nyuuzyou/wikis huggingface dataset:
Fourth Hyperoperation is Tetration Operation

== Theorem ==
The $4$th hyperoperation is the tetration operation restricted to the positive integers:
:$\forall x, y \in \Z_{\ge 0}: H_4 \left({x, y}\right) = x \uparrow \uparrow y$
where $\uparrow \uparrow$ denotes tetration:
:$x \uparrow \uparrow n := \begin{cases}
1 & : n = 0 \\
x \uparrow \left({x \uparrow \uparrow \left({n - 1}\right)}\right) & : n > 0 \\
\end{cases}$
In this context, $x \uparrow y$ is the Knuth notation for powers:
:$x \uparrow y := x^y$
== Proof ==
By definition of the hyperoperation sequence:
:$\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases}
y + 1 & : n = 0 \\
x & : n = 1, y = 0 \\
0 & : n = 2, y = 0 \\
1 & : n > 2, y = 0 \\
H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$
Thus the $4$th hyperoperation is defined as:
:$\forall x, y \in \Z_{\ge 0}: H_4 \left({x, y}\right) = \begin{cases}
1 & : y = 0 \\
H_3 \left({x, H_4 \left({x, y - 1}\right)}\right) & : y > 0 \end{cases}$
From Third Hyperoperation is Integer Power Operation:
:$(1): \quad \forall x, y \in \Z_{\ge 0}: H_4 \left({x, y}\right) = \begin{cases}
1 & : y = 0 \\
x \uparrow H_3 \left({x, y - 1}\right) & : y > 0 \end{cases}$
The proof proceeds by induction.
For all $y \in \Z_{\ge 0}$, let $P \left({y}\right)$ be the proposition:
:$\forall x \in \Z_{\ge 0}: H_4 \left({x, y}\right) = x \uparrow \uparrow y$
=== Basis for the Induction ===
$P \left({0}\right)$ is the case:
Thus $P \left({0}\right)$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the induction hypothesis:
:$\forall x \in \Z_{\ge 0}: H_4 \left({x, k}\right) = x \uparrow \uparrow k$
from which it is to be shown that:
:$\forall x \in \Z_{\ge 0}: H_4 \left({x, k + 1}\right) = x \uparrow \uparrow \left({k + 1}\right)$
=== Induction Step ===
This is the induction step:
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = x \uparrow \uparrow y$
Category:Hyperoperation
Fourth Power Modulo 5

== Theorem ==
Let $n \in \Z$ be an integer.
Then:
:$n^4 \equiv m \pmod 5$
where $m \in \set {0, 1}$.
== Proof ==
By Congruence of Powers:
:$a \equiv b \pmod 5 \iff a^4 \equiv b^4 \pmod 5$
so it is sufficient to demonstrate the result for $n \in \set {0, 1, 2, 3, 4}$.
Thus:
The result follows.
Category:Fourth Powers
Fourth Power as Summation of Groups of Consecutive Integers

== Theorem ==
Take the positive integers and group them in sets such that the $m$th set contains the next $m$ positive integers:
:$\set 1, \set {2, 3}, \set {4, 5, 6}, \set {7, 8, 9, 10}, \set {11, 12, 13, 14, 15}, \ldots$
Remove all the sets with an even number of elements.
Then the sum of all the integers in the first $n$ sets remaining equals $n^4$.
== Proof ==
Let $S_m$ be the $m$th set of $m$ consecutive integers.
Let $S_k$ be the $k$th set of $m$ consecutive integers after the sets with an even number of elements have been removed.
Then $S_k = S_m$ where $m = 2 k - 1$.
By the method of construction:
:the largest integer in $S_m$ is $T_m$, the $m$th triangular number
:there are $m$ integers in $S_m$.
We also have that the middle integer in $S_m$ is $T_m - \dfrac {m - 1} 2$ (by inspection).
Thus the sum of the elements of $S_k$ is:
We need to calculate the sum of all $S_k$ from $1$ to $n$.
Hence we have:
== Examples ==
== Sources ==
* 
* : $81$
* : $81$
Category:Fourth Powers
Category:Fourth Power as Summation of Groups of Consecutive Integers
Fourth Power as Summation of Groups of Consecutive Integers/Examples/3

