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-4700 | One-to-Many Image of Set Difference/Corollary 2 | Let $\RR \subseteq S \times T$ be a relation which is one-to-many.
Let $A$ be a subset of $S$.
Then:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
where $\complement$ (in this context) denotes relative complement.
In the language of direct image mappings this can be presented as:
:$\forall A \in \po... | By definition of the image of $\RR$:
:$\Img \RR = \RR \sqbrk S$
So, when $B = S$ in {{Corollary|One-to-Many Image of Set Difference|1}}:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \relcomp {\RR \sqbrk S} {\RR \sqbrk A}$
Hence:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
means exactly the same thing as... | Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] which is [[Definition:One-to-Many Relation|one-to-many]].
Let $A$ be a [[Definition:Subset|subset]] of $S$.
Then:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
where $\complement$ (in this context) denotes [[Definition:Relativ... | By definition of the [[Definition:Image of Relation|image of $\RR$]]:
:$\Img \RR = \RR \sqbrk S$
So, when $B = S$ in {{Corollary|One-to-Many Image of Set Difference|1}}:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \relcomp {\RR \sqbrk S} {\RR \sqbrk A}$
Hence:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \RR \sqbrk {\relcomp ... | One-to-Many Image of Set Difference/Corollary 2 | https://proofwiki.org/wiki/One-to-Many_Image_of_Set_Difference/Corollary_2 | https://proofwiki.org/wiki/One-to-Many_Image_of_Set_Difference/Corollary_2 | [
"Set Difference",
"Relative Complement",
"One-to-Many Image of Set Difference"
] | [
"Definition:Relation",
"Definition:One-to-Many Relation",
"Definition:Subset",
"Definition:Relative Complement",
"Definition:Direct Image Mapping"
] | [
"Definition:Image (Set Theory)/Relation/Relation",
"One-to-Many Image of Set Difference",
"Category:Set Difference",
"Category:Relative Complement",
"Category:One-to-Many Image of Set Difference"
] |
proofwiki-4701 | Integer Power Function is Bijective iff Index is Odd | Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $f_n: \R \to \R$ be the real function defined as:
:$\map {f_n} x = x^n$
Then $f_n$ is a bijection {{iff}} $n$ is odd. | === Even Index ===
Suppose $n$ is even.
Let $x \ne 0$.
Then $1^n = \paren {-1}^n$ by Power of Ring Negative, so $f_n$ is not injective.
Also, by Even Power is Non-Negative, $f_n$ is not surjective.
By definition, a bijection is both injective and surjective.
It follows that for even $n$, $f_n$ is not bijective.
{{qed|l... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Let $f_n: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\map {f_n} x = x^n$
Then $f_n$ is a [[Definition:Bijection|bijection]] {{iff}} $n$ is [[Definition:Odd Integer|odd]]. | === Even Index ===
Suppose $n$ is [[Definition:Even Integer|even]].
Let $x \ne 0$.
Then $1^n = \paren {-1}^n$ by [[Power of Ring Negative]], so $f_n$ is not [[Definition:Injection|injective]].
Also, by [[Even Power is Non-Negative]], $f_n$ is not [[Definition:Surjective|surjective]].
By definition, a [[Definition:... | Integer Power Function is Bijective iff Index is Odd | https://proofwiki.org/wiki/Integer_Power_Function_is_Bijective_iff_Index_is_Odd | https://proofwiki.org/wiki/Integer_Power_Function_is_Bijective_iff_Index_is_Odd | [
"Real Functions",
"Powers",
"Bijections"
] | [
"Definition:Positive/Integer",
"Definition:Real Function",
"Definition:Bijection",
"Definition:Odd Integer"
] | [
"Definition:Even Integer",
"Power of Ring Negative",
"Definition:Injection",
"Even Power is Non-Negative",
"Definition:Surjection",
"Definition:Bijection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Even Integer",
"Definition:Bijection",
"Definition:Injection",
"Definition:Su... |
proofwiki-4702 | Set of Infinite Sequences is Uncountable | Let $S$ be a set which contains more than one element.
Let $S^\infty$ denote the set of all sequences of elements of $S$.
Then $S^\infty$ is uncountable. | As $S$ has more than one element, it must have at least two.
So, let $a, b \in S$ be those two elements.
Let $Z$ be the set of all sequences from $\set {a, b}$.
Suppose $S^\infty$ were countable.
From Subset of Countably Infinite Set is Countable, $Z$ is likewise countable.
So by definition, it would be possible to set... | Let $S$ be a [[Definition:Set|set]] which contains more than one [[Definition:Element|element]].
Let $S^\infty$ denote the set of all [[Definition:Sequence|sequences]] of elements of $S$.
Then $S^\infty$ is [[Definition:Uncountable Set|uncountable]]. | As $S$ has more than one element, it must have at least two.
So, let $a, b \in S$ be those two elements.
Let $Z$ be the set of all sequences from $\set {a, b}$.
Suppose $S^\infty$ were [[Definition:Countable Set|countable]].
From [[Subset of Countably Infinite Set is Countable]], $Z$ is likewise [[Definition:Count... | Set of Infinite Sequences is Uncountable | https://proofwiki.org/wiki/Set_of_Infinite_Sequences_is_Uncountable | https://proofwiki.org/wiki/Set_of_Infinite_Sequences_is_Uncountable | [
"Countable Sets",
"Diagonal Arguments"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Sequence",
"Definition:Uncountable/Set"
] | [
"Definition:Countable Set",
"Subset of Countably Infinite Set is Countable",
"Definition:Countable Set",
"Definition:Bijection",
"Definition:Natural Numbers",
"Definition:Uncountable/Set",
"Rule of Transposition",
"Definition:Uncountable/Set"
] |
proofwiki-4703 | Union of Topologies is not necessarily Topology | Let $\tau_1$ and $\tau_2$ be topologies on a set $S$.
Then $\tau_1 \cup \tau_2$ is not necessarily also a topology on $S$. | Let $S := \set {0, 1, 2}$ be a set.
Let:
:$\tau_1 := \set {\O, \set 0, \set 1, \set {0, 1}, S}$
:$\tau_2 := \set {\O, \set 0, \set 2, \set {0, 2}, S}$
be topologies on $S$.
Then:
:$\tau := \tau_1 \cup \tau_2 = \set {\O, \set 0, \set 1, \set 2, \set {0, 1} \set {0, 2}, S}$
For $\tau$ to be a topology the union of any ... | Let $\tau_1$ and $\tau_2$ be [[Definition:Topology|topologies]] on a [[Definition:Set|set]] $S$.
Then $\tau_1 \cup \tau_2$ is not necessarily also a [[Definition:Topology|topology]] on $S$. | Let $S := \set {0, 1, 2}$ be a [[Definition:Set|set]].
Let:
:$\tau_1 := \set {\O, \set 0, \set 1, \set {0, 1}, S}$
:$\tau_2 := \set {\O, \set 0, \set 2, \set {0, 2}, S}$
be [[Definition:Topology|topologies]] on $S$.
Then:
:$\tau := \tau_1 \cup \tau_2 = \set {\O, \set 0, \set 1, \set 2, \set {0, 1} \set {0, 2}, S}$
... | Union of Topologies is not necessarily Topology | https://proofwiki.org/wiki/Union_of_Topologies_is_not_necessarily_Topology | https://proofwiki.org/wiki/Union_of_Topologies_is_not_necessarily_Topology | [
"Topologies",
"Set Union"
] | [
"Definition:Topology",
"Definition:Set",
"Definition:Topology"
] | [
"Definition:Set",
"Definition:Topology",
"Definition:Topology",
"Definition:Set Union",
"Definition:Topology"
] |
proofwiki-4704 | Niven's Theorem | Consider the angles $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$.
The only values of $\theta$ such that both $\dfrac \theta \pi$ and $\sin \theta$ are rational are:
:$\theta = 0: \sin \theta = 0$
:$\theta = \dfrac \pi 6: \sin \theta = \dfrac 1 2$
:$\theta = \dfrac \pi 2: \sin \theta = 1$ | We will prove that if both $\dfrac \theta \pi$ and $\cos \theta$ are rational then:
:$\theta \in \set{ {0, \dfrac \pi 3, \dfrac \pi 2} }$ | Consider the [[Definition:Angle|angles]] $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$.
The only values of $\theta$ such that both $\dfrac \theta \pi$ and $\sin \theta$ are [[Definition:Rational Number|rational]] are:
:$\theta = 0: \sin \theta = 0$
:$\theta = \dfrac \pi 6: \sin \theta = \dfrac 1 2$
:$\theta... | We will prove that if both $\dfrac \theta \pi$ and $\cos \theta$ are [[Definition:Rational Number|rational]] then:
:$\theta \in \set{ {0, \dfrac \pi 3, \dfrac \pi 2} }$ | Niven's Theorem | https://proofwiki.org/wiki/Niven's_Theorem | https://proofwiki.org/wiki/Niven's_Theorem | [
"Trigonometry",
"Niven's Theorem"
] | [
"Definition:Angle",
"Definition:Rational Number"
] | [
"Definition:Rational Number",
"Definition:Rational Number",
"Definition:Rational Number",
"Definition:Rational Number",
"Definition:Rational Number"
] |
proofwiki-4705 | Relative Frequency is Probability Measure | The relative frequency model is a probability measure. | We check all the Kolmogorov axioms in turn: | The [[Definition:Relative Frequency Model|relative frequency model]] is a [[Definition:Probability Measure|probability measure]]. | We check all the [[Axiom:Kolmogorov Axioms|Kolmogorov axioms]] in turn: | Relative Frequency is Probability Measure | https://proofwiki.org/wiki/Relative_Frequency_is_Probability_Measure | https://proofwiki.org/wiki/Relative_Frequency_is_Probability_Measure | [
"Probability Theory",
"Proofs by Induction",
"Statistics"
] | [
"Definition:Relative Frequency Model",
"Definition:Probability Measure"
] | [
"Axiom:Kolmogorov Axioms",
"Axiom:Kolmogorov Axioms"
] |
proofwiki-4706 | Sorgenfrey Line is Topology | The Sorgenfrey Line is a topological space. | We have to check that $\BB = \set {\hointr a b: a, b \in \R}$ fulfills the axioms of being a basis for a topology.
By definition of synthetic basis we only have to check that:
:$(1): \quad \bigcup \BB = \R$
:$(2): \quad \forall B_1, B_2 \in \BB: \exists V \in \BB: V \subseteq B_1 \cap B_2$
We have that:
:$\forall n \in... | The [[Definition:Sorgenfrey Line|Sorgenfrey Line]] is a [[Definition:Topological Space|topological space]]. | We have to check that $\BB = \set {\hointr a b: a, b \in \R}$ fulfills the axioms of being a [[Definition:Basis (Topology)|basis]] for a [[Definition:Topology|topology]].
By definition of [[Definition:Synthetic Basis|synthetic basis]] we only have to check that:
:$(1): \quad \bigcup \BB = \R$
:$(2): \quad \forall B_1... | Sorgenfrey Line is Topology | https://proofwiki.org/wiki/Sorgenfrey_Line_is_Topology | https://proofwiki.org/wiki/Sorgenfrey_Line_is_Topology | [
"Sorgenfrey Line"
] | [
"Definition:Sorgenfrey Line",
"Definition:Topological Space"
] | [
"Definition:Basis (Topology)",
"Definition:Topology",
"Definition:Basis (Topology)/Synthetic Basis"
] |
proofwiki-4707 | Sorgenfrey Line is Hausdorff | Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Then $T$ is Hausdorff. | Take $x, y \in \R$ such that $x \ne y$.
{{WLOG}}, assume that $x < y$.
From Real Numbers are Densely Ordered:
:$\exists t \in \R: x < t < y$
Then:
:$\hointr x t \cap \hointr t {y + 1} = \O$
We also have that $x \in \hointr x t$ by definition of half-open interval.
Also, as $t < y < y+1$ it is clear that $y \in \hointr ... | Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
Then $T$ is [[Definition:Hausdorff Space|Hausdorff]]. | Take $x, y \in \R$ such that $x \ne y$.
{{WLOG}}, assume that $x < y$.
From [[Real Numbers are Densely Ordered]]:
:$\exists t \in \R: x < t < y$
Then:
:$\hointr x t \cap \hointr t {y + 1} = \O$
We also have that $x \in \hointr x t$ by definition of [[Definition:Half-Open Real Interval|half-open interval]].
Also, a... | Sorgenfrey Line is Hausdorff | https://proofwiki.org/wiki/Sorgenfrey_Line_is_Hausdorff | https://proofwiki.org/wiki/Sorgenfrey_Line_is_Hausdorff | [
"Sorgenfrey Line",
"Examples of Hausdorff Spaces"
] | [
"Definition:Sorgenfrey Line",
"Definition:T2 Space"
] | [
"Real Numbers are Densely Ordered",
"Definition:Real Interval/Half-Open",
"Definition:Sorgenfrey Line",
"Definition:Open Set/Topology",
"Definition:Disjoint Sets",
"Definition:Open Set/Topology",
"Definition:Sorgenfrey Line",
"Definition:T2 Space"
] |
proofwiki-4708 | Convergent Sequence in Set of Integers | Let $\sequence{x_n}_{n \in \N}$ be a sequence in the set $\Z$ of integers considered as a subspace of the real number line $\R$ under the Euclidean metric.
Then $\sequence{x_n}_{n \in \N}$ converges in $\R$ to a limit {{iff}}:
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
That is, {{iff}} the sequence reaches... | Suppose $\sequence{x_n}_{n \in \N}$ converges to a limit $l$.
Consider the open set in $\R$:
:$U := \openint{l - \dfrac 1 2}{l + \dfrac 1 2}$
Then $\forall x \in \Z: x \in U \implies x = l$
It follows by definition of convergence that:
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
where $x_k = l$.
Now suppose... | Let $\sequence{x_n}_{n \in \N}$ be a [[Definition:Sequence|sequence]] in the set $\Z$ of [[Definition:Integer|integers]] considered as a [[Definition:Topological Subspace|subspace]] of the [[Definition:Real Number Line|real number line]] $\R$ under the [[Definition:Euclidean Metric on Real Number Line|Euclidean metric]... | Suppose $\sequence{x_n}_{n \in \N}$ [[Definition:Convergent Sequence (Topology)|converges]] to a [[Definition:Limit of Sequence (Topology)|limit]] $l$.
Consider the [[Definition:Open Set (Topology)|open set]] in $\R$:
:$U := \openint{l - \dfrac 1 2}{l + \dfrac 1 2}$
Then $\forall x \in \Z: x \in U \implies x = l$
It... | Convergent Sequence in Set of Integers | https://proofwiki.org/wiki/Convergent_Sequence_in_Set_of_Integers | https://proofwiki.org/wiki/Convergent_Sequence_in_Set_of_Integers | [
"Sequences",
"Convergence"
] | [
"Definition:Sequence",
"Definition:Integer",
"Definition:Topological Subspace",
"Definition:Real Number/Real Number Line",
"Definition:Euclidean Metric/Real Number Line",
"Definition:Convergent Sequence/Topology",
"Definition:Limit of Sequence/Topological Space"
] | [
"Definition:Convergent Sequence/Topology",
"Definition:Limit of Sequence/Topological Space",
"Definition:Open Set/Topology",
"Definition:Convergent Sequence/Topology",
"Definition:Convergent Sequence/Topology",
"Definition:Limit of Sequence/Topological Space",
"Category:Sequences",
"Category:Convergen... |
proofwiki-4709 | Sorgenfrey Line is Perfectly Normal | Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Then $T$ is perfectly normal. | From the definition of perfectly normal space, it is necessary to prove that:
:$T$ is a $T_1$ space
and that:
:an arbitrary closed set is a $G_\delta$ set.
From $T_2$ Space is $T_1$ Space and Sorgenfrey Line is Hausdorff:
:the Sorgenfrey line is a $T_1$ space.
From Complement of $F_\sigma$ Set is $G_\delta$ Set it is s... | Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
Then $T$ is [[Definition:Perfectly Normal Space|perfectly normal]]. | From the definition of [[Definition:Perfectly Normal Space|perfectly normal]] space, it is necessary to prove that:
:$T$ is a [[Definition:T1 Space|$T_1$ space]]
and that:
:an [[Definition:Arbitrary|arbitrary]] [[Definition:Closed Set (Topology)|closed set]] is a [[Definition:G-Delta Set|$G_\delta$ set]].
From [[T2 Sp... | Sorgenfrey Line is Perfectly Normal | https://proofwiki.org/wiki/Sorgenfrey_Line_is_Perfectly_Normal | https://proofwiki.org/wiki/Sorgenfrey_Line_is_Perfectly_Normal | [
"Sorgenfrey Line",
"Examples of Perfectly Normal Spaces"
] | [
"Definition:Sorgenfrey Line",
"Definition:Perfectly Normal Space"
] | [
"Definition:Perfectly Normal Space",
"Definition:T1 Space",
"Definition:Arbitrary",
"Definition:Closed Set/Topology",
"Definition:G-Delta Set",
"T2 Space is T1",
"Sorgenfrey Line is Hausdorff",
"Definition:Sorgenfrey Line",
"Definition:T1 Space",
"Complement of F-Sigma Set is G-Delta Set",
"Defi... |
proofwiki-4710 | Sorgenfrey Line is Expansion of Real Line | Let $\R = \struct {\R, d}$ be the metric space defined in Real Number Line is Metric Space.
Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Then $T$ is an expansion of $\R$ as a topological space. | It is enough to prove that any open set in $\R$ is open in $T$.
Let $a, b \in \R$.
Then:
:$\ds \openint a b = \bigcup_{\epsilon \mathop > 0} \hointr {a + \epsilon} b$
Since $\hointr {a + \epsilon} b$ are open in $T$, $\openint a b$ is also open in $T$.
{{qed}}
Category:Sorgenfrey Line
owmja931ica1kc7utbvp5gb94dcptg0 | Let $\R = \struct {\R, d}$ be the [[Definition:Metric Space|metric space]] defined in [[Real Number Line is Metric Space]].
Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
Then $T$ is an [[Definition:Expansion of Topology|expansion]] of $\R$ as a [[Definition:Topological Space|top... | It is enough to prove that any [[Definition:Open Set (Topology)|open set]] in $\R$ is [[Definition:Open Set (Topology)|open]] in $T$.
Let $a, b \in \R$.
Then:
:$\ds \openint a b = \bigcup_{\epsilon \mathop > 0} \hointr {a + \epsilon} b$
Since $\hointr {a + \epsilon} b$ are [[Definition:Open Set (Topology)|open]] in ... | Sorgenfrey Line is Expansion of Real Line | https://proofwiki.org/wiki/Sorgenfrey_Line_is_Expansion_of_Real_Line | https://proofwiki.org/wiki/Sorgenfrey_Line_is_Expansion_of_Real_Line | [
"Sorgenfrey Line"
] | [
"Definition:Metric Space",
"Real Number Line is Metric Space",
"Definition:Sorgenfrey Line",
"Definition:Expansion of Topology",
"Definition:Topological Space"
] | [
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Category:Sorgenfrey Line"
] |
proofwiki-4711 | Sorgenfrey Line satisfies all Separation Axioms | Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Then $T$ satisfies all separation axioms. | We have Sorgenfrey Line is Perfectly Normal.
The result follows from Sequence of Implications of Separation Axioms.
{{qed}}
Category:Sorgenfrey Line
Category:Separation Axioms
7utbg8yppc8vgkrn3ll9eeir5fu5de9 | Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
Then $T$ satisfies all [[Definition:Separation Axioms|separation axioms]]. | We have [[Sorgenfrey Line is Perfectly Normal]].
The result follows from [[Sequence of Implications of Separation Axioms]].
{{qed}}
[[Category:Sorgenfrey Line]]
[[Category:Separation Axioms]]
7utbg8yppc8vgkrn3ll9eeir5fu5de9 | Sorgenfrey Line satisfies all Separation Axioms | https://proofwiki.org/wiki/Sorgenfrey_Line_satisfies_all_Separation_Axioms | https://proofwiki.org/wiki/Sorgenfrey_Line_satisfies_all_Separation_Axioms | [
"Sorgenfrey Line",
"Separation Axioms"
] | [
"Definition:Sorgenfrey Line",
"Definition:Tychonoff Separation Axioms"
] | [
"Sorgenfrey Line is Perfectly Normal",
"Sequence of Implications of Separation Axioms",
"Category:Sorgenfrey Line",
"Category:Separation Axioms"
] |
proofwiki-4712 | Argument of Product equals Sum of Arguments | Let $z_1, z_2 \in \C$ be complex numbers.
Let $\arg$ be the argument operator.
Then:
:$\map \arg {z_1 z_2} = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$
where $k$ can be $0$, $1$ or $-1$. | Let $\theta_1 = \map \arg {z_1}, \theta_2 = \map \arg {z_2}$.
Then the polar forms of $z_1, z_2$ are:
{{begin-eqn}}
{{eqn | l = z_1
| r = \cmod {z_1} \paren {\cos \theta_1 + i \sin \theta_1}
}}
{{eqn | l = z_2
| r = \cmod {z_2} \paren {\cos \theta_2 + i \sin \theta_2}
}}
{{end-eqn}}
By the definition of com... | Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
Let $\arg$ be the [[Definition:Argument of Complex Number|argument operator]].
Then:
:$\map \arg {z_1 z_2} = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$
where $k$ can be $0$, $1$ or $-1$. | Let $\theta_1 = \map \arg {z_1}, \theta_2 = \map \arg {z_2}$.
Then the [[Definition:Polar Form of Complex Number|polar forms]] of $z_1, z_2$ are:
{{begin-eqn}}
{{eqn | l = z_1
| r = \cmod {z_1} \paren {\cos \theta_1 + i \sin \theta_1}
}}
{{eqn | l = z_2
| r = \cmod {z_2} \paren {\cos \theta_2 + i \sin \th... | Argument of Product equals Sum of Arguments | https://proofwiki.org/wiki/Argument_of_Product_equals_Sum_of_Arguments | https://proofwiki.org/wiki/Argument_of_Product_equals_Sum_of_Arguments | [
"Argument of Complex Number",
"Complex Multiplication"
] | [
"Definition:Complex Number",
"Definition:Argument of Complex Number"
] | [
"Definition:Complex Number/Polar Form",
"Definition:Multiplication/Complex Numbers",
"Sine of Sum",
"Cosine of Sum",
"Definition:Argument of Complex Number",
"Cosine of Angle plus Full Angle",
"Sine of Angle plus Full Angle",
"Definition:Argument of Complex Number",
"Definition:Argument of Complex N... |
proofwiki-4713 | Convergence of Sequence in Discrete Space | Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $S$.
Then $\sequence {x_n}_{n \mathop \in \N}$ converges in $T$ to a limit {{iff}}:
:$\exists N \in \N: \forall n \in \N: n > N \implies x_n = x_N$
That is, {{iff}} the sequence reaches some value of $... | === Sufficient Condition ===
Suppose $\sequence {x_n}_{n \mathop \in \N}$ converges to a limit $L$.
