epsilon > 0$.
Then $\dfrac \epsilon m > 0$.
We have that [[Real Number Line is Complete Metric Space]].
Therefore:
:$\exists y_1, y_2, \ldots, y_m \in \R$ and $N... | Euclidean Space is Complete Metric Space | https://proofwiki.org/wiki/Euclidean_Space_is_Complete_Metric_Space | https://proofwiki.org/wiki/Euclidean_Space_is_Complete_Metric_Space | [
"Real Euclidean Spaces",
"Examples of Complete Metric Spaces"
] | [
"Definition:Positive/Integer",
"Definition:Euclidean Space",
"Definition:Euclidean Metric/Real Vector Space",
"Definition:Complete Metric Space"
] | [
"Definition:Cauchy Sequence",
"Real Number Line is Complete Metric Space",
"Euclidean Space is Normed Vector Space",
"Definition:Euclidean Space",
"Definition:Complete Metric Space"
] |
proofwiki-5064 | Complex Plane is Banach Space | The complex plane, along with the complex modulus, forms a Banach space over $\C$. | The complex plane $\C$ is a vector space over $\C$.
That the norm axioms are satisfied is proven in Complex Modulus is Norm.
Then we have Complex Plane is Complete Metric Space.
Hence the result.
{{qed}} | The [[Definition:Complex Plane|complex plane]], along with the [[Definition:Modulus of Complex Number|complex modulus]], forms a [[Definition:Banach Space|Banach space]] over $\C$. | The [[Definition:Complex Plane|complex plane]] $\C$ is a [[Definition:Vector Space|vector space]] over $\C$.
That the [[Axiom:Vector Space Norm Axioms|norm axioms]] are satisfied is proven in [[Complex Modulus is Norm]].
Then we have [[Complex Plane is Complete Metric Space]].
Hence the result.
{{qed}} | Complex Plane is Banach Space | https://proofwiki.org/wiki/Complex_Plane_is_Banach_Space | https://proofwiki.org/wiki/Complex_Plane_is_Banach_Space | [
"Banach Spaces"
] | [
"Definition:Complex Number/Complex Plane",
"Definition:Complex Modulus",
"Definition:Banach Space"
] | [
"Definition:Complex Number/Complex Plane",
"Definition:Vector Space",
"Axiom:Vector Space Norm Axioms",
"Complex Modulus is Norm",
"Complex Plane is Complete Metric Space"
] |
proofwiki-5065 | Rational Number Space is not Complete Metric Space | Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean metric $d$.
Then $\struct {\Q, \tau_d}$ is not a complete metric space. | Proof by Counterexample:
First note that Rational Numbers form Metric Space.
It remains to be shown that $\struct {\Q, \tau_d}$ is not complete.
Consider the sequence $\sequence {a_n}$ given by:
:$a_n := \dfrac {f_{n + 1} } {f_n}$
where $\sequence {f_n}$ is the sequence of Fibonacci numbers.
By Ratio of Consecutive Fib... | Let $\struct {\Q, \tau_d}$ be the [[Definition:Rational Number Space|rational number space]] under the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]] $d$.
Then $\struct {\Q, \tau_d}$ is not a [[Definition:Complete Metric Space|complete metric space]]. | [[Proof by Counterexample]]:
First note that [[Rational Numbers form Metric Space]].
It remains to be shown that $\struct {\Q, \tau_d}$ is not [[Definition:Complete Metric Space|complete]].
Consider the [[Definition:Sequence|sequence]] $\sequence {a_n}$ given by:
:$a_n := \dfrac {f_{n + 1} } {f_n}$
where $\sequence ... | Rational Number Space is not Complete Metric Space | https://proofwiki.org/wiki/Rational_Number_Space_is_not_Complete_Metric_Space | https://proofwiki.org/wiki/Rational_Number_Space_is_not_Complete_Metric_Space | [
"Rational Number Space",
"Complete Metric Spaces"
] | [
"Definition:Rational Number Space",
"Definition:Euclidean Metric/Real Number Line",
"Definition:Complete Metric Space"
] | [
"Proof by Counterexample",
"Rational Numbers form Metric Space",
"Definition:Complete Metric Space",
"Definition:Sequence",
"Definition:Fibonacci Number",
"Ratio of Consecutive Fibonacci Numbers",
"Square Root of Prime is Irrational",
"Definition:Irrational Number",
"Definition:Cauchy Sequence",
"... |
proofwiki-5066 | Limit of Subsequence equals Limit of Sequence/Real Numbers | Let $\sequence {x_n}$ be a sequence in $\R$.
Let $l \in \R$ such that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\ds \lim_{r \mathop \to \infty} x_{n_r} = l$ | Let $\epsilon > 0$.
As $\ds \lim_{n \mathop \to \infty} x_n = l$, it follows that:
:$\exists N: \forall n > N: \size {x_n - l} < \epsilon$
Now let $R = N$.
Then from Strictly Increasing Sequence of Natural Numbers:
:$\forall r > R: n_r \ge r$
Thus $n_r > N$ and so:
:$\size {x_n - l} < \epsilon$
The result follows.
{{qe... | Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
Let $l \in \R$ such that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$.
Then:
:$\ds \lim_{r \mathop \to \infty} x_{n_r} = l$ | Let $\epsilon > 0$.
As $\ds \lim_{n \mathop \to \infty} x_n = l$, it follows that:
:$\exists N: \forall n > N: \size {x_n - l} < \epsilon$
Now let $R = N$.
Then from [[Strictly Increasing Sequence of Natural Numbers]]:
:$\forall r > R: n_r \ge r$
Thus $n_r > N$ and so:
:$\size {x_n - l} < \epsilon$
The result fo... | Limit of Subsequence equals Limit of Sequence/Real Numbers | https://proofwiki.org/wiki/Limit_of_Subsequence_equals_Limit_of_Sequence/Real_Numbers | https://proofwiki.org/wiki/Limit_of_Subsequence_equals_Limit_of_Sequence/Real_Numbers | [
"Convergence",
"Subsequences",
"Real Sequences"
] | [
"Definition:Real Sequence",
"Definition:Subsequence"
] | [
"Strictly Increasing Sequence of Natural Numbers"
] |
proofwiki-5067 | Distance Function of Metric Space is Continuous | Let $M = \struct {A, d}$ be a metric space.
Let $\tau_A$ be the topology on $A$ induced by $d$.
Let $\struct {A \times A, \tau}$ be the product space of $\struct {A, \tau_A}$ and itself.
Then the distance function $d: A \times A \to \R$ is a continuous mapping. | Let $d_\infty: \paren {A \times A} \times \paren {A \times A} \to \R$ be the metric on $A \times A$ defined by:
:$\map {d_\infty} {\tuple {x, y}, \tuple {x', y'} } = \max \set {\map d {x, x'}, \map d {y, y'} }$
By P-Product Metric Induces Product Topology, $\tau$ is the topology on $A \times A$ induced by $d_\infty$.
L... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $\tau_A$ be the [[Definition:Topology Induced by Metric|topology on $A$ induced by $d$]].
Let $\struct {A \times A, \tau}$ be the [[Definition:Product Space (Topology) of Two Factor Spaces|product space]] of $\struct {A, \tau_A}$ and itself.
... | Let $d_\infty: \paren {A \times A} \times \paren {A \times A} \to \R$ be the [[Definition:Metric|metric]] on $A \times A$ defined by:
:$\map {d_\infty} {\tuple {x, y}, \tuple {x', y'} } = \max \set {\map d {x, x'}, \map d {y, y'} }$
By [[P-Product Metric Induces Product Topology]], $\tau$ is the [[Definition:Topology ... | Distance Function of Metric Space is Continuous | https://proofwiki.org/wiki/Distance_Function_of_Metric_Space_is_Continuous | https://proofwiki.org/wiki/Distance_Function_of_Metric_Space_is_Continuous | [
"Continuous Mappings",
"Metric Spaces",
"Product Spaces"
] | [
"Definition:Metric Space",
"Definition:Topology Induced by Metric",
"Definition:Product Space (Topology)/Two Factor Spaces",
"Definition:Distance Function",
"Definition:Continuous Mapping (Topology)/Everywhere"
] | [
"Definition:Metric Space/Metric",
"P-Product Metric Induces Product Topology",
"Definition:Topology Induced by Metric",
"Definition:Strictly Positive/Real Number",
"Triangle Inequality/Real Numbers",
"Reverse Triangle Inequality",
"Definition:Continuous Mapping (Metric Space)",
"Category:Continuous Ma... |
proofwiki-5068 | Sequence Converges to Within Half Limit/Real Numbers | Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be convergent to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} x_n = l$.
