Exercise 1.25 (Scott 1981, PRG-19, ยง1) โ final segments of an ordinal #
Scott: let ฮ be a well-ordered set (ordinal) and let ๐ be the family of
non-empty final
segments of ฮ. Describe |๐|; are all elements finite; is every approximation
to a finite
element finite?
For a linear order, the final segments are exactly the non-empty upper sets
(IsUpperSet).
Any two upper sets of a linear order are nested (upperSet_subset_or), so the
family is a
neighbourhood system by Scott's nested-or-disjoint criterion
(finalSegmentSystem).
Classification of |๐|. For a well-order every non-empty upper set is
Set.Ici a with a
its minimum (exists_Ici_of_mem). Hence a filter x is completely described by
its trace
lowerSetOf x = {a โฃ Ici a โ x}, which is a non-empty lower set of ฮ, and
x โฆ lowerSetOf x
is an order isomorphism
|๐| โo {S : Set ฮ โฃ S non-empty lower set} (ordered by โ) โ
finalSegmentClassify.
So |๐| is the lattice of non-empty initial segments of ฮ (the "cuts" of ฮ).
Total / finite. The greatest element is the filter of all neighbourhoods
(topElement),
which is the unique total element (topElement_isTotal,
eq_topElement_of_isTotal). The finite
elements are the principal filters โ(Ici a). When ฮ has no greatest element
(e.g. โ = ฯ) the
total element is not finite (topElement_not_principal_of_noMax): not all
elements are finite.
The system and Ici lemmas are [propext, Quot.sound]; the classification's
surjectivity uses
the well-ordering (WellFounded.has_min), so it is classical.
In a linear order, any two upper sets are nested. (Hence final segments are nested-or-disjoint โ in fact always nested.)
Exercise 1.25 โ the final-segment neighbourhood system. Over a (non-empty)
linear order
ฮ, the non-empty upper sets (final segments) form a neighbourhood system. Built
via
ofNestedOrDisjoint since any two upper sets are nested (upperSet_subset_or).
Equations
- Domain.Neighborhood.FinalSegment.finalSegmentSystem ฮ = Domain.Neighborhood.NeighborhoodSystem.ofNestedOrDisjoint (fun (X : Set ฮ) => X.Nonempty โง IsUpperSet X) Set.univ โฏ โฏ โฏ
Instances For
Each Set.Ici a is a (non-empty, upper) neighbourhood.
The principal finite elements โ(Ici a); the correspondence a โฆ โ(Ici a) is
order-reversing (Ici a โ Ici b โ b โค a), one of Scott's "finite elements".
Well-order: every neighbourhood is Ici (its minimum). #
Well-order key lemma. In a well-order, every non-empty upper set is
Set.Ici a, where a
is its least element.
The classification of |๐| as the non-empty lower sets (initial #
segments).
The trace lowerSetOf x = {a โฃ Ici a โ x} of a filter x.
Instances For
The filter built from a non-empty lower set S: the neighbourhoods that
contain some
Ici a with a โ S.
Equations
- One or more equations did not get rendered due to their size.
Instances For
Exercise 1.25 โ classification of |๐|. For a well-order ฮ, the domain
of non-empty
final segments is order-isomorphic to the poset of non-empty initial segments
(lower sets) of
ฮ under inclusion: x โฆ {a โฃ Ici a โ x}. (So |๐| is the lattice of cuts of
ฮ.)
Equations
- One or more equations did not get rendered due to their size.
Instances For
Total elements and finiteness (Scott's prose questions). #
The greatest element: the filter of all neighbourhoods.
Equations
- Domain.Neighborhood.FinalSegment.topElement = { mem := fun (X : Set ฮ) => (Domain.Neighborhood.FinalSegment.finalSegmentSystem ฮ).mem X, sub := โฏ, master_mem := โฏ, inter_mem := โฏ, up_mem := โฏ }
Instances For
Exercise 1.25 โ the unique total element. The filter of all neighbourhoods
is total
(โ-maximal).
It is the unique total element: any total x equals topElement.
Exercise 1.25 โ not all elements are finite. If ฮ has no greatest
element (e.g.
ฮ = โ = ฯ), the total element topElement is not a principal filter โ(Ici a):
pick b > a,
then Ici b โ topElement but Ici a โ Ici b. So |๐| has a non-finite (limit)
element.