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Logicissuesintheconstructionofthetransfinite
sequenceofsubcomplexes$K^\bullet_\alpha$.Cases(1)
and(3)areindependentofprevioussteps,butitseems
(2)isnot:itrequirestomakeachoicesince
$N_\alpha^\bullet$isnotunique.Insteadoftransfinite
recursionon$\alpha$,maybewhatoneactuallyneedsis
thegeneralizedaxiomofdependentchoices?Let$\kappa$
beanaleph.Thisaxiomsays:
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Logic issues in the construction of the transfinite sequence of subcomplexes K^\bullet_\alpha. Cases (1) and (3) are independent of previous steps, but it seems (2) is not: it requires to make a choice since N_\alpha^\bullet is not unique. Instead of transfinite recursion on \alpha, maybe what one actually needs is the generalized axiom of dependent choices? Let \kappa be an aleph. This axiom says:

> \mathsf{DC}_\kappa [1, Sect. 8.1]. Let S be a nonempty set and let R be a binary relation such that for every \alpha<\kappa and every \alpha-sequence s=\left\langle x_{\xi}: \xi<\alpha\right\rangle of elements of S there exists y \in S such that sRy. Then there is a function f: \kappa \rightarrow S such that for every \alpha<\kappa,(f \mid \alpha) R f(\alpha).

(One has \mathsf{AC}\Leftrightarrow\forall\kappa,\mathsf{DC}_\kappa [1, Theorem 8.1].)

The axioms \mathsf{DC}_\kappa alone do not provide a transfinite sequence indexed on the class of all ordinals, but rather a transfinite sequence of length equal to an aleph. However, for our proof it suffices to consider some cardinal \kappa greater than \prod_{i\in\mathbb{Z}}|M^i| (as it is hinted in [#9511](https://stacks.math.columbia.edu/tag/079L#comment-9511), 4).

A disclaimer: All set theory I know I have learnt it in the last month, from scratch. I hadn't ever been into any kind of foundations of math, logical or set-theoretic ones.

### References ###

1. T. Jech, *The Axiom of Choice*