Suppose I have a sigma algebra M\mathcal{M} indexed by AA. Take α∈A,β∈A,α≠β\alpha \in A, \beta \in A, \alpha \neq \beta and intersect Dα∈M,Dβ∈MD_{\alpha} \in \mathcal{M}, D_{\beta} \in \mathcal{M}

Dα⋂Dβ=DrD_{\alpha} \bigcap D_{\beta} = D_r

where DrD_r is the result of this intersection. Therefore Dr=∅D_r = \emptyset in the case that DαD_{\alpha} and DβD_{\beta} are disjoint or Dr=HαβD_r = H_{\alpha \beta}, where HαβH_{\alpha \beta} contains the elements in both DαD_{\alpha} and DβD_{\beta}.

Repeat this process and put the results into the set RR and say each Hαβ={…}H_{\alpha \beta} = \{\dots\} (a set, we don’t care about the indices). My question is this:

Should

R={∅,{…},{…},…}R = \{\emptyset, \{\dots\}, \{\dots\}, …\}

or

R={{∅},{…},{…},…}R = \{\{\emptyset\}, \{\dots\}, \{\dots\}, …\}

I believe the first to be true because we are just sticking the empty set into RR. If not why?

=================

Let me ask something to clarify: Is HαβH_{\alpha \beta} equal to Dα∩DβD_\alpha\cap D_\beta? Also, is RR supposed to be the set all possible intersections?

– Wore

2 days ago

@Wore Yes and yes.

– Chris

2 days ago

Made a small edit. I believe the “first” to be true.

– Chris

2 days ago

In that case, if you have that Dα∩Dβ=∅D_\alpha\cap D_\beta=\emptyset (they are disjoint), then you should denote RR as in the first case: R={∅,…}R=\{\emptyset, …\}. However, given Dα≠XD_\alpha\neq X (where M\mathcal{M} is a sigma algebra of subsets of XX), you will have Dα∩X=DαD_\alpha\cap X=D_\alpha, for every α∈A\alpha\in A, so it seems that RR is simply equals to M=R∖{X}\mathcal{M}=R\setminus \{X\}. (Notice that XX is the only set that you cannot obtain as the intersection of two different sets in M\mathcal{M}).

– Wore

2 days ago

I am not sure I understand your construction completely, however I’ve got a feeling you mean that R⊆2ΩR \subseteq 2^\Omega where Ω\Omega is a set on which M\mathcal M is based (that is, Ω=⋃M∈MM\Omega = \bigcup_{M\in\mathcal M} M). In that case, the former is the correct way to write down RR, and the latter actually means that ∅∈Ω\emptyset \in \Omega which can be true when Ω\Omega is a collection of subsets of some other set, but I think it’s not what you mean here.

– Ilya

2 days ago

=================

=================