nyuuzyou/wikis huggingface dataset:
Sides of Orthic Triangle of Obtuse Triangle

== Theorem ==
Let $\triangle ABC$ be an obtuse triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $\angle A$ be the obtuse angle of $\triangle ABC$.
Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$.
Then the sides of $\triangle DEF$ are $-a \cos A$, $b \cos B$ and $c \cos C$.
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(51)$
Category:Sides of Orthic Triangle of Obtuse Triangle
Category:Orthic Triangles
Category:Obtuse Triangles
Sides of Orthic Triangle of Obtuse Triangle/Proof

== Theorem ==
== Proof ==
:
Let $H$ be the orthocenter of $\triangle ABC$.
On the other hand, 
Similar arguments can be applied to $DF$ by swapping $E$ with $F$ and $B$ with $C$.
Hence the result.
== Sources ==
* : $\text V$. Trigonometry: The pedal triangle
Category:Sides of Orthic Triangle of Obtuse Triangle
Sierpiński's Theorem

== Theorem ==
Let $\left({S, \tau}\right)$ be a compact connected Hausdorff space.
Let $\left\{{F_n: n \in \N}\right\}$ be a pairwise disjoint closed cover of $S$.
Then $F_n = S$ for some $n \in \N$.
== Proof ==
Category:Compact Spaces
Category:Connected Spaces
Category:Hausdorff Spaces
Category:Sierpiński's Theorem
Sierpiński's Theorem/Lemma 1

== Theorem ==
Let $\struct {S, \tau}$ be a compact connected Hausdorff space.
Let $A$ be a closed, non-empty proper subset of $S$.
Let $C$ be a component of $A$.
Then:
:$C \cap \partial A \ne \O$
where $\partial A$ denotes the boundary of $A$.
== Proof ==
Let $p \in C$.
Let $\VV$ be the set of all subsets of $A$ containing $p$ that are clopen relative to $A$..
By Quasicomponents and Components are Equal in Compact Hausdorff Space and Quasicomponent is Intersection of Clopen Sets:
:$C$ is the intersection of $\VV$.
:
:$C \cap \partial A = \O$
By Boundary of Set is Closed, $K \cap \partial A$ is closed for each $K \in \VV$.
Thus by Compact Space satisfies Finite Intersection Axiom, there exists a finite set $\VV' \subseteq \VV$ such that $\partial A \cap \bigcap \VV' = \O$.
But then:
:$\ds K = \bigcap \VV' \in \VV$
Therefore there exists a $K \in \VV$ such that $K \cap \partial A = \O$.
Since $A$ is closed in $S$, and $K$ is clopen in $A$, $K$ is closed in $S$.
We have that:
:$\partial A = A^- \setminus \map {\operatorname {Int} } A$
where $A^-$ is the closure of $A$ and $\map {\operatorname {Int} } A$ is the interior of $A$
Hence as $K \subseteq A$, it follows that $K \subseteq \map {\operatorname {Int} } A$.
Since $K$ is open relative to $A$, it is open relative to $\map {\operatorname {Int} } A$.
We have that $\map {\operatorname {Int} } A$ is open in $S$.
Therefore $K$ is open in $S$.
Thus $K$ is clopen in $S$.
We have that $p \in K \subseteq A \subsetneqq S$.
Therefore $S$ is not connected.
From this contradiction it follows that:
:$C \cap \partial A \ne \O$
Category:Sierpiński's Theorem
Sierpiński Problem/Historical Note