== Example of Fourth Power as Summation of Groups of Consecutive Integers ==
:$1 + \paren {4 + 5 + 6} + \paren {11 + 12 + 13 + 14 + 15} = 3^4$
== Sources ==
* : $81$
* : $81$
Category:Fourth Power as Summation of Groups of Consecutive Integers
Fourth Power expressible as Sum of 6 Fourth Powers

== Theorem ==
$28 \, 561$ can be expressed as the sum of $6$ fourth powers:
:$28 \, 561 = 13^4 = 12^4 + 8^4 + 7^4 + 6^4 + 2^4 + 2^4$
== Proof ==
== Sources ==
* : $28,561$
Category:Fourth Powers
Category:28,561
Fourth Power is Sum of 2 Triangular Numbers

== Theorem ==
Let $n \in \Z$ be an integer.
Then:
:$\exists a, b \in \Z_{\ge 0}: n^4 = T_a + T_b$
where $T_a$ and $T_b$ are triangular numbers.
That is, the $4$th power of an integer equals the sum of two triangular numbers.
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Historical Note ==
== Sources ==
* : $15$
* : $15$
Category:Fourth Powers
Category:Triangular Numbers
Category:Fourth Power is Sum of 2 Triangular Numbers
Fourth Power is Sum of 2 Triangular Numbers/Examples/T41 + T55

== Example of Use of Fourth Power is Sum of 2 Triangular Numbers ==
:$7^4 = T_{41} + T_{55}$
== Proof ==
We have:
:$41 = 7^2 - 7 - 1$
and:
:$55 = 7^2 + 7 - 1$
Hence from Fourth Power is Sum of 2 Triangular Numbers:
:$7^4 = T_{41} + T_{55}$
== Sources ==
* : $15$
* : $15$
Category:Fourth Power is Sum of 2 Triangular Numbers
Fourth Power is Sum of 2 Triangular Numbers/Proof 1

== Theorem ==
== Proof ==
Note first that:
:$\forall n \in \Z: \paren {-n}^4 = n^4$
Hence it is sufficient to consider the case where $n \ge 0$.
For $n = 0$:
:$0^4 = 0 = 0 + 0 = T_0 + T_0$
Let $n > 0$.
Then:
:$n^2 - 1 \ge 0$
and:
:$n^2 \ge 0$
Hence:
Category:Fourth Power is Sum of 2 Triangular Numbers
Fourth Power of 11 is Palindromic

== Theorem ==
::$11^4 = 14641$
== Proof ==
From Square of 11 is Palindromic:
:$11^2 = 121$
Then from Square of 121 is Palindromic:
:$121^2 = 14641 = \paren {11^2}^2 = 11^4$
== Sources ==
* : $121$
* : $121$
Category:Fourth Powers
Category:11
Category:14,641
Category:Palindromic Numbers
Fourth Power of Complex Number

== Theorem ==
Let $z = a + i b$ be a complex number.
Then its fourth power is given by:
:$z^4 = a^4 - 6 a^2 b^2 + b^4 + i \paren {4 a^3 b - 4 a b^3}$
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.20$
Category:Complex Numbers
Category:Fourth Powers
Fourth Powers which are Sum of 4 Fourth Powers

== Theorem ==
The following positive integers are such that their fourth powers can be expressed as the sum of the fourth powers of $4$ other positive integers with no common factors:
:$353, 651, 2487, 2501, 2829, \ldots$
== Proof ==
== Examples ==
== Sources ==
* 
* : $651$
* : $353$
* : $651$
* 
Category:Fourth Powers
Fourth Powers which are Sum of 4 Fourth Powers/Examples/353