Then:
{{begin-eqn}}
{{eqn | l = \set L
| o = \in
| r = \tau
| c = {{Defof|Discrete Space}}
}}
{{eqn | ll= \leadsto
| q = \exists N \in \N: \forall n \in \N
| lo= n > N \implies
| l = x... | Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Space|discrete topological space]].
Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] in $S$.
Then $\sequence {x_n}_{n \mathop \in \N}$ [[Definition:Convergent Sequence (Topology)|converges]] in $T$ to a [[Definition:Limit of Sequenc... | === Sufficient Condition ===
Suppose $\sequence {x_n}_{n \mathop \in \N}$ [[Definition:Convergent Sequence (Topology)|converges]] to a [[Definition:Limit of Sequence (Topology)|limit]] $L$.
Then:
{{begin-eqn}}
{{eqn | l = \set L
| o = \in
| r = \tau
| c = {{Defof|Discrete Space}}
}}
{{eqn | ll= \lea... | Convergence of Sequence in Discrete Space | https://proofwiki.org/wiki/Convergence_of_Sequence_in_Discrete_Space | https://proofwiki.org/wiki/Convergence_of_Sequence_in_Discrete_Space | [
"Convergence of Sequence in Discrete Space",
"Convergent Sequences",
"Discrete Topologies"
] | [
"Definition:Discrete Topology",
"Definition:Sequence",
"Definition:Convergent Sequence/Topology",
"Definition:Limit of Sequence/Topological Space"
] | [
"Definition:Convergent Sequence/Topology",
"Definition:Limit of Sequence/Topological Space",
"Universal Instantiation",
"Definition:Convergent Sequence/Topology"
] |
proofwiki-4714 | Discrete Subspace of Fortissimo Space | Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Let $T' = \struct {S \setminus \set p, \tau_p}$ be the topological subspace induced on $T$ by the subset $S \setminus \set p$.
Then $T'$ is a discrete topological space. | By the definition of Fortissimo space, every $A \subset S \setminus \set p$ is open in $T$, because $p \notin A$.
Thus by definition of topological subspace, $A \subset S \setminus \set p$ is open in $T'$.
The result follows by the definition of discrete space.
{{qed}}
Category:Fortissimo Spaces
Category:Discrete Topol... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
Let $T' = \struct {S \setminus \set p, \tau_p}$ be the [[Definition:Topological Subspace|topological subspace]] induced on $T$ by the [[Definition:Subset|subset]] $S \setminus \set p$.
Then $T'$ is a [[Definition:Discrete Space|disc... | By the definition of [[Definition:Fortissimo Space|Fortissimo space]], every $A \subset S \setminus \set p$ is [[Definition:Open Set (Topology)|open in $T$]], because $p \notin A$.
Thus by definition of [[Definition:Topological Subspace|topological subspace]], $A \subset S \setminus \set p$ is [[Definition:Open Set (T... | Discrete Subspace of Fortissimo Space | https://proofwiki.org/wiki/Discrete_Subspace_of_Fortissimo_Space | https://proofwiki.org/wiki/Discrete_Subspace_of_Fortissimo_Space | [
"Fortissimo Spaces",
"Discrete Topologies"
] | [
"Definition:Fortissimo Space",
"Definition:Topological Subspace",
"Definition:Subset",
"Definition:Discrete Topology"
] | [
"Definition:Fortissimo Space",
"Definition:Open Set/Topology",
"Definition:Topological Subspace",
"Definition:Open Set/Topology",
"Definition:Discrete Topology",
"Category:Fortissimo Spaces",
"Category:Discrete Topologies"
] |
proofwiki-4715 | Countable Space is Sigma-Compact | Let $T = \struct {S, \tau}$ be a countable topological space.
Then $T$ is $\sigma$-compact. | From Finite Space Satisfies All Compactness Properties, for every $p \in S$, $\set p$ is compact.
Then $\ds S = \bigcup_{p \mathop \in S} \set p$ is a countable union of compact sets.
Thus by definition $T$ is $\sigma$-compact.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Countable Topological Space|countable topological space]].
Then $T$ is [[Definition:Sigma-Compact Space|$\sigma$-compact]]. | From [[Finite Space Satisfies All Compactness Properties]], for every $p \in S$, $\set p$ is [[Definition:Compact Topological Space|compact]].
Then $\ds S = \bigcup_{p \mathop \in S} \set p$ is a [[Definition:Countable Union|countable union]] of [[Definition:Compact Topological Space|compact sets]].
Thus by definitio... | Countable Space is Sigma-Compact | https://proofwiki.org/wiki/Countable_Space_is_Sigma-Compact | https://proofwiki.org/wiki/Countable_Space_is_Sigma-Compact | [
"Sigma-Compact Spaces",
"Countable Topological Spaces"
] | [
"Definition:Countable Topological Space",
"Definition:Sigma-Compact Space"
] | [
"Finite Space Satisfies All Compactness Properties",
"Definition:Compact Topological Space",
"Definition:Set Union/Countable Union",
"Definition:Compact Topological Space",
"Definition:Sigma-Compact Space"
] |
proofwiki-4716 | Closed Sets of Fortissimo Space | Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Then $H \subseteq S$ is closed in $T$ {{iff}}:
:$p \in H$
or
:$H$ is countable
or both. | By definition of a Fortissimo space, $U \subseteq S$ is open in $T$ {{iff}}:
:$p \in \relcomp S U$
or
:$\relcomp S U$ is countable
or both.
The result follows from the definition of closed set.
{{qed}}
Category:Fortissimo Spaces
Category:Examples of Closed Sets
0xomz0ygotpxyb42i8wirdwnyjypfql | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
Then $H \subseteq S$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}}:
:$p \in H$
or
:$H$ is [[Definition:Countable Set|countable]]
or both. | By definition of a [[Definition:Fortissimo Space|Fortissimo space]], $U \subseteq S$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}}:
:$p \in \relcomp S U$
or
:$\relcomp S U$ is [[Definition:Countable Set|countable]]
or both.
The result follows from the definition of [[Definition:Closed Set (Topology)|closed... | Closed Sets of Fortissimo Space | https://proofwiki.org/wiki/Closed_Sets_of_Fortissimo_Space | https://proofwiki.org/wiki/Closed_Sets_of_Fortissimo_Space | [
"Fortissimo Spaces",
"Examples of Closed Sets"
] | [
"Definition:Fortissimo Space",
"Definition:Closed Set/Topology",
"Definition:Countable Set"
] | [
"Definition:Fortissimo Space",
"Definition:Open Set/Topology",
"Definition:Countable Set",
"Definition:Closed Set/Topology",
"Category:Fortissimo Spaces",
"Category:Examples of Closed Sets"
] |
proofwiki-4717 | Union of Topologies on Singleton or Doubleton is Topology | Let $S$ be a set which is either a singleton or a doubleton.
Let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty indexed set of topologies on $S$.
Then $\tau := \ds \bigcup_{i \mathop \in I} {\tau_i}$ is also a topology on $S$. | Let $S$ be a singleton.
Let $S = \set a$.
From Topology on Singleton is Indiscrete Topology, the only possible topology on $S$ is the indiscrete topology.
From Set Union is Idempotent, the union of any number of indiscrete topologies on $S$ is the indiscrete topology.
Thus the union of any number of topologies on a s... | Let $S$ be a [[Definition:Set|set]] which is either a [[Definition:Singleton|singleton]] or a [[Definition:Doubleton|doubleton]].
Let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary [[Definition:Non-Empty Set|non-empty]] [[Definition:Indexing Set|indexed set]] of [[Definition:Topology|topologies]] on $S$.
Then ... | Let $S$ be a [[Definition:Singleton|singleton]].
Let $S = \set a$.
From [[Topology on Singleton is Indiscrete Topology]], the only possible [[Definition:Topology|topology]] on $S$ is the [[Definition:Indiscrete Topology|indiscrete topology]].
From [[Set Union is Idempotent]], the [[Definition:Set Union|union]] of ... | Union of Topologies on Singleton or Doubleton is Topology | https://proofwiki.org/wiki/Union_of_Topologies_on_Singleton_or_Doubleton_is_Topology | https://proofwiki.org/wiki/Union_of_Topologies_on_Singleton_or_Doubleton_is_Topology | [
"Topologies",
"Singletons",
"Doubletons",
"Set Union"
] | [
"Definition:Set",
"Definition:Singleton",
"Definition:Doubleton",
"Definition:Non-Empty Set",
"Definition:Indexing Set",
"Definition:Topology",
"Definition:Topology"
] | [
"Definition:Singleton",
"Trivial Topological Space is Indiscrete",
"Definition:Topology",
"Definition:Indiscrete Topology",
"Set Union is Idempotent",
"Definition:Set Union",
"Definition:Indiscrete Topology",
"Definition:Indiscrete Topology",
"Definition:Set Union",
"Definition:Topology",
"Defin... |
proofwiki-4718 | Meager Sets in Arens-Fort Space | Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Let $A \subseteq S$.
Then $A$ is meager {{iff}} $A = \set {\tuple {0, 0} }$. | First let $A = \set {\tuple {0, 0} }$.
From the definition of Arens-Fort space, $\set {\tuple {0, 0} }$ is closed because $S \setminus \set {\tuple {0, 0} }$ is open.
From Closed Set Equals its Closure:
:$\set {\tuple {0, 0} }^- = \set {\tuple {0, 0} }$
where $\set {\tuple {0, 0} }^-$ denotes the closure of $\set {\tup... | Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
Let $A \subseteq S$.
Then $A$ is [[Definition:Meager Space|meager]] {{iff}} $A = \set {\tuple {0, 0} }$. | First let $A = \set {\tuple {0, 0} }$.
From the definition of [[Definition:Arens-Fort Space|Arens-Fort space]], $\set {\tuple {0, 0} }$ is [[Definition:Closed Set (Topology)|closed]] because $S \setminus \set {\tuple {0, 0} }$ is [[Definition:Open Set (Topology)|open]].
From [[Closed Set Equals its Closure]]:
:$\set ... | Meager Sets in Arens-Fort Space | https://proofwiki.org/wiki/Meager_Sets_in_Arens-Fort_Space | https://proofwiki.org/wiki/Meager_Sets_in_Arens-Fort_Space | [
"Arens-Fort Space",
"Examples of Meager Spaces"
] | [
"Definition:Arens-Fort Space",
"Definition:Meager Space"
] | [
"Definition:Arens-Fort Space",
"Definition:Closed Set/Topology",
"Definition:Open Set/Topology",
"Set is Closed iff Equals Topological Closure",
"Definition:Closure (Topology)",
"Definition:Arens-Fort Space",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Interior (Topol... |
proofwiki-4719 | Right and Left Regular Representations in Topological Group are Homeomorphisms | Let $\struct {G, \cdot, \tau}$ be a topological group.
Let $g \in G$ be an element of $G$.
Then the left and right regular representations with respect to $g$:
:$L_g: \struct {G, \tau} \to \struct {G, \tau}$
and:
:$R_g: \struct {G, \tau} \to \struct {G, \tau}$
are homeomorphisms. | Let:
:$m: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau}$
be the mapping defined as:
:$\forall \tuple {x, y} \in G \times G: \map m {x, y} = x \cdot y$
From the definition of topological group, $m$ is a continuous mapping.
Let $\phi_1: \struct {G, \tau} \to \struct {G, \tau} \times \struct {G, \tau}$ ... | Let $\struct {G, \cdot, \tau}$ be a [[Definition:Topological Group|topological group]].
Let $g \in G$ be an [[Definition:Element|element]] of $G$.
Then the [[Definition:Left Regular Representation|left]] and [[Definition:Right Regular Representation|right]] [[Definition:Regular Representations|regular representation... | Let:
:$m: \struct {G, \tau} \times \struct {G, \tau} \to \struct {G, \tau}$
be the [[Definition:Mapping|mapping]] defined as:
:$\forall \tuple {x, y} \in G \times G: \map m {x, y} = x \cdot y$
From the definition of [[Definition:Topological Group|topological group]], $m$ is a [[Definition:Continuous Mapping (Topology)... | Right and Left Regular Representations in Topological Group are Homeomorphisms | https://proofwiki.org/wiki/Right_and_Left_Regular_Representations_in_Topological_Group_are_Homeomorphisms | https://proofwiki.org/wiki/Right_and_Left_Regular_Representations_in_Topological_Group_are_Homeomorphisms | [
"Homeomorphisms (Topological Spaces)",
"Regular Representations",
"Topological Groups"
] | [
"Definition:Topological Group",
"Definition:Element",
"Definition:Regular Representations/Left Regular Representation",
"Definition:Regular Representations/Right Regular Representation",
"Definition:Regular Representations",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Definition:Mapping",
"Definition:Topological Group",
"Definition:Continuous Mapping (Topology)",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Injective Restriction",
"De... |
proofwiki-4720 | Center of Group is Subgroup | Let $G$ be a group.
The center $\map Z G$ of $G$ is a subgroup of $G$. | For brevity, suppress the symbol for the group operation (which may be $\circ$, or $+$).
Apply the Two-Step Subgroup Test:
=== Condition $(1)$ ===
By the definition of identity, $e g = g e = g$ for all $g \in G$.
So, $e \in \map Z G$, meaning $\map Z G$ is non-empty.
{{qed|lemma}}
=== Condition $(2)$ ===
Suppose $a, b ... | Let $G$ be a [[Definition:Group|group]].
The [[Definition:Center of Group|center]] $\map Z G$ of $G$ is a [[Definition:Subgroup|subgroup]] of $G$. | For brevity, suppress the symbol for the [[Definition:Group Operation|group operation]] (which may be $\circ$, or $+$).
Apply the [[Two-Step Subgroup Test]]:
=== Condition $(1)$ ===
By the definition of [[Definition:Identity Element|identity]], $e g = g e = g$ for all $g \in G$.
So, $e \in \map Z G$, meaning $\map... | Center of Group is Subgroup/Proof 1 | https://proofwiki.org/wiki/Center_of_Group_is_Subgroup | https://proofwiki.org/wiki/Center_of_Group_is_Subgroup/Proof_1 | [
"Subgroups",
"Direct Proofs",
"Centers of Groups",
"Center of Group is Subgroup"
] | [
"Definition:Group",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Subgroup"
] | [
"Definition:Group Product/Group Law",
"Two-Step Subgroup Test",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Non-Empty Set",
"Definition:Associative Operation",
"Definition:Center (Abstract Algebra)/Group"
] |
proofwiki-4721 | Center of Group is Subgroup | Let $G$ be a group.
The center $\map Z G$ of $G$ is a subgroup of $G$. | We have the result Center is Intersection of Centralizers.
That is, $\map Z G$ is the intersection of all the centralizers of $G$.
All of these are subgroups of $G$ by Centralizer of Group Element is Subgroup.
Thus from Intersection of Subgroups is Subgroup, $\map Z G$ is also a subgroup of $G$.
{{qed}} | Let $G$ be a [[Definition:Group|group]].
The [[Definition:Center of Group|center]] $\map Z G$ of $G$ is a [[Definition:Subgroup|subgroup]] of $G$. | We have the result [[Center is Intersection of Centralizers]].
That is, $\map Z G$ is the [[Definition:Set Intersection|intersection]] of all the [[Definition:Centralizer of Group Element|centralizers]] of $G$.
All of these are [[Definition:Subgroup|subgroups]] of $G$ by [[Centralizer of Group Element is Subgroup]].
... | Center of Group is Subgroup/Proof 2 | https://proofwiki.org/wiki/Center_of_Group_is_Subgroup | https://proofwiki.org/wiki/Center_of_Group_is_Subgroup/Proof_2 | [
"Subgroups",
"Direct Proofs",
"Centers of Groups",
"Center of Group is Subgroup"
] | [
"Definition:Group",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Subgroup"
] | [
"Center is Intersection of Centralizers",
"Definition:Set Intersection",
"Definition:Centralizer/Group Element",
"Definition:Subgroup",
"Centralizer of Group Element is Subgroup",
"Intersection of Subgroups is Subgroup",
"Definition:Subgroup"
] |
proofwiki-4722 | Center of Group is Abelian Subgroup | Let $G$ be a group.
Let $\map Z G$ be the center of $G$.
Then $\map Z G$ is an abelian subgroup of $G$. | By Center of Group is Subgroup, $\map Z G$ is a subgroup of $G$.
The definition of the center $\map Z G$ grants that all elements of $\map Z G)$ commute with all elements of $G$.
In particular, all elements of $\map Z G$ commute with all elements of $\map Z G$ as $\map Z G \subseteq G$.
Therefore $\map Z G$ is abelian.... | Let $G$ be a [[Definition:Group|group]].
Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
Then $\map Z G$ is an [[Definition:Abelian Group|abelian]] [[Definition:Subgroup|subgroup]] of $G$. | By [[Center of Group is Subgroup]], $\map Z G$ is a [[Definition:Subgroup|subgroup]] of $G$.
The definition of the [[Definition:Center of Group|center]] $\map Z G$ grants that all [[Definition:Element|elements]] of $\map Z G)$ [[Definition:Commute|commute]] with all [[Definition:Element|elements]] of $G$.
In particu... | Center of Group is Abelian Subgroup | https://proofwiki.org/wiki/Center_of_Group_is_Abelian_Subgroup | https://proofwiki.org/wiki/Center_of_Group_is_Abelian_Subgroup | [
"Subgroups",
"Abelian Groups",
"Centers of Groups"
] | [
"Definition:Group",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Abelian Group",
"Definition:Subgroup"
] | [
"Center of Group is Subgroup",
"Definition:Subgroup",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Element",
"Definition:Commutative/Elements",
"Definition:Element",
"Definition:Element",
"Definition:Commutative/Elements",
"Definition:Element",
"Definition:Abelian Group"
] |
proofwiki-4723 | Dihedral Group/Group Presentation | The dihedral group $D_n$ has the group presentation:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
That is, the dihedral group $D_n$ is generated by two elements $\alpha$ and $\beta$ such that:
:$(1): \quad \alpha^n = e$
:$(2): \quad \beta^2 = e$
:$(3): \quad \beta \alpha = \a... | By definition, the '''dihedral group''' $D_n$ of order $2 n$ is the group of symmetries of the regular $n$-gon.
So, let $P$ denote a regular polygon with $n$ sides.
Let $\alpha$ be a rotation of $P$ by $\dfrac {2 \pi} n$.
Let $\beta$ be a reflection $P$ whose axis of reflection is the $y$ axis.
It takes $n$ rotations b... | The [[Definition:Dihedral Group|dihedral group]] $D_n$ has the [[Definition:Group Presentation|group presentation]]:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
That is, the [[Definition:Dihedral Group|dihedral group]] $D_n$ is [[Definition:Generator of Group|generated]] b... | By definition, the '''[[Definition:Dihedral Group|dihedral group]]''' $D_n$ of order $2 n$ is the [[Definition:Group|group]] of [[Definition:Symmetry (Geometry)|symmetries]] of the [[Definition:Regular Polygon|regular $n$-gon]].
So, let $P$ denote a [[Definition:Regular Polygon|regular polygon]] with $n$ [[Definition:... | Dihedral Group/Group Presentation | https://proofwiki.org/wiki/Dihedral_Group/Group_Presentation | https://proofwiki.org/wiki/Dihedral_Group/Group_Presentation | [
"Dihedral Groups",
"Examples of Group Presentations"
] | [
"Definition:Dihedral Group",
"Definition:Group Presentation",
"Definition:Dihedral Group",
"Definition:Generator of Group"
] | [
"Definition:Dihedral Group",
"Definition:Group",
"Definition:Symmetry (Geometry)",
"Definition:Polygon/Regular",
"Definition:Polygon/Regular",
"Definition:Polygon/Side",
"Definition:Rotation (Geometry)/Plane",
"Definition:Reflection (Geometry)/Plane",
"Definition:Reflection (Geometry)/Plane/Axis",
... |
proofwiki-4724 | Intersection with Subgroup Product of Superset | Let $X, Y, Z$ be subgroups of a group $\struct {G, \circ}$.
Let $Y \subseteq X$.
Then:
:$X \cap \paren {Y \circ Z} = Y \circ \paren {X \cap Z}$
where $Y \circ Z$ denotes subset product. | By definition of set equality, it suffices to prove:
:$X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$
and:
:$Y \circ \paren {X \cap Z} \subseteq X \cap \paren {Y \circ Z}$ | Let $X, Y, Z$ be [[Definition:Subgroup|subgroups]] of a [[Definition:Group|group]] $\struct {G, \circ}$.
Let $Y \subseteq X$.
Then:
:$X \cap \paren {Y \circ Z} = Y \circ \paren {X \cap Z}$
where $Y \circ Z$ denotes [[Definition:Subset Product|subset product]]. | By definition of [[Definition:Set Equality/Definition 2|set equality]], it suffices to prove:
:$X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$
and:
:$Y \circ \paren {X \cap Z} \subseteq X \cap \paren {Y \circ Z}$ | Intersection with Subgroup Product of Superset | https://proofwiki.org/wiki/Intersection_with_Subgroup_Product_of_Superset | https://proofwiki.org/wiki/Intersection_with_Subgroup_Product_of_Superset | [
"Subset Products",
"Subgroups"
] | [
"Definition:Subgroup",
"Definition:Group",
"Definition:Subset Product"
] | [
"Definition:Set Equality/Definition 2"
] |
proofwiki-4725 | Properties of Semi-Inner Product | Let $V$ be a vector space over $\Bbb F \in \set {\R, \C}$.
Let $\innerprod \cdot \cdot$ be a semi-inner product on $V$.
Denote, for $x \in V$, $\norm x := \innerprod x x^{1 / 2}$.
Then, $\forall x, y \in V, a \in \Bbb F$:
:$(1): \quad \norm {x + y} \le \norm x + \norm y$
:$(2): \quad \norm {a x} = \size a \norm x$ | === Proof of $(1)$ ===
For $x, y \in V$, compute:
{{begin-eqn}}
{{eqn | l = \norm {x + y}^2
| r = \innerprod {x + y} {x + y}
| c = Definition of $\norm \cdot$
}}
{{eqn | r = \innerprod x x + \innerprod x y + \innerprod y x + \innerprod y y
| c = Linearity of $\innerprod \cdot \cdot$
}}
{{eqn | o = \le... | Let $V$ be a [[Definition:Vector Space|vector space]] over $\Bbb F \in \set {\R, \C}$.
Let $\innerprod \cdot \cdot$ be a [[Definition:Semi-Inner Product|semi-inner product]] on $V$.
Denote, for $x \in V$, $\norm x := \innerprod x x^{1 / 2}$.