== Historical Note on Sierpiński Problem ==
It was proved by in $1962$ that $78\ 557$ is a Sierpiński number of the second kind.
In 1967, and conjectured that $78\ 557$ is in fact the smallest Sierpiński number of the second kind.
To prove that this is the case, all odd positive integers smaller than $78\ 557$ are not Sierpiński numbers of the second kind.
That is, it must be shown that, for all $1 \le k \le 78\ 557$, where $k$ is odd, there exists some $n \in \Z$ such that:
:$k 2^n + 1$
is prime.
By March 2002, there were seventeen such $k$ whose status was still unknown.
That was when the distributed computing project Seventeen or Bust was established.
Its aim was to test all these remaining seventeen numbers by exhaustively checking all values of $n$ until finding a value of $n$ for which $k 2^n + 1$ is prime.
In April 2016 the project terminated as a result of the server going down, with the resulting loss of both the server and the backups. Work continues on PrimeGrid.
As of 18 January 2017, twelve of those remaining seventeen numbers have been found to be non-Sierpiński, by establishing a value of $n$ for which $k 2^n + 1$ is prime, as follows:
$k$ 	$n$ 	Number of digits in $k 2^n + 1$	Date discovered 	Discovered by
$\ 4847\ $ 	$\ 3\ 321\ 063\ $ 	$\ 999\ 744\ $ 	15 October 2005 	Richard Hassler
$\ 5359\ $ 	$\ 5\ 054\ 502\ $ 	$\ 1\ 521\ 561\ $ 	6 December 2003 	Randy Sundquist
$\ 10\ 223\ $	$\ 31\ 172\ 165\ $ 	$\ 9\ 383\ 761\ $ 	31 October 2016 	Szabolcs Peter
$\ 19\ 249\ $	$\ 13\ 018\ 586\ $ 	$\ 3\ 918\ 990\ $ 	5 May 2007 	Konstantin Agafonov
$\ 21\ 181\ $	$\ > 17\ 000\ 011\ $	 	Search ongoing 	
$\ 22\ 699\ $	$\ > 17\ 008\ 677\ $	 	Search ongoing 	
$\ 24\ 737\ $	$\ > 17\ 001\ 366\ $	 	Search ongoing 	
$\ 27\ 653\ $	$\ 9\ 167\ 433\ $ 	$\ 2\ 759\ 677\ $ 	8 June 2005 	Derek Gordon
$\ 28\ 433\ $	$\ 7\ 830\ 457\ $ 	$\ 2\ 357\ 207\ $ 	30 December 2004	anonymous
$\ 33\ 661\ $	$\ 7\ 031\ 232\ $ 	$\ 2\ 116\ 617\ $ 	17 October 2007 	Sturle Sunde
$\ 44\ 131\ $	$\ 995\ 972\ $ 	$\ 299\ 823\ $ 	6 December 2002 	anonymous
$\ 46\ 157\ $	$\ 698\ 207\ $ 	$\ 210\ 186\ $ 	27 November 2002	Stephen Gibson
$\ 54\ 767\ $	$\ 1\ 337\ 287\ $ 	$\ 402\ 569\ $ 	22 December 2002	Peter Coels
$\ 55\ 459\ $	$\ > 17\ 000\ 193\ $	 	Search ongoing 	
$\ 65\ 567\ $	$\ 1\ 013\ 803\ $ 	$\ 305\ 190\ $ 	3 December 2002 	James Burt
$\ 67\ 607\ $	$\ > 17\ 000\ 090\ $	 	Search ongoing 	
$\ 69\ 109\ $	$\ 1\ 157\ 446\ $ 	$\ 348\ 431\ $ 	7 December 2002 	Sean DiMichele
Category:Historical Notes
Sierpiński Space is Irreducible

== Theorem ==
Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is irreducible.
== Proof ==
A Sierpiński space is a particular point space by definition.
A Particular Point Space is Irreducible.
== Sources ==
* : Part $\text {II}$: Counterexamples: $11$. Sierpinski Space: $18$
Category:Irreducible Spaces
Category:Sierpiński Space
Sierpiński Space is Path-Connected

== Theorem ==
Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is path-connected.
== Proof ==
A Sierpiński space is a particular point space by definition.
A Particular Point Space is Path-Connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $11$. Sierpinski Space: $18$
Category:Path-Connected Spaces
Category:Sierpiński Space
Sierpiński Space is T4

== Theorem ==
Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is a $T_4$ space.
== Proof ==
We have that the Sierpiński Space is $T_5$.
Then we have that a $T_5$ Space is $T_4$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $11$. Sierpinski Space: $18$
Category:T4 Spaces
Category:Sierpiński Space
Sierpiński Space is T5

== Theorem ==
Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is a $T_5$ space.
== Proof ==
The only closed sets in $T$ are $\O$, $\set 1$ and $\set {0, 1}$.
So there are no two separated sets $A, B \subseteq \set {0, 1}$.
So $T$ is a $T_5$ space vacuously.
== Sources ==
* : Part $\text {II}$: Counterexamples: $11$. Sierpinski Space: $18$
Category:T5 Spaces
Category:Sierpiński Space
Sierpiński Space is Ultraconnected

