# Dynkin system

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Note: a Dynkin system is also called a "[ilmath]d[/ilmath]-system"[1] and the page d-system just redirects here.

## Definition

### First Definition


• [ilmath]X\in\mathcal{D} [/ilmath]
• For any [ilmath]D\in\mathcal{D} [/ilmath] we have [ilmath]D^c\in\mathcal{D} [/ilmath]
• For any [ilmath](D_n)_{n=1}^\infty\subseteq\mathcal{D}[/ilmath] is a sequence of pairwise disjoint sets we have [ilmath]\udot_{n=1}^\infty D_n\in\mathcal{D}[/ilmath]

### 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[3] 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[Note 1] and has [ilmath]\lim_{n\rightarrow\infty}(A_n)=A[/ilmath] we have [ilmath]A\in\mathcal{D}[/ilmath]

## Proof of equivalence of definitions

Claim: Definition 1 [ilmath]\iff[/ilmath] Definition 2


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 $\bigudot_{n=1}^\infty B_n=\bigcup_{n=1}^\infty A_n$
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]

TODO: That second half

## Immediate results

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

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.