François G. Dorais
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 Dec 5 answered Questions about generalizations of the Principle of Dependent Choices Dec 4 answered Proofs whose length depends on the input Sep 18 answered Does every nonempty definable finite set have a definable member? Aug 19 answered Borel linear order cannot have uncountable increasing chain Aug 19 comment Borel linear order cannot have uncountable increasing chain I had missed the link! The author cites Harrington and Shelah, who did prove the result I recalled earlier. I guess it would be best to check that reference. @William: No. There are no uncountable wellordered Borel chains at all so that weaker variant is vacuously true. Aug 19 comment Borel linear order cannot have uncountable increasing chain This is true if chain is replaced by wellordered chain. In other words, a Borel linear order cannot contain a copy of $\omega_1$ or its reverse. I would guess that's what is meant, otherwise "increasing or decreasing" is not very meaningful. Where is this from? Aug 16 comment What is actually “relatively consistent”? "If a system is not complete, then it is consistent." Seeing that complete usually means "proves $\sigma$ or $\lnot\sigma$ for every sentence $\sigma$" and that consistent usually means "does not prove every sentence $\sigma$," a system that is not complete must be consistent. There are variations but, in any case, I don't think your second sentence exactly says what you meant to say. Aug 14 awarded Quorum Aug 6 comment Axiom of choice - to use or not to use Also, the axiom of choice is not necessarily non-constructive. For example, full choice is valid in constructive type theory. Aug 3 awarded Revival Aug 3 comment Proving that the set of algebraic numbers is countable without AC And yes, countable means $\leq \aleph_0$ and, in particular, infinite Dedekind finite sets are "uncountable" in ZF. So one shouldn't think that "uncountable" means "large" in ZF... Aug 3 comment Proving that the set of algebraic numbers is countable without AC Yes, your answer is perfectly correct, my comment was just an addendum. Since I had a chance to look it up, the reference is: Hodges, Läuchli's algebraic closure of $\mathbb{Q}$, Math. Proc. Cambridge Philos. Soc. 79 (1976), 289-297. ams.org/mathscinet-getitem?mr=422022 Aug 3 comment Proving that the set of algebraic numbers is countable without AC This is correct if by "the algebraic numbers" you mean the algebraic closure of $\mathbb{Q}$ contained in $\mathbb{C}$. However, Hodges has shown that ZF does not prove that this is the only algebraic closure of $\mathbb{Q}$. In particular, since ZF proves that any two countable algebraic closures of $\mathbb{Q}$ are isomorphic, there is a very wild model of ZF where $\mathbb{Q}$ has an uncountable algebraic closure!!! Aug 1 awarded Scholar Aug 1 comment Solution space to a functional equation Perfect. Thanks! Aug 1 accepted Solution space to a functional equation Jul 31 revised Solution space to a functional equation minor correction Jul 31 awarded Student Jul 31 asked Solution space to a functional equation Jul 31 answered Is the Collatz conjecture in $\Sigma_1 / \Pi_1$?