Difference between revisions of "Dynkin system"

TODO: Flesh out the algebra (blue boxes)

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.

TODO: That second half

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.