== Theorem ==
Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is ultraconnected.
== Proof ==
The only closed sets of $T$ are $\O, \set 1$ and $\set {0, 1}$.
$\set 1$ and $\set {0, 1}$ are not disjoint.
Hence the result by definition of ultraconnected.
== Also see ==
* Particular Point Space is not Ultraconnected
== Sources ==
* : Part $\text {II}$: Counterexamples: $11$. Sierpinski Space: $18$
Category:Ultraconnected Spaces
Category:Sierpiński Space
Sierpiński Space is not Arc-Connected

== Theorem ==
Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
Then $T$ is not arc-connected.
== Proof ==
A Sierpiński space is a particular point space by definition.
A Particular Point Space is not Arc-Connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $11$. Sierpinski Space: $18$
Category:Arc-Connected Spaces
Category:Sierpiński Space
Sieve of Eratosthenes

== Algorithm ==
Take a set of natural numbers $S = \set {2, 3, 4, \ldots, n}$.
:$(1): \quad$ Set $k = 2$.
:$(2): \quad$ Delete from $S$ every multiple of $k$ (but not $k$ itself).
:$(3): \quad$ Set $k$ equal to the smallest number of those remaining in $S$ greater than the current value of $k$.
:$(4): \quad$ If $k \le \sqrt n$, go to step $(2)$. Otherwise, STOP.
What remains in $S$ is the complete set of all prime numbers less than or equal to $n$.
== Proof ==
The above constitutes an algorithm, for the following reasons:
=== Finiteness ===
For each iteration through the algorithm, step $(3)$ is executed, which increases $k$ by at least $1$.
The algorithm will terminate after at most $\sqrt n - 2$ iterations (and probably considerably less).
=== Definiteness ===
:Step 1: Trivially definite.
:Step 2: Trivially definite.
:Step 3: Trivially definite.
:Step 4: Trivially definite.
=== Inputs ===
The input to this algorithm is the set of natural numbers $S = \set {2, 3, 4, \ldots, n}$.
=== Outputs ===
The output to this algorithm is the set $S$ without any multiples of any numbers no greater than $\sqrt n$.
From Composite Number has Prime Factor not Greater Than its Square Root, it follows that all the numbers left in $S$ must be prime.
By the method of construction, no prime can have been deleted from $S$.
Hence the output consists of all, and only, the primes less than or equal to $n$.
=== Effective ===
Each step of the algorithm is basic enough to be done exactly and in a finite length of time.
:Sift the Two's and Sift the Three's,
:The Sieve of Eratosthenes.
:When the multiples sublime,
:The numbers that remain are Prime.
::-- Traditional: cited in .
== Historical Note ==
== Sources ==
* : $\text I$: The Series of Primes: $1.4$ The sequence of primes
* 
* : $33$
* 
* : Chapter $\text {B}.16$: The Sequence of Primes
* : $33$
* 
* 
* 
Category:Number Sieves
Category:Prime Numbers
Sigma-Algebra/Examples/Trivial Sigma-Algebra

