Bourbaki

Exercise 26, § 6

Tag 03O6, permanent. Cite it as [Bourbaki, Tag 03O6].

Theory of Sets, III, 248

Bourbaki marks this one as harder than the rest.

Let E be an infinite set and let F\mathfrak{F} be a set of subsets of E such that for each A∈FA \in \mathfrak{F} we have

Card (A)=Card (F)=Card (E)=a⩾ℵ0.\text{Card } (A) = \text{Card } (\mathfrak{F}) = \text{Card } (E) = \mathfrak{a} \geqslant \aleph_0.

Show that there exists a partition (Bι)ι∈I(B_\iota)_{\iota \in I} of E such that

Card (I)=Card (Bι)=a\text{Card } (I) = \text{Card } (B_\iota) = \mathfrak{a}

for all ι∈I\iota \in I, and such that A∩Bι≠∅A \cap B_\iota \neq \emptyset for all A∈FA \in \mathfrak{F} and all ι∈I\iota \in I.

(With the notation of Exercise 24 (a), consider first a surjective mapping ff of ωα\omega_\alpha into F\mathfrak{F} such that for each A∈F\mathrm{A} \in \mathfrak{F} the set of all ξ∈ωα\xi \in \omega_\alpha such that f(ξ)=Af(\xi) = \mathrm{A} has cardinal equal to a\mathfrak{a}. Then, by transfinite induction, define a bijection gg of ωα\omega_\alpha onto E such that g(ξ)∈f(ξ)g(\xi) \in f(\xi) for every ξ∈ωα\xi \in \omega_\alpha.)

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