# François G. Dorais

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bio website dorais.org location Hanover, NH age member for 4 years, 5 months seen 7 mins ago profile views 12,599

I like math and a few other things...

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 1d comment Can we construct cohomolgy theory on noetherian separated schemes without Axiom of Choice? David, there is a general problem with stalks without AC in that there might not be enough of them to ensure things like "surjective" = "surjective on stalks" make sense (depending on how things are set up). However, when suitably formulated, noetherianness should ensure enough stalks to make your assumptions work. Apr17 revised Suslin lines hereditarily Lindelof edited tags Apr15 comment A question about small sets of reals To complement the question, the recent paper by Goldstern, Kellner, Shelah, Wohofsky establishing the relative consistency of BC+dBC is here and on the arxiv. Also, Goldstern's overview of the proof is on the arxiv. Apr12 revised Showing that it is a matroid with axioms edited tags Apr12 comment Is there an intuitionistic generalized boolean algebra (of Stone)? The generalized Boolean algebras that the OP is talking about always have a bottom element. The difference is that they only have a relative complement operation $x - y$ rather than an absolute complement $-x$. Thus they always have a bottom $0 = x - x$ and they are Boolean algebras precisely when they have a top $1$, in which case the complement of $x$ is the relative complement $1-x$. Apr11 comment Is there an intuitionistic generalized boolean algebra (of Stone)? How about a Boolean ring without identity? en.wikipedia.org/wiki/Boolean_ring Apr8 comment $\aleph$ looks like $\mathbb N$? The $\aleph$ notation appears in Cantor's Contributions to the Founding of the Theory of Transfinite Numbers. As far as I can tell, N wasn't used there to denote the set of natural numbers. This suggests that the answer is 'no'. Apr7 comment Subposets of small Dushnik-Miller dimension Wonderful! Thank you both! Apr4 reviewed Approve suggested edit on Intuition on Lindeberg condition Apr2 awarded axiom-of-choice Apr1 comment Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory? @godelian: There is an $H$-valued model that separates all equivalence classes. Apr1 comment Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory? @godelian: In the classical case, that the theory is satisfiable is equivalent to BPI, but no choice is needed to see that the theory is consistent. Apr1 answered Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory? Mar31 revised On a modal correspondence deleted 27 characters in body Mar29 comment $n$th order arithmetic with predicates for orders One subtlety (that fortunately doesn't affect arithmetic theories much) are empty domains and their effect on the behavior of the existential quantifiers. In the usual interpretation of multi-sorted logic, $\exists x^1(x^1 =_1 x^1) \lor \cdots \lor \exists x^n(x^n =_n x^n)$ is not a tautology but $\exists x(x = x)$ is usually a tautology in classical single-sorted first-order logic. It's perhaps best to use free logic or similar when encoding using the single-sorted version, or to relax $\forall x(Z_1(x) \lor \cdots \lor Z_n(x))$ a bit to allow for all domains to be empty. Mar20 awarded Nice Answer Mar19 comment Order homomorphism functions on $\omega_1$ @MonroeEskew: The OP had originally written decreasing instead of regressive. Noah was just a bit too non-happy when he wrote his comment. Mar19 comment Order homomorphism functions on $\omega_1$ Since there are regressive functions with unbounded range but every function in $K$ has bounded range, you can't always have $f \leq h(f)$. (Assuming regressive is what you meant.) Mar19 comment Order homomorphism functions on $\omega_1$ @NoahS: That doesn't satisfy $f \leq h(f)$. Mar19 revised Order homomorphism functions on $\omega_1$ added 54 characters in body; edited tags