nyuuzyou/wikis huggingface dataset:
Sigma-Algebras with Independent Generators are Independent

== Theorem ==
Let $\struct {\Omega, \EE, \Pr}$ be a probability space.
Let $\Sigma, \Sigma'$ be sub-$\sigma$-algebras of $\EE$.
Suppose that $\GG, \HH$ are $\cap$-stable generators for $\Sigma, \Sigma'$, respectively.
Suppose that, for all $G \in \GG, H \in \HH$:
:$(1): \quad \map \Pr {G \cap H} = \map \Pr G \map \Pr H$
Then $\Sigma$ and $\Sigma'$ are $\Pr$-independent.
== Proof ==
Fix $H \in \HH$.
Define, for $E \in \Sigma$:
:$\map \mu E := \map \Pr {E \cap H}$
:$\map \nu E := \map \Pr E \map \Pr H$
Then by Intersection Measure is Measure and Restricted Measure is Measure, $\mu$ is a measure on $\Sigma$.
Namely, it is the intersection measure $\Pr_H$ restricted to $\Sigma$, that is $\Pr_H \restriction_\Sigma$.
Next, by Linear Combination of Measures and Restricted Measure is Measure, $\nu$ is also a measure on $\Sigma$.
Namely, it is the restricted measure $\map \Pr H \Pr \restriction_\Sigma$.
Let $\GG' := \GG \cup \set X$.
It is immediate that $\GG'$ is also a $\cap$-stable generator for $\Sigma$.
By assumption $(1)$, $\mu$ and $\nu$ coincide on $\GG$ (since $\map \Pr X = 1$).
From Restricting Measure Preserves Finiteness, $\mu$ and $\nu$ are also finite measures.
Hence, $\GG$ contains the exhausting sequence of which every term equals $X$.
Having verified all conditions, Uniqueness of Measures applies to yield $\mu = \nu$.
Now fix $E \in \Sigma$ and define, for $E' \in \Sigma'$:
:$\map {\mu'_E} {E'} := \map \Pr {E \cap E'}$
:$\map {\nu'_E} {E'} := \map \Pr E \map \Pr {E'}$
Mutatis mutandis, above consideration applies again, and we conclude by Uniqueness of Measures:
:$\mu'_E = \nu'_E$
for all $E \in \Sigma$. 
That is, expanding the definition of the measures $\mu'_E$ and $\nu'_E$:
:$\forall E \in \Sigma: \forall E' \in \Sigma': \map \Pr {E \cap E'} = \map \Pr E \map \Pr {E'}$
This is precisely the statement that $\Sigma$ and $\Sigma'$ are $\Pr$-independent $\sigma$-algebras.
== Sources ==
* : $\S 5$: Problem $10$
Category:Measure Theory
Sigma-Compact Space is Lindelöf

== Theorem ==
Every $\sigma$-compact space is a Lindelöf space.
== Proof ==
Let $T = \struct {S, \tau}$ be a $\sigma$-compact space.
By definition:
:$T$ is a Lindelöf space every open cover of $X$ has a countable subcover.
By definition of $\sigma$-compact space, $T = \bigcap \TT$ where $\TT$ is the union of countably many compact subspaces.
Let $\CC$ be an open cover of $T$.
Each element of $\TT$ is covered by a finite number of elements of $\CC$.
Hence $T$ is covered by a countable union of a finite number of elements of $\CC$.
Hence $\CC$ has a countable subcover.
Hence the result.
== Also see ==
* Weakly Sigma-Locally Compact iff Weakly Locally Compact and Lindelöf
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Sigma-Compact Spaces
Category:Lindelöf Spaces
Sigma-Compactness is Preserved under Continuous Surjection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is $\sigma$-compact, then $T_B$ is also $\sigma$-compact.
== Proof ==
Let $T_A$ be $\sigma$-compact.
Then:
:$\ds S_A = \bigcup_{i \mathop = 1}^\infty S_i$
where $S_i \subseteq S_A$ are compact.
Since $\phi$ is surjective, we have from Image of Union under Relation:
:$\ds \phi \sqbrk {S_A} = S_B = \phi \sqbrk {\bigcup_{i \mathop = 1}^\infty S_i} = \bigcup_{i \mathop = 1}^\infty \phi \sqbrk {S_i}$
From Compactness is Preserved under Continuous Surjection, we have that $\phi \sqbrk {S_i}$ is compact for all $i \in \N$.
So $S_B$ is the union of a countable number of compact subsets.
Thus, by definition, $T_B$ is also $\sigma$-compact.
== Also see ==
* Compactness Properties Preserved under Continuous Surjection
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Sigma-Compact Spaces
Category:Continuous Mappings
Category:Surjections
Sigma-Discrete Set of Subsets is Sigma-Locally Finite

