Exercise 1.24 (Scott 1981, PRG-19, §1) — every partial element extends to a #
total one (AC)
Scott (for set theorists): using the Axiom of Choice, prove that in every domain a partial element can always be extended to a total element. (Hint: the union of every transfinite chain of filters is again a filter.)
This file formalizes:
chainUnion C hne hchain— the union⋃ Cof a non-empty chainCof filters is itself a filter (V.Element); the only non-trivial axiom isinter_mem, which uses chain-directedness to find a single member ofCcontaining both factors. Plusle_chainUnion(every member of the chain is⊑the union).exists_total_ge x— with Zorn's lemma (zorn_le_nonempty_Ici₀), every elementxhas a total elementtabove it (x ⊑ t ∧ IsTotal t). The chains inIci xare bounded above by theirchainUnion, so Zorn produces a maximal — i.e. total — element.
Whether this is equivalent to AC is left as prose (Scott's question); we do not
formalize the
reversal. This is the explicitly classical exercise: exists_total_ge
legitimately uses
Classical.choice through zorn_le_nonempty_Ici₀. The chainUnion
construction is choice-free
([propext, Quot.sound]).
Exercise 1.24 — the union of a chain of filters is a filter. For a
non-empty chain C of
filters, ⋃ C = {X ∣ ∃ x ∈ C, X ∈ x} is again a filter. The intersection axiom
uses chain
totality: given X ∈ x and Y ∈ y with x, y ∈ C, one of x ⊑ y, y ⊑ x
holds, and the larger
filter contains both X and Y, hence X ∩ Y.
Equations
Instances For
Exercise 1.24 — partial elements extend to total ones (AC). Using Zorn's
lemma, every
element x admits a total element t with x ⊑ t. The upper bound of any
non-empty chain in
Ici x is its chainUnion, so a maximal element exists; maximality is exactly
totality
(IsMax = IsTotal).