Dynkin system

Note: a Dynkin system is also called a "$d$-system" and the page d-system just redirects here.

Definition

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

Second Definition

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

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

Proof of equivalence of definitions

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Proof of claim

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

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TODO: Flesh out the algebra (blue boxes)

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Definition 1 $\implies$ definition 2

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

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

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

   • we have $A-B\in\mathcal{D}$

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

   Let $(A_n)_{n=1}^\infty\subseteq\mathcal{D}$ be given.

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

   • $B_1=A_1$
   • $B_n=A_n-B_{n-1}$

   This is a pairwise disjoint sequence of sets.

   Now by hypothesis $\bigudot_{n=1}^\infty B_n\in\mathcal{D}$

   • Note that $$\bigudot_{n=1}^\infty B_n=\bigcup_{n=1}^\infty A_n$$

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

This completes the first half of the proof.

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

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TODO: That second half

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Immediate results

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• $\emptyset\in\mathcal{D}$

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Proof:

As $\mathcal{D}$ is closed under complements and $X\in\mathcal{D}$ by definition, $X^c\in\mathcal{D}$

$X^c=\emptyset$ so $\emptyset\in\mathcal{D}$

This completes the proof.

See also

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