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\FF$ be a $\sigma$-discrete set of subsets of $S$.
Then $\FF$ is a $\sigma$-locally finite set of subsets of $S$.
== Proof ==
This follows immediately from:
:Definition:Discrete Set of Subsets
:Definition:Locally Finite Set of Subsets
:Discrete Set of Subsets is Locally Finite
Category:Sigma-Discrete Sets of Subsets
Category:Sigma-Locally Finite Sets of Subsets
Sigma-Locally Finite Cover and Countable Locally Finite Cover have Common Locally Finite Refinement

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological space.
Let:
:$\SS = \ds \bigcup_{n \mathop = 0}^\infty \SS_n$ be a $\sigma$-locally finite cover of $X$
where each $\SS_n$ is locally finite for all $n \in \N$.
Let $\CC = \set {C_n : n \in \N}$ be a countable locally finite cover of $X$.
Then:
:there exists a common locally finite refinement $\AA$ of both $\SS$ and $\CC$
== Proof ==
Let:
:$\AA = \set {C_n \cap S : n \in \N, S \in \SS_n}$
=== $\AA$ is a Cover of $X$ ===
Let $x \in X$.
By definition of a cover:
:$\exists n \in \N : x \in C_n$
and
:$\exists S \in \SS : x \in S$
By definition of set union:
:$\exists n \in \N : S \in \SS_n$
By definition of set intersection:
:$x \in C_n \cap S$
By definition of $\AA$:
:$C_n \cap S \in \AA$
It follows that $\AA$ is a cover by definition.
=== $\AA$ is a Refinement of $\SS$ and $\CC$ ===
Let $A \in \AA$.
By definition of $\AA$:
:$\exists n \in \N, S \in SS_n : A = C_n \cap S$
From Intersection is Subset:
:$A \subseteq C_n$ and $A \subseteq S$
Hence $\AA$ is a refinement of both $\SS$ and $\CC$ by definition.
=== $\AA$ is Locally Finite ===
Let $x \in X$.
By definition of locally finite:
:$\exists U \in \tau : x \in U : \set {C_n \in \CC : C_n \cap U \ne \O}$ is finite.
From Subset of Naturals is Finite iff Bounded:
:$N = \set {n \in \N : C_n \cap U \ne \O}$ is bounded.
Hence $N$ has a greatest element $m$.
By definition of locally finite:
:$\forall n \le m : \exists V_n \in \tau : x \in V_n : \set {S \in \SS_n : S \cap V_n \ne \O}$ is finite.
Let:
:$W = U \cap \ds \bigcap_{n \mathop \le m} V_n$
By :
:$W \in \tau$
Let $A \in \AA : A \cap W \ne \O$.
Hence there exists $n \in \N, S \in \SS_n:$
:$A = C_n \cap S$
From Subsets of Disjoint Sets are Disjoint:
:$C_n \cap U \ne \O$
Hence $n \le m$.
From Subsets of Disjoint Sets are Disjoint:
:$S \cap V_n \ne \O$
We have:
From Finite Union of Finite Sets is Finite:
:$\ds \bigcup_{n \mathop \le m} \set {C_n \cap S : S \in \SS_n, S \cap V_n \ne \O}$ is finite.
From Subset of Finite Set is Finite:
:$\set {A \in \AA : A \cap W \ne \O}$ is finite.
Hence $\AA$ is locally finite by definition.
Category:Covers
Sigma-Locally Finite Cover has Locally Finite Refinement

