François G. Dorais
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 Aug3 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 Aug3 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!!! Aug1 awarded Scholar Aug1 comment Solution space to a functional equation Perfect. Thanks! Aug1 accepted Solution space to a functional equation Jul31 revised Solution space to a functional equation minor correction Jul31 awarded Student Jul31 asked Solution space to a functional equation Jul31 answered Is the Collatz conjecture in $\Sigma_1 / \Pi_1$? Jul31 comment Is the Collatz conjecture in $\Sigma_1 / \Pi_1$? The usual statement is $\Pi_2$, but since the Collatz conjecture is a sentence it is equivalent to either $0=1$ or $0=0$... (Assuming we're working in the standard model. If you're asking whether the conjecture is provably equivalent to a $\Pi_1$ or $\Sigma_1$ sentence over PA or ZFC, that's a different matter.) Jul29 comment All real functions are continuous A translation of Brouwer's original paper can be found in van Heijenoort's From Frege to Gödel: a source book in mathematical logic, 1879-1931 books.google.com/books/about/… Jul27 comment Intuitionistic Banach-Tarski Paradox Actually, the Vitali case is not simpler, this is exactly how Banach-Tarski use the Axiom of Choice. Instead of the group $\mathbb{Q}$ acting on $\mathbb{R}$ by translation, we have a free group $F$ generated by two rotations which acts on the unit sphere $S_2$ and we must pick one representative from each orbit (i.e. equivalence class). As in the Vitali case, each orbit is dense in $S_2$ so (a very significant fragment of) the Law of Excluded Middle is required to separate the points of $S_2$ into mutually disjoint orbits. Jul20 awarded Yearling Jul17 comment Existence of a prime ideal in an integral domain of finite type over a field without Axiom of Choice For future reference, Wilfrid Hodges wrote an excellent paper called Six impossible rings [J. Algebra 31 (1974)] where he examines the three Noetherian conditions and the three Artinian conditions. Using six pathological rings, he concludes that no implications between these six conditions other than the obvious ones are provable in ZF. Jul15 revised $\mu$-recursive definition of ulam (3n+1) function removed superfluous computations; added some backslashes Jul15 answered $\mu$-recursive definition of ulam (3n+1) function Jun8 awarded Constituent Jun8 awarded Caucus May27 comment A quick question about categoricity in model theory Categorical theory is often used to mean a theory that is $\kappa$-categorical for every infinite cardinal $\kappa$. Mar28 awarded Nice Answer