== Examples of $\sigma$-Algebra ==
Let $X$ be a set.
The trivial $\sigma$-algebra on $X$ is the $\sigma$-algebra defined as:
:$\set {\O, X}$
== Proof ==
Verify the axioms for a $\sigma$-algebra in turn:
=== Axiom $(\text {SA} 1)$ ===
From Intersection with Empty Set, there exists $\O \in \set {\O, X}$, such that:
:$\O \cap X = \O$
And from Intersection with Subset is Subset and Set is Subset of Itself, there exists $X \in \set {\O, X}$ such that:
:$ X \cap X = X$
So, we have:
:$\forall A \in \set {\O, X}: A \cap X = A$
Hence $X \in \Sigma$.
Hence $X$ is the unit of $\set {\O, X}$.
=== Axiom $(\text {SA} 2)$ ===
By Set Difference with Self is Empty Set again:
:$\relcomp X X = \O \in \Sigma$.
Where $\relcomp X X$, and is the relative complement of $X$ in $X$
Now suppose:
:$\O \in \Sigma$
Then by Set Difference with Empty Set is Self:
:$\relcomp X \O = X \in \Sigma$.
So $\forall A \in \Sigma$:
:$\relcomp X A \in \Sigma$
Hence $\Sigma$ is closed under complement.
=== Axiom $(\text {SA} 3)$ ===
Let the set of sets $\mathbb S$ be such that $\mathbb S$ is the empty set $\O$.
Then by Union is Empty iff Sets are Empty, the union of $\mathbb S$ is $\O$:
:$\mathbb S = \O \implies \bigcup \mathbb S = \O$
By Union with Empty Set we have:
:$X \cup \O = X$
By Union of Subsets is Subset and Set is Subset of Itself, we have:
:$\ds \bigcup \mathbb S \subseteq X$
So we have:
:$\forall A_n \in \Sigma: n = 1, 2, \ldots: \bigcup_{n \mathop = 1}^\infty A_n \in \Sigma$
Hence $\sigma$ is closed under countable unions.
Hence the result.
Category:Examples of Sigma-Algebras
Sigma-Algebra Closed under Countable Intersection

== Theorem ==
Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Suppose that $\sequence {E_n}_{n \mathop \in \N} \in \Sigma$ is a collection of measurable sets.
Then:
:$\ds \bigcap_{n \mathop \in \N} E_n \in \Sigma$,
where $\ds \bigcap$ denotes set intersection.
== Proof ==
From De Morgan's laws: Complement of Intersection:
:$\ds \bigcup_{n \mathop \in \N} \paren {X \setminus E_n} = X \setminus \paren {\bigcap_{n \mathop \in \N} E_n}$
Also, by Set Difference with Set Difference and Set Union Preserves Subsets:
:$\ds X \setminus \paren {X \setminus \paren {\bigcap_{n \mathop \in \N} E_n} } = \bigcap_{n \mathop \in \N} E_n$
Combining the previous equalities, it follows that:
:$\ds \bigcap_{n \mathop \in \N} E_n \in \Sigma$
== Sources ==
* : $3.2 \ \text{(iii)}$
Category:Sigma-Algebras
Sigma-Algebra Closed under Finite Intersection

== Theorem ==
Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $A_1, \ldots, A_n \in \Sigma$.
Then $\ds \bigcap_{k \mathop = 1}^n A_k \in \Sigma$.
== Proof ==
Define for $k \in \N, k > n: A_k = X$.
By axiom $(1)$ of a $\sigma$-algebra, it follows that $\forall k \in \N, k > n: A_k \in \Sigma$.
From Sigma-Algebra Closed under Countable Intersection, it follows that $\ds \bigcap_{k \mathop \in \N} A_k = \bigcap_{k \mathop = 1}^n A_k \in \Sigma$.
Category:Sigma-Algebras
Sigma-Algebra Closed under Set Difference

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $A, B \in \Sigma$.
Then the set difference $A \setminus B$ is contained in $\Sigma$. 
== Proof ==
Since $\sigma$-algebras are closed under relative complement, we have: 
:$\relcomp X B \in \Sigma$
By Sigma-Algebra Closed under Finite Intersection, we have: 
:$A \cap \relcomp X B \in \Sigma$
From Set Difference as Intersection with Relative Complement, we have: 
:$A \setminus B = A \cap \relcomp X B$
so:
:$A \setminus B \in \Sigma$
Category:Sigma-Algebras
Category:Set Difference
Sigma-Algebra Closed under Symmetric Difference

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $A, B \in \Sigma$.
Then the symmetric difference $A \Delta B$ is contained in $\Sigma$.
== Proof ==
From Sigma-Algebra Closed under Set Difference, we have: 
:$A \setminus B \in \Sigma$
and:
:$B \setminus A \in \Sigma$
Since $\sigma$-algebras are closed under countable union, we have: 
:$\paren {A \setminus B} \cup \paren {B \setminus A} \in \Sigma$
From the definition of symmetric difference, we have: 
:$A \Delta B = \paren {A \setminus B} \cup \paren {B \setminus A}$
so:
:$A \Delta B \in \Sigma$
Category:Sigma-Algebras
Category:Symmetric Difference
Sigma-Algebra Closed under Union

