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DefinitionDefinition: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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Small Dowker ladder topology

Definition

Assume AC. Put κ=ω1 and E={α<κ:cf(α)=ω}. Let (Sn)1n<ω be pairwise disjoint stationary subsets of E, equipped with a single-ladder two-target AD system (Aα)αn1Sn in clause 5 of Luzin sets, stick, and almost-disjoint guessing at omega one. In particular, Aαα is cofinal; if α<β, AαAβ is bounded in α; and for each pair of uncountable targets B0,B1κ and each n1, stationarily many αSn satisfy sup(AαBi)=α for both i<2.

Set S0=κn1Sn, let n(α) be the unique index with αSn(α), and put Wn=jnSj. If n(α)>0 and AαWn(α)1 is cofinal in α, set Lα=AαWn(α)1; in every other case set Lα=. Thus no undefined Aα or W1 is used on S0. Write Sˉ={α:Lα}.

The small Dowker ladder topology on κ declares Uκ open exactly when LαU is bounded in α for every αUSˉ. Here bounded means contained in some ordinal ε<α. Consequently, a set F is closed exactly when FLα is bounded in α for every αSˉF.

For an active αSˉ, write Nαε={α}(Lαε), ε<α. Every open set containing α contains one such set. These are weak neighborhood tests: they are not asserted to be open, or to form an open basis. The topology axioms, closedness test, and the simultaneous stationary accumulation property below are verified here.

Facts & Assumptions

Given: The preceding partition and single-ladder two-target AD system; AC; the definitions of Lα,Wn,Sˉ above.

[F1]

Clause 5 supplies cofinal ladders, bounded intersections at the smaller index, and simultaneous stationary guessing of two uncountable targets on each Sn (Luzin sets, stick, and almost-disjoint guessing at omega one).

[F2]

Under countable choice, a countable union of at most countable sets is at most countable (Countable unions of at most countable sets, assuming ACω).

[F4]

A topology contains the empty and whole sets and is closed under arbitrary unions and finite intersections (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[A1]

AC implies the countable choice used in F2 and F3 by applying a choice function to the nonempty enumeration-witness sets (The Axiom of Choice).

Verification

1.1

Both and κ satisfy the open-set test. If αUSˉ for a family U of open sets, take any member UU containing this one point. Then LαULαU is bounded. This pointwise existence argument selects no family of witnesses. At a point of a finite intersection, the omitted ladder points are a finite union of bounded subsets of the nonzero limit ordinal α, hence lie below the maximum of finitely many bounds. The empty intersection is κ. This proves the topology axioms.

givenF4
1.2

The complement of F is open precisely when, for each active αF, Lα(κF)=LαF is bounded. This proves both directions of the closedness test. Likewise, the open-set test gives NαεU for some ε<α whenever U is open and contains active α; it imposes no openness claim on Nαε.

givenalgebra
1.3

Each nonempty Lα is cofinal in α, has αE, and lies in strictly lower levels. If α<β are active, LαLβAαAβ is bounded in α. It is also bounded in β, because it is a subset of α<β. Thus intersections with any distinct active ladder are bounded at either index.

givenF1
1.4

Given uncountable B0,B1κ, F2 and A1 give indices mi such that BiSmi is uncountable: otherwise the level partition would express Bi as a countable union of countable sets. For every n>max(m0,m1) apply F1 to the two targets BiSmi. On a stationary subset of Sn, both AαBiSmi are cofinal in α. They lie in Wn1, so Lα is active and LαBi is cofinal for both i. This is the promised simultaneous two-target stationary accumulation property.

F1F2A1
2.1

Taking B0=B1=κ, which is uncountable by F3, step 1.4 gives a stationary subset of Sˉ on every sufficiently high level. Any set containing a stationary set meets every club, so Sˉ is stationary. Zero and successor ordinals are inactive, as they do not belong to E; empty ladders cause no open-set or closed-set condition. These observations include all endpoint cases of the definition. QED.

step 1.3step 1.4F3

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