Suppose $l > 0$.
Then:
:$\exists N: \forall n > N: x_n > \dfrac l 2$
Similarly, suppose $l < 0$.
Then:
:$\exists N: \forall n > N: x_n < \dfrac l 2$ | Suppose $l > 0$.
From the definition of convergence to a limit:
:$\forall \epsilon > 0: \exists N: \forall n > N: \size {x_n - l} < \epsilon$
That is, $l - \epsilon < x_n < l + \epsilon$.
As this is true for ''all'' $\epsilon > 0$, it is also true for $\epsilon = \dfrac l 2$ for some value of $N$.
Thus:
:$\exists N: \f... | Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
Let $\sequence {x_n}$ be [[Definition:Convergent Real Sequence|convergent]] to the [[Definition:Limit of Real Sequence|limit]] $l$.
That is, let $\ds \lim_{n \mathop \to \infty} x_n = l$.
Suppose $l > 0$.
Then:
:$\exists N: \forall n > N: x_... | Suppose $l > 0$.
From the definition of [[Definition:Convergent Real Sequence|convergence to a limit]]:
:$\forall \epsilon > 0: \exists N: \forall n > N: \size {x_n - l} < \epsilon$
That is, $l - \epsilon < x_n < l + \epsilon$.
As this is true for ''all'' $\epsilon > 0$, it is also true for $\epsilon = \dfrac l 2$ f... | Sequence Converges to Within Half Limit/Real Numbers | https://proofwiki.org/wiki/Sequence_Converges_to_Within_Half_Limit/Real_Numbers | https://proofwiki.org/wiki/Sequence_Converges_to_Within_Half_Limit/Real_Numbers | [
"Limits of Sequences"
] | [
"Definition:Real Sequence",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Limit of Sequence/Real Numbers"
] | [
"Definition:Convergent Sequence/Real Numbers"
] |
proofwiki-5069 | Sequence Converges to Within Half Limit/Complex Numbers | Let $\sequence {z_n}$ be a sequence in $\C$.
Let $\sequence {z_n}$ be convergent to the limit $l$.
That is, let $\ds \lim_{n \mathop \to \infty} z_n = l$ where $l \ne 0$.
Then:
:$\exists N: \forall n > N: \cmod {z_n} > \dfrac {\cmod l} 2$ | Suppose $l > 0$.
Let us choose $N$ such that:
:$\forall n > N: \cmod {z_n - l} < \dfrac {\cmod l} 2$
Then:
{{begin-eqn}}
{{eqn | l = \cmod {z_n - l}
| o = <
| r = \frac {\cmod l} 2
| c =
}}
{{eqn | ll= \leadsto
| l = \cmod l - \cmod {z_n}
| o = \le
| r = \cmod {z_n - l}
| c = ... | Let $\sequence {z_n}$ be a [[Definition:Complex Sequence|sequence in $\C$]].
Let $\sequence {z_n}$ be [[Definition:Convergent Sequence (Analysis)|convergent]] to the [[Definition:Limit of Sequence (Number Field)|limit]] $l$.
That is, let $\ds \lim_{n \mathop \to \infty} z_n = l$ where $l \ne 0$.
Then:
:$\exists N: ... | Suppose $l > 0$.
Let us choose $N$ such that:
:$\forall n > N: \cmod {z_n - l} < \dfrac {\cmod l} 2$
Then:
{{begin-eqn}}
{{eqn | l = \cmod {z_n - l}
| o = <
| r = \frac {\cmod l} 2
| c =
}}
{{eqn | ll= \leadsto
| l = \cmod l - \cmod {z_n}
| o = \le
| r = \cmod {z_n - l}
| ... | Sequence Converges to Within Half Limit/Complex Numbers | https://proofwiki.org/wiki/Sequence_Converges_to_Within_Half_Limit/Complex_Numbers | https://proofwiki.org/wiki/Sequence_Converges_to_Within_Half_Limit/Complex_Numbers | [
"Limits of Sequences"
] | [
"Definition:Complex Sequence",
"Definition:Convergent Sequence/Analysis",
"Definition:Limit of Sequence (Number Field)"
] | [
"Reverse Triangle Inequality",
"Category:Limits of Sequences"
] |
proofwiki-5070 | Number to Reciprocal Power is Decreasing | The real sequence $\sequence {n^{1/n} }$ is decreasing for $n \ge 3$. | We want to show that $\paren {n + 1}^{1 / \paren {n + 1} } \le n^{1/n}$.
Thus:
{{begin-eqn}}
{{eqn | l = \paren {n + 1}^{1 / \paren {n + 1} }
| o = \le
| r = n^{1/n}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {n + 1}^n
| o = \le
| r = n^{n + 1}
| c = raising both sides ... | The [[Definition:Real Sequence|real sequence]] $\sequence {n^{1/n} }$ is [[Definition:Decreasing Real Sequence|decreasing]] for $n \ge 3$. | We want to show that $\paren {n + 1}^{1 / \paren {n + 1} } \le n^{1/n}$.
Thus:
{{begin-eqn}}
{{eqn | l = \paren {n + 1}^{1 / \paren {n + 1} }
| o = \le
| r = n^{1/n}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {n + 1}^n
| o = \le
| r = n^{n + 1}
| c = raising both sides... | Number to Reciprocal Power is Decreasing | https://proofwiki.org/wiki/Number_to_Reciprocal_Power_is_Decreasing | https://proofwiki.org/wiki/Number_to_Reciprocal_Power_is_Decreasing | [
"Real Analysis",
"Reciprocals"
] | [
"Definition:Real Sequence",
"Definition:Decreasing/Sequence/Real Sequence"
] | [
"One Plus Reciprocal to the Nth",
"Definition:Decreasing/Sequence/Real Sequence"
] |
proofwiki-5071 | Inverse of Diagonal Matrix | Let:
:<nowiki>$\mathbf D = \begin{bmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn} \\
\end{bmatrix}$</nowiki>
be an $n \times n$ diagonal matrix.
Then its inverse is given by:
:<nowiki>$\mathbf D^{-1} = \begin{bmatrix}
\dfrac 1 {a_{11}} & 0 & \c... | {{WLOG}}, consider the right inverse of $\mathbf D$.
Suppose none of the diagonal elements are zero.
Then by the definition of inverse, our assertion is that the matrix product of the two matrices in question is the unit matrix of order $n$.
Now, observe that:
{{begin-eqn}}
{{eqn | l = \begin {bmatrix} a_{11} & 0 & \c... | Let:
:<nowiki>$\mathbf D = \begin{bmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn} \\
\end{bmatrix}$</nowiki>
be an $n \times n$ [[Definition:Diagonal Matrix|diagonal matrix]].
Then its [[Definition:Inverse Matrix|inverse]] is given by:
:<no... | {{WLOG}}, consider the [[Definition:Right Inverse Element|right inverse]] of $\mathbf D$.
Suppose none of the [[Definition:Diagonal Element|diagonal elements]] are zero.
Then by the definition of [[Definition:Inverse Matrix|inverse]], our assertion is that the [[Definition:Matrix Product (Conventional)|matrix produc... | Inverse of Diagonal Matrix | https://proofwiki.org/wiki/Inverse_of_Diagonal_Matrix | https://proofwiki.org/wiki/Inverse_of_Diagonal_Matrix | [
"Diagonal Matrices",
"Inverse Matrices"
] | [
"Definition:Diagonal Matrix",
"Definition:Inverse Matrix",
"Definition:Main Diagonal/Diagonal Elements",
"Definition:Main Diagonal/Diagonal Elements",
"Definition:Singular Matrix",
"Definition:Unit of Ring"
] | [
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Definition:Main Diagonal/Diagonal Elements",
"Definition:Inverse Matrix",
"Definition:Matrix Product (Conventional)",
"Definition:Unit Matrix",
"Definition:Main Diagonal/Diagonal Elements",
"Definition:Determinant/Matrix",
"Determinant of Inverse... |
proofwiki-5072 | Subspace of Complete Metric Space is Closed iff Complete | Let $\struct {M, d}$ be a complete metric space.