Then, $\forall x, y \in V, a \in \Bbb F$:
:$(1): \quad \norm {x + y} \le ... | === Proof of $(1)$ ===
For $x, y \in V$, compute:
{{begin-eqn}}
{{eqn | l = \norm {x + y}^2
| r = \innerprod {x + y} {x + y}
| c = Definition of $\norm \cdot$
}}
{{eqn | r = \innerprod x x + \innerprod x y + \innerprod y x + \innerprod y y
| c = [[Definition:Semi-Inner Product|Linearity]] of $\inner... | Properties of Semi-Inner Product | https://proofwiki.org/wiki/Properties_of_Semi-Inner_Product | https://proofwiki.org/wiki/Properties_of_Semi-Inner_Product | [
"Semi-Inner Product Spaces"
] | [
"Definition:Vector Space",
"Definition:Semi-Inner Product"
] | [
"Definition:Semi-Inner Product",
"Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces",
"Definition:Square Root",
"Definition:Semi-Inner Product",
"Definition:Semi-Inner Product",
"Definition:Semi-Inner Product",
"Definition:Square Root"
] |
proofwiki-4726 | Inner Product Norm is Norm | Let $V$ be an inner product space over a subfield $\Bbb F$ of $\C$.
Let $\norm {\, \cdot \,}$ denote the inner product norm on $V$.
Then $\norm {\, \cdot \,}$ is a norm on $V$. | Let us verify the norm axioms in turn. | Let $V$ be an [[Definition:Inner Product Space|inner product space]] over a [[Definition:Field (Abstract Algebra)|subfield]] $\Bbb F$ of $\C$.
Let $\norm {\, \cdot \,}$ denote the [[Definition:Inner Product Norm|inner product norm]] on $V$.
Then $\norm {\, \cdot \,}$ is a [[Definition:Norm on Vector Space|norm]] on ... | Let us verify the [[Definition:Norm on Vector Space|norm]] axioms in turn. | Inner Product Norm is Norm | https://proofwiki.org/wiki/Inner_Product_Norm_is_Norm | https://proofwiki.org/wiki/Inner_Product_Norm_is_Norm | [
"Examples of Norms",
"Inner Product Spaces"
] | [
"Definition:Inner Product Space",
"Definition:Field (Abstract Algebra)",
"Definition:Inner Product Norm",
"Definition:Norm/Vector Space"
] | [
"Definition:Norm/Vector Space",
"Definition:Norm/Vector Space"
] |
proofwiki-4727 | Metric Induced by Norm is Metric | Let $V$ be a normed vector space.
Let $\norm{\,\cdot\,}$ denote its norm.
Let $d$ be the metric induced by $\norm {\,\cdot\,}$.
Then $d$ is a metric. | === Proof of {{Metric-space-axiom|1}} and {{Metric-space-axiom|4}} ===
Let $x, y \in V$.
Then:
:$\map d {x, y} = \norm {x - y} \ge 0$
and furthermore:
{{begin-eqn}}
{{eqn | l = \map d {x, y}
| r = 0
}}
{{eqn | ll= \leadstoandfrom
| l = \norm {x - y}
| r = 0
}}
{{eqn | ll= \leadstoandfrom
| l = ... | Let $V$ be a [[Definition:Normed Vector Space|normed vector space]].
Let $\norm{\,\cdot\,}$ denote its [[Definition:Norm on Vector Space|norm]].
Let $d$ be the [[Definition:Metric Induced by Norm|metric induced by $\norm {\,\cdot\,}$]].
Then $d$ is a [[Definition:Metric|metric]]. | === Proof of {{Metric-space-axiom|1}} and {{Metric-space-axiom|4}} ===
Let $x, y \in V$.
Then:
:$\map d {x, y} = \norm {x - y} \ge 0$
and furthermore:
{{begin-eqn}}
{{eqn | l = \map d {x, y}
| r = 0
}}
{{eqn | ll= \leadstoandfrom
| l = \norm {x - y}
| r = 0
}}
{{eqn | ll= \leadstoandfrom
| ... | Metric Induced by Norm is Metric | https://proofwiki.org/wiki/Metric_Induced_by_Norm_is_Metric | https://proofwiki.org/wiki/Metric_Induced_by_Norm_is_Metric | [
"Norm Theory",
"Metric Spaces"
] | [
"Definition:Normed Vector Space",
"Definition:Norm/Vector Space",
"Definition:Metric Induced by Norm",
"Definition:Metric Space/Metric"
] | [] |
proofwiki-4728 | Product of Sums of Four Squares/Corollary | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n, c_1, c_2, \ldots, c_n, d_1, d_2, \ldots, d_n$ be integers.
Then:
:$\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
That is, the product of any number of sums of four squares is also a sum of four sq... | Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
$\map P 1$ is true, as this just says:
:$\exists w, x, y, z \in \Z: a^2 + b^2 + c^2 + d^2 = w^2 + x^2 + y^2 + z^2$
whic... | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n, c_1, c_2, \ldots, c_n, d_1, d_2, \ldots, d_n$ be [[Definition:Integer|integers]].
Then:
:$\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
That is, the [[Definition:Integer Multiplication|product]... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \exists w, x, y, z \in \Z: \prod_{j \mathop = 1}^n \paren {a_j^2 + b_j^2 + c_j^2 + d_j^2} = w^2 + x^2 + y^2 + z^2$
$\map P 1$ is true, as this just says:
:$\exists ... | Product of Sums of Four Squares/Corollary | https://proofwiki.org/wiki/Product_of_Sums_of_Four_Squares/Corollary | https://proofwiki.org/wiki/Product_of_Sums_of_Four_Squares/Corollary | [
"Product of Sums of Four Squares",
"Proofs by Induction"
] | [
"Definition:Integer",
"Definition:Multiplication/Integers",
"Definition:Addition/Integers",
"Definition:Square Number"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-4729 | Subgroup Subset of Subgroup Product | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Then:
:$H \subseteq H \circ K \supseteq K$
where $H \circ K$ denotes the subset product of $H$ and $K$. | By definition of subset product:
:$H \circ K = \set {h \circ k: h \in H, k \in K}$
So:
{{begin-eqn}}
{{eqn | l=x
| o=\in
| r=H
| c=
}}
{{eqn | ll=\leadsto
| l=x
| r=x \circ e
| c={{Defof|Identity Element}}
}}
{{eqn | o=\in
| r=H \circ K
| c=Identity of Subgroup
}}
{{eqn |... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
Then:
:$H \subseteq H \circ K \supseteq K$
where $H \circ K$ denotes the [[Definition:Subset Product|subset product]] of $H$ and $K$. | By definition of [[Definition:Subset Product|subset product]]:
:$H \circ K = \set {h \circ k: h \in H, k \in K}$
So:
{{begin-eqn}}
{{eqn | l=x
| o=\in
| r=H
| c=
}}
{{eqn | ll=\leadsto
| l=x
| r=x \circ e
| c={{Defof|Identity Element}}
}}
{{eqn | o=\in
| r=H \circ K
| c... | Subgroup Subset of Subgroup Product | https://proofwiki.org/wiki/Subgroup_Subset_of_Subgroup_Product | https://proofwiki.org/wiki/Subgroup_Subset_of_Subgroup_Product | [
"Subset Products",
"Subgroups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Subset Product"
] | [
"Definition:Subset Product",
"Identity of Subgroup",
"Identity of Subgroup",
"Category:Subset Products",
"Category:Subgroups"
] |
proofwiki-4730 | Product Space is Path-connected iff Factor Spaces are Path-connected | Let $\SS = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces for $i$ in some indexing set $I$ such that $\forall i \in I: S_i \ne \O$.
Let $\ds T = \struct {S, \tau} = \prod_{i \mathop \in I} \struct {S_i, \tau_i}$ be the product space of $\SS$.
Then $T$ is a path-connected ... | === Necessary Condition ===
Suppose $S_i$ is path-connected for each $i \in I$.
Let $x \ne y \in S$ be arbitrary.
Since each $S_i$ is path-connected, we have a continuous mapping:
:$f_i: \closedint 0 1 \to S_i$
such that $\map {f_i} 0 = x_i$ and $\map {f_i} 1 = y_i$ for all $i \in I$.
We then define:
: $f : \closedin... | Let $\SS = \family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] for $i$ in some [[Definition:Indexing Set|indexing set]] $I$ such that $\forall i \in I: S_i \ne \O$.
Let $\ds T = \struct {S, \tau} = \prod_{i \matho... | === Necessary Condition ===
Suppose $S_i$ is [[Definition:Path-Connected Space|path-connected]] for each $i \in I$.
Let $x \ne y \in S$ be arbitrary.
Since each $S_i$ is [[Definition:Path-Connected Space|path-connected]], we have a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]]:
:$f_i: \... | Product Space is Path-connected iff Factor Spaces are Path-connected | https://proofwiki.org/wiki/Product_Space_is_Path-connected_iff_Factor_Spaces_are_Path-connected | https://proofwiki.org/wiki/Product_Space_is_Path-connected_iff_Factor_Spaces_are_Path-connected | [
"Path-Connected Spaces",
"Product Spaces"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Path-Connected/Topological Space",
"Definition:Path-Connected/Topological Space"
] | [
"Definition:Path-Connected/Topological Space",
"Definition:Path-Connected/Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Continuous Mapping to Product Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Defin... |
proofwiki-4731 | Normal Subgroup of Subset Product of Subgroups | Let $G$ be a group whose identity is $e$.
Let:
: $H$ be a subgroup of $G$
: $N$ be a normal subgroup of $G$.
Then:
: $N \lhd N H$
where:
: $\lhd$ denotes normal subgroup
: $N H$ denotes subset product. | From Subset Product with Normal Subgroup is Subgroup:
:$N H = H N$ is a subgroup of $G$.
By definition of subset product all elements of $H N$ can be written in the form:
:$h n \in H N$
where $h \in H, n \in N$.
Let $h n \in H N$.
Let $n_1 \in N$.
From Inverse of Group Product:
:$\paren {h n} n_1 \paren {h n}^{-1} = h ... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let:
: $H$ be a [[Definition:Subgroup|subgroup]] of $G$
: $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Then:
: $N \lhd N H$
where:
: $\lhd$ denotes [[Definition:Normal Subgroup|normal subgroup]]
: $N H$... | From [[Subset Product with Normal Subgroup is Subgroup]]:
:$N H = H N$ is a [[Definition:Subgroup|subgroup]] of $G$.
By definition of [[Definition:Subset Product|subset product]] all [[Definition:Element|elements]] of $H N$ can be written in the form:
:$h n \in H N$
where $h \in H, n \in N$.
Let $h n \in H N$.
Let $... | Normal Subgroup of Subset Product of Subgroups | https://proofwiki.org/wiki/Normal_Subgroup_of_Subset_Product_of_Subgroups | https://proofwiki.org/wiki/Normal_Subgroup_of_Subset_Product_of_Subgroups | [
"Normal Subgroups",
"Subset Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup",
"Definition:Subset Product"
] | [
"Subset Product with Normal Subgroup is Subgroup",
"Definition:Subgroup",
"Definition:Subset Product",
"Definition:Element",
"Inverse of Group Product",
"Definition:Normal Subgroup"
] |
proofwiki-4732 | Subset Products of Normal Subgroup with Normal Subgroup of Subgroup | Let $G$ be a group.
Let:
:$(1): \quad H$ be a subgroup of $G$
:$(2): \quad K$ be a normal subgroup of $H$
:$(3): \quad N$ be a normal subgroup of $G$
Then:
:$N K \lhd N H$
where:
: $N K$ and $N H$ denote subset product
: $\lhd$ denotes the relation of being a normal subgroup. | Consider arbitrary $x_n \in N, x_h \in H$.
Thus:
:$x_n x_h \in N H$
We aim to show that:
:$x_n x_h N K \paren {x_n x_h}^{-1} \subseteq N K$
thus demonstrating $N K \lhd N H$ by the Normal Subgroup Test.
We have:
{{begin-eqn}}
{{eqn | l = x_n x_h N K \paren {x_n x_h}^{-1}
| r = x_n x_h N K {x_h}^{-1} {x_n}^{-1}
... | Let $G$ be a group.
Let:
:$(1): \quad H$ be a [[Definition:Subgroup|subgroup]] of $G$
:$(2): \quad K$ be a [[Definition:Normal Subgroup|normal subgroup]] of $H$
:$(3): \quad N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$
Then:
:$N K \lhd N H$
where:
: $N K$ and $N H$ denote [[Definition:Subset Product... | Consider arbitrary $x_n \in N, x_h \in H$.
Thus:
:$x_n x_h \in N H$
We aim to show that:
:$x_n x_h N K \paren {x_n x_h}^{-1} \subseteq N K$
thus demonstrating $N K \lhd N H$ by the [[Normal Subgroup Test]].
We have:
{{begin-eqn}}
{{eqn | l = x_n x_h N K \paren {x_n x_h}^{-1}
| r = x_n x_h N K {x_h}^{-1} {x_n... | Subset Products of Normal Subgroup with Normal Subgroup of Subgroup | https://proofwiki.org/wiki/Subset_Products_of_Normal_Subgroup_with_Normal_Subgroup_of_Subgroup | https://proofwiki.org/wiki/Subset_Products_of_Normal_Subgroup_with_Normal_Subgroup_of_Subgroup | [
"Normal Subgroups",
"Subset Products"
] | [
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup",
"Definition:Subset Product",
"Definition:Normal Subgroup"
] | [
"Normal Subgroup Test",
"Inverse of Group Product"
] |
proofwiki-4733 | Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group | Let $G$ be a group with the following properties:
:$(1): \quad G$ is non-abelian.
:$(2): \quad G$ is of order $8$.
:$(3): \quad G$ has precisely one element of order $2$.
Then $G$ is isomorphic to the quaternion group $Q$. | From Order of Element Divides Order of Finite Group all the elements in $G$ have order $1, 2, 4$ or $8$.
From Cyclic Group is Abelian, $\paren 1$ and $\paren 2$, no elements in $G$ have order $8$, i.e. they all have order $1, 2$ or $4$.
Let the identity element be $1$ and the one with order $2$ be $-1$.
Also denote $1 ... | Let $G$ be a [[Definition:Group|group]] with the following properties:
:$(1): \quad G$ is [[Definition:Abelian Group|non-abelian]].
:$(2): \quad G$ is of [[Definition:Order of Structure|order $8$]].
:$(3): \quad G$ has precisely one element of [[Definition:Order of Group Element|order $2$]].
Then $G$ is [[Definiti... | From [[Order of Element Divides Order of Finite Group]] all the [[Definition:Element|elements]] in $G$ have [[Definition:Order of Group Element|order]] $1, 2, 4$ or $8$.
From [[Cyclic Group is Abelian]], $\paren 1$ and $\paren 2$, no [[Definition:Element|elements]] in $G$ have order $8$, i.e. they all have order $1, 2... | Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Proof 1 | https://proofwiki.org/wiki/Non-Abelian_Order_8_Group_with_One_Order_2_Element_is_Quaternion_Group | https://proofwiki.org/wiki/Non-Abelian_Order_8_Group_with_One_Order_2_Element_is_Quaternion_Group/Proof_1 | [
"Quaternion Group",
"Groups of Order 8",
"Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group"
] | [
"Definition:Group",
"Definition:Abelian Group",
"Definition:Order of Structure",
"Definition:Order of Group Element",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Dicyclic Group/Quaternion Group"
] | [
"Order of Element Divides Order of Finite Group",
"Definition:Element",
"Definition:Order of Group Element",
"Cyclic Group is Abelian",
"Definition:Element",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 1",
"... |
proofwiki-4734 | Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group | Let $G$ be a group with the following properties:
:$(1): \quad G$ is non-abelian.
:$(2): \quad G$ is of order $8$.
:$(3): \quad G$ has precisely one element of order $2$.
Then $G$ is isomorphic to the quaternion group $Q$. | By Groups of Order 8, the only group satisfying all three properties is $Q = \Dic 2$.
{{qed}} | Let $G$ be a [[Definition:Group|group]] with the following properties:
:$(1): \quad G$ is [[Definition:Abelian Group|non-abelian]].
:$(2): \quad G$ is of [[Definition:Order of Structure|order $8$]].
:$(3): \quad G$ has precisely one element of [[Definition:Order of Group Element|order $2$]].
Then $G$ is [[Definiti... | By [[Groups of Order 8]], the only [[Definition:Group|group]] satisfying all three properties is $Q = \Dic 2$.
{{qed}} | Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Proof 2 | https://proofwiki.org/wiki/Non-Abelian_Order_8_Group_with_One_Order_2_Element_is_Quaternion_Group | https://proofwiki.org/wiki/Non-Abelian_Order_8_Group_with_One_Order_2_Element_is_Quaternion_Group/Proof_2 | [
"Quaternion Group",
"Groups of Order 8",
"Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group"
] | [
"Definition:Group",
"Definition:Abelian Group",
"Definition:Order of Structure",
"Definition:Order of Group Element",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Dicyclic Group/Quaternion Group"
] | [
"Groups of Order 8",
"Definition:Group"
] |
proofwiki-4735 | Quaternion Group not Dihedral Group | Let $Q$ be the quaternion group.
Then $Q$ is not isomorphic to the dihedral group $D_4$. | From Group Presentation of Dihedral Group, $D_4$ is generated by two elements of orders $4$ and $2$ respectively.
Let these generators be $\alpha$ and $\beta$ where:
:$\alpha^4 = e$
:$\beta^2 = e$
Hence $\alpha^2$ and $\beta$ are different elements of $D_4$ which both have order $2$.
But the quaternion group:
:$Q = \se... | Let $Q$ be the [[Definition:Quaternion Group|quaternion group]].
Then $Q$ is not [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Dihedral Group|dihedral group]] $D_4$. | From [[Group Presentation of Dihedral Group]], $D_4$ is [[Definition:Generator of Group|generated]] by two [[Definition:Element|elements]] of [[Definition:Order of Group Element|orders]] $4$ and $2$ respectively.
Let these [[Definition:Generator of Group|generators]] be $\alpha$ and $\beta$ where:
:$\alpha^4 = e$
:$\b... | Quaternion Group not Dihedral Group | https://proofwiki.org/wiki/Quaternion_Group_not_Dihedral_Group | https://proofwiki.org/wiki/Quaternion_Group_not_Dihedral_Group | [
"Quaternion Group",
"Dihedral Groups"
] | [
"Definition:Dicyclic Group/Quaternion Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Dihedral Group"
] | [
"Dihedral Group/Group Presentation",
"Definition:Generator of Group",
"Definition:Element",
"Definition:Order of Group Element",
"Definition:Generator of Group",
"Definition:Element",
"Definition:Order of Group Element",
"Quaternion Group/Complex Matrices",
"Definition:Element",
"Definition:Order ... |
proofwiki-4736 | Subset Product of Normal Subgroups with Trivial Intersection | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H, K$ be normal subgroups of $G$.
Let $H \cap K = e$.
Then $H K$ is isomorphic to $H \times K$ where:
: $H K$ denotes the subset product of $H$ and $K$
: $H \times K$ denotes the direct product of $H$ and $K$. | Let $G' = H K$.
From Subset Product of Normal Subgroups is Normal, $G'$ is a normal subgroup of $G$.
That is $G'$ is itself a group.
So by the Internal Direct Product Theorem, $G'$ is the internal group direct product of $H$ and $K$.
The result follows by definition of the internal group direct product.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H, K$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$.
Let $H \cap K = e$.
Then $H K$ is [[Definition:Group Isomorphism|isomorphic]] to $H \times K$ where:
: $H K$ denotes the [[Definition:... | Let $G' = H K$.
From [[Subset Product of Normal Subgroups is Normal]], $G'$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
That is $G'$ is itself a [[Definition:Group|group]].
So by the [[Internal Direct Product Theorem]], $G'$ is the [[Definition:Internal Group Direct Product|internal group direct prod... | Subset Product of Normal Subgroups with Trivial Intersection | https://proofwiki.org/wiki/Subset_Product_of_Normal_Subgroups_with_Trivial_Intersection | https://proofwiki.org/wiki/Subset_Product_of_Normal_Subgroups_with_Trivial_Intersection | [
"Normal Subgroups",
"Group Direct Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Normal Subgroup",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Subset Product",
"Definition:Group Direct Product"
] | [
"Subset Product of Normal Subgroups is Normal",
"Definition:Normal Subgroup",
"Definition:Group",
"Internal Direct Product Theorem",
"Definition:Internal Group Direct Product",
"Definition:Internal Group Direct Product"
] |
proofwiki-4737 | Quotient Group of Direct Products | Let $G$ and $G'$ be groups.
Let:
{{begin-eqn}}
{{eqn | l = H
| o = \lhd
| r = G
}}
{{eqn | l = H'
| o = \lhd
| r = G'
}}
{{end-eqn}}
where $\lhd$ denotes the relation of being a normal subgroup.
Then:
:$\paren {G \times G'} / \paren {H \times H'}$ is isomorphic to $\paren {G / H} \times \paren {... | Let $\phi_1: G \to G / H$ and $\phi_2: G' \to G' / H'$ be the quotient epimorphisms with $H$ and $H'$ as their kernels, respectively.
Now define a homomorphism $\phi: G \times G' \to \paren {G / H} \times \paren {G' / H'}$ by:
:$\phi = \phi_1 \times \phi_2$
so:
:$\map \phi {x, x'} = \tuple {\map {\phi_1} x, \map {\phi_... | Let $G$ and $G'$ be [[Definition:Group|groups]].
Let:
{{begin-eqn}}
{{eqn | l = H
| o = \lhd
| r = G
}}
{{eqn | l = H'
| o = \lhd
| r = G'
}}
{{end-eqn}}
where $\lhd$ denotes the relation of being a [[Definition:Normal Subgroup|normal subgroup]].
Then:
:$\paren {G \times G'} / \paren {H \ti... | Let $\phi_1: G \to G / H$ and $\phi_2: G' \to G' / H'$ be the [[Definition:Quotient Group Epimorphism|quotient epimorphisms]] with $H$ and $H'$ as their [[Definition:Kernel of Group Homomorphism|kernels]], respectively.
Now define a [[Definition:Group Homomorphism|homomorphism]] $\phi: G \times G' \to \paren {G / H} \... | Quotient Group of Direct Products | https://proofwiki.org/wiki/Quotient_Group_of_Direct_Products | https://proofwiki.org/wiki/Quotient_Group_of_Direct_Products | [
"Quotient Groups",
"Group Direct Products"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Direct Product",
"Definition:Quotient Group"
] | [
"Definition:Quotient Epimorphism/Group",
"Definition:Kernel of Group Homomorphism",
"Definition:Group Homomorphism",
"Definition:Kernel of Group Homomorphism",
"Definition:Surjection",
"First Isomorphism Theorem/Groups"
] |
proofwiki-4738 | Completion Theorem (Inner Product Space) | Let $\GF \in \set {\R, \C}$.
Let $\struct {V, \innerprod \cdot \cdot_V}$ be an inner product space over a subfield $\GF$.
Let $\norm {\, \cdot \,}_V$ be the inner product norm induced by $\innerprod \cdot \cdot_V$.