== Theorem ==
Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $A, B \in \Sigma$ be measurable sets.
Then $A \cup B \in \Sigma$, where $\cup$ denotes set union.
=== Corollary ===
== Proof ==
Define $A_1 = A, A_2 = B$, and for $n \in \N, n \ge 2: A_n = \O$.
Then by Sigma-Algebra Contains Empty Set, axiom $(3)$ of a $\sigma$-algebra applies.
Hence:
:$\ds \bigcup_{n \mathop \in \N} A_n = A \cup B \in \Sigma$
== Sources ==
* : $3.2 \ \text{(ii)}$
Category:Sigma-Algebras
Sigma-Algebra Closed under Union/Corollary

== Theorem ==
Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $A_1, \ldots, A_n \in \Sigma$.
Then $\ds \bigcup_{k \mathop = 1}^n A_k \in \Sigma$.
== Proof ==
Define for $k \in \N, k > n: A_k = \O$.
Then by Sigma-Algebra Contains Empty Set, axiom $(3)$ of a $\sigma$-algebra applies.
Hence:
:$\ds \bigcup_{k \mathop \in \N} A_k = \bigcup_{k \mathop = 1}^n A_k \in \Sigma$
Category:Sigma-Algebras
Sigma-Algebra Contains Empty Set

== Theorem ==
Let $X$ be a set.
Let $\Sigma$ be a $\sigma$-algebra on $X$.
Then:
:$\O \in \Sigma$
== Proof ==
Axiom $(1)$ of a $\sigma$-algebra grants:
:$X \in \Sigma$
By axiom $(2)$ and Set Difference with Self is Empty Set, it follows that:
:$\O = X \setminus X \in \Sigma$
== Sources ==
* : $3.2 \ \text{(i)}$
Category:Sigma-Algebras
Sigma-Algebra Contains Generated Sigma-Algebra of Subset

== Theorem ==
Let $\sigma_\FF$ be a be a $\sigma$-algebra on a set $\FF$.
Let $\sigma_\FF$ contain a set of sets $\EE$.
Let $\map \sigma \EE$ be the $\sigma$-algebra generated by $\EE$.
Then $\map \sigma \EE \subseteq \sigma_\FF$
== Proof ==
$\sigma_\FF$ is a $\sigma$-algebra containing $\EE$.
$\map \sigma \EE$ is a subset of all $\sigma$-algebras containing $\EE$, by definition of a generated $\sigma$-algebra.
Therefore it contains $\map \sigma \EE$.
== Sources ==
	
* : $\S 1.2$
Category:Sigma-Algebras
Sigma-Algebra Extended by Single Set

== Theorem ==
Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Let $S \subseteq X$ be a subset of $X$.
For subsets $T \subseteq X$ of $X$, denote $T^\complement$ for the set difference $X \setminus T$.
Then:
:$\map \sigma {\Sigma \cup \set S} = \set {\paren {E_1 \cap S} \cup \paren {E_2 \cap S^\complement}: E_1, E_2 \in \Sigma}$
where $\sigma$ denotes generated $\sigma$-algebra.
== Proof ==
Define $\Sigma'$ as follows:
:$\Sigma' := \set {\paren {E_1 \cap S} \cup \paren {E_2 \cap S^\complement}: E_1, E_2 \in \Sigma}$
Picking $E_1 = X$ and $E_2 = \O$ (allowed by Sigma-Algebra Contains Empty Set), it follows that $S \in \Sigma'$.
On the other hand, for any $E_1 \in \Sigma$, have by Intersection Distributes over Union and Union with Relative Complement:
:$\paren {E_1 \cap S} \cup \paren {E_1 \cap S^\complement} = E_1 \cap \paren {S \cup S^\complement} = E_1 \cap X = E_1$
Hence $E_1 \in \Sigma'$ for all $E_1$, hence $\Sigma \subseteq \Sigma'$.
Therefore, $\Sigma \cup \set S \subseteq \Sigma'$.
Moreover, from Sigma-Algebra Closed under Union, Sigma-Algebra Closed under Intersection and axiom $(2)$ for a $\sigma$-algebra, it is necessarily the case that:
:$\Sigma' \subseteq \map \sigma {\Sigma \cup \set S}$
It will thence suffice to demonstrate that $\Sigma'$ is a $\sigma$-algebra.
Since $X \in \Sigma$, also $X \in \Sigma'$.
Next, for any $E_1, E_2 \in \Sigma$, observe:
As $\Sigma$ is a $\sigma$-algebra, $E_1^\complement, E_2^\complement \in \Sigma$ and so indeed:
:$\paren {\paren {E_1 \cap S} \cup \paren {E_2 \cap S^\complement} }^\complement \in \Sigma'$
Finally, let $\sequence {E_{1, n} }_{n \mathop \in \N}$ and $\sequence {E_{2, n} }_{n \mathop \in \N}$ be sequences in $\Sigma$.
Then:
Since $\ds \bigcup_{n \mathop \in \N} E_{1, n}, \bigcup_{n \mathop \in \N} E_{2, n} \in \Sigma$, it follows that:
:$\ds \bigcup_{n \mathop \in \N} \paren {E_{1, n} \cap S} \cup \paren {E_{2, n} \cap S^\complement} \in \Sigma'$
Hence it is established that $\Sigma'$ is a $\sigma$-algebra.
It follows that:
:$\ds \map \sigma {\Sigma \cup \set S} = \Sigma'$
Category:Sigma-Algebras
Sigma-Algebra Generated by Complements of Generators