Let $\struct {S, d}$ be a subspace of $\struct {M, d}$.
Then $S$ is closed {{iff}} $S$ is complete. | This will be proved by demonstrating the contrapositive:
:$S$ is not complete {{iff}} $S$ is not closed. | Let $\struct {M, d}$ be a [[Definition:Complete Metric Space|complete metric space]].
Let $\struct {S, d}$ be a [[Definition:Metric Subspace|subspace]] of $\struct {M, d}$.
Then $S$ is [[Definition:Closed Set (Metric Space)|closed]] {{iff}} $S$ is [[Definition:Complete Metric Space|complete]]. | This will be proved by demonstrating the [[Definition:Contrapositive Statement|contrapositive]]:
:$S$ is not [[Definition:Complete Metric Space|complete]] {{iff}} $S$ is not [[Definition:Closed Set (Metric Space)|closed]]. | Subspace of Complete Metric Space is Closed iff Complete | https://proofwiki.org/wiki/Subspace_of_Complete_Metric_Space_is_Closed_iff_Complete | https://proofwiki.org/wiki/Subspace_of_Complete_Metric_Space_is_Closed_iff_Complete | [
"Metric Subspaces",
"Complete Metric Spaces"
] | [
"Definition:Complete Metric Space",
"Definition:Metric Subspace",
"Definition:Closed Set/Metric Space",
"Definition:Complete Metric Space"
] | [
"Definition:Contrapositive Statement",
"Definition:Complete Metric Space",
"Definition:Closed Set/Metric Space",
"Definition:Complete Metric Space",
"Definition:Complete Metric Space",
"Definition:Closed Set/Metric Space",
"Definition:Complete Metric Space",
"Definition:Complete Metric Space"
] |
proofwiki-5073 | Set of Chains is Closed under Chain Unions under Subset Relation | Let $\struct {P, \le}$ be an ordered set.
Let $\map {\operatorname {ch} } P$ be the set of chains in $P$.
Then $\map {\operatorname {ch} } P$ is closed under chain unions.
That is, if $\CC$ is a chain in $\map {\operatorname {ch} } P$ with respect to the subset ordering, the union $\ds \bigcup \CC$ of $\CC$ is also a c... | Let $\ds D = \bigcup \CC$.
Observe that any $C \in \CC$ is a chain in $P$, hence $C \subseteq P$.
Therefore, $D \subseteq P$, by Set Union Preserves Subsets.
Now suppose that $a, b \in D$.
By definition of union, there exist $A, B \in \CC$ such that $a \in A$ and $b \in B$.
Since $\CC$ is a chain in $\struct {\map {\... | Let $\struct {P, \le}$ be an [[Definition:Ordered Set|ordered set]].
Let $\map {\operatorname {ch} } P$ be the [[Definition:Set of Sets|set]] of [[Definition:Chain (Order Theory)|chains]] in $P$.
Then $\map {\operatorname {ch} } P$ is [[Definition:Closure under Chain Unions|closed under chain unions]].
That is, if ... | Let $\ds D = \bigcup \CC$.
Observe that any $C \in \CC$ is a [[Definition:Chain of Sets|chain]] in $P$, hence $C \subseteq P$.
Therefore, $D \subseteq P$, by [[Set Union Preserves Subsets]].
Now suppose that $a, b \in D$.
By definition of [[Definition:Set Union|union]], there exist $A, B \in \CC$ such that $a \in... | Set of Chains is Closed under Chain Unions under Subset Relation | https://proofwiki.org/wiki/Set_of_Chains_is_Closed_under_Chain_Unions_under_Subset_Relation | https://proofwiki.org/wiki/Set_of_Chains_is_Closed_under_Chain_Unions_under_Subset_Relation | [
"Order Theory",
"Closure under Chain Unions",
"Subset Relation"
] | [
"Definition:Ordered Set",
"Definition:Set of Sets",
"Definition:Chain (Order Theory)",
"Definition:Closure under Chain Unions",
"Definition:Chain (Order Theory)/Subset Relation",
"Subset Relation is Ordering",
"Definition:Set Union",
"Definition:Chain (Order Theory)"
] | [
"Definition:Chain (Order Theory)/Subset Relation",
"Set Union Preserves Subsets",
"Definition:Set Union",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Chain (Order Theory)",
"Definition:Totally Ordered Set",
"Definition:Chain (Order Theory)",
"Category:Order Theory",
"Category:Clos... |
proofwiki-5074 | Jensen's Inequality (Real Analysis) | Let $I$ be a real interval.
Let $\phi: I \to \R$ be a convex function.
Let $x_1, x_2, \ldots, x_n \in I$.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$ be real numbers, at least one of which is non-zero.
Then:
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \... | The proof proceeds by mathematical induction on $n$.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lambda_k \map \phi {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
$\map P 1... | Let $I$ be a [[Definition:Real Interval|real interval]].
Let $\phi: I \to \R$ be a [[Definition:Convex Real Function|convex function]].
Let $x_1, x_2, \ldots, x_n \in I$.
Let $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$ be [[Definition:Real Number|real numbers]], at least one of which is non-zero.
Then:
:$\ds \... | The proof proceeds by [[Principle of Mathematical Induction|mathematical induction]] on $n$.
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \map \phi {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \le \frac {\sum_{k \mathop = 1}^n \lamb... | Jensen's Inequality (Real Analysis) | https://proofwiki.org/wiki/Jensen's_Inequality_(Real_Analysis) | https://proofwiki.org/wiki/Jensen's_Inequality_(Real_Analysis) | [
"Real Analysis",
"Inequalities",
"Jensen's Inequality (Real Analysis)"
] | [
"Definition:Real Interval",
"Definition:Convex Real Function",
"Definition:Real Number",
"Definition:Strictly Convex Real Function"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Principle of Mathematical Induction"
] |
proofwiki-5075 | Inequality of Hölder Means | Let $p, q \in \overline \R$ be extended real numbers such that $p < q$.
Let $x_1, x_2, \ldots, x_n \ge 0$ be real numbers.
If $p < 0$, then we require that $x_1, x_2, \ldots, x_n > 0$.
Then the Hölder mean satisfies the inequality:
:$\map {M_p} {x_1, x_2, \ldots, x_n} \le \map {M_q} {x_1, x_2, \ldots, x_n}$
Equality ho... | For real $p \ne 0$, the Hölder mean is defined as:
:$\ds \map {M_p} {x_1, x_2, \ldots, x_n} = \paren {\frac 1 n \sum_{k \mathop = 1}^n {x_k}^p}^{1 / p}$
whenever the above expression is defined.
First we note that by definition of Hölder mean with $p = \infty$:
:$\map {M_\infty} {x_1, x_2, \ldots, x_n} = \max \set {x_1... | Let $p, q \in \overline \R$ be [[Definition:Extended Real Number Line|extended real numbers]] such that $p < q$.
Let $x_1, x_2, \ldots, x_n \ge 0$ be [[Definition:Real Number|real numbers]].
If $p < 0$, then we require that $x_1, x_2, \ldots, x_n > 0$.
Then the [[Definition:Hölder Mean|Hölder mean]] satisfies the i... | For [[Definition:Real Number|real]] $p \ne 0$, the [[Definition:Hölder Mean|Hölder mean]] is defined as:
:$\ds \map {M_p} {x_1, x_2, \ldots, x_n} = \paren {\frac 1 n \sum_{k \mathop = 1}^n {x_k}^p}^{1 / p}$
whenever the above expression is defined.