Let $\struct {H, \norm {\, \cdot \,}_H}$ be a completion of $V$ with respect to $\norm {\, \cdot \,}$ giv... | We first show that $\norm {\, \cdot \,}_H$ is induced by an inner product by using Norm satisfying Parallelogram Law induced by Inner Product.
From Completion Theorem (Normed Vector Space), there exists a linear isometry $\phi : V \to H$ such that $\phi \sqbrk V$ is dense in $H$.
From Parallelogram Law (Inner Product ... | Let $\GF \in \set {\R, \C}$.
Let $\struct {V, \innerprod \cdot \cdot_V}$ be an [[Definition:Inner Product Space|inner product space]] over a [[Definition:Subfield|subfield]] $\GF$.
Let $\norm {\, \cdot \,}_V$ be the [[Definition:Inner Product Norm|inner product norm]] induced by $\innerprod \cdot \cdot_V$.
Let $\str... | We first show that $\norm {\, \cdot \,}_H$ is [[Definition:Inner Product Norm|induced by an inner product]] by using [[Norm satisfying Parallelogram Law induced by Inner Product]].
From [[Completion Theorem (Normed Vector Space)]], there exists a [[Definition:Linear Isometry|linear isometry]] $\phi : V \to H$ such tha... | Completion Theorem (Inner Product Space) | https://proofwiki.org/wiki/Completion_Theorem_(Inner_Product_Space) | https://proofwiki.org/wiki/Completion_Theorem_(Inner_Product_Space) | [
"Hilbert Spaces",
"Inner Product Spaces",
"Named Theorems"
] | [
"Definition:Inner Product Space",
"Definition:Subfield",
"Definition:Inner Product Norm",
"Definition:Completion (Metric Space)",
"Completion Theorem (Normed Vector Space)",
"Definition:Inner Product",
"Definition:Hilbert Space",
"Definition:Inner Product Space",
"Definition:Completion (Metric Space... | [
"Definition:Inner Product Norm",
"Norm satisfying Parallelogram Law induced by Inner Product",
"Completion Theorem (Normed Vector Space)",
"Definition:Linear Isometry",
"Definition:Everywhere Dense",
"Parallelogram Law (Inner Product Space)",
"Definition:Linear Isometry",
"Definition:Everywhere Dense"... |
proofwiki-4739 | Identity Mapping is Group Endomorphism | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then $I_G: \struct {G, \circ} \to \struct {G, \circ}$ is a group endomorphism. | From Identity Mapping is Group Automorphism, $G$ is a group automorphism.
The result follows from Group Automorphism is Endomorphism.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Then $I_G: \struct {G, \circ} \to \struct {G, \circ}$ is a [[Definition:Group Endomorphism|group endomorphism]]. | From [[Identity Mapping is Group Automorphism]], $G$ is a [[Definition:Group Automorphism|group automorphism]].
The result follows from [[Group Automorphism is Endomorphism]].
{{qed}} | Identity Mapping is Group Endomorphism/Proof 1 | https://proofwiki.org/wiki/Identity_Mapping_is_Group_Endomorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Group_Endomorphism/Proof_1 | [
"Identity Mapping is Group Endomorphism",
"Identity Mappings",
"Group Endomorphisms"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Endomorphism"
] | [
"Identity Mapping is Automorphism/Groups",
"Definition:Group Automorphism",
"Group Automorphism is Endomorphism"
] |
proofwiki-4740 | Pythagoras's Theorem (Inner Product Space) | Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $\norm \cdot$ be the inner product norm on $\struct {V, \innerprod \cdot \cdot}$.
Let $f_1, \ldots, f_n \in V$ form an orthogonal set.
Then:
:$\ds \norm {\sum_{i \mathop = 1}^n f_i}^2 = \sum_{i \mathop = 1}^n \norm {f_i}^2$ | {{begin-eqn}}
{{eqn | l = \norm {\sum_{i \mathop = 1}^n f_i}^2
| r = \innerprod {\sum_{i \mathop = 1}^n f_i} {\sum_{j \mathop = 1}^n f_j}
| c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \innerprod {f_i} {f_j}
| c = since the Inner Product is linear in th... | Let $\struct {V, \innerprod \cdot \cdot}$ be an [[Definition:Inner Product Space|inner product space]].
Let $\norm \cdot$ be the [[Definition:Inner Product Norm|inner product norm]] on $\struct {V, \innerprod \cdot \cdot}$.
Let $f_1, \ldots, f_n \in V$ form an [[Definition:Orthogonal (Linear Algebra)/Set|orthogonal ... | {{begin-eqn}}
{{eqn | l = \norm {\sum_{i \mathop = 1}^n f_i}^2
| r = \innerprod {\sum_{i \mathop = 1}^n f_i} {\sum_{j \mathop = 1}^n f_j}
| c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \innerprod {f_i} {f_j}
| c = since the [[Definition:Inner Product|In... | Pythagoras's Theorem (Inner Product Space) | https://proofwiki.org/wiki/Pythagoras's_Theorem_(Inner_Product_Space) | https://proofwiki.org/wiki/Pythagoras's_Theorem_(Inner_Product_Space) | [
"Inner Product Spaces",
"Pythagoras's Theorem"
] | [
"Definition:Inner Product Space",
"Definition:Inner Product Norm",
"Definition:Orthogonal (Linear Algebra)/Set"
] | [
"Definition:Inner Product",
"Definition:Linear Functional",
"Definition:Orthogonal (Linear Algebra)"
] |
proofwiki-4741 | Primitive of Reciprocal of x squared plus a squared/Arctangent Form | :$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$ | Let:
:$a \tan \theta = x$
for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
From Shape of Tangent Function, this substitution is valid for all real $x$.
Then:
{{begin-eqn}}
{{eqn | l = x
| r = a \tan \theta
| c = from above
}}
{{eqn | ll= \leadsto
| l = \frac {\d x} {\d \theta}
| r = a ... | :$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$ | Let:
:$a \tan \theta = x$
for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
From [[Shape of Tangent Function]], this substitution is valid for all [[Definition:Real Number|real]] $x$.
Then:
{{begin-eqn}}
{{eqn | l = x
| r = a \tan \theta
| c = from above
}}
{{eqn | ll= \leadsto
| l = \fra... | Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Proof 1 | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_x_squared_plus_a_squared/Arctangent_Form | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_x_squared_plus_a_squared/Arctangent_Form/Proof_1 | [
"Primitive of Reciprocal of x squared plus a squared",
"Expressions whose Primitives are Inverse Trigonometric Functions",
"Primitives involving x squared plus a squared",
"Arctangent Function",
"Primitives involving Reciprocals"
] | [] | [
"Shape of Tangent Function",
"Definition:Real Number",
"Derivative of Tangent Function",
"Integration by Substitution",
"Primitive of Constant Multiple of Function",
"Sum of Squares of Sine and Cosine/Corollary 1",
"Integral of Constant",
"Definition:Real Interval/Open",
"Definition:Primitive (Calcu... |
proofwiki-4742 | Primitive of Reciprocal of x squared plus a squared/Arctangent Form | :$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$ | We have that $x^2 + a^2$ is in the form $a x^2 + b x + c$, where $b^2 - 4 a c < 0$.
Thus from Primitive of $\dfrac 1 {a x^2 + b x + c}$ for $b^2 - 4 a c > 0$:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
setting $a := 1, b := 0,... | :$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$ | We have that $x^2 + a^2$ is in the form $a x^2 + b x + c$, where $b^2 - 4 a c < 0$.
Thus from [[Primitive of Reciprocal of a x squared plus b x plus c|Primitive of $\dfrac 1 {a x^2 + b x + c}$]] for $b^2 - 4 a c > 0$:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x... | Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Proof 2 | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_x_squared_plus_a_squared/Arctangent_Form | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_x_squared_plus_a_squared/Arctangent_Form/Proof_2 | [
"Primitive of Reciprocal of x squared plus a squared",
"Expressions whose Primitives are Inverse Trigonometric Functions",
"Primitives involving x squared plus a squared",
"Arctangent Function",
"Primitives involving Reciprocals"
] | [] | [
"Primitive of Reciprocal of a x squared plus b x plus c",
"Primitive of Reciprocal of a x squared plus b x plus c"
] |
proofwiki-4743 | Primitive of Reciprocal of x squared plus a squared/Arctangent Form | :$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$ | {{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 + a^2}
| r = \frac 1 a \int \frac {\d t} {t^2 + 1}
| c = Substitution of $x \to a t$
}}
{{eqn | r = \frac 1 a \int \frac {\d t} {\paren {1 + i t} \paren {1 - i t} }
| c = factoring
}}
{{eqn | r = \frac 1 {2 a} \paren {\int \frac {\d t} {1 + i t} + \int ... | :$\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$ | {{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 + a^2}
| r = \frac 1 a \int \frac {\d t} {t^2 + 1}
| c = [[Integration by Substitution|Substitution of $x \to a t$]]
}}
{{eqn | r = \frac 1 a \int \frac {\d t} {\paren {1 + i t} \paren {1 - i t} }
| c = factoring
}}
{{eqn | r = \frac 1 {2 a} \paren {\in... | Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Proof 3 | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_x_squared_plus_a_squared/Arctangent_Form | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_x_squared_plus_a_squared/Arctangent_Form/Proof_3 | [
"Primitive of Reciprocal of x squared plus a squared",
"Expressions whose Primitives are Inverse Trigonometric Functions",
"Primitives involving x squared plus a squared",
"Arctangent Function",
"Primitives involving Reciprocals"
] | [] | [
"Integration by Substitution",
"Primitive of Reciprocal",
"Sum of Logarithms",
"Arctangent Logarithmic Formulation"
] |
proofwiki-4744 | Endomorphism from Integers to Multiples | Let $\struct {\Z, +}$ be the additive group of integers.
Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be a mapping.
Then $\phi$ is a group endomorphism {{iff}}:
:$\exists k \in \Z: \forall n \in \Z: \map \phi n = k n$ | === Necessary Condition ===
Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be an endomorphism.
Let $k = \map \phi 1$.
We have that $n = \overbrace {1 + \cdots + 1}^n$ for any positive integer $n$.
Thus:
{{begin-eqn}}
{{eqn | l = \map \phi n
| r = \map \phi {\overbrace {1 + \cdots + 1}^n}
}}
{{eqn | r = \overbrac... | Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be a [[Definition:Mapping|mapping]].
Then $\phi$ is a [[Definition:Group Endomorphism|group endomorphism]] {{iff}}:
:$\exists k \in \Z: \forall n \in \Z: \map \phi n = k ... | === Necessary Condition ===
Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be an [[Definition:Group Endomorphism|endomorphism]].
Let $k = \map \phi 1$.
We have that $n = \overbrace {1 + \cdots + 1}^n$ for any [[Definition:Strictly Positive Integer|positive integer]] $n$.
Thus:
{{begin-eqn}}
{{eqn | l = \map \phi ... | Endomorphism from Integers to Multiples | https://proofwiki.org/wiki/Endomorphism_from_Integers_to_Multiples | https://proofwiki.org/wiki/Endomorphism_from_Integers_to_Multiples | [
"Integers",
"Group Endomorphisms"
] | [
"Definition:Additive Group of Integers",
"Definition:Mapping",
"Definition:Group Endomorphism"
] | [
"Definition:Group Endomorphism",
"Definition:Strictly Positive/Integer",
"Definition:Group Endomorphism"
] |
proofwiki-4745 | Finite Cyclic Group is Isomorphic to Integers under Modulo Addition | Let $\struct {G, \circ}$ be a finite group whose identity element is $e$.
Then $\struct {G, \circ}$ is cyclic of order $n$ {{iff}} $\struct {G, \circ}$ is isomorphic with the additive group of integers modulo $n$ $\struct {\Z_n, +_n}$. | === Necessary Condition ===
Let $\struct {G, \circ}$ be a cyclic group of order $n$.
From List of Elements in Finite Cyclic Group:
:$G = \set {a^0, a^1, a^2, \ldots, a^n}$
where $a^0 = e, a^1 = a$.
From the definition of integers modulo $n$, $\Z_n$ can be expressed as:
:$\Z_n = \set {\eqclass 0 n, \eqclass 1 n, \ldots,... | Let $\struct {G, \circ}$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity element]] is $e$.
Then $\struct {G, \circ}$ is [[Definition:Cyclic Group|cyclic]] of [[Definition:Order of Structure|order $n$]] {{iff}} $\struct {G, \circ}$ is [[Definition:Group Isomorphism|isomorphic... | === Necessary Condition ===
Let $\struct {G, \circ}$ be a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order $n$]].
From [[List of Elements in Finite Cyclic Group]]:
:$G = \set {a^0, a^1, a^2, \ldots, a^n}$
where $a^0 = e, a^1 = a$.
From the definition of [[Definition:Integers Modulo m... | Finite Cyclic Group is Isomorphic to Integers under Modulo Addition | https://proofwiki.org/wiki/Finite_Cyclic_Group_is_Isomorphic_to_Integers_under_Modulo_Addition | https://proofwiki.org/wiki/Finite_Cyclic_Group_is_Isomorphic_to_Integers_under_Modulo_Addition | [
"Cyclic Groups",
"Modulo Arithmetic"
] | [
"Definition:Finite Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Additive Group of Integers Modulo m"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"List of Elements in Finite Cyclic Group",
"Definition:Integers Modulo m",
"Definition:Residue Class",
"Definition:Mapping",
"Definition:Bijection",
"Powers of Group Elements/Sum of Indices",
"Definition:Morphism Property",
"Definition:Gr... |
proofwiki-4746 | Parallelogram Law (Inner Product Space) | Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $\norm \cdot$ be the inner product norm of $\struct {V, \innerprod \cdot \cdot}$.
Let $f, g \in V$ be arbitrary.
Then:
:$\norm {f + g}^2 + \norm {f - g}^2 = 2 \paren {\norm f^2 + \norm g^2}$ | {{begin-eqn}}
{{eqn | l = \norm {f + g}^2 + \norm {f - g}^2
| r = \innerprod {f + g} {f + g} + \innerprod {f - g} {f - g}
| c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \innerprod f f + \innerprod f g + \innerprod g f + \innerprod g g + \innerprod f f - \innerprod f g - \innerprod g f + \innerprod g g
... | Let $\struct {V, \innerprod \cdot \cdot}$ be an [[Definition:Inner Product Space|inner product space]].
Let $\norm \cdot$ be the [[Definition:Inner Product Norm|inner product norm]] of $\struct {V, \innerprod \cdot \cdot}$.
Let $f, g \in V$ be arbitrary.
Then:
:$\norm {f + g}^2 + \norm {f - g}^2 = 2 \paren {\norm ... | {{begin-eqn}}
{{eqn | l = \norm {f + g}^2 + \norm {f - g}^2
| r = \innerprod {f + g} {f + g} + \innerprod {f - g} {f - g}
| c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \innerprod f f + \innerprod f g + \innerprod g f + \innerprod g g + \innerprod f f - \innerprod f g - \innerprod g f + \innerprod g g
... | Parallelogram Law (Inner Product Space) | https://proofwiki.org/wiki/Parallelogram_Law_(Inner_Product_Space) | https://proofwiki.org/wiki/Parallelogram_Law_(Inner_Product_Space) | [
"Inner Product Spaces"
] | [
"Definition:Inner Product Space",
"Definition:Inner Product Norm"
] | [
"Definition:Inner Product"
] |
proofwiki-4747 | Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space | Let $H$ be a Hilbert space, and let $h \in H$.
Let $K \subseteq H$ be a closed, convex, non-empty subset of $H$.
Then there is a unique point $k_0 \in K$ such that:
:$\norm {h - k_0} = \map d {h, K}$
where $d$ denotes distance to a set. | Let $\mathbf 0_H$ be the zero of $H$.
Since for every $k \in K$, we have:
:$\map d {h, k} = \norm {h - k} = \map d {\mathbf 0_H, k - h}$
it follows that:
:$\map d {h, K} = \map d {\mathbf 0_H, K - h}$
{{WLOG}}, we may therefore assume that $h = \mathbf 0_H$.
The problem has therefore reduced to finding $k_0 \in K$ such... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]], and let $h \in H$.
Let $K \subseteq H$ be a [[Definition:Closed Set (Topology)|closed]], [[Definition:Convex Set (Vector Space)|convex]], [[Definition:Non-Empty Set|non-empty]] subset of $H$.
Then there is a unique point $k_0 \in K$ such that:
:$\norm {h - k_... | Let $\mathbf 0_H$ be the [[Definition:Zero Vector|zero]] of $H$.
Since for every $k \in K$, we have:
:$\map d {h, k} = \norm {h - k} = \map d {\mathbf 0_H, k - h}$
it follows that:
:$\map d {h, K} = \map d {\mathbf 0_H, K - h}$
{{WLOG}}, we may therefore assume that $h = \mathbf 0_H$.
The problem has therefore r... | Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space | https://proofwiki.org/wiki/Unique_Point_of_Minimal_Distance_to_Closed_Convex_Subset_of_Hilbert_Space | https://proofwiki.org/wiki/Unique_Point_of_Minimal_Distance_to_Closed_Convex_Subset_of_Hilbert_Space | [
"Hilbert Spaces",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Hilbert Space",
"Definition:Closed Set/Topology",
"Definition:Convex Set (Vector Space)",
"Definition:Non-Empty Set",
"Definition:Distance/Sets/Metric Spaces"
] | [
"Definition:Zero Vector",
"Definition:Infimum of Set/Real Numbers",
"Definition:Sequence",
"Parallelogram Law (Inner Product Space)",
"Definition:Convex Set (Vector Space)",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Hilbert Space",
"Definition:Closed Set/Metric Space",
"Norm is Continuo... |
proofwiki-4748 | Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form | :$\ds \int \frac {\d x} {\sqrt {a^2 - x^2} } = \arcsin \frac x a + C$ | {{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\sqrt {a^2 - x^2} }
| r = \int \frac {\d x} {\sqrt {a^2 \paren {1 - \frac {x^2} {a^2} } } }
| c = factor $a^2$ out of the radicand
}}
{{eqn | r = \int \frac {\d x} {\sqrt {a^2} \sqrt {1 - \paren {\frac x a}^2} }
| c =
}}
{{eqn | r = \frac 1 a \int \frac {... | :$\ds \int \frac {\d x} {\sqrt {a^2 - x^2} } = \arcsin \frac x a + C$ | {{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\sqrt {a^2 - x^2} }
| r = \int \frac {\d x} {\sqrt {a^2 \paren {1 - \frac {x^2} {a^2} } } }
| c = factor $a^2$ out of the [[Definition:Radicand|radicand]]
}}
{{eqn | r = \int \frac {\d x} {\sqrt {a^2} \sqrt {1 - \paren {\frac x a}^2} }
| c =
}}
{{eqn | r ... | Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_Root_of_a_squared_minus_x_squared/Arcsine_Form | https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_Root_of_a_squared_minus_x_squared/Arcsine_Form | [
"Primitive of Reciprocal of Root of a squared minus x squared",
"Arcsine Function"
] | [] | [
"Definition:Radicand",
"Integration by Substitution",
"Real Sine Function is Bounded",
"Shape of Sine Function",
"Powers of Group Elements",
"Negative of Absolute Value",
"Definition:Differentiation",
"Derivative of Sine Function",
"Derivative of Composite Function",
"Integration by Substitution",... |
proofwiki-4749 | Inner Automorphism is Automorphism | Let $G$ be a group.
Let $x \in G$.
Let $\kappa_x$ be the inner automorphism of $x$ in $G$.
Then $\kappa_x$ is an automorphism of $G$. | By definition, $\kappa_x: G \to G$ is a mapping defined as:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
We need to show that $\kappa_x$ is an automorphism.
First we show $\kappa_x$ is a homomorphism.
{{begin-eqn}}
{{eqn | q = \forall g, h \in G
| l = \map {\kappa_x} g \map {\kappa_x} h
| r = \paren {... | Let $G$ be a [[Definition:Group|group]].
Let $x \in G$.
Let $\kappa_x$ be the [[Definition:Inner Automorphism|inner automorphism]] of $x$ in $G$.
Then $\kappa_x$ is an [[Definition:Group Automorphism|automorphism]] of $G$. | By definition, $\kappa_x: G \to G$ is a [[Definition:Mapping|mapping]] defined as:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
We need to show that $\kappa_x$ is an [[Definition:Group Automorphism|automorphism]].
First we show $\kappa_x$ is a [[Definition:Group Homomorphism|homomorphism]].
{{begin-eqn}}
{{e... | Inner Automorphism is Automorphism | https://proofwiki.org/wiki/Inner_Automorphism_is_Automorphism | https://proofwiki.org/wiki/Inner_Automorphism_is_Automorphism | [
"Inner Automorphisms",
"Group Automorphisms"
] | [
"Definition:Group",
"Definition:Inner Automorphism",
"Definition:Group Automorphism"
] | [
"Definition:Mapping",
"Definition:Group Automorphism",
"Definition:Group Homomorphism",
"Definition:Morphism Property",
"Definition:Injection",
"Cancellation Laws",
"Definition:Injection",
"Definition:Surjection",
"Definition:Group",
"Definition:Closure (Abstract Algebra)",
"Definition:Surjectio... |
proofwiki-4750 | Order of Automorphism Group | Let $G$ be a finite group whose order is greater than $2$.
Let $\Aut G$ be the automorphism group of $G$.
Then the order of $\Aut G$ is greater than $1$. | There are $3$ cases, and for each case we find an automorphism that is not the identity automorphism. | Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Order of Structure|order]] is greater than $2$.
Let $\Aut G$ be the [[Definition:Automorphism Group of Group|automorphism group]] of $G$.
Then the [[Definition:Order of Structure|order]] of $\Aut G$ is greater than $1$. | There are $3$ cases, and for each case we find an [[Definition:Group Automorphism|automorphism]] that is not the [[Definition:Identity Mapping|identity automorphism]]. | Order of Automorphism Group | https://proofwiki.org/wiki/Order_of_Automorphism_Group | https://proofwiki.org/wiki/Order_of_Automorphism_Group | [
"Automorphism Groups"
] | [
"Definition:Finite Group",
"Definition:Order of Structure",
"Definition:Automorphism Group/Group",
"Definition:Order of Structure"
] | [
"Definition:Group Automorphism",
"Definition:Identity Mapping",
"Definition:Group Automorphism",
"Definition:Identity Mapping",
"Definition:Group Automorphism",
"Definition:Identity Mapping",
"Definition:Group Automorphism",
"Definition:Group Automorphism",
"Definition:Identity Mapping"
] |
proofwiki-4751 | Characteristic Subgroup is Transitive | Let $G$ be a group.
Let $H$ be a characteristic subgroup of $G$.
Let $K$ be a characteristic subgroup of $H$.
Then $K$ is a characteristic subgroup of $G$. | Let $\phi: G \to G$ be a group automorphism.
