Types of set algebras/Type table

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$\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }$ $\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}$ $\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }$

Type table

System Type	Definition	Deductions
Ring^[1]^[2]	$\mathcal{A}$ is $\backslash$ -closed^{[Note 1]} $\mathcal{A}$ is $\cup$ -closed^{[Note 2]}	$\emptyset\in\mathcal{A}$ ^{[Note 3]}
$\sigma$ -ring^[1]^[2]	$\mathcal{A}$ is a ring $\mathcal{A}$ is $\sigma$ - $\cup$ -closed^{[Note 4]}	$\sigma$ - $\cap$ -closed also^{[Theorem 1]}
Algebra^[1]^[2]	$\mathcal{A}$ is closed under complements $\mathcal{A}$ is $\cup$ -closed	$\mathcal{A}$ is $\backslash$ -closed^{[Note 5]} $\emptyset\in\mathcal{A}$ ^{[Note 6]} $\Omega\in\mathcal{A}$ ^{[Note 7]} $\mathcal{A}$ is $\cap$ -closed^{[Theorem 2]}
$\sigma$ -algebra^[1]^[2]	$\mathcal{A}$ is an algebra $\mathcal{A}$ is $\sigma$ - $\cup$ -closed	also $\sigma$ - $\cap$ -closed^{[Theorem 1]} Is a Dynkin system^[3]^{[Note 8]}
Semiring^[1]	TODO: Page 3 in^[1]
Dynkin system^[1]^[3]	$\Omega\in\mathcal{A}$ $\mathcal{A}$ is closed under complements $\sigma$ - $\udot$ -closed	$\emptyset\in\mathcal{A}$ ^{[Note 9]}

Theorems

↑ ^{Jump up to: 1.0} ^1.1
Using Class of sets closed under set-subtraction properties we know that if \mathcal{A} is closed under Set subtraction then:
- $\mathcal{A}$ is $\cap$ -closed
- $\sigma$ - $\cup$ -closed $\implies$ $\sigma$ - $\cap$ -closed
Jump up ↑
Using Class of sets closed under complements properties we see that if \mathcal{A} is closed under complements then:
- $\mathcal{A}$ is $\cap$ -closed $\iff$ $\mathcal{A}$ is $\cup$ -closed
- $\mathcal{A}$ is $\sigma$ - $\cap$ -closed $\iff$ $\mathcal{A}$ is $\sigma$ - $\cup$ -closed

Notes

Jump up ↑ Closed under finite Set subtraction
Jump up ↑ Closed under finite Union
Jump up ↑ As given $A\in\mathcal{A}$ we must have $A-A\in\mathcal{A}$ and $A-A=\emptyset$
Jump up ↑ closed under finite or countably infinite union
Jump up ↑ Note that $A-B=A\cap B^c=(A^c\cup B)^c$ - or that $A-B=(A^c\cup B)^c$ - so we see that being closed under union and complement means we have closure under set subtraction.
Jump up ↑ As we are closed under set subtraction we see $A-A=\emptyset$ so $\emptyset\in\mathcal{A}$
Jump up ↑ As we are closed under set subtraction we see that $A-A\in\mathcal{A}$ and $A-A=\emptyset$ , so $\emptyset\in\mathcal{A}$ - but we are also closed under complements, so $\emptyset^c\in\mathcal{A}$ and $\emptyset^c=\Omega\in\mathcal{A}$
Jump up ↑ Trivial - satisfies the definitions
Jump up ↑ As $\Omega^c=\emptyset$ by being closed of complements, $\emptyset\in\mathcal{A}$

References

↑ ^{Jump up to: 1.0} ^1.1 ^1.2 ^1.3 ^1.4 ^1.5 ^1.6 Probability Theory - A comprehensive course - second edition - Achim Klenke
↑ ^{Jump up to: 2.0} ^2.1 ^2.2 ^2.3 Measure Theory - Paul R. Halmos
↑ ^{Jump up to: 3.0} ^3.1 Measures Integrals and Martingales - Rene L. Schilling

[.5C-closed-consequences-7] {Jump up to: 1.0} ^1.1 Using Class of sets closed under set-subtraction properties we know that if $\mathcal{A}$ is closed under Set subtraction then:
$\mathcal{A}$ is $\cap$ -closed

$\sigma$ - $\cup$ -closed $\implies$ $\sigma$ - $\cap$ -closed

[2] $\mathcal{A}$ is $\cap$ -closed

[3] $\sigma$ - $\cup$ -closed $\implies$ $\sigma$ - $\cap$ -closed

[complement-closed-consequences-11] Jump up ↑ Using Class of sets closed under complements properties we see that if $\mathcal{A}$ is closed under complements then:
$\mathcal{A}$ is $\cap$ -closed $\iff$ $\mathcal{A}$ is $\cup$ -closed

$\mathcal{A}$ is $\sigma$ - $\cap$ -closed $\iff$ $\mathcal{A}$ is $\sigma$ - $\cup$ -closed

[5] $\mathcal{A}$ is $\cap$ -closed $\iff$ $\mathcal{A}$ is $\cup$ -closed

[6] $\mathcal{A}$ is $\sigma$ - $\cap$ -closed $\iff$ $\mathcal{A}$ is $\sigma$ - $\cup$ -closed

[.5C-closed-3] Jump up ↑ Closed under finite Set subtraction

[U-closed-4] Jump up ↑ Closed under finite Union

[5] Jump up ↑ As given $A\in\mathcal{A}$ we must have $A-A\in\mathcal{A}$ and $A-A=\emptyset$

[sigma-U-closed-6] Jump up ↑ closed under finite or countably infinite union

[8] Jump up ↑ Note that $A-B=A\cap B^c=(A^c\cup B)^c$ - or that $A-B=(A^c\cup B)^c$ - so we see that being closed under union and complement means we have closure under set subtraction.

[9] Jump up ↑ As we are closed under set subtraction we see $A-A=\emptyset$ so $\emptyset\in\mathcal{A}$

[10] Jump up ↑ As we are closed under set subtraction we see that $A-A\in\mathcal{A}$ and $A-A=\emptyset$ , so $\emptyset\in\mathcal{A}$ - but we are also closed under complements, so $\emptyset^c\in\mathcal{A}$ and $\emptyset^c=\Omega\in\mathcal{A}$

[13] Jump up ↑ Trivial - satisfies the definitions

[14] Jump up ↑ As $\Omega^c=\emptyset$ by being closed of complements, $\emptyset\in\mathcal{A}$

[PTACC-1] {Jump up to: 1.0} ^1.1 ^1.2 ^1.3 ^1.4 ^1.5 ^1.6 Probability Theory - A comprehensive course - second edition - Achim Klenke

[MTH-2] {Jump up to: 2.0} ^2.1 ^2.2 ^2.3 Measure Theory - Paul R. Halmos

[MIM-12] {Jump up to: 3.0} ^3.1 Measures Integrals and Martingales - Rene L. Schilling

[1]

[2]

[Note 1]

[Note 2]

[Note 3]

[Note 4]

[Theorem 1]

[Note 5]

[Note 6]

[Note 7]

[Theorem 2]

[3]

[Note 8]

[Note 9]

Types of set algebras/Type table

Contents

Type table

Theorems

Notes

References

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