Bourbaki

Exercise 31, § 6

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

Theory of Sets, III, 250

Bourbaki marks this one as harder than the rest.

Let A be an infinite set and let E be the set of all infinite subsets of A, ordered by inclusion. Show that E is completely ramified (§ 2, Exercise 8) but not antidirected (§ 1, Exercise 23) and that E has an antidirected cofinal subset F. (Consider first the set D(A)\mathfrak{D}(A) of countable infinite subsets of A (which is cofinal in E) and let Z=R0(D(A))Z = R_0(\mathfrak{D}(A)) (§ 1, Exercise 23). Write Z in the form (zλ)λ∈L(z_\lambda)_{\lambda \in L}, where L is a well-ordered set, and take F to be a set of countable subsets XnλX_n^\lambda, where λ\lambda runs through a suitable subset of L, n∈Nn \in \mathbf{N}, Xmλ⊃XnλX_m^\lambda \supset X_n^\lambda whenever m⩽nm \leqslant n, Xnλ−Xn+1λX_n^\lambda - X_{n+1}^\lambda is infinite for all n⩾0n \geqslant 0, and

⋂n∈NXnλ=∅;\bigcap_{n \in \mathbf{N}} X_n^\lambda = \emptyset;

the XnλX_n^\lambda are to be defined by transfinite induction in such a way that the images of the sets XnλX_n^\lambda under the canonical mapping r:D(A)→Zr : \mathfrak{D}(A) \to Z (§ 1, Exercise 23) are mutually disjoint and form a cofinal subset of Z.)

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