Dynkin system/Proof that definitions 1 and 2 are equivalent

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

Claim

The following definitions are the same:

Definition 1	Definition 2
Given a set $X$ and a family of subsets of $X$, which we shall denote $\mathcal{D}\subseteq\mathcal{P}(X)$ is a Dynkin system if: $X\in\mathcal{D}$ For any $D\in\mathcal{D}$ we have $D^c\in\mathcal{D}$ For any $(D_n)_{n=1}^\infty\subseteq\mathcal{D}$ is a sequence of pairwise disjoint sets we have $\udot_{n=1}^\infty D_n\in\mathcal{D}$	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

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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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