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological space.
Let $\SS = \ds \bigcup_{n \mathop = 0}^\infty \SS_n$ be a $\sigma$-locally finite cover of $X$, where each $\SS_n$ is locally finite for all $n \in \N$.
Then:
:there exists a locally finite refinement $\AA$ of $\SS$.
== Proof ==
For each $n \in \N$, let:
:$B_n = \bigcup \SS_n$
Let:
:$\BB = \set {B_n : n \in \N}$
=== Lemma 1 ===
For each $n \in \N$, let:
:$C_n = B_n \setminus \ds \bigcup_{i \mathop < n} B_i$
Let:
:$\CC = \set {C_n : n \in \N}$
=== Lemma 2 ===
From Sigma-Locally Finite Cover and Countable Locally Finite Cover have Common Locally Finite Refinement:
:there exists a common locally finite refinement $\AA$ of both $\CC$ and $\SS$.
Category:Covers
Category:Sigma-Locally Finite Cover has Locally Finite Refinement
Sigma-Locally Finite Cover has Locally Finite Refinement/Lemma 1

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological Space.
Let $\SS = \ds \bigcup_{n = 0}^\infty \SS_n$ be a cover of $X$, where each $\SS_n$ is a set of subsets of $X$ for each $n \in \N$.
For each $n \in \N$, let:
:$B_n = \ds \bigcup \SS_n$
Let:
:$\BB = \set{B_n : n \in \N}$
Then:
:$\BB$ is a cover of $X$
== Proof ==
Let $x \in X$.
By definition of cover of set:
:$\exists S \in \SS : x \in S$
By definition of union:
:$\exists n \in \N : S \in \SS_n$
From Set is Subset of Union:
:$S \subseteq B_n$
By definition of subset:
:$x \in B_n$
It follows by definition, $\BB$ is a cover of $X$.
Category:Sigma-Locally Finite Cover has Locally Finite Refinement
Sigma-Locally Finite Cover has Locally Finite Refinement/Lemma 2

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological Space.
Let $\SS = \ds \bigcup_{n = 0}^\infty \SS_n$ be a cover of $X$, where each $\SS_n$ is a set of subsets of $X$ for each $n \in \N$.
For each $n \in \N$, let:
:$B_n = \ds \bigcup \SS_n$
Let:
:$\BB = \set{B_n : n \in \N}$
For each $n \in \N$, let:
:$C_n = B_n \setminus \ds \bigcup_{i < n} B_i$
Let:
:$\CC = \set{C_n : n \in \N}$
Then:
:$\CC$ is a locally finite refinement of $\BB$
== Proof ==
==== Lemma 1 ====
==== $\CC$ is a Cover of $X$ ====
Let $x \in X$.
By definition of cover of set:
:$\exists n \in \N : x \in B_n$
From Well-Ordering Principle:
:$\exists m \in \N : x \in B_m : \forall n < m : x \notin B_n$
By definition of set difference:
:$\forall n < m : x \in B_m \setminus B_n$
By definition of set intersection:
:$x \in \ds \bigcap_{i < m} \paren{B_m \setminus B_i}$
From De Morgan's Laws for Set Difference:
:$\ds \bigcap_{i < m} \paren{B_m \setminus B_i} = B_m \setminus \paren{\bigcup_{i < m} B_i}$
Hence:
:$x \in C_m$
It follows that $\CC$ is a cover of $X$.
==== $\CC$ is a Refinement of $\WW$ ====
From Set Difference is Subset:
:$\forall n \in \N : C_n \subseteq B_n$
It follows that $\CC$ is a refinement of $\BB$ by definition.
==== $\CC$ is Locally Finite ====
Let $x \in X$.
By definition of cover of set:
:$\exists S \in \SS : x \in S$
By definition of union:
:$\exists m \in \N : S \in \SS_m$
From Set is Subset of Union:
:$V \subseteq B_m$
From Set is Subset of Union:
:$\forall n > m : B_m \subseteq \ds \bigcup_{i < n} B_i$
From Set Difference with Subset is Superset of Set Difference:
:$\forall n > m : C_n = B_n \setminus \ds \bigcup_{i < n} B_i \subseteq B_n \setminus B_m$
From Subset of Set Difference iff Disjoint Set:
:$\forall n > m : C_n \cap B_m = \O$
From Subsets of Disjoint Sets are Disjoint:
:$\forall n > m : C_n \cap V = \O$
Hence:
:$\set{C_n \in \CC : C_n \cap V \ne \O} \subseteq \set{C_n \in \CC : n \le m}$
The set $\set{C_n \in \CC : n \le m}$ is finite.
From Subset of Finite Set is Finite:
:$\set{C_n \in \CC : C_n \cap V \ne \O}$ is finite
It follows that $\CC$ is a locally finite refinement of $\BB$ by definition.
Category:Sigma-Locally Finite Cover has Locally Finite Refinement
Sigma-Ring contains Limit Superior of Sequence of Sets

