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 Yearling
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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 $\Sigma_1 / \Pi_1$?
Jul
31
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.)
Jul
29
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/…
Jul
27
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.
Jul
20
awarded  Yearling
Jul
17
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.
Jul
15
revised $\mu$-recursive definition of ulam (3n+1) function
removed superfluous computations; added some backslashes
Jul
15
answered $\mu$-recursive definition of ulam (3n+1) function
Jun
8
awarded  Constituent
Jun
8
awarded  Caucus