First we note that by definition of [[Definition:Hölder Mean with Pos... | Inequality of Hölder Means | https://proofwiki.org/wiki/Inequality_of_Hölder_Means | https://proofwiki.org/wiki/Inequality_of_Hölder_Means | [
"Inequalities",
"Hölder Mean"
] | [
"Definition:Extended Real Number Line",
"Definition:Real Number",
"Definition:Hölder Mean"
] | [
"Definition:Real Number",
"Definition:Hölder Mean",
"Definition:Hölder Mean/Positive Infinite Exponent",
"Limit of Hölder Mean as Exponent tends to Infinity",
"Maximum is Greater than or Equal to Hölder Mean",
"Definition:Hölder Mean/Negative Infinite Exponent",
"Limit of Hölder Mean as Exponent tends t... |
proofwiki-5076 | Real Linear Subspace Contains Zero Vector | Let $\mathbb W \subseteq \R^n$ such that $\mathbb W$ is a linear subspace of $\R^n$.
Then $\mathbb W$ contains the zero vector:
:$\mathbf 0_{n \times 1} = \begin{bmatrix}
0 \\ 0 \\ \vdots \\ 0
\end{bmatrix} \in \mathbb W$ | This is a consequence of Vector Subspace of Real Vector Space.
{{qed}} | Let $\mathbb W \subseteq \R^n$ such that $\mathbb W$ is a [[Definition:Vector Subspace|linear subspace of $\R^n$]].
Then $\mathbb W$ contains the [[Definition:Zero Vector|zero vector]]:
:$\mathbf 0_{n \times 1} = \begin{bmatrix}
0 \\ 0 \\ \vdots \\ 0
\end{bmatrix} \in \mathbb W$ | This is a consequence of [[Vector Subspace of Real Vector Space]].
{{qed}} | Real Linear Subspace Contains Zero Vector | https://proofwiki.org/wiki/Real_Linear_Subspace_Contains_Zero_Vector | https://proofwiki.org/wiki/Real_Linear_Subspace_Contains_Zero_Vector | [
"Linear Algebra"
] | [
"Definition:Vector Subspace",
"Definition:Zero Vector"
] | [
"Vector Subspace of Real Vector Space"
] |
proofwiki-5077 | Zero Subspace is Subspace | Let $V$ be a vector space over $K$ with zero vector $\mathbf 0$.
The zero subspace $\set {\mathbf 0}$ is a subspace of $V$. | We use the Two-Step Vector Subspace Test.
$\set {\mathbf 0}$ is not empty, because it contains $\mathbf 0$.
$\set {\mathbf 0}$ is closed under $+$ because:
:$\forall \mathbf x, \mathbf y \in \set {\mathbf 0}, \mathbf x + \mathbf y = \mathbf 0 + \mathbf 0 = \mathbf 0 \in \set {\mathbf 0}$
$\set {\mathbf 0}$ is closed un... | Let $V$ be a [[Definition:Vector Space|vector space]] over $K$ with [[Definition:Zero Vector|zero vector]] $\mathbf 0$.
The [[Definition:Zero Subspace|zero subspace]] $\set {\mathbf 0}$ is a [[Definition:Vector Subspace|subspace]] of $V$. | We use the [[Two-Step Vector Subspace Test]].
$\set {\mathbf 0}$ is [[Definition:Non-Empty Set|not empty]], because it contains $\mathbf 0$.
$\set {\mathbf 0}$ is [[Definition:Closure (Abstract Algebra)|closed]] under $+$ because:
:$\forall \mathbf x, \mathbf y \in \set {\mathbf 0}, \mathbf x + \mathbf y = \mathbf ... | Zero Subspace is Subspace | https://proofwiki.org/wiki/Zero_Subspace_is_Subspace | https://proofwiki.org/wiki/Zero_Subspace_is_Subspace | [
"Linear Algebra"
] | [
"Definition:Vector Space",
"Definition:Zero Vector",
"Definition:Zero Subspace",
"Definition:Vector Subspace"
] | [
"Two-Step Vector Subspace Test",
"Definition:Non-Empty Set",
"Definition:Closure (Abstract Algebra)",
"Definition:Closure (Abstract Algebra)",
"Two-Step Vector Subspace Test"
] |
proofwiki-5078 | Cartesian Product of Natural Numbers with Itself is Countable | The Cartesian product $\N \times \N$ of the set of natural numbers $\N$ with itself is countable. | This is simply a special case of Cartesian Product of Countable Sets is Countable.
{{qed}} | The [[Definition:Cartesian Product|Cartesian product]] $\N \times \N$ of the set of [[Definition:Natural Numbers|natural numbers]] $\N$ with itself is [[Definition:Countable|countable]]. | This is simply a special case of [[Cartesian Product of Countable Sets is Countable]].
{{qed}} | Cartesian Product of Natural Numbers with Itself is Countable | https://proofwiki.org/wiki/Cartesian_Product_of_Natural_Numbers_with_Itself_is_Countable | https://proofwiki.org/wiki/Cartesian_Product_of_Natural_Numbers_with_Itself_is_Countable | [
"Set Theory",
"Cartesian Product",
"Countable Sets"
] | [
"Definition:Cartesian Product",
"Definition:Natural Numbers",
"Definition:Countable Set"
] | [
"Cartesian Product of Countable Sets is Countable"
] |
proofwiki-5079 | Minkowski's Inequality for Sums | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.
Let $p \in \R$, $p \ne 0$ be a real number.
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be strictly positive.
If $p > 1$, then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{... | === Proof for $p = 2$ ===
$p = 2$ is an easily proved special case:
{{:Minkowski's Inequality for Sums/Index 2}}{{qed|lemma}} | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be [[Definition:Non-Negative Real Number|non-negative real numbers]].
Let $p \in \R$, $p \ne 0$ be a [[Definition:Real Number|real number]].
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be [[Definition:Strictly Positi... | === [[Minkowski's Inequality for Sums/Index 2|Proof for $p = 2$]] ===
$p = 2$ is an easily proved special case:
{{:Minkowski's Inequality for Sums/Index 2}}{{qed|lemma}} | Minkowski's Inequality for Sums | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums | [
"Minkowski's Inequality for Sums",
"Minkowski's Inequality",
"Algebra",
"Analysis"
] | [
"Definition:Positive/Real Number",
"Definition:Real Number",
"Definition:Strictly Positive/Real Number"
] | [
"Minkowski's Inequality for Sums/Index 2"
] |
proofwiki-5080 | Minkowski's Inequality for Sums | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.
Let $p \in \R$, $p \ne 0$ be a real number.
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be strictly positive.
If $p > 1$, then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{... | Define:
:$q = \dfrac p {p - 1}$
Then:
:$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
It follows that:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \paren {a_k + b_k}^p
| r = \sum_{k \mathop = 1}^n a_k \paren {a_k + b_k}^{p - 1} + \sum_{k \mathop = 1}^n b_k \paren {a_k + b_k}^{p - 1}
| c ... | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be [[Definition:Non-Negative Real Number|non-negative real numbers]].
Let $p \in \R$, $p \ne 0$ be a [[Definition:Real Number|real number]].
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be [[Definition:Strictly Positi... | Define:
:$q = \dfrac p {p - 1}$
Then:
:$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
It follows that:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \paren {a_k + b_k}^p
| r = \sum_{k \mathop = 1}^n a_k \paren {a_k + b_k}^{p - 1} + \sum_{k \mathop = 1}^n b_k \paren {a_k + b_k}^{p - 1}
|... | Minkowski's Inequality for Sums/Index Greater than 1/Proof 1 | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums/Index_Greater_than_1/Proof_1 | [
"Minkowski's Inequality for Sums",
"Minkowski's Inequality",
"Algebra",
"Analysis"
] | [
"Definition:Positive/Real Number",
"Definition:Real Number",
"Definition:Strictly Positive/Real Number"
] | [
"Hölder's Inequality for Sums",
"Exponent Combination Laws/Power of Power"
] |
proofwiki-5081 | Minkowski's Inequality for Sums | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be non-negative real numbers.
Let $p \in \R$, $p \ne 0$ be a real number.
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be strictly positive.
If $p > 1$, then:
:$\ds \paren {\sum_{k \mathop = 1}^n \paren {a_k + b_k}^p}^{... | Let $\mathbf a$ and $\mathbf b$ be real finite sequences:
{{begin-eqn}}
{{eqn | l = \mathbf a
| r = \sequence {a_k}_{1 \mathop \le k \mathop \le n}
}}
{{eqn | l = \mathbf b
| r = \sequence {b_k}_{1 \mathop \le k \mathop \le n}
}}
{{end-eqn}}
Let $\norm {\mathbf a}_p$ denote the $p$-norm of $\mathbf a$:
:$\n... | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in \R_{\ge 0}$ be [[Definition:Non-Negative Real Number|non-negative real numbers]].