Since $H$ is a characteristic subgroup of $G$, we have:
:$\phi \sqbrk H = H$
Thus, from Group Homomorphism Preserves Subgroups, $\phi {\restriction_H}$, the restriction of $\phi$ to $H$, is an automorphism of $H$.
Now since $K$ is a characteristic subgroup of $H$, we have th... | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Characteristic Subgroup|characteristic subgroup]] of $G$.
Let $K$ be a [[Definition:Characteristic Subgroup|characteristic subgroup]] of $H$.
Then $K$ is a [[Definition:Characteristic Subgroup|characteristic subgroup]] of $G$. | Let $\phi: G \to G$ be a [[Definition:Group Automorphism|group automorphism]].
Since $H$ is a [[Definition:Characteristic Subgroup|characteristic subgroup]] of $G$, we have:
:$\phi \sqbrk H = H$
Thus, from [[Group Homomorphism Preserves Subgroups]], $\phi {\restriction_H}$, the [[Definition:Restriction of Mapping|re... | Characteristic Subgroup is Transitive | https://proofwiki.org/wiki/Characteristic_Subgroup_is_Transitive | https://proofwiki.org/wiki/Characteristic_Subgroup_is_Transitive | [
"Characteristic Subgroups"
] | [
"Definition:Group",
"Definition:Characteristic Subgroup",
"Definition:Characteristic Subgroup",
"Definition:Characteristic Subgroup"
] | [
"Definition:Group Automorphism",
"Definition:Characteristic Subgroup",
"Group Homomorphism Preserves Subgroups",
"Definition:Restriction/Mapping",
"Definition:Group Automorphism",
"Definition:Characteristic Subgroup",
"Definition:Restriction/Mapping",
"Definition:Characteristic Subgroup"
] |
proofwiki-4752 | Quotient Theorem for Group Homomorphisms | Let $\phi: G \to G'$ be a (group) homomorphism between two groups $G$ and $G'$.
Then $\phi$ can be decomposed into the form:
:$\phi = \alpha \beta \gamma$
where:
:$\alpha: \Img \phi \to G'$ is a monomorphism
:$\beta: G / \map \ker \phi \to \Img \phi$ is an isomorphism
:$\gamma: G \to G / \map \ker \phi$ is an epimorphi... | === Monomorphism ===
The mapping $\alpha$ is identified with the inclusion mapping $i: \Img \phi \to G'$ defined as:
:$\forall x \in \Img \phi: \map i x = x$
From Inclusion Mapping is Monomorphism, it follows that $\alpha$ is a monomorphism.
{{qed|lemma}} | Let $\phi: G \to G'$ be a [[Definition:Group Homomorphism|(group) homomorphism]] between two [[Definition:Group|groups]] $G$ and $G'$.
Then $\phi$ can be decomposed into the form:
:$\phi = \alpha \beta \gamma$
where:
:$\alpha: \Img \phi \to G'$ is a [[Definition:Group Monomorphism|monomorphism]]
:$\beta: G / \map \ker... | === Monomorphism ===
The [[Definition:Mapping|mapping]] $\alpha$ is identified with the [[Definition:Inclusion Mapping|inclusion mapping]] $i: \Img \phi \to G'$ defined as:
:$\forall x \in \Img \phi: \map i x = x$
From [[Inclusion Mapping is Monomorphism]], it follows that $\alpha$ is a [[Definition:Group Monomorphis... | Quotient Theorem for Group Homomorphisms | https://proofwiki.org/wiki/Quotient_Theorem_for_Group_Homomorphisms | https://proofwiki.org/wiki/Quotient_Theorem_for_Group_Homomorphisms | [
"Group Homomorphisms",
"Quotient Groups",
"Quotient Theorems",
"Quotient Theorem for Group Homomorphisms"
] | [
"Definition:Group Homomorphism",
"Definition:Group",
"Definition:Group Monomorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Epimorphism"
] | [
"Definition:Mapping",
"Definition:Inclusion Mapping",
"Inclusion Mapping is Monomorphism",
"Definition:Group Monomorphism",
"Definition:Mapping"
] |
proofwiki-4753 | Quotient Theorem for Group Homomorphisms | Let $\phi: G \to G'$ be a (group) homomorphism between two groups $G$ and $G'$.
Then $\phi$ can be decomposed into the form:
:$\phi = \alpha \beta \gamma$
where:
:$\alpha: \Img \phi \to G'$ is a monomorphism
:$\beta: G / \map \ker \phi \to \Img \phi$ is an isomorphism
:$\gamma: G \to G / \map \ker \phi$ is an epimorphi... | From {{Corollary|Quotient Theorem for Group Homomorphisms|1}}:
:$N \subseteq K$
{{iff}}:
:there exists a group homomorphism $\psi: G / N \to H$ such that $\phi = \psi \circ q_N$
From Surjection if Composite is Surjection, it follows that the group homomorphism $\psi$ is a surjection.
Hence by definition, $\psi$ is an e... | Let $\phi: G \to G'$ be a [[Definition:Group Homomorphism|(group) homomorphism]] between two [[Definition:Group|groups]] $G$ and $G'$.
Then $\phi$ can be decomposed into the form:
:$\phi = \alpha \beta \gamma$
where:
:$\alpha: \Img \phi \to G'$ is a [[Definition:Group Monomorphism|monomorphism]]
:$\beta: G / \map \ker... | From {{Corollary|Quotient Theorem for Group Homomorphisms|1}}:
:$N \subseteq K$
{{iff}}:
:there exists a [[Definition:Group Homomorphism|group homomorphism]] $\psi: G / N \to H$ such that $\phi = \psi \circ q_N$
From [[Surjection if Composite is Surjection]], it follows that the [[Definition:Group Homomorphism|group ... | Quotient Theorem for Group Homomorphisms/Corollary 2/Proof 1 | https://proofwiki.org/wiki/Quotient_Theorem_for_Group_Homomorphisms | https://proofwiki.org/wiki/Quotient_Theorem_for_Group_Homomorphisms/Corollary_2/Proof_1 | [
"Group Homomorphisms",
"Quotient Groups",
"Quotient Theorems",
"Quotient Theorem for Group Homomorphisms"
] | [
"Definition:Group Homomorphism",
"Definition:Group",
"Definition:Group Monomorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Epimorphism"
] | [
"Definition:Group Homomorphism",
"Surjection if Composite is Surjection",
"Definition:Group Homomorphism",
"Definition:Surjection",
"Definition:Quotient Epimorphism/Group"
] |
proofwiki-4754 | Quotient Theorem for Group Homomorphisms | Let $\phi: G \to G'$ be a (group) homomorphism between two groups $G$ and $G'$.
Then $\phi$ can be decomposed into the form:
:$\phi = \alpha \beta \gamma$
where:
:$\alpha: \Img \phi \to G'$ is a monomorphism
:$\beta: G / \map \ker \phi \to \Img \phi$ is an isomorphism
:$\gamma: G \to G / \map \ker \phi$ is an epimorphi... | Let $e$ be the identity element of $G$.
Let $\RR$ be the congruence relation defined by $N$ in $G$.
Let $\RR_\phi$ be the equivalence relation induced by $\phi$.
From Condition for Existence of Epimorphism from Quotient Structure to Epimorphic Image:
:there exists an epimorphism $\psi$ from $G / N$ to $H$ which satisfi... | Let $\phi: G \to G'$ be a [[Definition:Group Homomorphism|(group) homomorphism]] between two [[Definition:Group|groups]] $G$ and $G'$.
Then $\phi$ can be decomposed into the form:
:$\phi = \alpha \beta \gamma$
where:
:$\alpha: \Img \phi \to G'$ is a [[Definition:Group Monomorphism|monomorphism]]
:$\beta: G / \map \ker... | Let $e$ be the [[Definition:Identity Element|identity element]] of $G$.
Let $\RR$ be the [[Definition:Congruence Modulo Subgroup|congruence relation defined by $N$]] in $G$.
Let $\RR_\phi$ be the [[Definition:Equivalence Relation Induced by Mapping|equivalence relation induced by $\phi$]].
From [[Condition for Exi... | Quotient Theorem for Group Homomorphisms/Corollary 2/Proof 2 | https://proofwiki.org/wiki/Quotient_Theorem_for_Group_Homomorphisms | https://proofwiki.org/wiki/Quotient_Theorem_for_Group_Homomorphisms/Corollary_2/Proof_2 | [
"Group Homomorphisms",
"Quotient Groups",
"Quotient Theorems",
"Quotient Theorem for Group Homomorphisms"
] | [
"Definition:Group Homomorphism",
"Definition:Group",
"Definition:Group Monomorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Epimorphism"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Congruence Modulo Subgroup",
"Definition:Equivalence Relation Induced by Mapping",
"Condition for Existence of Epimorphism from Quotient Structure to Epimorphic Image",
"Definition:Epimorphism (Abstract Algebra)"
] |
proofwiki-4755 | Inclusion Mapping is Monomorphism | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \circ}$ be an algebraic substructure of $S$.
Let $\iota: T \to S$ be the inclusion mapping from $T$ to $S$.
Then $\iota$ is a monomorphism. | We have that the inclusion mapping is an injection.
Now let $x, y \in T$:
{{begin-eqn}}
{{eqn | l = \map \iota {x \circ y}
| r = x \circ y
| c = {{Defof|Inclusion Mapping}}
}}
{{eqn | r = \map \iota x \circ \map \iota y
| c = {{Defof|Inclusion Mapping}}
}}
{{end-eqn}}
demonstrating that $\iota$ has th... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $\struct {T, \circ}$ be an [[Definition:Algebraic Substructure|algebraic substructure]] of $S$.
Let $\iota: T \to S$ be the [[Definition:Inclusion Mapping|inclusion mapping]] from $T$ to $S$.
Then $\iota$ is a [[Definition:Mo... | We have that the [[Inclusion Mapping is Injection|inclusion mapping is an injection]].
Now let $x, y \in T$:
{{begin-eqn}}
{{eqn | l = \map \iota {x \circ y}
| r = x \circ y
| c = {{Defof|Inclusion Mapping}}
}}
{{eqn | r = \map \iota x \circ \map \iota y
| c = {{Defof|Inclusion Mapping}}
}}
{{end-eq... | Inclusion Mapping is Monomorphism | https://proofwiki.org/wiki/Inclusion_Mapping_is_Monomorphism | https://proofwiki.org/wiki/Inclusion_Mapping_is_Monomorphism | [
"Monomorphisms (Abstract Algebra)",
"Inclusion Mappings"
] | [
"Definition:Algebraic Structure",
"Definition:Algebraic Substructure",
"Definition:Inclusion Mapping",
"Definition:Monomorphism (Abstract Algebra)"
] | [
"Inclusion Mapping is Injection",
"Definition:Morphism Property",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Injection",
"Definition:Monomorphism (Abstract Algebra)",
"Category:Monomorphisms (Abstract Algebra)",
"Category:Inclusion Mappings"
] |
proofwiki-4756 | Function is Odd Iff Inverse is Odd | Let $f$ be an odd real function with an inverse $f^{-1}$.
Then $f^{-1}$ is also odd. | First note that we have:
{{begin-eqn}}
{{eqn | l = y
| r = \map f x
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = \map {f^{-1} } y
| c = Image of Element under Inverse Mapping
}}
{{eqn | ll= \leadsto
| l = -x
| r = -\map {f^{-1} } y
| c = Multiply both sides by $-1$
}}
{{e... | Let $f$ be an [[Definition:Odd Function|odd]] [[Definition:Real Function|real function]] with an [[Definition:Inverse Mapping|inverse]] $f^{-1}$.
Then $f^{-1}$ is also [[Definition:Odd Function|odd]]. | First note that we have:
{{begin-eqn}}
{{eqn | l = y
| r = \map f x
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = \map {f^{-1} } y
| c = [[Image of Element under Inverse Mapping]]
}}
{{eqn | ll= \leadsto
| l = -x
| r = -\map {f^{-1} } y
| c = Multiply both sides by $-1$
}... | Function is Odd Iff Inverse is Odd | https://proofwiki.org/wiki/Function_is_Odd_Iff_Inverse_is_Odd | https://proofwiki.org/wiki/Function_is_Odd_Iff_Inverse_is_Odd | [
"Inverse Mappings",
"Odd Functions"
] | [
"Definition:Odd Function",
"Definition:Real Function",
"Definition:Inverse Mapping",
"Definition:Odd Function"
] | [
"Image of Element under Inverse Mapping",
"Definition:Mapping",
"Composite of Bijection with Inverse is Identity Mapping",
"Definition:Odd Function",
"Category:Inverse Mappings",
"Category:Odd Functions"
] |
proofwiki-4757 | Zassenhaus Lemma | Let $G$ be a group.
Let $H_1$ and $H_2$ be subgroups of $G$ such that:
{{begin-eqn}}
{{eqn | l = N_1
| o = \lhd
| r = H_1
}}
{{eqn | l = N_2
| o = \lhd
| r = H_2
}}
{{end-eqn}}
where $\lhd$ denotes the relation of being a normal subgroup.
Then:
{{begin-eqn}}
{{eqn | l = \frac {N_1 \paren {H_1 \c... | In order for the expressions to make sense, the denominators need to be normal subgroups.
This is demonstrated in the following lemmata: | Let $G$ be a [[Definition:Group|group]].
Let $H_1$ and $H_2$ be [[Definition:Subgroup|subgroups]] of $G$ such that:
{{begin-eqn}}
{{eqn | l = N_1
| o = \lhd
| r = H_1
}}
{{eqn | l = N_2
| o = \lhd
| r = H_2
}}
{{end-eqn}}
where $\lhd$ denotes the relation of being a [[Definition:Normal Subgroup... | In order for the [[Definition:Expression|expressions]] to make sense, the [[Definition:Denominator|denominators]] need to be [[Definition:Normal Subgroup|normal subgroups]].
This is demonstrated in the following [[Definition:Lemma|lemmata]]: | Zassenhaus Lemma | https://proofwiki.org/wiki/Zassenhaus_Lemma | https://proofwiki.org/wiki/Zassenhaus_Lemma | [
"Zassenhaus Lemma",
"Group Isomorphisms",
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Subset Product",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Expression",
"Definition:Fraction/Denominator",
"Definition:Normal Subgroup",
"Definition:Lemma"
] |
proofwiki-4758 | Integral Resulting in Arcsecant | :<nowiki>$\ds \int \frac 1 {x \sqrt {x^2 - a^2} } \rd x = \begin{cases}
\dfrac 1 {\size a} \arcsec \dfrac x {\size a} + C & : x > \size a \\
-\dfrac 1 {\size a} \arcsec \dfrac x {\size a} + C & : x < -\size a
\end{cases}$</nowiki>
where $a$ is a constant. | {{begin-eqn}}
{{eqn | l = \int \frac 1 {x \sqrt {x^2 - a^2} } \rd x
| r = \int \frac 1 {x \sqrt {a^2 \paren {\frac {x^2} {a^2} - 1} } } \rd x
| c = factor $a^2$ out of the radicand
}}
{{eqn | r = \int \frac 1 {x \sqrt {a^2} \sqrt {\paren {\frac x a}^2 - 1} } \rd x
| c =
}}
{{eqn | r = \frac 1 {\size ... | :<nowiki>$\ds \int \frac 1 {x \sqrt {x^2 - a^2} } \rd x = \begin{cases}
\dfrac 1 {\size a} \arcsec \dfrac x {\size a} + C & : x > \size a \\
-\dfrac 1 {\size a} \arcsec \dfrac x {\size a} + C & : x < -\size a
\end{cases}$</nowiki>
where $a$ is a [[Definition:Constant|constant]]. | {{begin-eqn}}
{{eqn | l = \int \frac 1 {x \sqrt {x^2 - a^2} } \rd x
| r = \int \frac 1 {x \sqrt {a^2 \paren {\frac {x^2} {a^2} - 1} } } \rd x
| c = factor $a^2$ out of the [[Definition:Radicand|radicand]]
}}
{{eqn | r = \int \frac 1 {x \sqrt {a^2} \sqrt {\paren {\frac x a}^2 - 1} } \rd x
| c =
}}
{{e... | Integral Resulting in Arcsecant | https://proofwiki.org/wiki/Integral_Resulting_in_Arcsecant | https://proofwiki.org/wiki/Integral_Resulting_in_Arcsecant | [
"Integral Calculus"
] | [
"Definition:Constant"
] | [
"Definition:Radicand",
"Integration by Substitution",
"Definition:By Hypothesis",
"Definition:Integration/Integrand",
"Definition:Differentiation",
"Derivative of Secant Function",
"Derivative of Composite Function",
"Integration by Substitution",
"Shape of Tangent Function",
"Integral of Constant... |
proofwiki-4759 | Schreier-Zassenhaus Theorem | Let $G$ be a finite group.
Let $\HH_1$ and $\HH_2$ be two normal series for $G$.
Then $\HH_1$ and $\HH_2$ have refinements of equal length whose factors are isomorphic. | Suppose that:
:$(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$
and:
:$(2): \quad \set e = H_0 \lhd H_1 \lhd \cdots \lhd H_{m - 1} \lhd H_m = G$
are two normal series for $G$.
Let a new series be formed:
:$(3): \quad \set e = \hat G_0 \subseteq \hat G_1 \subseteq \cdots \subseteq \hat G_{n m -... | Let $G$ be a [[Definition:Finite Group|finite group]].
Let $\HH_1$ and $\HH_2$ be two [[Definition:Normal Series|normal series]] for $G$.
Then $\HH_1$ and $\HH_2$ have [[Definition:Refinement of Normal Series|refinements]] of equal [[Definition:Length of Normal Series|length]] whose [[Definition:Factor of Normal Ser... | Suppose that:
:$(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$
and:
:$(2): \quad \set e = H_0 \lhd H_1 \lhd \cdots \lhd H_{m - 1} \lhd H_m = G$
are two [[Definition:Normal Series|normal series]] for $G$.
Let a new series be formed:
:$(3): \quad \set e = \hat G_0 \subseteq \hat G_1 \subseteq... | Schreier-Zassenhaus Theorem | https://proofwiki.org/wiki/Schreier-Zassenhaus_Theorem | https://proofwiki.org/wiki/Schreier-Zassenhaus_Theorem | [
"Normal Series"
] | [
"Definition:Finite Group",
"Definition:Normal Series",
"Definition:Refinement of Normal Series",
"Definition:Normal Series/Length",
"Definition:Normal Series/Factor Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Normal Series",
"Definition:Well-Defined/Mapping",
"Definition:Normal Subgroup",
"Definition:Normal Series",
"Zassenhaus Lemma",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Normal Series/Factor Group",
"Definition:Normal Series/Factor Group",
"Definition:Is... |
proofwiki-4760 | Jordan-Hölder Theorem | Let $G$ be a finite group.
Let $\HH_1$ and $\HH_2$ be two composition series for $G$.
Then:
:$\HH_1$ and $\HH_2$ have the same length
:Corresponding factors of $\HH_1$ and $\HH_2$ are isomorphic. | By the Schreier-Zassenhaus Theorem, two normal series have refinements of equal length whose factors are isomorphic.
But from the definition of composition series, $\HH_1$ and $\HH_2$ have no proper refinements.
Hence any such refinements must be identical to $\HH_1$ and $\HH_2$ themselves.
{{qed}}
{{Namedfor|Marie Enn... | Let $G$ be a [[Definition:Finite Group|finite group]].
Let $\HH_1$ and $\HH_2$ be two [[Definition:Composition Series|composition series]] for $G$.
Then:
:$\HH_1$ and $\HH_2$ have the same [[Definition:Length of Normal Series|length]]
:Corresponding [[Definition:Factor of Normal Series|factors]] of $\HH_1$ and $\HH_2... | By the [[Schreier-Zassenhaus Theorem]], two [[Definition:Normal Series|normal series]] have [[Definition:Refinement of Normal Series|refinements]] of equal [[Definition:Length of Normal Series|length]] whose [[Definition:Factor of Normal Series|factors]] are [[Definition:Group Isomorphism|isomorphic]].
But from the de... | Jordan-Hölder Theorem | https://proofwiki.org/wiki/Jordan-Hölder_Theorem | https://proofwiki.org/wiki/Jordan-Hölder_Theorem | [
"Composition Series"
] | [
"Definition:Finite Group",
"Definition:Composition Series",
"Definition:Normal Series/Length",
"Definition:Normal Series/Factor Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Schreier-Zassenhaus Theorem",
"Definition:Normal Series",
"Definition:Refinement of Normal Series",
"Definition:Normal Series/Length",
"Definition:Normal Series/Factor Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Composition Series",
"Definition:Refinement of Nor... |
proofwiki-4761 | Double Orthocomplement is Closed Linear Span | Let $H$ be a Hilbert space.
Let $A \subseteq H$ be a subset of $H$.
Then the following identity holds:
:$\paren {A^\perp}^\perp = \vee A$
Here $A^\perp$ denotes orthocomplementation, and $\vee A$ denotes the closed linear span. | From Orthocomplement is Closed Linear Subspace:
:$\paren {A^\perp}^\perp$ is a closed linear subspace of $H$.
Also:
:$\paren {A^\perp}^\perp \supseteq A$
Indeed, for each $a \in A$, we have:
:$\forall a' \in A^\perp : \innerprod a {a'} = 0$
by definition of $A^\perp$.
This also means that:
:$a \in \paren{A^\perp}^\perp... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $A \subseteq H$ be a [[Definition:Subset|subset]] of $H$.
Then the following identity holds:
:$\paren {A^\perp}^\perp = \vee A$
Here $A^\perp$ denotes [[Definition:Orthocomplement|orthocomplementation]], and $\vee A$ denotes the [[Definition:Closed Linea... | From [[Orthocomplement is Closed Linear Subspace]]:
:$\paren {A^\perp}^\perp$ is a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
Also:
:$\paren {A^\perp}^\perp \supseteq A$
Indeed, for each $a \in A$, we have:
:$\forall a' \in A^\perp : \innerprod a {a'} = 0$
by definition of $A^\perp$.
This a... | Double Orthocomplement is Closed Linear Span | https://proofwiki.org/wiki/Double_Orthocomplement_is_Closed_Linear_Span | https://proofwiki.org/wiki/Double_Orthocomplement_is_Closed_Linear_Span | [
"Hilbert Spaces",
"Orthocomplements"
] | [
"Definition:Hilbert Space",
"Definition:Subset",
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement",
"Definition:Closed Linear Span"
] | [
"Orthocomplement is Closed Linear Subspace",
"Definition:Closed Linear Subspace",
"Orthocomplement Reverses Subset",
"Double Orthocomplement of Closed Linear Subspace"
] |
proofwiki-4762 | Orthocomplement is Closed Linear Subspace | Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $A \subseteq V$ be a subset of $V$.
Then the orthocomplement $A^\perp$ of $A$ is a closed linear subspace of $H$. | Let $\sequence {x_n} \subset A^\perp$ be a convergent sequence.
Let $x$ be the limit of $\sequence {x_n}$.