== Examples of Fourth Powers which are Sum of 4 Fourth Powers ==
:$353^4 = 30^4 + 120^4 + 272^4 + 315^4$
== Proof ==
Now we have that:
Hence:
But:
:$3 \times 17 = 2 \times 26 - 1$
So:
Hence:
:$353^2 + 272^2 = 2^4 + 8^4 + 21^4$
but:
So:
:$353^4 = 30^4 + 120^4 + 272^4 + 315^4$
== Historical Note ==
== Sources ==
* 
* 
* : $353$
* : $353$
* 
Category:Fourth Powers
Category:353
Fourth Powers which are Sum of 4 Fourth Powers/Mistake

== Source Work ==
:
:The Dictionary
::$353$
== Mistake ==
:$353^4$ is the smallest number that is the sum of $4$ other $4$th powers ... The sequence of such numbers continues: $651 \quad 2487 \quad 2501 \quad 2829 \ldots$
It needs to be specified that the $4$th powers themselves must have no common divisor, otherwise, for example, $706$ would be included in this list:
== Sources ==
* : $353$
Category:Mistakes/Curious and Interesting Numbers
Fourth Sylow Theorem

== Theorem ==
The number of Sylow $p$-subgroups of a finite group is congruent to $1 \pmod p$.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
Some sources call this the second Sylow theorem.
Others merge this result with what we call the Fifth Sylow Theorem and call it the third Sylow theorem.
== Also see ==
* Sylow Theorems
== Historical Note ==
Category:Sylow Theorems
Category:Fourth Sylow Theorem
Fourth Sylow Theorem/Proof 1

== Theorem ==
== Proof ==
Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$ and $n > 0$.
Let $r$ be the number of Sylow $p$-subgroups of $G$.
We want to show that $r \equiv 1 \pmod p$.
Let $\mathbb S = \set {S \subseteq G: \card S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.
From the reasoning in the First Sylow Theorem, we have:
:$\size {\mathbb S} = \dbinom {p^n k} {p^n}$
Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements::
:$\forall S \in \mathbb S: g \wedge S = g S = \set {x \in G: x = g s: s \in S}$
From Orbits of Group Action on Sets with Power of Prime Size:
:there are exactly as many Sylow $p$-subgroups as there are orbits whose length is not divisible by $p$.
Also by Orbits of Group Action on Sets with Power of Prime Size:
:all the terms in the Partition Equation are divisible by $k$, perhaps also divisible by $p$.
We can write the Partition Equation as:
:$\size {\mathbb S} = \size {\Orb {S_1} } + \size {\Orb {S_2} } + \cdots + \size {\Orb {S_r} } + \size {\Orb {S_{r + 1} } } + \cdots + \size {\Orb {S_s} }$
where the first $r$ terms are the orbits containing the Sylow $p$-subgroups:
:$\Stab {S_i}$
For each of these:
:$\order G = \size {\Orb {S_i} } \times \size {\Stab {S_i} } = p^n \size {\Orb {S_i} }$
Thus:
:$\size {\Orb {S_i} } = k$
for $1 \le i \le r$.
Each of the rest of the orbits are divisible by both $p$ and $k$, as we have seen.
So:
:$\size {\mathbb S} = k r + m p k$
where:
:the first term corresponds to the $r$ orbits containing the Sylow $p$-subgroups
:the second term corresponds to all the rest of the orbits
:$m$ is some unspecified integer.
That is, there exists some integer $m$ such that:
:$\size {\mathbb S} = \dbinom {p^n k} {p^n} = k r + m p k$
Now this of course applies to the special case of the cyclic group $C_{p^n k}$.
In this case, there is exactly one subgroup for each divisor of $p^n k$.
In particular, there is exactly one subgroup of order $p^n$.
Hence, in this case:
:$r = 1$
So we have an integer $m'$ such that $\dbinom {p^n k} {p^n} = k + m' p k$.
We can now equate these expressions:
and the proof is complete.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Theorem $11.6$
Category:Fourth Sylow Theorem
Fourth Sylow Theorem/Proof 2

