Difference between revisions of "Dynkin system"

[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]

TODO: Flesh out the algebra (blue boxes)

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.

TODO: That second half

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.