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Countable intersections of club subsets of omega_1 are club
Statement
Assume . If is closed and unbounded for every , then is closed and unbounded in .
Facts & Assumptions
Given: Each is closed and unbounded, and holds.
Under , every countable subset of is bounded below . (Assuming countable choice: every at most countable subset of is bounded below , so no at most countable subset of is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable)
Proof
Finite intersections of clubs are club: closedness is immediate, and for two clubs one alternately chooses larger points in them; the supremum of the resulting omega-sequence is below by [L1] and belongs to both by closedness. Induction handles finitely many.
Given , recursively choose with , starting above ; step 1.1 supplies such a point. Let by [L1]. For each fixed , the tail lies in , so closedness gives . Also .
Thus the intersection is unbounded. It is closed as an arbitrary intersection of closed sets, hence is club.
Depends on
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- The first uncountable ordinal $\omega_1 := \aleph(\omega)$
- The order topology on an ordinal, with the half-open intervals $(\alpha, \beta]$ and the initial segments $[0, \beta]$ as a basis
- Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$, so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable
Used by
Dependency tree · two levels
24 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
- Donald L. Cohn, Measure Theory, 2nd ed., Chapter 7 (standard reference, not scraped)