Dynkin system

Note: a Dynkin system is also called a "[ilmath]d[/ilmath]-system" and the page d-system just redirects here.

Definition

===[[Dynkin systeyesnkin system/Definition 1}}

Second Definition

Given a set [ilmath]X[/ilmath] and a family of subsets of [ilmath]X[/ilmath] we denote [ilmath]\mathcal{D}\subseteq\mathcal{P}(X)[/ilmath] is a Dynkin system on [ilmath]X[/ilmath] if:

• [ilmath]X\in\mathcal{D} [/ilmath]
• [ilmath]\forall A,B\in\mathcal{D}[B\subseteq A\implies A-B\in\mathcal{D}][/ilmath]
• Given a sequence [ilmath](A_n)_{n=1}^\infty\subseteq\mathcal{D}[/ilmath] that is increasing and has [ilmath]\lim_{n\rightarrow\infty}(A_n)=A[/ilmath] we have [ilmath]A\in\mathcal{D}[/ilmath]

Proof of equivalence of definitions

Proof of claim

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[math]\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}} }[/math][math]\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}[/math][math]\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }[/math]

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TODO: Flesh out the algebra (blue boxes)

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Definition 1 [ilmath]\implies[/ilmath] definition 2

Let [ilmath]\mathcal{D} [/ilmath] be a subgroup satisfying definition 1, then I claim it satisfies definition 2. Let us check the conditions.

1. [ilmath]X\in\mathcal{D} [/ilmath] is satisfied by definition
2. For [ilmath]A,B\in\mathcal{D} [/ilmath] with [ilmath]B\subseteq A[/ilmath] then [ilmath]A-B\in\mathcal{D} [/ilmath]
   • Note that [ilmath]A-B=(A^c\udot B)^c[/ilmath] (this is not true in general, it requires [ilmath]B\subseteq A[/ilmath]Include ven diagram

   As by hypothesis [ilmath]\mathcal{D} [/ilmath] is closed under complements and disjoint unions, we see that [ilmath](A^c\udot B)^c\in\mathcal{D} [/ilmath] thus

   • we have [ilmath]A-B\in\mathcal{D} [/ilmath]

3. Given [ilmath](A_n)_{n=1}^\infty\subseteq\mathcal{D}[/ilmath] being an increasing sequence of subsets, we have [ilmath]\lim_{n\rightarrow\infty}(A_n)=A[/ilmath] where [ilmath]A:=\bigcup_{n=1}^\infty A_n[/ilmath] (See limit of an increasing sequence of sets for more information)

   Let [ilmath](A_n)_{n=1}^\infty\subseteq\mathcal{D}[/ilmath] be given.

   Define a new sequence of sets, [ilmath](B_n)_{n=1}^\infty[/ilmath] by:

   • [ilmath]B_1=A_1[/ilmath]
   • [ilmath]B_n=A_n-B_{n-1}[/ilmath]

   This is a pairwise disjoint sequence of sets.

   Now by hypothesis [ilmath]\bigudot_{n=1}^\infty B_n\in\mathcal{D}[/ilmath]

   • Note that [math]\bigudot_{n=1}^\infty B_n=\bigcup_{n=1}^\infty A_n[/math]

   So we have [ilmath]\bigcup_{n=1}^\infty A_n\in\mathcal{D} := A[/ilmath], thus the limit is in [ilmath]\mathcal{D} [/ilmath] - as required.

This completes the first half of the proof.

The second half isn't tricky, the only bit I recommend knowing is [ilmath]A\udot B=(A^c-B)^c[/ilmath]

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TODO: That second half

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Immediate results

• [ilmath]\emptyset\in\mathcal{D} [/ilmath]

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Proof:

As [ilmath]\mathcal{D} [/ilmath] is closed under complements and [ilmath]X\in\mathcal{D} [/ilmath] by definition, [ilmath]X^c\in\mathcal{D} [/ilmath]

[ilmath]X^c=\emptyset[/ilmath] so [ilmath]\emptyset\in\mathcal{D} [/ilmath]

This completes the proof.

See also

• Dynkin system generated by
• Types of set algebras
• [ilmath]p[/ilmath]-system
• Conditions for a [ilmath]d[/ilmath]-system to be a [ilmath]\sigma[/ilmath]-algebra