Let $p \in \R$, $p \ne 0$ be a [[Definition:Real Number|real number]].
If $p < 0$, then we require that $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be [[Definition:Strictly Positi... | Let $\mathbf a$ and $\mathbf b$ be [[Definition:Real Sequence|real]] [[Definition:Finite Sequence|finite sequences]]:
{{begin-eqn}}
{{eqn | l = \mathbf a
| r = \sequence {a_k}_{1 \mathop \le k \mathop \le n}
}}
{{eqn | l = \mathbf b
| r = \sequence {b_k}_{1 \mathop \le k \mathop \le n}
}}
{{end-eqn}}
Let ... | Minkowski's Inequality for Sums/Index Greater than 1/Proof 2 | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Sums/Index_Greater_than_1/Proof_2 | [
"Minkowski's Inequality for Sums",
"Minkowski's Inequality",
"Algebra",
"Analysis"
] | [
"Definition:Positive/Real Number",
"Definition:Real Number",
"Definition:Strictly Positive/Real Number"
] | [
"Definition:Real Sequence",
"Definition:Finite Sequence",
"Definition:P-Norm/Real",
"Hölder's Inequality for Sums",
"Transformation of P-Norm"
] |
proofwiki-5082 | Minkowski's Inequality for Integrals | Let $f, g$ be (Darboux) integrable functions.
Let $p \in \R$ such that $p > 1$.
Then:
:$\ds \paren {\int_a^b \size {\map f x + \map g x}^p \rd x}^{1/p} \le \paren {\int_a^b \size {\map f x}^p \rd x}^{1 / p} + \paren {\int_a^b \size {\map g x}^p \rd x}^{1 / p}$ | Define:
:$q = \dfrac p {p - 1}$
Then:
:$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
It follows that:
{{begin-eqn}}
{{eqn | l = \int_a^b \size {\map f x + \map g x}^p \rd x
| r = \int_a^b \size {\map f x} \size {\map f x + \map g x}^{p - 1} \rd x + \int_a^b \size {\map g x} \size {\map f x + \map ... | Let $f, g$ be [[Definition:Darboux Integrable Function|(Darboux) integrable functions]].
Let $p \in \R$ such that $p > 1$.
Then:
:$\ds \paren {\int_a^b \size {\map f x + \map g x}^p \rd x}^{1/p} \le \paren {\int_a^b \size {\map f x}^p \rd x}^{1 / p} + \paren {\int_a^b \size {\map g x}^p \rd x}^{1 / p}$ | Define:
:$q = \dfrac p {p - 1}$
Then:
:$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p - 1} p = 1$
It follows that:
{{begin-eqn}}
{{eqn | l = \int_a^b \size {\map f x + \map g x}^p \rd x
| r = \int_a^b \size {\map f x} \size {\map f x + \map g x}^{p - 1} \rd x + \int_a^b \size {\map g x} \size {\map f x + \m... | Minkowski's Inequality for Integrals | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Integrals | https://proofwiki.org/wiki/Minkowski's_Inequality_for_Integrals | [
"Minkowski's Inequality for Integrals",
"Minkowski's Inequality",
"Integral Calculus"
] | [
"Definition:Darboux Integrable Function"
] | [
"Hölder's Inequality for Integrals",
"Exponent Combination Laws/Power of Power"
] |
proofwiki-5083 | Characteristic Function Determined by 1-Fiber | Let $A \subseteq S$.
Let $f:S \to \left\{{0, 1}\right\}$ be a mapping.
Denote by $\chi_A$ the characteristic function on $A$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f {{=}} \chi_A$}}
{{item|(2):|$\forall s \in S: \map f s {{=}} 1 \iff s \in A$}}
{{end-itemize}}
Using the notion of a fiber, $(2)$ may also be expressed ... | === $(1)$ implies $(2)$ ===
Follows directly from the definition of characteristic function.
{{qed|lemma}} | Let $A \subseteq S$.
Let $f:S \to \left\{{0, 1}\right\}$ be a [[Definition:Mapping|mapping]].
Denote by $\chi_A$ the [[Definition:Characteristic Function of Set|characteristic function]] on $A$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$f {{=}} \chi_A$}}
{{item|(2):|$\forall s \in S: \map f s {{=}} 1 \iff s \in A$}}
... | === $(1)$ implies $(2)$ ===
Follows directly from the definition of [[Definition:Characteristic Function of Set|characteristic function]].
{{qed|lemma}} | Characteristic Function Determined by 1-Fiber | https://proofwiki.org/wiki/Characteristic_Function_Determined_by_1-Fiber | https://proofwiki.org/wiki/Characteristic_Function_Determined_by_1-Fiber | [
"Characteristic Functions"
] | [
"Definition:Mapping",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Preimage/Mapping/Element"
] | [
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Characteristic Function (Set Theory)/Set"
] |
proofwiki-5084 | Characteristic Function of Set Difference | Let $A, B \subseteq S$.
Then:
:$\chi_{A \mathop \setminus B} = \chi_A - \chi_{A \cap B}$
where:
:$A \setminus B$ denotes set difference
:$\chi$ denotes characteristic function. | Suppose that $\map {\chi_{A \mathop \setminus B} } s = 1$.
Then by definition of characteristic function:
:$s \in A \setminus B$
That is, by definition of set difference, $s \in A$ and $s \notin B$; in particular, $s \notin A \cap B$.
Hence:
:$\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$
Conclude that:
:$\m... | Let $A, B \subseteq S$.
Then:
:$\chi_{A \mathop \setminus B} = \chi_A - \chi_{A \cap B}$
where:
:$A \setminus B$ denotes [[Definition:Set Difference|set difference]]
:$\chi$ denotes [[Definition:Characteristic Function of Set|characteristic function]]. | Suppose that $\map {\chi_{A \mathop \setminus B} } s = 1$.
Then by definition of [[Definition:Characteristic Function of Set|characteristic function]]:
:$s \in A \setminus B$
That is, by definition of [[Definition:Set Difference|set difference]], $s \in A$ and $s \notin B$; in particular, $s \notin A \cap B$.
Hence:... | Characteristic Function of Set Difference | https://proofwiki.org/wiki/Characteristic_Function_of_Set_Difference | https://proofwiki.org/wiki/Characteristic_Function_of_Set_Difference | [
"Characteristic Functions"
] | [
"Definition:Set Difference",
"Definition:Characteristic Function (Set Theory)/Set"
] | [
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Set Difference",
"Definition:Set Intersection",
"Definition:Set Difference",
"Characteristic Function Determined by 1-Fiber"
] |
proofwiki-5085 | Characteristic Function of Symmetric Difference | Let $A, B \subseteq S$.
Then:
:$\chi_{A \symdif B} = \chi_A + \chi_B - 2 \chi_{A \cap B}$
where:
:$\chi$ denotes characteristic function
:$\symdif$ denotes symmetric difference. | By definition of symmetric difference:
:$A \symdif B = \paren {A \cup B} \setminus \paren {A \cap B}$
Thus:
:$\chi_{A \symdif B} = \chi_{A \mathop \cup B} - \chi_{\paren {A \mathop \cup B} \mathop \cap \paren {A \mathop \cap B} }$
by Characteristic Function of Set Difference.
But by Intersection is Subset of Union and ... | Let $A, B \subseteq S$.