== Theorem ==
Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Let $\GG$ be a generator for $\Sigma$.
Then:
:$\GG' := \set {X \setminus G: G \in \GG}$
the set of relative complements of $\GG$, is also a generator for $\Sigma$.
== Proof ==
Hence the result.
Category:Sigma-Algebras Generated by Collection of Subsets
Sigma-Algebra Generated by Finite Partition is Finite Sub-Sigma-Algebra

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\xi$ be a finite partition of $\Omega$.
Let $\map \sigma \xi$ the generated $\sigma$-algebra by $\xi$.
Then, $\map \sigma \xi$ is a finite sub-$\sigma$-algebra of $\Sigma$.
Furthermore:
:$\ds \map \sigma \xi = \set {\bigcup S: S \subseteq \xi}$
== Proof ==
Let:
:$\ds \Gamma := \set {\bigcup S: S \subseteq \xi}$
By Power Set of Finite Set is Finite, $\Gamma$ is finite.
Therefore, it suffices to show:
:$\map \sigma \xi = \Gamma$
As $\xi$ is finite, from (SA3) follows:
:$\map \sigma \xi \supseteq \Gamma$
To conclude the equality, by definition of $\map\sigma\xi$, we need to show that $\Gamma$ is a $\sigma$-algebra.
(SA1) follows from:
:$\ds X = \bigcup \xi \in \Gamma$
(SA2) follows from:
:$\ds \forall S\subseteq\xi : X\setminus\bigcup S = \bigcup \paren {\xi\setminus S} \in \Gamma$
(SA3) follows from:
:$\ds \forall S_1, S_2,\ldots\subseteq\xi : \bigcup _{i=1}^\infty \paren {\bigcup S_i} = \bigcup \paren {\bigcup _{i=1}^\infty S_i} \in \Gamma$
Category:Probability Theory
Category:Ergodic Theory
Sigma-Algebra as Magma of Sets

== Theorem ==
The concept of $\sigma$-algebra is an instance of a magma of sets.
== Proof ==
It will suffice to define partial mappings such that the axiom for a magma of sets crystallises into the axioms for a $\sigma$-algebra.
Let $X$ be any set, and let $\powerset X$ be its power set.
Let $I$ be an index set.
For $S \in \powerset X$, define:
:$\phi_1: \powerset X \to \powerset X: \map {\phi_1} S := X$
:$\phi_2: \powerset X \to \powerset X: \map {\phi_2} S := X \setminus S$
:$\phi_3: \powerset X^\N \to \powerset X: \map {\phi_3} {\sequence {S_n}_{n \mathop \in \N} } := \ds \bigcup_{n \mathop \in \N} S_n$
It is blatantly obvious that $\phi_1, \phi_2$ and $\phi_3$ capture the axioms for a $\sigma$-algebra.
Let $I$ be the closed (real) interval $[1, 3]$.
For every $i \in I = [1, 3]$, let $J_i = \N$ be the index set, and let:
:$\phi_i: \powerset X^\N \to \powerset X$
be a partial mapping.
We have:
:$X \in \SS$
Also:
:$X \setminus S \in \SS$
From Union of Subsets is Subset/Subset of Power Set, we have:
:$\ds \bigcup_{n \mathop \in \N} S_n \in \SS$
It follows that for every indexed family $\family {S_n}_{n \mathop \in \N} \in \SS^{\N}$ in the domain of $\phi_i$ for every $i \in I$:
:$\forall i \in [1, 3]: \map {\phi_i} {\family {S_n}_{n \mathop \in \N} } \in \SS$
It follows that $\SS$ is closed under $\phi_i$ for all $i \in I$.
So $\SS$ is a magma of sets for $\set {\phi_i: i \in I}$ on $X$
Hence the result.
Category:Sigma-Algebras
Category:Magmas of Sets
Sigma-Algebra is Delta-Algebra