== Theorem ==
== Proof ==
Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$.
Let $r$ be the number of Sylow $p$-subgroups of $G$.
Let $H$ be a Sylow $p$-subgroup of $G$.
We have that:
:$\order H = p^n$
:$\index G H = m$
Let $S_1, S_2, \ldots, S_m$ denote the elements of the left coset space of $G / H$.
We have that $H$ acts on $G / H$ by the rule:
:$g * S_i = g S_i$
for $S_i \in G / H$.
Unless $H = G$ and $r = 1$, there is more than $1$ orbit.
We have that $H$ is the stabilizer of the coset $H$, which must be one of $S_1, S_2, \ldots, S_m$.
Let $S_1, S_2, \ldots, S_k$ be the elements of $G / H$ whose stabilizer is $H$.
From the Orbit-Stabilizer Theorem and from $\order H = p^n$ we see there are $2$ cases:
:$(1): \quad$ The orbit of $S_i$ contains $p^t$ elements where $0 < t < n$
:$(2): \quad$ The orbit of $S_i$ contains only the element $S_i$.
$(2)$ occurs $S_i$ is one of the cosets $S_1, S_2, \ldots, S_k$ whose stabilizer is $H$.
So counting the elements of $G / H$, we see that:
:$m = k + u p$
or:
:$m \equiv k \pmod p$
From the Fifth Sylow Theorem, we have:
:$m \equiv k r \pmod p$
and so:
:$k r \equiv k \pmod p$
from which it follows:
:$r \equiv 1 \pmod p$
because $k \not \equiv 0 \pmod p$.
Hence the result.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 58$: Third Sylow Theorem
Category:Fourth Sylow Theorem
Fractal Dimension of Cantor Set

== Theorem ==
The Cantor set is a fractal with similarity dimension of $\dfrac {\ln 2} {\ln 3}$.
== Proof ==
Let $C$ denote the Cantor set.
$C$ has the following self-similarities:
Thus we have:
Hence the result.
== Sources ==
* 
* 
* 
* 
Category:Cantor Set
Category:Similarity Dimensions
Fraction/Examples

== Examples of Fractions ==
:$(1): \quad \dfrac 1 2$ is a proper fraction.
:$(2): \quad \dfrac 5 2$ is an improper fraction.
It can be expressed as a mixed fraction as follows:
:$\dfrac 5 2 = \dfrac {4 + 1} 2 = \dfrac 4 2 + \dfrac 1 2 = 2 \frac 1 2$
:$(3): \quad \dfrac {24} {36}$ is a proper fraction, although not in canonical form.
It is found that when $\dfrac {24} {36}$ is expressed in canonical form:
:$\dfrac {24} {36} = \dfrac {12 \times 2} {12 \times 3} = \dfrac 2 3$
its denominator is not $1$.
Hence $\dfrac {24} {36}$ is indeed a vulgar fraction.
Category:Examples of Fractions
Fractional Sobolev Embedding Theorem

== Theorem ==
Let $S'$ denote the space of tempered distributions.
Let $\FF : S' \to S'$ denote the Fourier transform.
For each $s \in \R$ and $p \in \closedint 1 \infty$, let:
:$\map {W^{s, p} } {\R^n} = \set {u \in S': \sequence \xi^s \hat u \in \map {L^p} {\R^n} }$
where:
:$\sequence \xi = \paren {1 + \size \xi^2}^\frac 1 2$
Then:
:$(1): \quad$ If $s > t$ then $\map {W^{s, p} } {\R^n}$ embeds continuously into $\map {W^{t, q} } {\R^n}$ where $q$ is given by $\dfrac 1 q = \dfrac 1 p - \dfrac {s - t} n$.
Category:Fourier Analysis
Frame-Homogeneous Riemannian Manifold is Homogeneous and Isotropic

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Suppose $M$ is frame-homogeneous.
Then $M$ is homogeneous and isotropic.
== Proof ==
== Sources ==
* : $\S 3$: Model Riemannian Manifolds. Symmetries of Riemannian Manifolds
Category:Riemannian Geometry
Frame Homomorphism Induced by Completely Prime Filter Induced by Frame Homomorphism