== Theorem ==
Let $\RR$ be a $\sigma$-ring.
Let $\sequence {A_n}_{n \mathop \in \N} \in \RR$ be a sequence of sets in $\RR$.
Then:
:$\ds \limsup_{n \mathop \to \infty} A_n \in \RR$
== Proof ==
Define the sequence of sets:
:$B_i := \ds \bigcup_{n \mathop = i}^\infty A_n$
From $\sigma$-ring axiom $(\text {SR} 3)$: Closure under Countable Unions:
:$\forall i: B_i \in \RR$
From Sigma-Ring is Closed under Countable Intersections:
:$\forall n: \ds \bigcap_{i \mathop = 0}^\infty B_i \in \RR$
Thus:
== Sources ==
* : $\S 1.1$: Rings and Algebras: Problem $1.1.3$
Category:Sigma-Rings
Sigma-Ring is Closed under Countable Intersections

== Theorem ==
Let $\RR$ be a $\sigma$-ring.
Let $\sequence {A_n}_{n \mathop \in \N} \in \RR$ be a sequence of sets in $\RR$.
Then:
:$\ds \bigcap_{n \mathop = 1}^\infty A_n \in \RR$
== Proof ==
From De Morgan's laws: Difference with Intersection:
:$\ds \bigcup_{n \mathop = 2}^\infty \paren {A_1 \setminus A_n} = A_1 \setminus \paren {\bigcap_{n \mathop = 2}^\infty A_n}$
From Set Difference with Set Difference:
Combining the previous equalities, it follows that:
:$\ds \bigcap_{n \mathop = 1}^\infty A_n \in \RR$
== Sources ==
* : $\S 1.1$: Rings and Algebras: Problem $1.1.3$
Category:Sigma-Rings
Sigmoid Function is Strictly Increasing

== Theorem ==
The real sigmoid function $\map S x$ is strictly increasing.
== Proof ==
By Cumulative Distribution Function of Logistic Distribution, $S$ is the cumulative distribution function of a logistic distribution, with $\mu = 0$ and $s = 1$.
By Cumulative Distribution Function is Increasing, $S$ is an increasing real function.
, suppose that $S$ is not strictly increasing.
Then there are $a < b$ such that $\map S a = \map S b$.
Thus:
which contradicts $a < b$.
Therefore, by Proof by Contradiction, the real sigmoid function is strictly increasing.
Category:Sigmoid Function
Sign of Composition of Permutations

