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Ccc and countably closed forcings are proper
Statement
In ZFC, every ccc forcing preorder and every countably closed forcing preorder is proper. No converse is asserted.
Facts & Assumptions
Given: A nonempty forcing preorder and a sufficiently large well-ordered structure containing it.
Properness may be checked by producing an -master below each . Master-condition characterizations
Ccc means that every antichain is countable. Compatibility, ccc and Knaster for posets
Countable closure means that every countable descending chain has a common lower bound. Closure, distributivity, and chain conditions for forcing orders
AC supplies maximal antichains, enumerations of the countable family of dense sets in , and the recursive choices in the closed case. The Axiom of Choice
Proof
Suppose first that is ccc, let be a relevant countable elementary model, and fix . For each dense , elementarity and A1 give a maximal antichain with . By F2, is externally countable. Any externally countable set is a subset of : elementarity supplies in a surjection from onto , and every natural number belongs to . Hence is predense below every condition, so itself is -generic and is a master below . F1 proves that is proper.
Suppose instead that is countably closed. Enumerate all dense subsets of belonging to as , repeating one if the family is finite. Starting with , use elementarity and A1 to choose with ; every stays in . By F3 there is for all . For every , the condition lies above , so is predense below . Thus is an -master below , and F1 again makes proper.
The two arguments cover the ccc and countably closed hypotheses independently and use no converse. AC is spent exactly in the maximal-antichain and enumeration/recursive-choice operations identified in steps 1.1 and 1.2. Therefore every forcing in either class is proper.
Depends on
Used by
- PFA implies MA(aleph-one) and the Suslin Hypothesis Corollary
- Ccc posets are proper by maximal antichains Example
- PFA specializes an Aronszajn tree Example
- Ccc and proper are not equivalent False statement
- PFA implies the simple ideal dichotomy Theorem
Dependency tree · two levels
14 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
- Karagila, Forcing & Symmetric Extensions, Propositions 8.5 and 8.7 with complete proofs, printed p. 39 (standard reference, not scraped)