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a frame.
Let $\struct{\mathbf 2, \vee, \wedge, \preceq}$ denote the (Boolean Lattice) $\mathbf 2$.
For every frame homomorphism $\phi : L \to \mathbf 2$, let:
:$p_\phi = \map {\phi^{-1}} \top$ 
where 
:$\map {\phi^{-1}} \top$ denotes the preimage of $\top \in \mathbf 2$ under the mapping $\phi$
which is the completely prime filter of $L$ induced by $\phi$ 
For every completely prime filter $p$ of $L$, let:
:$\phi_p : L \to \mathbf 2$ be the frame homomorphism defined by:
::$\forall a \in L : \map {\phi_p} a = \begin{cases}
\top & : a \in p\\
\bot & : a \notin p 
\end{cases}$
Let $\phi : L \to \mathbf 2$ be a frame homomorphism of $L$ into $\mathbf 2$.
Then:
:$\phi_{p_\phi} = \phi$
== Proof ==
From Completely Prime Filter Induced by Frame Homomorphism Onto Two:
:For every frame homomorphism $\phi : L \to \mathbf 2$, $p_\phi = \map {\phi^{-1}} \top$ is a completely prime filter of $L$
From Frame Homomorphism Onto Two Induced by Completely Prime Filter:
:For every completely prime filter $p$ of $L$, $\phi_p : L \to \mathbf 2$ is a frame homomorphism
We have:
From Equality of Mappings:
:$\phi_{p_\phi} = \phi$
Category:Completely Prime Filters
Category:Two (Boolean Lattice)
Category:Frame Homomorphisms
Frame Homomorphism Onto Two Induced by Completely Prime Filter

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a frame.
Let $\struct{\mathbf 2, \vee, \wedge, \preceq}$ denote the (Boolean Lattice) $\mathbf 2$.
Let $p$ be a completely prime filter of $L$.
Let $\phi_p : L \to \mathbf 2$ be the mapping defined by:
::$\forall a \in L : \map {\phi_p} a = \begin{cases}
\top & : a \in p\\
\bot & : a \notin p 
\end{cases}$
Then:
:$\phi_p$ is a frame homomorphism
== Proof ==
=== $\phi_p$ is Arbitrary Join Preserving ===
Let $\set{a_i : i \in I}$ be an indexed subset of elements of $L$.
We have:
It follows that $\phi_p$ is arbitrary join preserving.
=== $\phi_p$ is Finite Meet Preserving ===
Let $\set{a_i : i \in F}$ be a finite indexed subset of elements of $L$.
We have:
It follows that $\phi_p$ is finite meet preserving.
Hence $\phi_p$ is a frame homomorphim by definition.
Category:Completely Prime Filters
Category:Two (Boolean Lattice)
Category:Frame Homomorphisms
Frame Homomorphism Preserves Greatest Element

== Theorem ==
Let $L_1 = \struct{S_1, \preceq_1}$ and $L_2 = \struct{S_2, \preceq_2}$ be frames.
Let $\phi: L_1 \to L_2$ be a frame homomophism.
Let $\top_1$ and $\top_2$ denote the greatest elelment of $L_1$ and $L_2$ respectively.
Then:
:$\map \phi \top_1 = \top_2$
== Proof ==
we have:
== Also see==
* Frame Homomorphism Preserves Smallest Element
Category:Frame Homomorphisms
Category:Greatest Elements
Frame Homomorphism Preserves Smallest Element

== Theorem ==
Let $L_1 = \struct{S_1, \preceq_1}$ and $L_2 = \struct{S_2, \preceq_2}$ be frames.
Let $\phi: L_1 \to L_2$ be a frame homomophism.
Let $\bot_1$ and $\bot_2$ denote the smallest elelment of $L_1$ and $L_2$ respectively.
Then:
:$\map \phi \bot_1 = \bot_2$
== Proof ==
We have:
== Also see==
* Frame Homomorphism Preserves Greatest Element
Category:Frame Homomorphisms
Category:Smallest Elements
Frame Homomorphism is Lower Adjoint of Galois Connection