== Theorem ==
Let $n \in \N$ be a natural number.
Let $N_n$ denote the set of natural numbers $\set {1, 2, \ldots, n}$.
Let $S_n$ denote the set of permutations on $N_n$.
Let $\map \sgn \pi$ denote the sign of $\pi$ of a permutation $\pi$ of $N_n$.
Let $\pi_1, \pi_2 \in S_n$.
Then:
:$\map \sgn {\pi_1} \map \sgn {\pi_2} = \map \sgn {\pi_1 \circ \pi_2}$
where $\pi_1 \circ \pi_2$ denotes the composite of $\pi_1$ and $\pi_2$.
== Proof ==
From Sign of Permutation on n Letters is Well-Defined, it is established that the sign each of $\pi_1$, $\pi_2$ and $\pi_1 \circ \pi_2$ is either $+1$ and $-1$.
By Existence and Uniqueness of Cycle Decomposition, each of $\pi_1$ and $\pi_2$ has a unique cycle decomposition.
Thus each of $\pi_1$ and $\pi_2$ can be expressed as the composite of $p_1$ and $p_2$ transpositions respectively.
Thus $\pi_1 \circ \pi_2$ can be expressed as the composite of $p_1 + p_2$ transpositions.
From Sum of Even Integers is Even, if $p_1$ and $p_2$ are both even then $p_1 + p_2$ is even.
In this case:
:$\map \sgn {\pi_1} = 1$
:$\map \sgn {\pi_2} = 1$
:$\map \sgn {\pi_1} \map \sgn {\pi_2} = 1 = 1 \times 1$
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 81$
* : Appendix: Elementary set and number theory
Category:Sign of Permutation
Sign of Cosecant

== Theorem ==
Let $x$ be a real number.
Then:
where $\csc$ is the real cosecant function.
== Proof ==
For the first part:
For the second part:
== Also see ==
* Sign of Sine
* Sign of Cosine
* Sign of Tangent
* Sign of Cotangent
* Sign of Secant
Category:Cosecant Function
Sign of Cosine

== Theorem ==
Let $x$ be a real number.
Then:
where $\cos$ is the real cosine function.
== Proof ==
Proof by induction:
=== Base case ===
For $n = 0$, it follows from Sine and Cosine are Periodic on Reals/Corollary.
=== Induction Hypothesis ===
This is our induction hypothesis:
Now we need to show true for $n=k+1$:
=== Induction Step ===
This is our induction step:
Also:
The result follows by induction.
For negative $n$:
=== Induction Hypothesis ===
This is our induction hypothesis:
Now we need to show true for $n=k-1$:
=== Induction Step ===
This is our induction step:
Also:
The result follows by induction.
== Also see ==
* Sign of Sine
* Sign of Tangent
* Sign of Cotangent
* Sign of Secant
* Sign of Cosecant
Category:Cosine Function
Sign of Cotangent

== Theorem ==
Let $x$ be a real number.
Then:
where $\cot$ is the real cotangent function.
== Proof ==
For the first part:
For the second part:
== Also see ==
* Sign of Sine
* Sign of Cosine
* Sign of Tangent
* Sign of Secant
* Sign of Cosecant
Category:Cotangent Function
Sign of Difference between Prime-Counting Function and Eulerian Logarithmic Integral Changes Infinitely Often

== Theorem ==
The sign of $\map \pi n - \ds \int_2^n \frac {\d x} {\ln x}$ changes infinitely often as $n$ increases indefinitely.
== Proof ==
== Historical Note ==
== Sources ==
* 
* : $10^{10^{10^{34}}}$
* : $10^{10^{10^{34}}}$
Category:Prime Number Theorem
Sign of Difference between Prime-Counting Function and Eulerian Logarithmic Integral Changes Infinitely Often/Historical Note