Then:
:$\chi_{A \symdif B} = \chi_A + \chi_B - 2 \chi_{A \cap B}$
where:
:$\chi$ denotes [[Definition:Characteristic Function of Set|characteristic function]]
:$\symdif$ denotes [[Definition:Symmetric Difference|symmetric difference]]. | By definition of [[Definition:Symmetric Difference/Definition 2|symmetric difference]]:
:$A \symdif B = \paren {A \cup B} \setminus \paren {A \cap B}$
Thus:
:$\chi_{A \symdif B} = \chi_{A \mathop \cup B} - \chi_{\paren {A \mathop \cup B} \mathop \cap \paren {A \mathop \cap B} }$
by [[Characteristic Function of Set Di... | Characteristic Function of Symmetric Difference | https://proofwiki.org/wiki/Characteristic_Function_of_Symmetric_Difference | https://proofwiki.org/wiki/Characteristic_Function_of_Symmetric_Difference | [
"Characteristic Functions"
] | [
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Symmetric Difference"
] | [
"Definition:Symmetric Difference/Definition 2",
"Characteristic Function of Set Difference",
"Intersection is Subset of Union",
"Intersection with Subset is Subset",
"Characteristic Function of Union/Variant 2"
] |
proofwiki-5086 | Reverse Young's Inequality for Products | Let $p, q \in \R_{> 0}$ be strictly positive real numbers satisfying:
:$\dfrac 1 p - \dfrac 1 q = 1$
Let $a \in \R_{\ge 0}$ be a positive real number.
Let $b \in \R_{> 0}$ be a strictly positive real number.
Then:
:$a b \ge \dfrac {a^p} p - \dfrac {b^{-q} } q$ | Define:
{{begin-eqn}}
{{eqn | n = 1
| l = u
| r = \frac 1 p
}}
{{eqn | l = v
| r = \frac q p
}}
{{eqn | l = x
| r = \paren {a b}^p
}}
{{eqn | l = y
| r = b^{-p}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \frac 1 u + \frac 1 v
| r = p + \frac p q
}}
{{eqn | r = p \paren {1 + \frac... | Let $p, q \in \R_{> 0}$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]] satisfying:
:$\dfrac 1 p - \dfrac 1 q = 1$
Let $a \in \R_{\ge 0}$ be a [[Definition:Positive Real Number|positive real number]].
Let $b \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|strictly positive... | Define:
{{begin-eqn}}
{{eqn | n = 1
| l = u
| r = \frac 1 p
}}
{{eqn | l = v
| r = \frac q p
}}
{{eqn | l = x
| r = \paren {a b}^p
}}
{{eqn | l = y
| r = b^{-p}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \frac 1 u + \frac 1 v
| r = p + \frac p q
}}
{{eqn | r = p \paren {1 + \... | Reverse Young's Inequality for Products | https://proofwiki.org/wiki/Reverse_Young's_Inequality_for_Products | https://proofwiki.org/wiki/Reverse_Young's_Inequality_for_Products | [
"Young's Inequality for Products"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Positive/Real Number",
"Definition:Strictly Positive/Real Number"
] | [
"Young's Inequality for Products"
] |
proofwiki-5087 | Reverse Hölder's Inequality for Sums | Let $p, q \in \R_{>0}$ be strictly positive real numbers such that $\dfrac 1 p - \dfrac 1 q = 1$.
Suppose that the sequences $\mathbf x = \sequence {x_n}$ and $\mathbf y = \sequence {y_n}$ in $\C$ (or $\R$) are such that the series
:$\ds \paren {\sum_{n \mathop = 1}^\infty \size {x_n}^p}^{1/p}$
and
:$\ds \paren {\sum_{... | {{WLOG}}, assume that $\mathbf x$ and $\mathbf y$ are non-zero.
Let:
:$\mathbf u = \sequence {u_n} = \dfrac {\mathbf x} {\norm {\mathbf x}_p}$
and:
:$\mathbf v = \sequence {v_n} = \dfrac {\mathbf y} {\norm {\mathbf y}_{-q} }$
Then:
:$\ds \norm {\mathbf u}_p = \dfrac 1 {\norm {\mathbf x}_p} \paren {\sum_{n \mathop = 1}^... | Let $p, q \in \R_{>0}$ be [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real numbers]] such that $\dfrac 1 p - \dfrac 1 q = 1$.
Suppose that the [[Definition:Sequence|sequences]] $\mathbf x = \sequence {x_n}$ and $\mathbf y = \sequence {y_n}$ in [[Definition:Complex Number|$\C$]] (or [[De... | {{WLOG}}, assume that $\mathbf x$ and $\mathbf y$ are non-zero.
Let:
:$\mathbf u = \sequence {u_n} = \dfrac {\mathbf x} {\norm {\mathbf x}_p}$
and:
:$\mathbf v = \sequence {v_n} = \dfrac {\mathbf y} {\norm {\mathbf y}_{-q} }$
Then:
:$\ds \norm {\mathbf u}_p = \dfrac 1 {\norm {\mathbf x}_p} \paren {\sum_{n \mathop = 1... | Reverse Hölder's Inequality for Sums | https://proofwiki.org/wiki/Reverse_Hölder's_Inequality_for_Sums | https://proofwiki.org/wiki/Reverse_Hölder's_Inequality_for_Sums | [
"Functional Analysis"
] | [
"Definition:Strictly Positive",
"Definition:Real Number",
"Definition:Sequence",
"Definition:Complex Number",
"Definition:Real Number",
"Definition:Series",
"Definition:Convergent Series",
"Definition:Norm/Vector Space",
"Definition:P-Norm",
"Definition:Lebesgue Space"
] | [
"Reverse Young's Inequality for Products"
] |
proofwiki-5088 | Real Vector Space is Vector Space | Let $\R$ be the set of real numbers.
Then the real vector space $\R^n$ is a vector space. | === Construction of Real Vector Space ===
From the definition, a vector space is a unitary module whose scalar ring is a field.
In order to call attention to the precise scope of the operators, let real addition and real multiplication be expressed as $+_\R$ and $\times_\R$ respectively.
Then we can express the field o... | Let $\R$ be the set of [[Definition:Real Number|real numbers]].
Then the [[Definition:Real Vector Space|real vector space $\R^n$]] is a [[Definition:Vector Space|vector space]]. | === Construction of Real Vector Space ===
From the definition, a [[Definition:Vector Space|vector space]] is a [[Definition:Unitary Module|unitary module]] whose [[Definition:Scalar Ring of Unitary Module|scalar ring]] is a [[Definition:Field (Abstract Algebra)|field]].
In order to call attention to the precise scop... | Real Vector Space is Vector Space | https://proofwiki.org/wiki/Real_Vector_Space_is_Vector_Space | https://proofwiki.org/wiki/Real_Vector_Space_is_Vector_Space | [
"Examples of Vector Spaces",
"Real Numbers",
"Analytic Geometry"
] | [
"Definition:Real Number",
"Definition:Real Vector Space",
"Definition:Vector Space"
] | [
"Definition:Vector Space",
"Definition:Unitary Module over Ring",
"Definition:Scalar Ring/Unitary Module",
"Definition:Field (Abstract Algebra)",
"Definition:Addition/Real Numbers",
"Definition:Multiplication/Real Numbers",
"Definition:Field of Real Numbers",
"Real Numbers under Addition form Group",
... |
proofwiki-5089 | Symmetric Difference is Subset of Union of Symmetric Differences | Let $R, S, T$ be sets.
Then:
:$R \symdif S \subseteq \paren {R \symdif T} \cup \paren {S \symdif T}$
where $R \symdif S$ denotes the symmetric difference between $R$ and $S$. | From the definition of symmetric difference, we have:
:$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$
Then from Set Difference is Subset of Union of Differences, we have:
:$R \setminus S \subseteq \paren {R \setminus T} \cup \paren {T \setminus S}$
:$S \setminus R \subseteq \paren {S \setminus T} \c... | Let $R, S, T$ be [[Definition:Set|sets]].
Then:
:$R \symdif S \subseteq \paren {R \symdif T} \cup \paren {S \symdif T}$
where $R \symdif S$ denotes the [[Definition:Symmetric Difference|symmetric difference]] between $R$ and $S$. | From the definition of [[Definition:Symmetric Difference|symmetric difference]], we have:
:$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$
Then from [[Set Difference is Subset of Union of Differences]], we have:
:$R \setminus S \subseteq \paren {R \setminus T} \cup \paren {T \setminus S}$
:$S \se... | Symmetric Difference is Subset of Union of Symmetric Differences | https://proofwiki.org/wiki/Symmetric_Difference_is_Subset_of_Union_of_Symmetric_Differences | https://proofwiki.org/wiki/Symmetric_Difference_is_Subset_of_Union_of_Symmetric_Differences | [
"Symmetric Difference",
"Set Union"
] | [
"Definition:Set",
"Definition:Symmetric Difference"
] | [
"Definition:Symmetric Difference",
"Set Difference is Subset of Union of Differences",
"Set Union Preserves Subsets",
"Union is Commutative",
"Category:Symmetric Difference",
"Category:Set Union"
] |
proofwiki-5090 | Sigma-Algebra Contains Empty Set | Let $X$ be a set.
