The intersection of an arbitrary family of Dynkin systems is itself a Dynkin system
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Contents
Statement
Let [ilmath]\{\mathcal{D}_\alpha\}_{\alpha\in I} [/ilmath] be an arbitrary family of Dynkin systems on a set [ilmath]X[/ilmath], then[1] we claim:
- [math]\mathcal{D}:\eq\bigcap_{\alpha\in I}\mathcal{D}_\alpha[/math] is itself a Dynkin system on [ilmath]X[/ilmath].
Proof
Recall the definition of a Dynkin system:
Given a set [ilmath]X[/ilmath] and a family of subsets of [ilmath]X[/ilmath], which we shall denote [ilmath]\mathcal{D}\subseteq\mathcal{P}(X)[/ilmath] is a Dynkin system[1] if:
- [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]
Proof
- There are 3 properties we must show, we will show them in the order above:
- Show that [ilmath]X\in\mathcal{D} [/ilmath]
- [ilmath]X\in\mathcal{D}\iff\forall\alpha\in I[X\in\mathcal{D}_\alpha][/ilmath] by definition of intersection
- Well let [ilmath]\alpha\in I[/ilmath] be given
- [ilmath]X\in\mathcal{D}_\alpha[/ilmath] as [ilmath]\mathcal{D}_\alpha[/ilmath] is itself a Dynkin system on [ilmath]X[/ilmath].
- Since [ilmath]\alpha\in I[/ilmath] was arbitrary we have shown it for all
- Well let [ilmath]\alpha\in I[/ilmath] be given
- Thus [ilmath]X\in\mathcal{D} [/ilmath] - as required
- [ilmath]X\in\mathcal{D}\iff\forall\alpha\in I[X\in\mathcal{D}_\alpha][/ilmath] by definition of intersection
- Show: [ilmath]\forall D\in\mathcal{D}[X-D\in\mathcal{D}][/ilmath] (recall the set complement of a set in another set can be defined in terms of the relative complement as done here)
- Let [ilmath]D\in\mathcal{D} [/ilmath] be given, we wish to show [ilmath]X-D\in\mathcal{D} [/ilmath]
- [ilmath]X-D\in\mathcal{D}\iff\forall\alpha\in I[X-D\in\mathcal{D}_\alpha][/ilmath] by definition of intersection
- We will show the RHS of this
- Let [ilmath]\alpha\in I[/ilmath] be given
- As [ilmath]D\in\mathcal{D} [/ilmath] by definition, we see that [ilmath]\forall\beta\in I[D\in\mathcal{D}_\beta][/ilmath] by definition of intersection again
- Thus [ilmath]D\in\mathcal{D}_\alpha[/ilmath] for the given [ilmath]\alpha[/ilmath]
- As [ilmath]D\in\mathcal{D}_\alpha[/ilmath] and [ilmath]\mathcal{D}_\alpha[/ilmath] is a Dynkin system on [ilmath]X[/ilmath]
- [ilmath]X-D\in\mathcal{D}_\alpha[/ilmath]
- As [ilmath]D\in\mathcal{D}_\alpha[/ilmath] and [ilmath]\mathcal{D}_\alpha[/ilmath] is a Dynkin system on [ilmath]X[/ilmath]
- Since [ilmath]\alpha\in I[/ilmath] was arbitrary we have shown it for all
- Thus [ilmath]\forall\alpha\in I[X-D\in\mathcal{D}_\alpha][/ilmath]
- [ilmath]X-D\in\mathcal{D}\iff\forall\alpha\in I[X-D\in\mathcal{D}_\alpha][/ilmath] by definition of intersection
- This is the very definition of [ilmath]X-D\in\mathcal{D} [/ilmath] - as required
- Let [ilmath]D\in\mathcal{D} [/ilmath] be given, we wish to show [ilmath]X-D\in\mathcal{D} [/ilmath]
- Let [math](D_n)_{n\in\mathbb{N} }\subseteq\mathcal{D} [/math] be a family of pairwise disjoint sets in [ilmath]\mathcal{D} [/ilmath] - we must show [ilmath]\bigudot_{n\in\mathbb{N} }D_n\in\mathcal{D} [/ilmath]
- Again [math]\left[\bigudot_{n\in\mathbb{N} }D_n\in\mathcal{D}\right]\iff\left[\forall\alpha\in I\left(\bigudot_{n\in\mathbb{N} }D_n\in\mathcal{D}_\alpha\right)\right][/math] - we will show the RHS
- Let [ilmath]\alpha\in I[/ilmath] be given
- By hypothesis [ilmath]\forall n\in\mathbb{N}[D_n\in\mathcal{D}][/ilmath], so:
- [ilmath]\forall\beta\in I\forall n\in\mathbb{N}[D_n\in\mathcal{D}_\beta][/ilmath]
- Specifically: let [ilmath]\beta:\eq\alpha[/ilmath], then:
- [ilmath]\forall n\in\mathbb{N}[D_n\in\mathcal{D}_\beta][/ilmath] - otherwise written as [ilmath](D_n)_{n\in\mathbb{N} }\subseteq\mathcal{D}_\alpha[/ilmath]
- As [ilmath]\mathcal{D}_\alpha[/ilmath] is itself a Dynkin system:
- [ilmath]\bigudot_{n\in\mathbb{N} }D_n\in\mathcal{D}_\alpha[/ilmath]
- As [ilmath]\mathcal{D}_\alpha[/ilmath] is itself a Dynkin system:
- Since [ilmath]\alpha\in I[/ilmath] was arbitrary we have shown it for all.
- Let [ilmath]\alpha\in I[/ilmath] be given
- This completes the proof
- Again [math]\left[\bigudot_{n\in\mathbb{N} }D_n\in\mathcal{D}\right]\iff\left[\forall\alpha\in I\left(\bigudot_{n\in\mathbb{N} }D_n\in\mathcal{D}_\alpha\right)\right][/math] - we will show the RHS
- Show that [ilmath]X\in\mathcal{D} [/ilmath]
References