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Small Dowker ladder shrinking obstruction

Statement

Assume AC and the hypotheses of Small Dowker ladder topology. Its ladder space $X$ has cardinality $ℵ_{1}$ and is not countably paracompact. The decreasing closed sets $D_{n} = ω_{1} ∖ W_{n}$ , $n < ω$ , have empty intersection, but every sequence of open expansions $U_{n} \supseteq D_{n}$ has nonempty intersection. The product $X \times [0, 1]$ , with the usual interval and product topology, is not normal.

Facts & Assumptions

Given: The ladder space $X$ on $ω_{1}$ and AC; $W_{n} = ⋃_{j \leq n} S_{j}$ and $D_{n} = ω_{1} ∖ W_{n}$ .

[F1]

The sets $(S_{n})_{n < ω}$ partition $ω_{1}$ , and each $S_{n}$ for $n \geq 1$ is stationary (Small Dowker ladder topology).

[F2]

This space is normal Hausdorff, each $W_{n}$ is open, and no two disjoint closed sets are both uncountable (Small Dowker ladder normality).

[F3]

Under countable choice, countable unions of countable sets are countable and countable subsets of $ω_{1}$ are bounded; $ω_{1}$ is the least uncountable ordinal and is a cardinal (Countable unions of at most countable sets, assuming ${A C}_{ω}$ , 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, $ω_{1}$ 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).

[F4]

In a normal space, countable paracompactness is equivalent to the existence of open expansions with empty intersection for every decreasing closed sequence with empty intersection (Increasing-cover and decreasing-closed-set criteria, condition (iv)).

[F5]

For a $T_{1}$ space, normality of its product with $[0, 1]$ is equivalent to normality and countable paracompactness of the space (Dowker product characterization).

[A1]

AC supplies countable choice in F3 and the assumptions of F4–F5 (The Axiom of Choice).

Proof

1.1

Since $W_{n}$ is open and increasing, its complements $D_{n}$ are closed and decreasing. A point belongs to a unique level $S_{j}$ and is excluded from $D_{n}$ for all $n \geq j$ , so $⋂_{n} D_{n} = \emptyset$ . Each $D_{n}$ contains $S_{n + 1}$ . A stationary subset of $ω_{1}$ is uncountable: a countable set is bounded by F3 and therefore misses a club tail. Hence each $D_{n}$ is uncountable, including $D_{0}$ .

F1F2F3A1

2.1

Let $(U_{n})_{n < ω}$ be any sequence of open sets with $D_{n} \subseteq U_{n}$ . The closed complement $F_{n} = ω_{1} ∖ U_{n}$ is disjoint from the uncountable closed $D_{n}$ . Thus $F_{n}$ is countable by F2. AC, through F3, makes $⋃_{n} F_{n}$ countable, so it cannot equal $ω_{1}$ . Consequently $⋂_{n} U_{n} = ω_{1} ∖ ⋃_{n} F_{n} \neq \emptyset$ . No monotonicity of the expansions is needed; empty $F_{n}$ cause no exception.

step 1.1F2F3A1

3.1

The normal space supplied by F2 fails condition (iv) of F4 by steps 1.1–2.1, and hence is not countably paracompact. It is $T_{1}$ , since the normality lemma proves singleton closedness. F5 therefore shows $X \times [0, 1]$ cannot be normal. Finally, its underlying set is literally $ω_{1}$ , whose cardinality is $ℵ_{1}$ by F3: the identity is the required bijection, with no quotient or choice of representatives. QED.

step 1.1step 2.1F2F3F4F5A1

Depends on

Used by

Conditional Dowker constructions of cardinality aleph one Theorem

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Rinot–Shalev, A guessing principle from a Souslin tree, with applications to topology, Lemma 3.1 p. 16 and Corollaries 3.9, 3.11 p. 19 (standard reference, not scraped)