Then, by the definition of orthocomplement:
:$\forall n \in \N, y \in A: \innerprod {x_n} y = 0$
Passing to the limit, from Inner Product is Continuous we have:
:$\forall y \in A: \innerprod x y = 0$
So:
:$x \in A... | Let $\struct {V, \innerprod \cdot \cdot}$ be an [[Definition:Inner Product Space|inner product space]].
Let $A \subseteq V$ be a [[Definition:Subset|subset]] of $V$.
Then the [[Definition:Orthocomplement|orthocomplement]] $A^\perp$ of $A$ is a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$. | Let $\sequence {x_n} \subset A^\perp$ be a [[Definition:Convergent Sequence in Normed Vector Space|convergent]] [[Definition:Sequence|sequence]].
Let $x$ be the [[Definition:Limit of Sequence|limit]] of $\sequence {x_n}$.
Then, by the definition of [[Definition:Orthocomplement|orthocomplement]]:
:$\forall n \in \N, y... | Orthocomplement is Closed Linear Subspace | https://proofwiki.org/wiki/Orthocomplement_is_Closed_Linear_Subspace | https://proofwiki.org/wiki/Orthocomplement_is_Closed_Linear_Subspace | [
"Inner Product Spaces",
"Orthocomplements"
] | [
"Definition:Inner Product Space",
"Definition:Subset",
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement",
"Definition:Closed Linear Subspace"
] | [
"Definition:Convergent Sequence/Normed Vector Space",
"Definition:Sequence",
"Definition:Limit of Sequence",
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement",
"Definition:Limit of Sequence",
"Inner Product is Continuous",
"Definition:Closed Linear Subspace"
] |
proofwiki-4763 | Linear Subspace Dense iff Zero Orthocomplement | Let $H$ be a Hilbert space.
Let $K$ be a linear subspace of $H$.
Then $K$ is everywhere dense {{iff}} $K^\perp = \paren 0$, where $K^\perp$ is the orthocomplement of $K$, and $\paren 0$ denotes the zero subspace. | === Sufficient Condition ===
Assume that $K$ is everywhere dense.
Let $x \in K^\perp$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = K^\perp
}}
{{eqn | ll= \leadsto
| q = \forall y \in K
| l = x
| o = \perp
| r = y
}}
{{eqn | ll= \leadsto
| q = \forall y \in H
| l = x... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $K$ be a [[Definition:Vector Subspace|linear subspace]] of $H$.
Then $K$ is [[Definition:Everywhere Dense|everywhere dense]] {{iff}} $K^\perp = \paren 0$, where $K^\perp$ is the [[Definition:Orthocomplement|orthocomplement]] of $K$, and $\paren 0$ denotes ... | === Sufficient Condition ===
Assume that $K$ is [[Definition:Everywhere Dense|everywhere dense]].
Let $x \in K^\perp$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = K^\perp
}}
{{eqn | ll= \leadsto
| q = \forall y \in K
| l = x
| o = \perp
| r = y
}}
{{eqn | ll= \leadsto
... | Linear Subspace Dense iff Zero Orthocomplement | https://proofwiki.org/wiki/Linear_Subspace_Dense_iff_Zero_Orthocomplement | https://proofwiki.org/wiki/Linear_Subspace_Dense_iff_Zero_Orthocomplement | [
"Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Vector Subspace",
"Definition:Everywhere Dense",
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement",
"Definition:Zero Subspace"
] | [
"Definition:Everywhere Dense",
"Inner Product is Continuous",
"Definition:Everywhere Dense"
] |
proofwiki-4764 | Finite Group has Composition Series | Let $G$ be a finite group.
Then $G$ has a composition series. | Let $G$ be a finite group whose identity is $e$.
Either $G$ has a proper non-trivial normal subgroup or it does not.
If not, then:
:$\set e \lhd G$
is the composition series for $G$.
Otherwise, $G$ has one or more proper non-trivial normal subgroup.
Of these, one or more will have a maximum order.
Select one of these a... | Let $G$ be a [[Definition:Finite Group|finite group]].
Then $G$ has a [[Definition:Composition Series|composition series]]. | Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e$.
Either $G$ has a [[Definition:Proper Subgroup|proper]] [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]] or it does not.
If not, then:
:$\set e \lhd G$
is the [[Defini... | Finite Group has Composition Series/Proof 1 | https://proofwiki.org/wiki/Finite_Group_has_Composition_Series | https://proofwiki.org/wiki/Finite_Group_has_Composition_Series/Proof_1 | [
"Finite Group has Composition Series",
"Composition Series",
"Finite Groups"
] | [
"Definition:Finite Group",
"Definition:Composition Series"
] | [
"Definition:Finite Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Proper Subgroup",
"Definition:Non-Trivial Group",
"Definition:Normal Subgroup",
"Definition:Composition Series",
"Definition:Proper Subgroup",
"Definition:Non-Trivial Group",
"Definition:Normal Subgro... |
proofwiki-4765 | Finite Group has Composition Series | Let $G$ be a finite group.
Then $G$ has a composition series. | {{WIP|Bring up to site standards}}
Let $G$ be a finite group whose identity is $e_G$.
We shall use induction on $|G|$.
If $G$ is trivial ($|G|=1$), then its composition series is
:$G=\{e_G\}$.
Suppose $G$ has a composition series if $|G|<n$, then it suffices to construct a composition series for $G$ with order $n$.... | Let $G$ be a [[Definition:Finite Group|finite group]].
Then $G$ has a [[Definition:Composition Series|composition series]]. | {{WIP|Bring up to site standards}}
Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e_G$.
We shall use [[Principle_of_Mathematical_Induction|induction]] on $|G|$.
If $G$ is trivial ($|G|=1$), then its composition series is
:$G=\{e_G\}$.
Suppose $G$ has a comp... | Finite Group has Composition Series/Proof 2 | https://proofwiki.org/wiki/Finite_Group_has_Composition_Series | https://proofwiki.org/wiki/Finite_Group_has_Composition_Series/Proof_2 | [
"Finite Group has Composition Series",
"Composition Series",
"Finite Groups"
] | [
"Definition:Finite Group",
"Definition:Composition Series"
] | [
"Definition:Finite Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Principle_of_Mathematical_Induction",
"Definition:Simple_Group",
"Definition:Proper Subgroup",
"Definition:Non-Trivial Group",
"Definition:Normal Subgroup",
"Zorn's Lemma"
] |
proofwiki-4766 | Length of Subgroup Plus Length of Quotient Group | Let $G$ be a finite group.
Let $H$ be a normal subgroup of $G$.
Then:
:$\map l G = \map l H + \map l {G / H}$
where:
:$\map l G$ denotes the length of $G$
:$G / H$ denotes the quotient group of $G$ by $H$. | Let $\HH$ be a composition series of $H$:
:$\set e = H_0 \subset H_1 \subset \dots \subset H_m = H$,
We can construct a composition series $\GG$ for $G$ which contains $H$:
:$\set e = H_0 \subset H_1 \subset \dots \subset H_m = H = G_0 \subset G_1 \subset \dots \subset G_n = G$
where, for $k = 0, \dots, n - 1$:
:$G_k \... | Let $G$ be a [[Definition:Finite Group|finite group]].
Let $H$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Then:
:$\map l G = \map l H + \map l {G / H}$
where:
:$\map l G$ denotes the [[Definition:Length of Group|length]] of $G$
:$G / H$ denotes the [[Definition:Quotient Group|quotient group]] of $G$... | Let $\HH$ be a [[Definition:Composition Series|composition series]] of $H$:
:$\set e = H_0 \subset H_1 \subset \dots \subset H_m = H$,
We can construct a [[Definition:Composition Series|composition series]] $\GG$ for $G$ which contains $H$:
:$\set e = H_0 \subset H_1 \subset \dots \subset H_m = H = G_0 \subset G_1 ... | Length of Subgroup Plus Length of Quotient Group | https://proofwiki.org/wiki/Length_of_Subgroup_Plus_Length_of_Quotient_Group | https://proofwiki.org/wiki/Length_of_Subgroup_Plus_Length_of_Quotient_Group | [
"Normal Subgroups",
"Quotient Groups"
] | [
"Definition:Finite Group",
"Definition:Normal Subgroup",
"Definition:Length of Group",
"Definition:Quotient Group"
] | [
"Definition:Composition Series",
"Definition:Composition Series",
"Definition:Normal Subgroup",
"Definition:Sequence",
"Definition:Normal Series",
"Definition:Normal Series",
"Third Isomorphism Theorem",
"Definition:Normal Subgroup",
"Third Isomorphism Theorem",
"Definition:Proper Subgroup",
"De... |
proofwiki-4767 | Condition for Composition Series | Let $G$ be a finite group.
Then:
:a normal series $\HH$ for $G$ is a composition series for $G$
{{iff}}:
:every factor group of $\HH$ is a simple group. | Let $G$ be a finite group whose identity is $e$.
Let:
:$(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$
be a normal series for $G$. | Let $G$ be a [[Definition:Finite Group|finite group]].
Then:
:a [[Definition:Normal Series|normal series]] $\HH$ for $G$ is a [[Definition:Composition Series|composition series]] for $G$
{{iff}}:
:every [[Definition:Factor of Normal Series|factor group]] of $\HH$ is a [[Definition:Simple Group|simple group]]. | Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e$.
Let:
:$(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$
be a [[Definition:Normal Series|normal series]] for $G$. | Condition for Composition Series | https://proofwiki.org/wiki/Condition_for_Composition_Series | https://proofwiki.org/wiki/Condition_for_Composition_Series | [
"Composition Series"
] | [
"Definition:Finite Group",
"Definition:Normal Series",
"Definition:Composition Series",
"Definition:Normal Series/Factor Group",
"Definition:Simple Group"
] | [
"Definition:Finite Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Normal Series",
"Definition:Normal Series",
"Definition:Normal Series",
"Definition:Normal Series"
] |
proofwiki-4768 | Factors of Composition Series for Prime Power Group | Let $G$ be a group such that $\order G = p^n$ where $p$ is a prime number.
Then $G$ has a composition series in which each factor group is cyclic of order $p$. | From Composition Series of Group of Prime Power Order, $G$ has a sequence of subgroups:
:$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$
such that $\order {G_k} = p^k$, $G_k \lhd G_{k + 1}$ and $G_{k + 1} / G_k$ is cyclic and of order $p$.
From Prime Group is Simple it follows that $G_{k + 1} / G_k$ is a simp... | Let $G$ be a [[Definition:Group|group]] such that $\order G = p^n$ where $p$ is a [[Definition:Prime Number|prime number]].
Then $G$ has a [[Definition:Composition Series|composition series]] in which each [[Definition:Factor of Normal Series|factor group]] is [[Definition:Cyclic Group|cyclic]] of [[Definition:Order ... | From [[Composition Series of Group of Prime Power Order]], $G$ has a [[Definition:Sequence|sequence]] of [[Definition:Subgroup|subgroups]]:
:$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$
such that $\order {G_k} = p^k$, $G_k \lhd G_{k + 1}$ and $G_{k + 1} / G_k$ is [[Definition:Cyclic Group|cyclic]] and of [... | Factors of Composition Series for Prime Power Group | https://proofwiki.org/wiki/Factors_of_Composition_Series_for_Prime_Power_Group | https://proofwiki.org/wiki/Factors_of_Composition_Series_for_Prime_Power_Group | [
"Composition Series",
"Cyclic Groups"
] | [
"Definition:Group",
"Definition:Prime Number",
"Definition:Composition Series",
"Definition:Normal Series/Factor Group",
"Definition:Cyclic Group",
"Definition:Order of Structure"
] | [
"Composition Series of Group of Prime Power Order",
"Definition:Sequence",
"Definition:Subgroup",
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Prime Group is Simple",
"Definition:Simple Group",
"Prime Group is Cyclic",
"Definition:Cyclic Group",
"Condition for Composition Series"
] |
proofwiki-4769 | Continuity of Linear Functionals | Let $V$ be a normed vector space, and let $L$ be a linear functional on $V$.
Then the following four statements are equivalent:
:$(1): \quad L$ is continuous
:$(2): \quad L$ is continuous at $\mathbf 0_V$
:$(3): \quad L$ is continuous at some point
:$(4): \quad L$ is bounded: $\exists c > 0: \forall v \in H: \size {L v... | === $(1)$ iff $(2)$ === | Let $V$ be a [[Definition:Normed Vector Space|normed vector space]], and let $L$ be a [[Definition:Linear Functional|linear functional]] on $V$.
Then the following four statements are equivalent:
:$(1): \quad L$ is [[Definition:Continuous Mapping (Normed Vector Space)|continuous]]
:$(2): \quad L$ is [[Definition:Con... | === $(1)$ iff $(2)$ === | Continuity of Linear Functionals | https://proofwiki.org/wiki/Continuity_of_Linear_Functionals | https://proofwiki.org/wiki/Continuity_of_Linear_Functionals | [
"Linear Functionals",
"Normed Vector Spaces"
] | [
"Definition:Normed Vector Space",
"Definition:Linear Functional",
"Definition:Continuous Mapping (Normed Vector Space)",
"Definition:Continuous Mapping (Normed Vector Space)/Point",
"Definition:Continuous Mapping (Normed Vector Space)/Point",
"Definition:Bounded Linear Functional"
] | [] |
proofwiki-4770 | Equivalence of Definitions of Closed Linear Span | Let $H$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$, and let $A \subseteq H$ be a subset.
{{TFAE|def = Closed Linear Span|view = closed linear span of $A$}}
:$(1): \quad \ds \vee A = \bigcap \Bbb M$, where $\Bbb M$ consists of all closed linear subspaces $M$ of $H$ with $A \subseteq M$
:$(2): \quad \vee A$ is th... | Let the proposition $(1)$ hold:
Assume the closed linear subspace $M'$ contains the set $A$.
Then because $M' \in \Bbb M$:
:$\vee A \subseteq M'$
The intersection of arbitrary family of subspaces is a subspace.
For suppose $\CC$ is a family of subspaces.
Denote $\bigcap \CC = \set {f\in H: \forall V \in \CC: \exists f ... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] over $\Bbb F \in \set {\R, \C}$, and let $A \subseteq H$ be a [[Definition:Subset|subset]].
{{TFAE|def = Closed Linear Span|view = closed linear span of $A$}}
:$(1): \quad \ds \vee A = \bigcap \Bbb M$, where $\Bbb M$ consists of all [[Definition:Closed Linear Su... | Let the proposition $(1)$ hold:
Assume the [[Definition:Closed Linear Subspace|closed linear subspace]] $M'$ contains the set $A$.
Then because $M' \in \Bbb M$:
:$\vee A \subseteq M'$
The intersection of arbitrary family of subspaces is a subspace.
For suppose $\CC$ is a family of subspaces.
Denote $\bigcap \CC =... | Equivalence of Definitions of Closed Linear Span | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Closed_Linear_Span | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Closed_Linear_Span | [
"Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Subset",
"Definition:Closed Linear Subspace",
"Definition:Closed Linear Subspace",
"Definition:Closure (Topology)"
] | [
"Definition:Closed Linear Subspace",
"Point in Closure of Subset of Metric Space iff Limit of Sequence",
"Definition:Closed Linear Subspace"
] |
proofwiki-4771 | Equivalence of Definitions of Norm of Linear Functional | Let $V$ be a normed vector space, and let $L$ be a bounded linear functional on $V$.
Define the following norms of $L$:
{{begin-eqn}}
{{eqn | n = 1
| l = \norm L_1
| r = \sup \set {\size {L h}: \norm h \le 1}
}}
{{eqn | n = 2
| l = \norm L_2
| r = \sup \set {\size {L h}: \norm h = 1}
}}
{{eqn | ... | We have:
:$\set {v \in V : \norm v = 1} \subseteq \set {v \in V : \norm v \le 1} \subseteq V$
So it follows from the definition of the supremum that
:$\norm L_2 \le \norm L_1 \le \norm L_3$
Next we show that $\norm L_2 = \norm L_3$:
{{begin-eqn}}
{{eqn | l = \norm L_2
| r = \sup \set {\norm {L v}: \norm h = 1}
}... | Let $V$ be a [[Definition:Normed Vector Space|normed vector space]], and let $L$ be a [[Definition:Bounded Linear Functional|bounded linear functional]] on $V$.
Define the following [[Definition:Norm on Bounded Linear Functional|norms]] of $L$:
{{begin-eqn}}
{{eqn | n = 1
| l = \norm L_1
| r = \sup \set {\... | We have:
:$\set {v \in V : \norm v = 1} \subseteq \set {v \in V : \norm v \le 1} \subseteq V$
So it follows from the definition of the [[Definition:Supremum of Set|supremum]] that
:$\norm L_2 \le \norm L_1 \le \norm L_3$
Next we show that $\norm L_2 = \norm L_3$:
{{begin-eqn}}
{{eqn | l = \norm L_2
| r = \sup... | Equivalence of Definitions of Norm of Linear Functional | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Norm_of_Linear_Functional | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Norm_of_Linear_Functional | [
"Bounded Linear Functionals",
"Definition Equivalences"
] | [
"Definition:Normed Vector Space",
"Definition:Bounded Linear Functional",
"Definition:Norm/Bounded Linear Functional"
] | [
"Definition:Supremum of Set",
"Definition:Linear Functional",
"Definition:Infimum of Set",
"Definition:Contradiction"
] |
proofwiki-4772 | Riesz Representation Theorem (Hilbert Spaces) | Let $H$ be a Hilbert space.
Let $L$ be a bounded linear functional on $H$.
Then there is a unique $h_0 \in H$ such that:
:$\forall h \in H: L h = \innerprod h {h_0}$ | If $L \equiv 0$ identically, then:
:$L h = 0 = \innerprod h 0$
and the theorem holds.
By Kernel of Bounded Linear Transformation is Closed Linear Subspace:
:$M := \ker L$ is a closed linear subspace of $H$.
Then we can decompose $H$ as a direct sum:
:$H \cong M \oplus M^\perp$
where $M^\perp$ denotes the orthocomplemen... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $L$ be a [[Definition:Bounded Linear Functional|bounded linear functional]] on $H$.
Then there is a unique $h_0 \in H$ such that:
:$\forall h \in H: L h = \innerprod h {h_0}$ | If $L \equiv 0$ identically, then:
:$L h = 0 = \innerprod h 0$
and the theorem holds.
By [[Kernel of Bounded Linear Transformation is Closed Linear Subspace]]:
:$M := \ker L$ is a [[Definition:Closed Linear Subspace|closed linear subspace]] of $H$.
Then we can [[Direct Sum of Subspace and Orthocomplement|decompose $H... | Riesz Representation Theorem (Hilbert Spaces) | https://proofwiki.org/wiki/Riesz_Representation_Theorem_(Hilbert_Spaces) | https://proofwiki.org/wiki/Riesz_Representation_Theorem_(Hilbert_Spaces) | [
"Hilbert Spaces",
"Bounded Linear Functionals",
"Representation Theorems"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Functional"
] | [
"Kernel of Bounded Linear Transformation is Closed Linear Subspace",
"Definition:Closed Linear Subspace",
"Direct Sum of Subspace and Orthocomplement",
"Definition:Orthogonal (Linear Algebra)/Orthogonal Complement",
"Definition:Positive Definite",
"Definition:Inner Product"
] |
proofwiki-4773 | Orthonormal Subset of Hilbert Space Extends to Basis | Let $H$ be a Hilbert space, and let $S$ be an orthonormal subset of $H$.
Then there exists a basis for $H$ that contains $S$ as a subset. | Consider the set $\OO$ of orthonormal subsets $S'$ of $H$ that contain $S$:
:$\OO := \set{ S' \subseteq H: S \subseteq S', \text{$S'$ is orthonormal} }$
In particular, $S \in \OO$ so that $\OO$ is non-empty.
Give $\OO$ the subset ordering.
For a chain $\CC \subseteq \OO$, we assert that $\bigcup \CC \in \OO$.
For each ... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]], and let $S$ be an [[Definition:Orthonormal Subset|orthonormal subset]] of $H$.
Then there exists a [[Definition:Basis (Hilbert Space)|basis]] for $H$ that contains $S$ as a [[Definition:Subset|subset]]. | Consider the set $\OO$ of [[Definition:Orthonormal Subset|orthonormal subsets]] $S'$ of $H$ that contain $S$:
:$\OO := \set{ S' \subseteq H: S \subseteq S', \text{$S'$ is orthonormal} }$
In particular, $S \in \OO$ so that $\OO$ is [[Definition:Non-Empty Set|non-empty]].
Give $\OO$ the [[Definition:Subset Ordering|su... | Orthonormal Subset of Hilbert Space Extends to Basis | https://proofwiki.org/wiki/Orthonormal_Subset_of_Hilbert_Space_Extends_to_Basis | https://proofwiki.org/wiki/Orthonormal_Subset_of_Hilbert_Space_Extends_to_Basis | [
"Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Orthonormal Subset",
"Definition:Basis (Hilbert Space)",
"Definition:Subset"
] | [
"Definition:Orthonormal Subset",
"Definition:Non-Empty Set",
"Definition:Set Ordered by Subset Relation",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Chain (Order Theory)/Subset Relation",
"Zorn's Lemma",
"Definition:Maximal/Element",
"Definition:Maximal/Set",
"Definition:Orthonor... |
proofwiki-4774 | Gram-Schmidt Orthogonalization | Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\R$ or $\C$.
Let $S = \set {v_n: n \in \N_{>0} }$ be a linearly independent subset of $V$.
Then there exists an orthonormal subset $E = \set {e_n: n \in \N_{>0} }$ of $V$ such that:
:$\forall k \in \N: \span \set {v_1 , \ldots , v_k} = \span \se... | For all $k \in \N_{>0}$, define $u_k , e_k \in V$ inductively as:
{{begin-eqn}}
{{eqn | o = :=
| l = u_k
| r = \ds v_k - \sum_{i \mathop= 1}^{k-1} \innerprod {v_k}{e_i} e_i
}}
{{eqn | o = :=
| l = e_k
| r = \dfrac {1}{\norm {u_k} } u_k
}}
{{end-eqn}}
where $\norm \cdot$ denotes the inner produc... | Let $\struct {V, \innerprod \cdot \cdot}$ be an [[Definition:Inner Product Space|inner product space]] over $\R$ or $\C$.
Let $S = \set {v_n: n \in \N_{>0} }$ be a [[Definition:Linearly Independent Set|linearly independent]] [[Definition:Subset|subset]] of $V$.