== Historical Note on Sign of Difference between Prime-Counting Function and Eulerian Logarithmic Integral Changes Infinitely Often ==
For values of $n$ lower than some large number, $\map \pi n - \ds \int_2^n \frac {\d x} {\ln x}$ is greater than $0$.
However, in $1914$, proved that at some point $\map \pi n - \ds \int_2^n \frac {\d x} {\ln x}$ switches to being less than $0$.
Furthermore, he proved that the lead changes an infinite number of times, if $n$ becomes large enough.
== Sources ==
* : $10^{10^{10^{34}}}$
* : $10^{10^{10^{34}}}$
Category:Historical Notes
Sign of Function Matches Sign of Definite Integral

== Theorem ==
Let $f$ be a real function continuous on some closed interval $\closedint a b$, where $a < b$.
Then:
:If $\forall x \in \closedint a b: \map f x \ge 0$ then $\ds \int_a^b \map f x \rd x \ge 0$
:If $\forall x \in \closedint a b: \map f x > 0$ then $\ds \int_a^b \map f x \rd x > 0$
:If $\forall x \in \closedint a b: \map f x \le 0$ then $\ds \int_a^b \map f x \rd x \le 0$
:If $\forall x \in \closedint a b: \map f x < 0$ then $\ds \int_a^b \map f x \rd x < 0$
== Proof ==
From Continuous Real Function is Darboux Integrable, the definite integrals under discussion are guaranteed to exist.
Consider the case where $\forall x \in \closedint a b: \map f x \ge 0$.
Define a constant mapping:
:$f_0: \closedint a b \to \R$:
:$\map {f_0} x = 0$
Then:
The proofs of the other cases are similar.
Category:Integral Calculus
Sign of Half-Plane is Well-Defined

== Theorem ==
Let $\LL$ be a straight line embedded in a cartesian plane $\CC$, given by the equation:
:$l x + m y + n = 0$
Let $\HH_1$ and $\HH_2$ be the half-planes into which $\LL$ divides $\CC$.
Let the sign of a point $P = \tuple {x_1, y_1}$ in $\CC$ be defined as the sign of the expression $l x_1 + m y_1 + n$.
Then the sign of $\HH_1$ and $\HH_2$ is well-defined in the sense that:
:all points in one half-plane $\HH \in \set {\HH_1, \HH_2}$ have the same sign
:all points in $\HH_1$ are of the opposite sign from the points in $\HH_2$
:all points on $\LL$ itself have sign $0$.
== Proof ==
By definition of $\LL$, if $P$ is on $\LL$ then $l x_1 + m y_1 + n = 0$.
Similarly, if $P$ is not on $\LL$ then $l x_1 + m y_1 + n \ne 0$.
Let $P = \tuple {x_1, y_1}$ and $Q = \tuple {x_2, y_2}$ be two points not on $\LL$ such that the line $PQ$ intersects $\LL$ at $R = \tuple {x, y}$.
Let $PR : RQ = k$.
Then from Joachimsthal's Section-Formulae:
If $u_1$ and $u_2$ have the same sign, then $k$ is negative.
By definition of the position-ratio of $R$, it then follows that $R$ is not on the ine segment $PQ$.
Hence $P$ and $Q$ are in the same one of the half-planes defined by $\LL$.
Similarly, if $u_1$ and $u_2$ have the opposite signs, then $k$ is positive.
Again by definition of the position-ratio of $R$, it then follows that $R$ is on the ine segment $PQ$.
That is, $\LL$ intersects the ine segment $PQ$.
That is, $P$ and $Q$ are on opposite sides of $\LL$.
Hence $P$ and $Q$ are in opposite half-planes.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $7$.
Category:Half-Planes
Sign of Haversine

== Theorem ==
The haversine is non-negative for all $\theta \in \R$.
== Proof ==
The haversine is conventionally defined on the real numbers only.
We have that:
:$\forall \theta \in \R: -1 < \cos \theta < 1$
and so:
:$\forall \theta \in \R: 0 < 1 - \cos \theta < 2$
from which the result follows by definition of haversine.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $13$. The haversine formula.
Category:Haversines
Sign of Odd Power

