Let (Xι)ι∈I be a family of sets. Show that if (Yι)ι∈I is a family of sets such that Yι⊂Xι for each ι∈I, then
ι∈I∏Yι=ι∈I⋂prι−1(Yι).
Solution
This solution was written by nemotron-3-ultra-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.
Let (Xι)ι∈I be a family of sets and let (Yι)ι∈I be a family of sets such that Yι⊂Xι for each ι∈I.
By Definition 1 of §3 (tag 03Q8), the product ∏ι∈IXι is the set of all functional graphs F with domain I such that F(ι)∈Xι for every ι∈I. For each α∈I, the projection prα is defined in the same Definition 1 as the mapping of ∏ι∈IXι into Xα that sends F to F(α).
For a fixed ι∈I, the inverse image prι−1(Yι) is therefore
prι−1(Yι)={F∈ι∈I∏Xι∣F(ι)∈Yι}. Taking the intersection over all ι∈I gives ι∈I⋂prι−1(Yι)={F∈ι∈I∏Xι∣∀ι∈I,F(ι)∈Yι}.
On the other hand, the product ∏ι∈IYι is, by the same Definition 1, the set of all functional graphs F with domain I such that F(ι)∈Yι for every ι∈I. Because Yι⊂Xι for all ι, Corollary 3 of Proposition 6 (tag 03QD) implies ∏ι∈IYι⊂∏ι∈IXι. Hence an element of ∏ι∈IYι is exactly a functional graph F with domain I that belongs to ∏ι∈IXι and satisfies F(ι)∈Yι for all ι∈I.
Therefore the two sets are equal:
ι∈I∏Yι=ι∈I⋂prι−1(Yι).
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