Then there exists an [[Definition:Orthonormal Subset|o... | For all $k \in \N_{>0}$, define $u_k , e_k \in V$ inductively as:
{{begin-eqn}}
{{eqn | o = :=
| l = u_k
| r = \ds v_k - \sum_{i \mathop= 1}^{k-1} \innerprod {v_k}{e_i} e_i
}}
{{eqn | o = :=
| l = e_k
| r = \dfrac {1}{\norm {u_k} } u_k
}}
{{end-eqn}}
where $\norm \cdot$ denotes the [[Definiti... | Gram-Schmidt Orthogonalization | https://proofwiki.org/wiki/Gram-Schmidt_Orthogonalization | https://proofwiki.org/wiki/Gram-Schmidt_Orthogonalization | [
"Gram-Schmidt Orthogonalization",
"Inner Product Spaces"
] | [
"Definition:Inner Product Space",
"Definition:Linearly Independent/Set",
"Definition:Subset",
"Definition:Orthonormal Subset",
"Definition:Generated Submodule/Linear Span"
] | [
"Definition:Inner Product Norm",
"Definition:Inner Product",
"Definition:Inner Product"
] |
proofwiki-4775 | Orthogonal Projection onto Closed Linear Span | Let $H$ be a Hilbert space with inner product $\innerprod \cdot \cdot$ and inner product norm $\norm \cdot$.
Let $E = \set {e_1, \ldots, e_n}$ be an orthonormal subset of $H$.
Let $M = \vee E$, where $\vee E$ is the closed linear span of $E$.
Let $P$ be the orthogonal projection onto $M$.
Then:
:$\forall h \in H: P h... | Let $h \in H$.
Let:
:$\ds u = \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k$
We have that:
:$u \in \map \span E$
and from the definition of closed linear span:
:$M = \paren {\map \span E}^-$
We therefore have, by the definition of closure:
:$u \in M$
Let $v = h - u$
We want to show that $v \in M^\bot$.
From Inte... | Let $H$ be a [[Definition:Hilbert Space|Hilbert space]] with [[Definition:Inner Product|inner product]] $\innerprod \cdot \cdot$ and [[Definition:Inner Product Norm|inner product norm]] $\norm \cdot$.
Let $E = \set {e_1, \ldots, e_n}$ be an [[Definition:Orthonormal Subset|orthonormal subset]] of $H$.
Let $M = \vee E... | Let $h \in H$.
Let:
:$\ds u = \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k$
We have that:
:$u \in \map \span E$
and from the definition of [[Definition:Closed Linear Span|closed linear span]]:
:$M = \paren {\map \span E}^-$
We therefore have, by the definition of [[Definition:Topological Closure|closure]]:
... | Orthogonal Projection onto Closed Linear Span | https://proofwiki.org/wiki/Orthogonal_Projection_onto_Closed_Linear_Span | https://proofwiki.org/wiki/Orthogonal_Projection_onto_Closed_Linear_Span | [
"Orthogonal Projections"
] | [
"Definition:Hilbert Space",
"Definition:Inner Product",
"Definition:Inner Product Norm",
"Definition:Orthonormal Subset",
"Definition:Closed Linear Span",
"Definition:Orthogonal Projection"
] | [
"Definition:Closed Linear Span",
"Definition:Closure (Topology)",
"Intersection of Orthocomplements is Orthocomplement of Closed Linear Span",
"Definition:Inner Product",
"Definition:Linear Transformation",
"Definition:Linear Transformation",
"Definition:Inner Product",
"Definition:Linear Transformati... |
proofwiki-4776 | Bessel's Inequality | Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $\norm \cdot$ be the inner product norm for $\struct {V, \innerprod \cdot \cdot}$.
Let $E = \set {e_n: n \in \N}$ be a countably infinite orthonormal subset of $V$.
Then, for all $h \in V$:
:$\ds \sum_{n \mathop = 1}^\infty \size {\innerprod h {e_... | Note that for any natural number $n$ we have, applying sesquilinearity of the inner product:
{{begin-eqn}}
{{eqn | l = \norm {h - \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k}^2
| r = \innerprod {h - \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k} {h - \sum_{j \mathop = 1}^n \innerprod h {e_j} e_j}
| c = {... | Let $\struct {V, \innerprod \cdot \cdot}$ be an [[Definition:Inner Product Space|inner product space]].
Let $\norm \cdot$ be the [[Definition:Inner Product Norm|inner product norm]] for $\struct {V, \innerprod \cdot \cdot}$.
Let $E = \set {e_n: n \in \N}$ be a [[Definition:Countably Infinite Set|countably infinite]] ... | Note that for any [[Definition:Natural Number|natural number]] $n$ we have, applying [[Definition:Sesquilinear Form|sesquilinearity]] of the [[Definition:Inner Product|inner product]]:
{{begin-eqn}}
{{eqn | l = \norm {h - \sum_{k \mathop = 1}^n \innerprod h {e_k} e_k}^2
| r = \innerprod {h - \sum_{k \mathop = 1... | Bessel's Inequality | https://proofwiki.org/wiki/Bessel's_Inequality | https://proofwiki.org/wiki/Bessel's_Inequality | [
"Bessel's Inequality",
"Inner Product Spaces"
] | [
"Definition:Inner Product Space",
"Definition:Inner Product Norm",
"Definition:Countably Infinite/Set",
"Definition:Orthonormal Subset"
] | [
"Definition:Natural Numbers",
"Definition:Sesquilinear Form",
"Definition:Inner Product",
"Definition:Inner Product",
"Pythagoras's Theorem (Inner Product Space)",
"Definition:Sesquilinear Form",
"Definition:Inner Product",
"Definition:Inner Product",
"Definition:Inner Product",
"Product of Comple... |
proofwiki-4777 | Between two Rational Numbers exists Irrational Number | Let $a, b \in \Q$ where $a < b$.
Then:
:$\exists \xi \in \R \setminus \Q: a < \xi < b$ | Let $d = b - a$.
As $a, b \in \Q: a < b$ it follows from Rational Numbers form Ordered Integral Domain that $d \in \Q: d > 0$.
From Square Root of 2 is Irrational, $\sqrt 2$ is not a rational number, so it is an element of $\R \setminus \Q$.
From Square Number Less than One, for any given real number $x$:
:$x^2 < 1 \im... | Let $a, b \in \Q$ where $a < b$.
Then:
:$\exists \xi \in \R \setminus \Q: a < \xi < b$ | Let $d = b - a$.
As $a, b \in \Q: a < b$ it follows from [[Rational Numbers form Ordered Integral Domain]] that $d \in \Q: d > 0$.
From [[Square Root of 2 is Irrational]], $\sqrt 2$ is not a [[Definition:Rational Number|rational number]], so it is an [[Definition:Element|element]] of $\R \setminus \Q$.
From [[Squar... | Between two Rational Numbers exists Irrational Number/Proof 1 | https://proofwiki.org/wiki/Between_two_Rational_Numbers_exists_Irrational_Number | https://proofwiki.org/wiki/Between_two_Rational_Numbers_exists_Irrational_Number/Proof_1 | [
"Real Analysis",
"Between two Rational Numbers exists Irrational Number"
] | [] | [
"Rational Numbers form Ordered Integral Domain",
"Square Root of 2 is Irrational",
"Definition:Rational Number",
"Definition:Element",
"Square Number Less than One",
"Definition:Real Number",
"Between two Rational Numbers exists Irrational Number/Lemma 1",
"Between two Rational Numbers exists Irration... |
proofwiki-4778 | Between two Rational Numbers exists Irrational Number | Let $a, b \in \Q$ where $a < b$.
Then:
:$\exists \xi \in \R \setminus \Q: a < \xi < b$ | From Between two Real Numbers exists Rational Number, there exists $x \in \Q$ such that:
:$a - \sqrt2 < x < b - \sqrt2$
Since Square Root of 2 is Irrational, by the Lemma:
:$x + \sqrt 2$ is irrational.
But:
:$a < x + \sqrt 2 < b$
which is what we wanted to show.
{{qed}} | Let $a, b \in \Q$ where $a < b$.
Then:
:$\exists \xi \in \R \setminus \Q: a < \xi < b$ | From [[Between two Real Numbers exists Rational Number]], there exists $x \in \Q$ such that:
:$a - \sqrt2 < x < b - \sqrt2$
Since [[Square Root of 2 is Irrational]], by the [[Between two Rational Numbers exists Irrational Number/Lemma 2|Lemma]]:
:$x + \sqrt 2$ is [[Definition:Irrational Number|irrational]].
But:
:$a ... | Between two Rational Numbers exists Irrational Number/Proof 2 | https://proofwiki.org/wiki/Between_two_Rational_Numbers_exists_Irrational_Number | https://proofwiki.org/wiki/Between_two_Rational_Numbers_exists_Irrational_Number/Proof_2 | [
"Real Analysis",
"Between two Rational Numbers exists Irrational Number"
] | [] | [
"Between two Real Numbers exists Rational Number",
"Square Root of 2 is Irrational",
"Between two Rational Numbers exists Irrational Number/Lemma 2",
"Definition:Irrational Number"
] |
proofwiki-4779 | Abelian Group is Simple iff Prime | Let $G$ be a non-trivial abelian group.
Then $G$ is simple {{iff}} $G$ is a prime group. | First we note that the trivial group is (trivially) simple, but not prime because $1$ is not prime.
Hence the specification of $G$ as being non-trivial. | Let $G$ be a [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Abelian Group|abelian group]].
Then $G$ is [[Definition:Simple Group|simple]] {{iff}} $G$ is a [[Definition:Prime Group|prime group]]. | First we note that the [[Definition:Trivial Group|trivial group]] is (trivially) [[Definition:Simple Group|simple]], but not [[Definition:Prime Group|prime]] because [[One is not Prime|$1$ is not prime]].
Hence the specification of $G$ as being [[Definition:Non-Trivial Group|non-trivial]]. | Abelian Group is Simple iff Prime | https://proofwiki.org/wiki/Abelian_Group_is_Simple_iff_Prime | https://proofwiki.org/wiki/Abelian_Group_is_Simple_iff_Prime | [
"Simple Groups",
"Abelian Groups"
] | [
"Definition:Non-Trivial Group",
"Definition:Abelian Group",
"Definition:Simple Group",
"Definition:Prime Group"
] | [
"Definition:Trivial Group",
"Definition:Simple Group",
"Definition:Prime Group",
"One is not Prime",
"Definition:Non-Trivial Group",
"Definition:Simple Group",
"Definition:Non-Trivial Group",
"Definition:Simple Group",
"Definition:Simple Group",
"Definition:Simple Group",
"Definition:Prime Group... |
proofwiki-4780 | Cyclic Group is Simple iff Prime | Let $G$ be a cyclic group.
Then $G$ is simple {{iff}} $G$ is a prime group. | Let $G$ be a cyclic group.
From Cyclic Group is Abelian it follows that $G$ is an abelian group.
The result follows from Abelian Group is Simple iff Prime.
{{qed}}
Category:Simple Groups
Category:Cyclic Groups
ihiodpl0ircj9fl5b5ya623qbow3qh2 | Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
Then $G$ is [[Definition:Simple Group|simple]] {{iff}} $G$ is a [[Definition:Prime Group|prime group]]. | Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
From [[Cyclic Group is Abelian]] it follows that $G$ is an [[Definition:Abelian Group|abelian group]].
The result follows from [[Abelian Group is Simple iff Prime]].
{{qed}}
[[Category:Simple Groups]]
[[Category:Cyclic Groups]]
ihiodpl0ircj9fl5b5ya623qbow3qh2 | Cyclic Group is Simple iff Prime | https://proofwiki.org/wiki/Cyclic_Group_is_Simple_iff_Prime | https://proofwiki.org/wiki/Cyclic_Group_is_Simple_iff_Prime | [
"Simple Groups",
"Cyclic Groups"
] | [
"Definition:Cyclic Group",
"Definition:Simple Group",
"Definition:Prime Group"
] | [
"Definition:Cyclic Group",
"Cyclic Group is Abelian",
"Definition:Abelian Group",
"Abelian Group is Simple iff Prime",
"Category:Simple Groups",
"Category:Cyclic Groups"
] |
proofwiki-4781 | Factors of Solvable Group are Prime | Let $G$ be a solvable group.
Let $\HH$ be a composition series of $G$.
Then all factor groups of $\HH$ must be prime. | By definition, a composition series is a normal series whose factor groups are all simple.
A solvable group, by definition, is one which has a composition series whose factor groups are all cyclic.
From Cyclic Group is Simple iff Prime, it follows that all the factor groups of a composition series of a solvable group m... | Let $G$ be a [[Definition:Solvable Group|solvable group]].
Let $\HH$ be a [[Definition:Composition Series|composition series]] of $G$.
Then all [[Definition:Factor of Normal Series|factor groups]] of $\HH$ must be [[Definition:Prime Group|prime]]. | By definition, a [[Definition:Composition Series|composition series]] is a [[Definition:Normal Series|normal series]] whose [[Definition:Factor of Normal Series|factor groups]] are all [[Definition:Simple Group|simple]].
A [[Definition:Solvable Group|solvable group]], by definition, is one which has a [[Definition:Com... | Factors of Solvable Group are Prime | https://proofwiki.org/wiki/Factors_of_Solvable_Group_are_Prime | https://proofwiki.org/wiki/Factors_of_Solvable_Group_are_Prime | [
"Solvable Groups",
"Prime Groups"
] | [
"Definition:Solvable Group",
"Definition:Composition Series",
"Definition:Normal Series/Factor Group",
"Definition:Prime Group"
] | [
"Definition:Composition Series",
"Definition:Normal Series",
"Definition:Normal Series/Factor Group",
"Definition:Simple Group",
"Definition:Solvable Group",
"Definition:Composition Series",
"Definition:Normal Series/Factor Group",
"Definition:Cyclic Group",
"Cyclic Group is Simple iff Prime",
"De... |
proofwiki-4782 | Linear Subspace is Convex Set | Let $V$ be a vector space over $\R$ or $\C$, and let $L$ be a linear subspace of $V$.
Then $L$ is a convex set. | Let $x, y \in L, t \in \closedint 0 1$ be arbitrary.
Then, as $L$ is by definition closed under addition and scalar multiplication, it follows immediately that
:$t x + \paren {1 - t} y \in L$
Hence $L$ is convex.
{{qed}} | Let $V$ be a vector space over $\R$ or $\C$, and let $L$ be a [[Definition:Vector Subspace|linear subspace]] of $V$.
Then $L$ is a [[Definition:Convex Set (Vector Space)|convex set]]. | Let $x, y \in L, t \in \closedint 0 1$ be arbitrary.
Then, as $L$ is by definition closed under addition and scalar multiplication, it follows immediately that
:$t x + \paren {1 - t} y \in L$
Hence $L$ is [[Definition:Convex Set (Vector Space)|convex]].
{{qed}} | Linear Subspace is Convex Set | https://proofwiki.org/wiki/Linear_Subspace_is_Convex_Set | https://proofwiki.org/wiki/Linear_Subspace_is_Convex_Set | [
"Vector Subspaces",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Vector Subspace",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Convex Set (Vector Space)"
] |
proofwiki-4783 | Singleton is Convex Set | Let $V$ be a vector space over $\R$ or $\C$, and let $v \in V$.
Then the singleton $S = \set v$ is a convex set. | For any $x, y \in S$, we have $x = y = v$.
It follows that:
:$\forall t \in \closedint 0 1: t x + \paren {1 - t} y = v \in S$
Hence $S$ is a convex set.
{{qed}} | Let $V$ be a [[Definition:Vector Space|vector space]] over $\R$ or $\C$, and let $v \in V$.
Then the [[Definition:Singleton|singleton]] $S = \set v$ is a [[Definition:Convex Set (Vector Space)|convex set]]. | For any $x, y \in S$, we have $x = y = v$.
It follows that:
:$\forall t \in \closedint 0 1: t x + \paren {1 - t} y = v \in S$
Hence $S$ is a [[Definition:Convex Set (Vector Space)|convex set]].
{{qed}} | Singleton is Convex Set | https://proofwiki.org/wiki/Singleton_is_Convex_Set | https://proofwiki.org/wiki/Singleton_is_Convex_Set | [
"Vector Spaces",
"Singletons",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Vector Space",
"Definition:Singleton",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Convex Set (Vector Space)"
] |
proofwiki-4784 | Group is Solvable iff Normal Subgroup and Quotient are Solvable | Let $G$ be a finite group.
Let $H$ be a normal subgroup of $G$.
Then $G$ is solvable {{iff}}:
:$(1): \quad H$ is solvable
and
:$(2): \quad G / H$ is solvable
where $G / H$ is the quotient group of $G$ by $H$. | As $H \lhd G$ we can construct the normal series:
:$(A): \quad \set e \lhd H \lhd G$
By Finite Group has Composition Series, $(A)$ can be refined to a composition series for $G$:
:$(B): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$
Suppose $G_k = H$.
Then we can construct the composition series:
:$(C): \quad \s... | Let $G$ be a [[Definition:Finite Group|finite group]].
Let $H$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Then $G$ is [[Definition:Solvable Group|solvable]] {{iff}}:
:$(1): \quad H$ is [[Definition:Solvable Group|solvable]]
and
:$(2): \quad G / H$ is [[Definition:Solvable Group|solvable]]
where $G /... | As $H \lhd G$ we can construct the [[Definition:Normal Series|normal series]]:
:$(A): \quad \set e \lhd H \lhd G$
By [[Finite Group has Composition Series]], $(A)$ can be [[Definition:Refinement of Normal Series|refined]] to a [[Definition:Composition Series|composition series]] for $G$:
:$(B): \quad \set e = G_0 \lhd... | Group is Solvable iff Normal Subgroup and Quotient are Solvable | https://proofwiki.org/wiki/Group_is_Solvable_iff_Normal_Subgroup_and_Quotient_are_Solvable | https://proofwiki.org/wiki/Group_is_Solvable_iff_Normal_Subgroup_and_Quotient_are_Solvable | [
"Normal Subgroups",
"Quotient Groups",
"Solvable Groups"
] | [
"Definition:Finite Group",
"Definition:Normal Subgroup",
"Definition:Solvable Group",
"Definition:Solvable Group",
"Definition:Solvable Group",
"Definition:Quotient Group"
] | [
"Definition:Normal Series",
"Finite Group has Composition Series",
"Definition:Refinement of Normal Series",
"Definition:Composition Series",
"Definition:Composition Series",
"Third Isomorphism Theorem",
"Definition:Normal Series/Factor Group",
"Definition:Composition Series",
"Definition:Normal Ser... |
proofwiki-4785 | Finite Abelian Group is Solvable | Let $G$ be a finite abelian group.
Then $G$ is solvable. | Proof by strong induction on the order of $G$:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:All abelian groups of order $n$ and below are solvable. | Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
Then $G$ is [[Definition:Solvable Group|solvable]]. | Proof by [[Second Principle of Mathematical Induction|strong induction]] on the [[Definition:Order of Structure|order]] of $G$:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:All [[Definition:Abelian Group|abelian groups]] of [[Definition:Order of Structure|order]] $n$ and belo... | Finite Abelian Group is Solvable | https://proofwiki.org/wiki/Finite_Abelian_Group_is_Solvable | https://proofwiki.org/wiki/Finite_Abelian_Group_is_Solvable | [
"Abelian Groups",
"Solvable Groups"
] | [
"Definition:Finite Group",
"Definition:Abelian Group",
"Definition:Solvable Group"
] | [
"Second Principle of Mathematical Induction",
"Definition:Order of Structure",
"Definition:Proposition",
"Definition:Abelian Group",
"Definition:Order of Structure",
"Definition:Solvable Group",
"Definition:Solvable Group",
"Definition:Abelian Group",
"Definition:Order of Structure",
"Definition:S... |
proofwiki-4786 | Intersection of Convex Sets is Convex Set (Vector Spaces) | Let $V$ be a vector space over $\R$ or $\C$.
Let $\CC$ be a family of convex subsets of $V$.
Then the intersection $\ds \bigcap \CC$ is also a convex subset of $V$. | Let $x, y \in \ds \bigcap \CC$.
Then by definition of set intersection, $\forall C \in \CC: x, y \in C$.
The convexity of each $C$ yields:
:$\forall t \in \closedint 0 1: t x + \paren {1 - t} y \in C$
Therefore, these elements are also in $\ds \bigcap \CC$, by definition of set intersection.
Hence $\ds \bigcap \CC$ is ... | Let $V$ be a [[Definition:Vector Space|vector space]] over $\R$ or $\C$.
Let $\CC$ be a family of [[Definition:Convex Set (Vector Space)|convex subsets]] of $V$.
Then the [[Definition:Set Intersection|intersection]] $\ds \bigcap \CC$ is also a [[Definition:Convex Set (Vector Space)|convex subset]] of $V$. | Let $x, y \in \ds \bigcap \CC$.
Then by definition of [[Definition:Set Intersection|set intersection]], $\forall C \in \CC: x, y \in C$.
The [[Definition:Convex Set (Vector Space)|convexity]] of each $C$ yields:
:$\forall t \in \closedint 0 1: t x + \paren {1 - t} y \in C$
Therefore, these [[Definition:Element|elem... | Intersection of Convex Sets is Convex Set (Vector Spaces) | https://proofwiki.org/wiki/Intersection_of_Convex_Sets_is_Convex_Set_(Vector_Spaces) | https://proofwiki.org/wiki/Intersection_of_Convex_Sets_is_Convex_Set_(Vector_Spaces) | [
"Vector Spaces",
"Convex Sets (Vector Spaces)"
] | [
"Definition:Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Set Intersection",
"Definition:Convex Set (Vector Space)"
] | [
"Definition:Set Intersection",
"Definition:Convex Set (Vector Space)",
"Definition:Element",
"Definition:Set Intersection",
"Definition:Convex Set (Vector Space)"
] |
proofwiki-4787 | Prime Power Group is Solvable | Let $G$ be a group whose order is $p^n$ where $p$ is a prime number and $n$ is a positive integer.
Then $G$ is solvable. | A direct consequence of Factors of Composition Series for Prime Power Group and the definition of solvable group.
{{qed}}
Category:P-Groups
Category:Solvable Groups
31xinsqnol2xxwyvwlv1v8vg7gcgpq1 | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of Group|order]] is $p^n$ where $p$ is a [[Definition:Prime Number|prime number]] and $n$ is a [[Definition:Positive Integer|positive integer]].
Then $G$ is [[Definition:Solvable Group|solvable]]. | A direct consequence of [[Factors of Composition Series for Prime Power Group]] and the definition of [[Definition:Solvable Group|solvable group]].
{{qed}}
[[Category:P-Groups]]
[[Category:Solvable Groups]]
31xinsqnol2xxwyvwlv1v8vg7gcgpq1 | Prime Power Group is Solvable | https://proofwiki.org/wiki/Prime_Power_Group_is_Solvable | https://proofwiki.org/wiki/Prime_Power_Group_is_Solvable | [
"P-Groups",
"Solvable Groups"
] | [
"Definition:Group",
"Definition:Order of Structure",
"Definition:Prime Number",
"Definition:Positive/Integer",
"Definition:Solvable Group"
] | [
"Factors of Composition Series for Prime Power Group",
"Definition:Solvable Group",
"Category:P-Groups",
"Category:Solvable Groups"
] |
proofwiki-4788 | Subgroup of Solvable Group is Solvable | Let $G$ be a solvable group.