== Theorem ==
Let $L_1 = \struct{S_1, \preceq_1}, L_2 = \struct{S_2, \preceq_2}$ be frames.
Let $f : L_1 \to L_2$ be a frame homomorphism.
Then there exists a Galois connection $g = \tuple {\upperadjoint g, \loweradjoint g}$:
:the lower adjoint $\loweradjoint g = f$ 
== Proof ==
By definition of frame homomorphism:
:$f$ is arbitrary join preserving
From All Suprema Preserving Mapping is Lower Adjoint of Galois Connection:
:there exists a Galois connection $g = \tuple {\upperadjoint g, \loweradjoint g} : \loweradjoint g = f$
Category:Frame Homomorphisms
Category:Galois Connections
Frame Homomorphism of Continuous Mapping is Frame Homomorphism

== Theorem ==
Let $T_1 = \struct{S_1, \tau_1}, T_2 = \struct{S_2, \tau_2}$ be topological spaces.
Let $f : T_1 \to T_2$ be a continuous mapping.
Let $\map \Omega {T_1} = \struct{\tau_1, \subseteq}$ and $\map \Omega {T_2} = \struct{\tau_2, \subseteq}$ denote the frames of $T_1$ and $T_2$ respectively.
Let $\map \Omega f : \map \Omega {T_2} \to \map \Omega {T_1}$ be the frame homomorphism of $f$
Then:
:$\map \Omega f : \map \Omega {T_2} \to \map \Omega {T_1}$ is a frame homomorphism
== Proof ==
Recall the definition of continuous mapping:
:$f$ is continuous $U \in \tau_2 \implies f^{-1} \sqbrk U \in \tau_1$
By definition of frame homomorphism of $f$:
:$\map \Omega f$ is the inverse image mapping restricted to $\tau_2 \times \tau_1$
By definition of inverse image mapping:
:$\forall U \in \tau_2 : \map {\map \Omega f} U = f^{-1} \sqbrk U$
From Inverse Image Mapping is Mapping:
:$\map \Omega f : \tau_2 \to \tau_1$ is a well-defined mapping.
=== $\map \Omega f$ Preserves All Suprema ===
Let $\TT \subseteq \tau_2$.
By :
:$\bigcup \TT \in \tau_2$
:$\bigcup \set{\map {\map \Omega f} U : U \in \TT} \in \tau_1$
We have
Hence $\map \Omega f$ is arbitrary join preserving.
=== $\map \Omega f$ Preserves Finite Infima ===
Let $\FF \subseteq \tau_2$ be finite.
By :
:$\bigcap \FF \in \tau_2$
:$\bigcap \set{\map {\map \Omega f} U : U \in \FF} \in \tau_1$
We have
Hence $\map \Omega f$ is finite meet preserving.
It follows that $\map \Omega f$ is a frame homomorphism by definition.
== Sources ==
* : Chapter II: Introduction to Locales, $\S1.1$
* : Chapter II: Frames and Locales. Spectra, $\S 1.3$
Category:Continuous Mappings
Category:Frame Homomorphisms
Frame of Open Sets Functor is Contravariant

== Theorem ==
Let $\mathbf{Top}$ denote the category of topological spaces.
Let $\mathbf{Frm}$ denote the category of frames.
Then:
:the open sets functor $\mathbf \Omega : \mathbf{Top} \to \mathbf{Frm}$ is a contravariant functor
== Proof ==
Recall the open sets functor $\mathbf \Omega : \mathbf{Top} \to \mathbf{Frm}$ is defined by:
=== Object Functor is Well-Defined ===
From Frame of Topological Space is Frame:
:given a topological space $T$:
::$\map \Omega T$ is a frame.
Hence the object functor is well-defined.
=== Arrow Functor is Well-Defined ===
From Frame Homomorphism of Continuous Mapping is Frame Homomorphism:
:given a continuous mapping $f : T_1 \to T_2$ between topological spaces $T_1$ and $T_2$:
::$\map \Omega f : \map \Omega {T_2} \to \map \Omega {T_1}$ is a frame homomorphism.
Hence the arrow functor is well-defined.
=== Arrow Functor Preserves Composition ===
Let $T_1 = \struct{S_1, \tau_1}, T_2 = \struct{S_2, \tau_2}, T_3 = \struct{S_3, \tau_3}$ be topological spaces.
Let $f : T_1 \to T_2$ and $g : T_2 \to T_3$ be continuous mappings.
We have
It follows that:
:$\forall f, g : \map \Omega {g \circ f} = {\map \Omega f} \circ {\map \Omega g}$
=== Arrow Functor Preserves Identities ===
Let $T = \struct{S, \tau}$ be a topological space.
By definition of frame of topological space:
:$\map \Omega T = \struct{\tau, \subseteq}$
We have
It follows that:
:$\forall T : \map \Omega {\operatorname{id}_T } = \operatorname{id}_{\map \Omega T}$
It follows that $\Omega$ is a contravariant functor by definition.
== Sources ==
* : Chapter II: Frames and Locales. Spectra, $\S 1.3$
Category:Functors
Frame of Reference moving at Constant Velocity relative to Inertial Frame is Inertial Frame