== Theorem ==
Let $x \in \R$ be a real number.
Let $n \in \Z$ be an odd integer.
Then:
:$x^n = 0 \iff x = 0$
:$x^n > 0 \iff x > 0$
:$x^n < 0 \iff x < 0$
That is, the sign of an odd power matches the number it is a power of.
=== Corollary ===
== Proof ==
If $n$ is an odd integer, then $n = 2 k + 1$ for some $k \in \N$.
Thus $x^n = x \cdot x^{2 k}$.
But $x^{2 k} \ge 0$ from Even Power is Non-Negative.
The result follows.
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.12 \ (1)$
Category:Real Analysis
Sign of Odd Power/Corollary

== Corollary to Sign of Odd Power ==
Let $x \in \R$ be a real number.
Let $n \in \Z$ be an odd integer.
Then:
:$\paren {-x}^n = -\paren {x^n}$
== Proof ==
We have:
:$\size {-x}^n = \size {-\paren {x^n} }$
and the result follows from Sign of Odd Power.
Category:Real Analysis
Sign of Permutation is Plus or Minus Unity

== Theorem ==
Let $n \in \N$ be a natural number.
Let $\N_n$ denote the set of natural numbers $\set {1, 2, \ldots, n}$.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\sequence {x_k}_{k \mathop \in \N_n}$ be a finite sequence in $\R$.
Let $\pi \in S_n$.
Let $\map {\Delta_n} {x_1, x_2, \ldots, x_n}$ be the product of differences of $\tuple {x_1, x_2, \ldots, x_n}$.
Let $\map \sgn \pi$ be the sign of $\pi$.
Let $\pi \cdot \map {\Delta_n} {x_1, x_2, \ldots, x_n}$ be defined as:
:$\pi \cdot \map {\Delta_n} {x_1, x_2, \ldots, x_n} := \map {\Delta_n} {x_{\map \pi 1}, x_{\map \pi 2}, \ldots, x_{\map \pi n} }$
Then either:
:$\pi \cdot \Delta_n = \Delta_n$
or:
:$\pi \cdot \Delta_n = -\Delta_n$
That is:
:$\map \sgn \pi = \begin {cases}
1 & :\pi \cdot \Delta_n = \Delta_n \\
-1 & : \pi \cdot \Delta_n = -\Delta_n
\end {cases}$
Thus:
:$\pi \cdot \Delta_n = \map \sgn \pi \Delta_n$
== Proof ==
If $\exists i, j \in \N_n$ such that $x_i = x_j$, then $\map {\Delta_n} {x_1, x_2, \ldots, x_n} = 0$ and the result follows trivially.
So, suppose all the elements $x_k$ are distinct.
Let us use $\Delta_n$ to denote $\map {\Delta_n} {x_1, x_2, \ldots, x_n}$.
Let $1 \le a < b \le n$.
Then $x_a - x_b$ is a divisor of $\Delta_n$.
Then $x_{\map \pi a} - x_{\map \pi b}$ is a factor of $\pi \cdot \Delta_n$.
There are two possibilities for the ordering of $\map \pi a$ and $\map \pi b$:
Either $\map \pi a < \map \pi b$ or $\map \pi a > \map \pi b$.
If the former, then $x_{\map \pi a} - x_{\map \pi b}$ is a factor of $\Delta_n$.
If the latter, then $-\paren {x_{\map \pi a} - x_{\map \pi b} }$ is a factor of $\Delta_n$.
The same applies to all factors of $\Delta_n$.
Thus:
== Sources ==
* : $\S 7.4$. Kernel and image: Example $142$
* : Chapter $2$: The Symmetric Groups: $\S 81$
* : Chapter $9$: Permutations: Definition $9.15$
Category:Symmetric Groups
Category:Sign of Permutation