Let $H$ be a subgroup of $G$.
Then $H$ is solvable. | Let $\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$ be a normal series for $G$ with abelian quotients.
For every $i = 1, 2, \ldots, n$ we have:
:$\paren {H \cap G_i} \cap G_{i - 1} = H \cap G_{i - 1}$
From the Second Isomorphism Theorem for Groups:
:$\dfrac {\paren {H \cap G_i} G_{i - 1} } {G_{i - 1} } \cong \dfrac {H... | Let $G$ be a [[Definition:Solvable Group|solvable group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $H$ is [[Definition:Solvable Group|solvable]]. | Let $\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$ be a [[Definition:Normal Series|normal series]] for $G$ with [[Definition:Abelian_Group|abelian]] quotients.
For every $i = 1, 2, \ldots, n$ we have:
:$\paren {H \cap G_i} \cap G_{i - 1} = H \cap G_{i - 1}$
From the [[Second Isomorphism Theorem for Groups]]:
:$\df... | Subgroup of Solvable Group is Solvable/Proof 1 | https://proofwiki.org/wiki/Subgroup_of_Solvable_Group_is_Solvable | https://proofwiki.org/wiki/Subgroup_of_Solvable_Group_is_Solvable/Proof_1 | [
"Subgroups",
"Solvable Groups",
"Subgroup of Solvable Group is Solvable"
] | [
"Definition:Solvable Group",
"Definition:Subgroup",
"Definition:Solvable Group"
] | [
"Definition:Normal Series",
"Definition:Abelian_Group",
"Second Isomorphism Theorem/Groups",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Normal Subgroup",
"Correspondence Theorem (Group Theory)",
"Definition:Abelian Group",
"Subgroup of Abelian Group is Abelian",
"Defi... |
proofwiki-4789 | Subgroup of Solvable Group is Solvable | Let $G$ be a solvable group.
Let $H$ be a subgroup of $G$.
Then $H$ is solvable. | Firstly we, know that a group is solvable {{iff}} its derived series:
:$\map D G = \sqbrk {G, G} \ , \ \map {D^i} G = \sqbrk {\map {D^{i - 1} } G, \map {D^{i - 1} } G}$
becomes trivial after finite iteration.
Meaning:
{{handwaving|trivial}}
$\map {D^j} G = \set 1$
for some finite $j$.
Now it is trivial that:
{{handwav... | Let $G$ be a [[Definition:Solvable Group|solvable group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $H$ is [[Definition:Solvable Group|solvable]]. | Firstly we, know that a [[Definition:Group|group]] is [[Definition:Solvable Group|solvable]] {{iff}} its [[Definition:Derived Series|derived series]]:
:$\map D G = \sqbrk {G, G} \ , \ \map {D^i} G = \sqbrk {\map {D^{i - 1} } G, \map {D^{i - 1} } G}$
becomes [[Definition:Trivial Subgroup|trivial]] after finite iterati... | Subgroup of Solvable Group is Solvable/Proof 2 | https://proofwiki.org/wiki/Subgroup_of_Solvable_Group_is_Solvable | https://proofwiki.org/wiki/Subgroup_of_Solvable_Group_is_Solvable/Proof_2 | [
"Subgroups",
"Solvable Groups",
"Subgroup of Solvable Group is Solvable"
] | [
"Definition:Solvable Group",
"Definition:Subgroup",
"Definition:Solvable Group"
] | [
"Definition:Group",
"Definition:Solvable Group",
"Definition:Derived Series",
"Definition:Trivial Subgroup",
"Definition:Trivial Subgroup"
] |
proofwiki-4790 | Subgroup of Solvable Group is Solvable | Let $G$ be a solvable group.
Let $H$ be a subgroup of $G$.
Then $H$ is solvable. | Let $H \le G$ and $G$ be solvable with normal series:
:$\set e = G_0 \lhd G_1 \lhd \dots \lhd G_m = G$
such that $G_{i + 1} / G_i$ is abelian for all $i$.
Define $N_i = G_i \cap H$.
We show that these $N_i$ will form a normal series with abelian factors.
;Normality:
Let $x \in N_i$ and $y \in N_{i + 1}$.
Then $y x y^{-... | Let $G$ be a [[Definition:Solvable Group|solvable group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $H$ is [[Definition:Solvable Group|solvable]]. | Let $H \le G$ and $G$ be [[Definition:Solvable Group|solvable]] with [[Definition:Normal Series|normal series]]:
:$\set e = G_0 \lhd G_1 \lhd \dots \lhd G_m = G$
such that $G_{i + 1} / G_i$ is [[Definition:Abelian Group|abelian]] for all $i$.
Define $N_i = G_i \cap H$.
We show that these $N_i$ will form a [[Definit... | Subgroup of Solvable Group is Solvable/Proof 3 | https://proofwiki.org/wiki/Subgroup_of_Solvable_Group_is_Solvable | https://proofwiki.org/wiki/Subgroup_of_Solvable_Group_is_Solvable/Proof_3 | [
"Subgroups",
"Solvable Groups",
"Subgroup of Solvable Group is Solvable"
] | [
"Definition:Solvable Group",
"Definition:Subgroup",
"Definition:Solvable Group"
] | [
"Definition:Solvable Group",
"Definition:Normal Series",
"Definition:Abelian Group",
"Definition:Normal Series",
"Definition:Abelian Group",
"Definition:Normal Series/Factor Group",
"Definition:Group",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Normal Subgroup/Definitio... |
proofwiki-4791 | Group with Normal Series with Solvable Factor Groups is Solvable | Let $G$ be a solvable group.
Let $\HH$ be a normal series for $G$.
Let all the factor groups of $\HH$ be solvable.
Then $G$ is solvable. | To be proved by construction of a composition series of $G$ from $\HH$.
Suppose $\HH$ is a normal series for $G$:
:$\HH: \quad \set e = G_0 \subset G_1 \subset \dots \subset G_n = G$,
where $G_{k + 1} / G_k$ is solvable for $k = 0, 1, \dots, n - 1$.
If every factor of $\HH$ is cyclic of prime order then we are done.
Ot... | Let $G$ be a [[Definition:Solvable Group|solvable group]].
Let $\HH$ be a [[Definition:Normal Series|normal series]] for $G$.
Let all the [[Definition:Factor of Normal Series|factor groups]] of $\HH$ be [[Definition:Solvable Group|solvable]].
Then $G$ is [[Definition:Solvable Group|solvable]]. | To be proved by construction of a [[Definition:Composition Series|composition series]] of $G$ from $\HH$.
Suppose $\HH$ is a [[Definition:Normal Series|normal series]] for $G$:
:$\HH: \quad \set e = G_0 \subset G_1 \subset \dots \subset G_n = G$,
where $G_{k + 1} / G_k$ is [[Definition:Solvable Group|solvable]] for... | Group with Normal Series with Solvable Factor Groups is Solvable | https://proofwiki.org/wiki/Group_with_Normal_Series_with_Solvable_Factor_Groups_is_Solvable | https://proofwiki.org/wiki/Group_with_Normal_Series_with_Solvable_Factor_Groups_is_Solvable | [
"Subgroups",
"Solvable Groups"
] | [
"Definition:Solvable Group",
"Definition:Normal Series",
"Definition:Normal Series/Factor Group",
"Definition:Solvable Group",
"Definition:Solvable Group"
] | [
"Definition:Composition Series",
"Definition:Normal Series",
"Definition:Solvable Group",
"Definition:Normal Series/Factor Group",
"Definition:Cyclic Group",
"Definition:Prime Group",
"Definition:Normal Series/Factor Group",
"Definition:Cyclic Group",
"Definition:Prime Group",
"Definition:Composit... |
proofwiki-4792 | Group with Order Less than 60 is Solvable | Every group whose order is less than $60$ is solvable. | Let $n \in \Z$ be the smallest integer such that there is a non-solvable group of order $n$.
{{AimForCont}} $n < 60$.
Let $G$ be a non-solvable group of order $n$.
Since $\order G < 60$, it follows from Simple Group of Order Less than 60 is Prime that either:
:$G$ is prime
or:
:$G$ is not simple.
Since {{hypothesis}} $... | Every [[Definition:Group|group]] whose [[Definition:Order of Group|order]] is less than $60$ is [[Definition:Solvable Group|solvable]]. | Let $n \in \Z$ be the smallest [[Definition:Integer|integer]] such that there is a [[Definition:Solvable Group|non-solvable group]] of [[Definition:Order of Group|order]] $n$.
{{AimForCont}} $n < 60$.
Let $G$ be a [[Definition:Solvable Group|non-solvable group]] of [[Definition:Order of Group|order]] $n$.
Since $\or... | Group with Order Less than 60 is Solvable | https://proofwiki.org/wiki/Group_with_Order_Less_than_60_is_Solvable | https://proofwiki.org/wiki/Group_with_Order_Less_than_60_is_Solvable | [
"Solvable Groups"
] | [
"Definition:Group",
"Definition:Order of Structure",
"Definition:Solvable Group"
] | [
"Definition:Integer",
"Definition:Solvable Group",
"Definition:Order of Structure",
"Definition:Solvable Group",
"Definition:Order of Structure",
"Simple Group of Order Less than 60 is Prime",
"Definition:Prime Group",
"Definition:Simple Group",
"Definition:Solvable Group",
"Definition:Prime Group... |
proofwiki-4793 | Direct Product of Solvable Groups is Solvable | Let $G$ and $H$ be groups which are solvable.
Then their (external) direct product $G \times H$ is also solvable. | By Image of Canonical Injection is Normal Subgroup, $G \times \set {e_H}$ is a normal subgroup of $G \times H$.
Also, by Quotient Group of Direct Products, $\paren {G \times H} / \paren {G \times \set{e_H} }$ is isomorphic to $H$.
The result then follows from Group is Solvable iff Normal Subgroup and Quotient are Solva... | Let $G$ and $H$ be [[Definition:Group|groups]] which are [[Definition:Solvable Group|solvable]].
Then their [[Definition:Group Direct Product|(external) direct product]] $G \times H$ is also [[Definition:Solvable Group|solvable]]. | By [[Image of Canonical Injection is Normal Subgroup]], $G \times \set {e_H}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G \times H$.
Also, by [[Quotient Group of Direct Products]], $\paren {G \times H} / \paren {G \times \set{e_H} }$ is [[Definition:Group Isomorphism|isomorphic]] to $H$.
The result then... | Direct Product of Solvable Groups is Solvable | https://proofwiki.org/wiki/Direct_Product_of_Solvable_Groups_is_Solvable | https://proofwiki.org/wiki/Direct_Product_of_Solvable_Groups_is_Solvable | [
"Solvable Groups",
"Group Direct Products"
] | [
"Definition:Group",
"Definition:Solvable Group",
"Definition:Group Direct Product",
"Definition:Solvable Group"
] | [
"Image of Canonical Injection is Normal Subgroup",
"Definition:Normal Subgroup",
"Quotient Group of Direct Products",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Group is Solvable iff Normal Subgroup and Quotient are Solvable"
] |
proofwiki-4794 | Group with One Sylow Subgroup per Prime Divisor is Solvable | Let $G$ be a group of order $n$.
Suppose that, for each prime number $p$ which divides $n$, $G$ has exactly one $p$-Sylow subgroup.
Then $G$ is solvable. | From Finite Group with One Sylow p-Subgroup per Prime Divisor is Isomorphic to Direct Product,
:$G$ is isomorphic to the direct product of its $p$-Sylow subgroups.
From Prime Power Group is Solvable, each $p$-Sylow subgroup is solvable.
From Direct Product of Solvable Groups is Solvable, $G$ is solvable.
{{qed}} | Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $n$.
Suppose that, for each [[Definition:Prime Number|prime number]] $p$ which [[Definition:Divisor of Integer|divides]] $n$, $G$ has exactly one [[Definition:Sylow p-Subgroup|$p$-Sylow subgroup]].
Then $G$ is [[Definition:Solvable Gr... | From [[Finite Group with One Sylow p-Subgroup per Prime Divisor is Isomorphic to Direct Product]],
:$G$ is [[Definition:Isomorphism (Abstract Algebra)/Group Isomorphism|isomorphic]] to the [[Definition:Group Direct Product|direct product]] of its [[Definition:Sylow p-Subgroup|$p$-Sylow subgroups]].
From [[Prime Powe... | Group with One Sylow Subgroup per Prime Divisor is Solvable | https://proofwiki.org/wiki/Group_with_One_Sylow_Subgroup_per_Prime_Divisor_is_Solvable | https://proofwiki.org/wiki/Group_with_One_Sylow_Subgroup_per_Prime_Divisor_is_Solvable | [
"Solvable Groups",
"Prime Groups"
] | [
"Definition:Group",
"Definition:Order of Structure",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Sylow p-Subgroup",
"Definition:Solvable Group"
] | [
"Finite Group with One Sylow p-Subgroup per Prime Divisor is Isomorphic to Direct Product",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Direct Product",
"Definition:Sylow p-Subgroup",
"Prime Power Group is Solvable",
"Definition:Sylow p-Subgroup",
"Definition:Solvabl... |
proofwiki-4795 | Characterization of Bases (Hilbert Spaces) | Let $\HH$ be a Hilbert space, and let $\EE$ be an orthonormal subset of $\HH$.
Then the following six statements are equivalent:
:$(1): \quad \EE$ is a basis for $\HH$
:$(2): \quad h \in \HH, h \perp \EE \implies h = \mathbf 0_\HH$, where $\perp$ denotes orthogonality
:$(3): \quad \vee \EE = \HH$, where $\vee \EE$ deno... | === $(1)$ implies $(2)$ ===
Suppose that $\EE$ is a basis for $\HH$.
Suppose that $h \perp \EE$ for some $h \in \HH$ with $h \ne {\mathbf 0}_\HH$.
Then $\innerprod h x = 0$ for all $x \in \EE$.
So $\ds \innerprod {\frac h {\norm h} } x = 0$ for all $x \in \EE$.
Hence:
:$\ds \EE \cup \set {\frac h {\norm h} }$ is a ort... | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]], and let $\EE$ be an [[Definition:Orthonormal Subset|orthonormal subset]] of $\HH$.
Then the following six statements are equivalent:
:$(1): \quad \EE$ is a [[Definition:Basis (Hilbert Space)|basis]] for $\HH$
:$(2): \quad h \in \HH, h \perp \EE \implies h = ... | === $(1)$ implies $(2)$ ===
Suppose that $\EE$ is a [[Definition:Basis (Hilbert Space)|basis]] for $\HH$.
Suppose that $h \perp \EE$ for some $h \in \HH$ with $h \ne {\mathbf 0}_\HH$.
Then $\innerprod h x = 0$ for all $x \in \EE$.
So $\ds \innerprod {\frac h {\norm h} } x = 0$ for all $x \in \EE$.
Hence:
:$\ds \E... | Characterization of Bases (Hilbert Spaces) | https://proofwiki.org/wiki/Characterization_of_Bases_(Hilbert_Spaces) | https://proofwiki.org/wiki/Characterization_of_Bases_(Hilbert_Spaces) | [
"Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Orthonormal Subset",
"Definition:Basis (Hilbert Space)",
"Definition:Orthogonal (Linear Algebra)",
"Definition:Closed Linear Span",
"Definition:Generalized Sum"
] | [
"Definition:Basis (Hilbert Space)",
"Definition:Orthonormal Subset",
"Definition:Orthonormal Subset",
"Definition:Basis (Hilbert Space)",
"Definition:Orthonormal Subset",
"Definition:Basis (Hilbert Space)"
] |
proofwiki-4796 | Square Number Less than One | Let $x$ be a real number such that $x^2 < 1$.
Then:
:$x \in \openint {-1} 1$
where $\openint {-1} 1$ is the open interval $\left\{{x \in \R: -1 < x < 1}\right\}$. | First note that from Square of Real Number is Non-Negative:
:$x^2 \ge 0$
From Ordering of Squares in Reals:
:$(1): \quad x > 1 \implies x^2 > 1$
:$(2): \quad x < 1 \implies x^2 < 1$
From Identity Element of Multiplication on Numbers:
:$1^2 = 1$
so it is clear that the strict inequalities apply above.
For clarity, there... | Let $x$ be a [[Definition:Real Number|real number]] such that $x^2 < 1$.
Then:
:$x \in \openint {-1} 1$
where $\openint {-1} 1$ is the [[Definition:Open Real Interval|open interval]] $\left\{{x \in \R: -1 < x < 1}\right\}$. | First note that from [[Square of Real Number is Non-Negative]]:
:$x^2 \ge 0$
From [[Ordering of Squares in Reals]]:
:$(1): \quad x > 1 \implies x^2 > 1$
:$(2): \quad x < 1 \implies x^2 < 1$
From [[Identity Element of Multiplication on Numbers]]:
:$1^2 = 1$
so it is clear that the strict inequalities apply above.
Fo... | Square Number Less than One | https://proofwiki.org/wiki/Square_Number_Less_than_One | https://proofwiki.org/wiki/Square_Number_Less_than_One | [
"Real Analysis"
] | [
"Definition:Real Number",
"Definition:Real Interval/Open"
] | [
"Square of Real Number is Non-Negative",
"Ordering of Squares in Reals",
"Identity Element of Multiplication on Numbers",
"Square of Real Number is Non-Negative",
"Rule of Transposition",
"Category:Real Analysis"
] |
proofwiki-4797 | Uniform Limit Theorem | Let $\struct {M, d_M}$ and $\struct {N, d_N}$ be metric spaces.
Let $\sequence {f_n}$ be a sequence of mappings from $M$ to $N$ such that:
:$(1): \quad \forall n \in \N: f_n$ is continuous at every point of $M$
:$(2): \quad \sequence {f_n}$ converges uniformly to $f$
Then:
:$f$ is continuous at every point of $M$. | Let $a \in M$.
We are given that $d_N$ is a metric on $N$.
By applying {{Metric-space-axiom|2}} twice:
{{begin-eqn}}
{{eqn | n = 3
| q = \forall n \in \N, \forall x \in M
| l = \map {d_N} {\map f x, \map f a}
| o = \le
| r = \map {d_N} {\map f x, \map {f_n} x} + \map {d_N} {\map {f_n} x, \map {f... | Let $\struct {M, d_M}$ and $\struct {N, d_N}$ be [[Definition:Metric Space|metric spaces]].
Let $\sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Mapping|mappings]] from $M$ to $N$ such that:
:$(1): \quad \forall n \in \N: f_n$ is [[Definition:Continuous at Point of Metric Space|continuous at eve... | Let $a \in M$.
We are given that $d_N$ is a [[Definition:Metric|metric]] on $N$.
By applying {{Metric-space-axiom|2}} twice:
{{begin-eqn}}
{{eqn | n = 3
| q = \forall n \in \N, \forall x \in M
| l = \map {d_N} {\map f x, \map f a}
| o = \le
| r = \map {d_N} {\map f x, \map {f_n} x} + \map {d_... | Uniform Limit Theorem | https://proofwiki.org/wiki/Uniform_Limit_Theorem | https://proofwiki.org/wiki/Uniform_Limit_Theorem | [
"Functional Analysis",
"Continuous Mappings",
"Metric Spaces",
"Uniform Convergence",
"Named Theorems"
] | [
"Definition:Metric Space",
"Definition:Sequence",
"Definition:Mapping",
"Definition:Continuous Mapping (Metric Space)/Point",
"Definition:Uniform Convergence",
"Definition:Continuous Mapping (Metric Space)/Point"
] | [
"Definition:Metric Space/Metric",
"Definition:Uniform Convergence",
"Universal Instantiation",
"Definition:Continuous Mapping (Metric Space)/Point",
"Universal Instantiation",
"Definition:Continuous Mapping (Metric Space)/Point"
] |
proofwiki-4798 | Equality is Transitive | :$\forall a, b, c: \paren {a = b} \land \paren {b = c} \implies a = c$ | {{begin-eqn}}
{{eqn | l = a
| r = b
| c =
}}
{{eqn | ll= \vdash
| l = \map P a
| o = \iff
| r = \map P b
| c = Leibniz's law
}}
{{eqn | l = b
| r = c
| c =
}}
{{eqn | ll= \vdash
| l = \map P b
| o = \iff
| r = \map P c
| c = Leibniz's law
}}
{{eq... | :$\forall a, b, c: \paren {a = b} \land \paren {b = c} \implies a = c$ | {{begin-eqn}}
{{eqn | l = a
| r = b
| c =
}}
{{eqn | ll= \vdash
| l = \map P a
| o = \iff
| r = \map P b
| c = [[Axiom:Leibniz's Law|Leibniz's law]]
}}
{{eqn | l = b
| r = c
| c =
}}
{{eqn | ll= \vdash
| l = \map P b
| o = \iff
| r = \map P c
| c... | Equality is Transitive | https://proofwiki.org/wiki/Equality_is_Transitive | https://proofwiki.org/wiki/Equality_is_Transitive | [
"Logic",
"Equality"
] | [] | [
"Axiom:Leibniz's Law",
"Axiom:Leibniz's Law",
"Biconditional is Transitive",
"Axiom:Leibniz's Law"
] |
proofwiki-4799 | Axiom of Pairing from Infinity and Replacement | The Axiom of Pairing is a consequence of:
:the Axiom of Infinity
and
:the Axiom of Replacement. | The set $2 = \set {\O, \set \O}$ is used with the Axiom of Replacement as the domain for a mapping whose image is $\set {A, B}$.
A suitable mapping would be:
:$\paren {y = \O \land z = A} \lor \paren {y = \set \O \land z = B}$
The set $2$ is shown to exist as a member of the infinite set whose existence is asserted by ... | The [[Axiom:Axiom of Pairing (Set Theory)|Axiom of Pairing]] is a consequence of:
:the [[Axiom:Axiom of Infinity|Axiom of Infinity]]
and
:the [[Axiom:Axiom of Replacement|Axiom of Replacement]]. | The set $2 = \set {\O, \set \O}$ is used with the [[Axiom:Axiom of Replacement|Axiom of Replacement]] as the [[Definition:Domain of Mapping|domain]] for a [[Definition:Mapping|mapping]] whose [[Definition:Image of Mapping|image]] is $\set {A, B}$.
A suitable [[Definition:Mapping|mapping]] would be:
:$\paren {y = \O \l... | Axiom of Pairing from Infinity and Replacement | https://proofwiki.org/wiki/Axiom_of_Pairing_from_Infinity_and_Replacement | https://proofwiki.org/wiki/Axiom_of_Pairing_from_Infinity_and_Replacement | [
"Axiom of Pairing",
"Doubletons"
] | [
"Axiom:Axiom of Pairing/Set Theory",
"Axiom:Axiom of Infinity",
"Axiom:Axiom of Replacement"
] | [
"Axiom:Axiom of Replacement",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping",
"Definition:Infinite Set",
"Axiom:Axiom of Infinity",
"Category:Axiom of Pairing",
"Category:Doubletons"
] |
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