== Theorem ==
Let $F_1$ be a frame of reference.
Let $F_2$ be an inertial frame of reference.
Let $F_1$ be moving at constant velocity relative to $F_2$.
Then $F_1$ is an inertial frame of reference.
== Proof ==
== Sources ==
* 
* 
Category:Inertial Frames of Reference
Frame of Topological Space is Frame

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $\map \Omega T$ denote the frame of topological space $T$.
Then:
:$\map \Omega T$ is a frame
== Proof ==
Recall that the frame of topological space $T$ is defined to be:
:$\map \Omega T = \struct{\tau, \subseteq}$
where $\subseteq$ denotes the subset relation.
From Topology forms Complete Lattice:
:$\map \Omega T$ is a complete lattice
Furthermore:
:$\forall \TT \subseteq \tau : \sup \TT = \bigcup \TT, \inf \TT = \paren{\bigcap \TT}^\circ$
where $\paren{\bigcap \TT}^\circ$ denotes the interior of $\bigcap \TT$
From Interior of Open Set and :
:$\forall U, V \in \tau : \paren{U \cap V}^\circ = U \cap V$
Hence:
:$\forall U, V \in \tau : U \wedge V = U \cap V$
From Intersection Distributes over Union (General Result):
:$\map \Omega T$ satisfies the infinite join distributive law
Hence $\map \Omega T$ is a frame by definition.
== Also see ==
* Frame Homomorphism of Continuous Mapping is Frame Homomorphism
== Sources ==
* : Chapter II: Introduction to Locales, $\S1.1$
* : Chapter II: Frames and Locales. Spectra, $\S 1.3$
Category:Topological Spaces
Category:Frames
Frattini's Argument

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $K$ be a finite normal subgroup of $G$, and $p$ a prime which divides the order of $K$.
Let $P$ be a Sylow $p$-subgroup of $K$, and $\map {N_G} P$ the normalizer of $P$ in $G$.
Then:
:$G = \map {N_G} P \circ K = K \circ \map {N_G} P$
== Proof ==
Let $g \in G$.
Since $K$ is normal in $G$, and $P \subset K$, the conjugate $g \circ P \circ g^{-1}$ of $P$ is also a subset of $K$.
From Inner Automorphism is Automorphism, $g \circ P \circ g^{-1}$ is a subgroup of $K$ of the same order as $P$.
Thus $g \circ P \circ g^{-1}$ is also a Sylow $p$-subgroup of $K$.
By Second Sylow Theorem:
:$\exists k \in K: g \circ P \circ g^{-1} = k \circ P \circ k^{-1}$
Then:
:$g^{-1} \circ k \circ P \circ k^{-1} \circ g = P$
and so by definition of normalizer:
:$k^{-1} \circ g \in \map {N_G} P$
We can then write $g$ as
:$g = k \circ \paren {k^{-1} \circ g} \in K \circ \map {N_G} P$
Since $K$ is normal:
:$\map {N_G} P \circ K = K \circ \map {N_G} P$
by Subset Product with Normal Subgroup as Generator.
== Sources ==
* : Chapter $5.2$ Theorem $5.2.14$
Category:Group Theory
Category:Normalizers