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Small Dowker ladder topology

Definition

Assume AC. Put $κ = ω_{1}$ and $E = {α < κ : cf (α) = ω}$ . Let $(S_{n})_{1 \leq n < ω}$ be pairwise disjoint stationary subsets of $E$ , equipped with a single-ladder two-target AD system $(A_{α})_{α \in ⋃_{n \geq 1} S_{n}}$ in clause 5 of Luzin sets, stick, and almost-disjoint guessing at omega one. In particular, $A_{α} \subseteq α$ is cofinal; if $α < β$ , $A_{α} \cap A_{β}$ is bounded in $α$ ; and for each pair of uncountable targets $B_{0}, B_{1} \subseteq κ$ and each $n \geq 1$ , stationarily many $α \in S_{n}$ satisfy $\sup (A_{α} \cap B_{i}) = α$ for both $i < 2$ .

Set $S_{0} = κ ∖ ⋃_{n \geq 1} S_{n}$ , let $n (α)$ be the unique index with $α \in S_{n (α)}$ , and put $W_{n} = ⋃_{j \leq n} S_{j}$ . If $n (α) > 0$ and $A_{α} \cap W_{n (α) - 1}$ is cofinal in $α$ , set $L_{α} = A_{α} \cap W_{n (α) - 1}$ ; in every other case set $L_{α} = \emptyset$ . Thus no undefined $A_{α}$ or $W_{- 1}$ is used on $S_{0}$ . Write $\overset{ˉ}{S} = {α : L_{α} \neq \emptyset}$ .

The small Dowker ladder topology on $κ$ declares $U \subseteq κ$ open exactly when $L_{α} ∖ U$ is bounded in $α$ for every $α \in U \cap \overset{ˉ}{S}$ . Here bounded means contained in some ordinal $ε < α$ . Consequently, a set $F$ is closed exactly when $F \cap L_{α}$ is bounded in $α$ for every $α \in \overset{ˉ}{S} ∖ F$ .

For an active $α \in \overset{ˉ}{S}$ , write $N_{α}^{ε} = {α} \cup (L_{α} ∖ ε)$ , $ε < α$ . 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 $L_{α}, W_{n}, \overset{ˉ}{S}$ above.

[F1]

Clause 5 supplies cofinal ladders, bounded intersections at the smaller index, and simultaneous stationary guessing of two uncountable targets on each $S_{n}$ (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 ${A C}_{ω}$ ).

[F3]

Every ordinal below $ω_{1}$ is countable and $ω_{1}$ is uncountable; under countable choice its countable subsets are bounded ( $ω_{1}$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF, Assuming countable choice: every at most countable subset of $ω_{1}$ is bounded below $ω_{1}$ , so no at most countable subset of $ω_{1}$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable).

[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

1.1givenF4

Both $\emptyset$ and $κ$ satisfy the open-set test. If $α \in ⋃ U \cap \overset{ˉ}{S}$ for a family $U$ of open sets, take any member $U \in U$ containing this one point. Then $L_{α} ∖ ⋃ U \subseteq L_{α} ∖ U$ 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.

1.2givenalgebra

The complement of $F$ is open precisely when, for each active $α \notin F$ , $L_{α} ∖ (κ ∖ F) = L_{α} \cap F$ is bounded. This proves both directions of the closedness test. Likewise, the open-set test gives $N_{α}^{ε} \subseteq U$ for some $ε < α$ whenever $U$ is open and contains active $α$ ; it imposes no openness claim on $N_{α}^{ε}$ .

1.3givenF1

Each nonempty $L_{α}$ is cofinal in $α$ , has $α \in E$ , and lies in strictly lower levels. If $α < β$ are active, $L_{α} \cap L_{β} \subseteq A_{α} \cap A_{β}$ 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.

1.4F1F2A1

Given uncountable $B_{0}, B_{1} \subseteq κ$ , F2 and A1 give indices $m_{i}$ such that $B_{i} \cap S_{m_{i}}$ is uncountable: otherwise the level partition would express $B_{i}$ as a countable union of countable sets. For every $n > \max (m_{0}, m_{1})$ apply F1 to the two targets $B_{i} \cap S_{m_{i}}$ . On a stationary subset of $S_{n}$ , both $A_{α} \cap B_{i} \cap S_{m_{i}}$ are cofinal in $α$ . They lie in $W_{n - 1}$ , so $L_{α}$ is active and $L_{α} \cap B_{i}$ is cofinal for both $i$ . This is the promised simultaneous two-target stationary accumulation property.

2.1step 1.3step 1.4F3

Taking $B_{0} = B_{1} = κ$ , which is uncountable by F3, step 1.4 gives a stationary subset of $\overset{ˉ}{S}$ on every sufficiently high level. Any set containing a stationary set meets every club, so $\overset{ˉ}{S}$ 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.

Depends on

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Sources

Rinot–Shalev, A guessing principle from a Souslin tree, with applications to topology, §3, space definition and Lemma 3.2, p. 17 (standard reference, not scraped)