Let $\Sigma$ be a $\sigma$-algebra on $X$.
Then:
:$\O \in \Sigma$ | Axiom $(1)$ of a $\sigma$-algebra grants:
:$X \in \Sigma$
By axiom $(2)$ and Set Difference with Self is Empty Set, it follows that:
:$\O = X \setminus X \in \Sigma$
{{qed}} | Let $X$ be a [[Definition:Set|set]].
Let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
Then:
:$\O \in \Sigma$ | Axiom $(1)$ of a [[Definition:Sigma-Algebra|$\sigma$-algebra]] grants:
:$X \in \Sigma$
By axiom $(2)$ and [[Set Difference with Self is Empty Set]], it follows that:
:$\O = X \setminus X \in \Sigma$
{{qed}} | Sigma-Algebra Contains Empty Set | https://proofwiki.org/wiki/Sigma-Algebra_Contains_Empty_Set | https://proofwiki.org/wiki/Sigma-Algebra_Contains_Empty_Set | [
"Sigma-Algebras"
] | [
"Definition:Set",
"Definition:Sigma-Algebra"
] | [
"Definition:Sigma-Algebra",
"Set Difference with Self is Empty Set"
] |
proofwiki-5091 | Sigma-Algebra Closed under Union | Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $A, B \in \Sigma$ be measurable sets.
Then $A \cup B \in \Sigma$, where $\cup$ denotes set union. | Define $A_1 = A, A_2 = B$, and for $n \in \N, n \ge 2: A_n = \O$.
Then by Sigma-Algebra Contains Empty Set, axiom $(3)$ of a $\sigma$-algebra applies.
Hence:
:$\ds \bigcup_{n \mathop \in \N} A_n = A \cup B \in \Sigma$
{{qed}} | Let $X$ be a [[Definition:Set|set]], and let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
Let $A, B \in \Sigma$ be [[Definition:Measurable Set|measurable sets]].
Then $A \cup B \in \Sigma$, where $\cup$ denotes [[Definition:Set Union|set union]]. | Define $A_1 = A, A_2 = B$, and for $n \in \N, n \ge 2: A_n = \O$.
Then by [[Sigma-Algebra Contains Empty Set]], axiom $(3)$ of a [[Definition:Sigma-Algebra|$\sigma$-algebra]] applies.
Hence:
:$\ds \bigcup_{n \mathop \in \N} A_n = A \cup B \in \Sigma$
{{qed}} | Sigma-Algebra Closed under Union | https://proofwiki.org/wiki/Sigma-Algebra_Closed_under_Union | https://proofwiki.org/wiki/Sigma-Algebra_Closed_under_Union | [
"Sigma-Algebras"
] | [
"Definition:Set",
"Definition:Sigma-Algebra",
"Definition:Measurable Set",
"Definition:Set Union"
] | [
"Sigma-Algebra Contains Empty Set",
"Definition:Sigma-Algebra"
] |
proofwiki-5092 | Sigma-Algebra Closed under Countable Intersection | Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Suppose that $\sequence {E_n}_{n \mathop \in \N} \in \Sigma$ is a collection of measurable sets.
Then:
:$\ds \bigcap_{n \mathop \in \N} E_n \in \Sigma$,
where $\ds \bigcap$ denotes set intersection. | {{begin-eqn}}
{{eqn | q = \forall n \in \N
| l = E_n
| o = \in
| r = \Sigma
}}
{{eqn | ll= \leadsto
| l = X \setminus E_n
| o = \in
| r = \Sigma
| c = Axiom $(2)$ for $\sigma$-algebras
}}
{{eqn | ll= \leadsto
| l = \bigcup_{n \mathop \in \N} \paren {X \setminus E_n}
... | Let $X$ be a [[Definition:Set|set]], and let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
Suppose that $\sequence {E_n}_{n \mathop \in \N} \in \Sigma$ is a [[Definition:Indexed Set|collection]] of [[Definition:Measurable Set|measurable sets]].
Then:
:$\ds \bigcap_{n \mathop \in \N} E_n \in \Si... | {{begin-eqn}}
{{eqn | q = \forall n \in \N
| l = E_n
| o = \in
| r = \Sigma
}}
{{eqn | ll= \leadsto
| l = X \setminus E_n
| o = \in
| r = \Sigma
| c = Axiom $(2)$ for [[Definition:Sigma-Algebra/Definition 1|$\sigma$-algebras]]
}}
{{eqn | ll= \leadsto
| l = \bigcup_{n \mat... | Sigma-Algebra Closed under Countable Intersection | https://proofwiki.org/wiki/Sigma-Algebra_Closed_under_Countable_Intersection | https://proofwiki.org/wiki/Sigma-Algebra_Closed_under_Countable_Intersection | [
"Sigma-Algebras"
] | [
"Definition:Set",
"Definition:Sigma-Algebra",
"Definition:Indexing Set/Indexed Set",
"Definition:Measurable Set",
"Definition:Set Intersection"
] | [
"Definition:Sigma-Algebra/Definition 1",
"Definition:Sigma-Algebra/Definition 1",
"Definition:Sigma-Algebra/Definition 1",
"De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Intersection",
"Set Difference with Set Difference",
"Set Union Preserves Subsets"
] |
proofwiki-5093 | Sigma-Algebra of Countable Sets | Let $X$ be a set.
Let $\Sigma$ be the set of countable and co-countable subsets of $X$.
Then $\Sigma$ is a $\sigma$-algebra. | Let us verify in turn the axioms of a $\sigma$-algebra. | Let $X$ be a [[Definition:Set|set]].
Let $\Sigma$ be the [[Definition:Set|set]] of [[Definition:Countable Set|countable]] and [[Definition:Co-Countable Set|co-countable]] [[Definition:Subset|subsets]] of $X$.
Then $\Sigma$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]]. | Let us verify in turn the [[Definition:Axiom|axioms]] of a [[Definition:Sigma-Algebra|$\sigma$-algebra]]. | Sigma-Algebra of Countable Sets | https://proofwiki.org/wiki/Sigma-Algebra_of_Countable_Sets | https://proofwiki.org/wiki/Sigma-Algebra_of_Countable_Sets | [
"Sigma-Algebras",
"Examples of Sigma-Algebras"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Countable Set",
"Definition:Co-Countable Set",
"Definition:Subset",
"Definition:Sigma-Algebra"
] | [
"Definition:Axiom",
"Definition:Sigma-Algebra"
] |
proofwiki-5094 | Trace Sigma-Algebra is Sigma-Algebra | Let $X$ be a set.
Let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $E \subseteq X$ be a subset of $X$.
Then the trace $\sigma$-algebra $\Sigma_E$ is a $\sigma$-algebra on $E$. | Verifying the axioms for a $\sigma$-algebra in turn: | Let $X$ be a [[Definition:Set|set]].
Let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
Let $E \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
Then the [[Definition:Trace Sigma-Algebra|trace $\sigma$-algebra]] $\Sigma_E$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $E$. | Verifying the axioms for a [[Definition:Sigma-Algebra|$\sigma$-algebra]] in turn: | Trace Sigma-Algebra is Sigma-Algebra | https://proofwiki.org/wiki/Trace_Sigma-Algebra_is_Sigma-Algebra | https://proofwiki.org/wiki/Trace_Sigma-Algebra_is_Sigma-Algebra | [
"Sigma-Algebras",
"Trace Sigma-Algebras"
] | [
"Definition:Set",
"Definition:Sigma-Algebra",
"Definition:Subset",
"Definition:Trace Sigma-Algebra",
"Definition:Sigma-Algebra"
] | [
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra"
] |
proofwiki-5095 | Pre-Image Sigma-Algebra on Domain is Sigma-Algebra | Let $X, X'$ be sets, and let $f: X \to X'$ be a mapping.