== Theorem ==
A $\sigma$-algebra is also a $\delta$-algebra.
== Proof ==
Let $\SS$ be a $\sigma$-algebra whose unit is $\mathbb U$.
Let $A_1, A_2, \ldots$ be a countably infinite collection of elements of $\SS$.
Then:
Thus $\SS$ is a $\delta$-algebra.
== Also see ==
* Delta-Algebra is Sigma-Algebra
Category:Sigma-Algebras
Sigma-Algebra is Dynkin System

== Theorem ==
Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Then $\Sigma$ is a Dynkin system on $X$.
== Proof ==
The axioms $(1)$ and $(2)$ for both $\sigma$-algebras and Dynkin systems are identical.
Dynkin system axiom $(3)$ is seen to be a specification of $\sigma$-algebra axiom $(3)$ to pairwise disjoint sequences.
Hence $\Sigma$ is trivially a Dynkin system on $X$.
== Sources ==
* : $5.2$, $\S 5$: Problem $1$
Category:Sigma-Algebras
Category:Dynkin Systems
Sigma-Algebra is Monotone Class

== Theorem ==
Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Then $\Sigma$ is also a monotone class.
== Proof ==
By definition, $\Sigma$, being a $\sigma$-algebra, is closed under countable unions.
From Sigma-Algebra Closed under Countable Intersection, it is also closed under countable intersections.
Thence, by definition, $\Sigma$ is a monotone class.
Category:Sigma-Algebras
Category:Monotone Classes
Sigma-Algebra of Countable Sets

== Theorem ==
Let $X$ be a set.
Let $\Sigma$ be the set of countable and co-countable subsets of $X$.
Then $\Sigma$ is a $\sigma$-algebra.
== Proof ==
Let us verify in turn the axioms of a $\sigma$-algebra.
=== Axiom $(1)$ ===
By Relative Complement with Self is Empty Set:
:$\relcomp X X = \O$
In particular, it is countable.
Hence $X$ is co-countable, and so $X \in \Sigma$.
=== Axiom $(2)$ ===
The relative complement of a countable set is by definition co-countable.
The converse holds by Relative Complement of Relative Complement.
Hence:
:$E \in \Sigma \implies \relcomp X E \in \Sigma$
=== Axiom $(3)$ ===
Let $\family {E_n}_{n \mathop \in \N} \in \Sigma$ be a family of elements of $\Sigma$.
Suppose that all the $E_n$ are countable.
Then by Countable Union of Countable Sets is Countable, $\ds \bigcup_{n \mathop \in \N} E_n$ is also countable.
Hence in $\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$.
Now suppose that at least one $E_n$ is only co-countable.
Then by Set Complement inverts Subsets and Set is Subset of Union: General Result:
:$\ds \relcomp X {\bigcup_{n \mathop \in \N} E_n} \subseteq \relcomp X {E_n}$
By definition of co-countable, the latter is countable.
Thus, by Subset of Countably Infinite Set is Countable, it follows that $\ds \bigcup_{n \mathop \in \N} E_n$ is co-countable.
Hence:
:$\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$
Hence the result, from Proof by Cases.
== Sources ==
* : $\S 1.2$
* : $3.3 \ \text{(v)}$
Category:Sigma-Algebras
Category:Examples of Sigma-Algebras