Let $\Sigma'$ be a $\sigma$-algebra on $X'$.
Then $f^{-1} \sqbrk {\Sigma'}$, the pre-image $\sigma$-algebra on the domain of $f$, is a $\sigma$-algebra on $X$. | Verify the axioms for a $\sigma$-algebra in turn: | Let $X, X'$ be [[Definition:Set|sets]], and let $f: X \to X'$ be a [[Definition:Mapping|mapping]].
Let $\Sigma'$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X'$.
Then $f^{-1} \sqbrk {\Sigma'}$, the [[Definition:Pre-Image Sigma-Algebra on Domain|pre-image $\sigma$-algebra]] on the [[Definition:Domain of Ma... | Verify the axioms for a [[Definition:Sigma-Algebra|$\sigma$-algebra]] in turn: | Pre-Image Sigma-Algebra on Domain is Sigma-Algebra | https://proofwiki.org/wiki/Pre-Image_Sigma-Algebra_on_Domain_is_Sigma-Algebra | https://proofwiki.org/wiki/Pre-Image_Sigma-Algebra_on_Domain_is_Sigma-Algebra | [
"Sigma-Algebras"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Sigma-Algebra",
"Definition:Pre-Image Sigma-Algebra/Domain",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Sigma-Algebra"
] | [
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra"
] |
proofwiki-5096 | Intersection of Sigma-Algebras | Let $X$ be a set.
Let $\family {\Sigma_i}_{i \mathop \in I}$ be a indexed set of $\sigma$-algebras on $X$.
Then $\Sigma := \ds \bigcap_{i \mathop \in I} \Sigma_i$ is also a $\sigma$-algebra on $X$.
Here, $\ds \bigcap$ denotes set intersection. | Verify the axioms for a $\sigma$-algebra in turn: | Let $X$ be a [[Definition:Set|set]].
Let $\family {\Sigma_i}_{i \mathop \in I}$ be a [[Definition:Indexed Set|indexed set]] of [[Definition:Sigma-Algebra|$\sigma$-algebras]] on $X$.
Then $\Sigma := \ds \bigcap_{i \mathop \in I} \Sigma_i$ is also a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $X$.
Here, $\ds \bi... | Verify the axioms for a [[Definition:Sigma-Algebra|$\sigma$-algebra]] in turn: | Intersection of Sigma-Algebras | https://proofwiki.org/wiki/Intersection_of_Sigma-Algebras | https://proofwiki.org/wiki/Intersection_of_Sigma-Algebras | [
"Sigma-Algebras",
"Set Intersection"
] | [
"Definition:Set",
"Definition:Indexing Set/Indexed Set",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra",
"Definition:Set Intersection"
] | [
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra"
] |
proofwiki-5097 | Generated Sigma-Algebra Preserves Subset | Let $X$ be a set.
Let $\FF, \GG \subseteq \powerset X$ be collections of subsets of $X$.
Suppose that:
:$\FF \subseteq \GG$
Then:
:$\map \sigma \FF \subseteq \map \sigma \GG$
where $\map \sigma \GG$ denotes the $\sigma$-algebra generated by $\GG$ | By definition of $\sigma$-algebra generated by $\GG$:
:$\GG \subseteq \map \sigma \GG$
It follows that also:
:$\FF \subseteq \map \sigma \GG$
By definition of $\sigma$-algebra generated by $\FF$:
:$\FF \subseteq \map \sigma \FF$
We are given that:
:$\FF \subseteq \GG$.
Hence:
:$\map \sigma \FF \subseteq \map \sigma \GG... | Let $X$ be a [[Definition:Set|set]].
Let $\FF, \GG \subseteq \powerset X$ be collections of [[Definition:Subset|subsets]] of $X$.
Suppose that:
:$\FF \subseteq \GG$
Then:
:$\map \sigma \FF \subseteq \map \sigma \GG$
where $\map \sigma \GG$ denotes the [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\... | By definition of [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebra generated by $\GG$]]:
:$\GG \subseteq \map \sigma \GG$
It follows that also:
:$\FF \subseteq \map \sigma \GG$
By definition of [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebra generated by $\F... | Generated Sigma-Algebra Preserves Subset | https://proofwiki.org/wiki/Generated_Sigma-Algebra_Preserves_Subset | https://proofwiki.org/wiki/Generated_Sigma-Algebra_Preserves_Subset | [
"Sigma-Algebras Generated by Collection of Subsets"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Given"
] |
proofwiki-5098 | Characterization of Euclidean Borel Sigma-Algebra | Let $\OO^n$, $\CC^n$ and $\KK^n$ be the collections of open, closed and compact subsets of the Euclidean space $\struct {\R^n, \tau}$, respectively.
Let $\JJ_{ho}^n$ be the collection of half-open rectangles in $\R^n$.
Let $\JJ^n_{ho, \text {rat} }$ be the collection of half-open rectangles in $\R^n$ with rational endp... | By definition of Borel $\sigma$-algebra, $\map \BB {\R^n} = \map \sigma {\OO^n}$.
The rest of the proof will be split in proving the following equalities:
:$(1): \quad \map \sigma {\OO^n} = \map \sigma {\CC^n}$
:$(2): \quad \map \sigma {\CC^n} = \map \sigma {\KK^n}$
:$(3): \quad \map \sigma {\OO^n} = \map \sigma {\JJ_{... | Let $\OO^n$, $\CC^n$ and $\KK^n$ be the collections of [[Definition:Open Set (Topology)|open]], [[Definition:Closed Set (Topology)|closed]] and [[Definition:Compact (Real Analysis)|compact]] [[Definition:Subset|subsets]] of the [[Definition:Euclidean Space|Euclidean space]] $\struct {\R^n, \tau}$, respectively.
Let $\... | By definition of [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]], $\map \BB {\R^n} = \map \sigma {\OO^n}$.
The rest of the proof will be split in proving the following equalities:
:$(1): \quad \map \sigma {\OO^n} = \map \sigma {\CC^n}$
:$(2): \quad \map \sigma {\CC^n} = \map \sigma {\KK^n}$
:$(3): \quad \ma... | Characterization of Euclidean Borel Sigma-Algebra | https://proofwiki.org/wiki/Characterization_of_Euclidean_Borel_Sigma-Algebra | https://proofwiki.org/wiki/Characterization_of_Euclidean_Borel_Sigma-Algebra | [
"Sigma-Algebras",
"Characterization of Euclidean Borel Sigma-Algebra"
] | [
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Compact Space/Real Analysis",
"Definition:Subset",
"Definition:Euclidean Space",
"Definition:Half-Open Rectangle",
"Definition:Half-Open Rectangle",
"Definition:Rational Number",
"Definition:Borel Sigma-Algebra",
"Defini... | [
"Definition:Borel Sigma-Algebra"
] |
proofwiki-5099 | Generated Sigma-Algebra Preserves Finiteness | Let $X$ be a set, and let $A_1, \ldots, A_n \subseteq X$ be subsets of $X$.
Then $\map \sigma {\set {A_1, \ldots, A_n} }$ is a finite set, where $\sigma$ denotes generated $\sigma$-algebra. | Proceed by induction on $n$, that is, on the number of generators. | Let $X$ be a [[Definition:Set|set]], and let $A_1, \ldots, A_n \subseteq X$ be [[Definition:Subset|subsets]] of $X$.
Then $\map \sigma {\set {A_1, \ldots, A_n} }$ is a [[Definition:Finite Set|finite set]], where $\sigma$ denotes [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]... | Proceed by [[Principle of Mathematical Induction|induction]] on $n$, that is, on the number of [[Definition:Sigma-Algebra Generated by Collection of Subsets/Generator|generators]]. | Generated Sigma-Algebra Preserves Finiteness | https://proofwiki.org/wiki/Generated_Sigma-Algebra_Preserves_Finiteness | https://proofwiki.org/wiki/Generated_Sigma-Algebra_Preserves_Finiteness | [
"Sigma-Algebras Generated by Collection of Subsets"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Finite Set",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Principle of Mathematical Induction",
"Definition:Sigma-Algebra Generated by Collection of Subsets/Generator",
"Principle of Mathematical Induction"
] |
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