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bio website dorais.org
location Hanover, NH
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visits member for 4 years, 5 months
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I like math and a few other things...


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.
Apr
17
revised Suslin lines hereditarily Lindelof
edited tags
Apr
15
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.
Apr
12
revised Showing that it is a matroid with axioms
edited tags
Apr
12
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 $1-x$.
Apr
11
comment Is there an intuitionistic generalized boolean algebra (of Stone)?
How about a Boolean ring without identity? en.wikipedia.org/wiki/Boolean_ring
Apr
8
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'.
Apr
7
comment Subposets of small Dushnik-Miller dimension
Wonderful! Thank you both!
Apr
4
reviewed Approve suggested edit on Intuition on Lindeberg condition
Apr
2
awarded  axiom-of-choice
Apr
1
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.
Apr
1
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.
Apr
1
answered Is every Heyting algebra the Lindenbaum algebra of an intuitionistic first order theory?
Mar
31
revised On a modal correspondence
deleted 27 characters in body
Mar
29
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.
Mar
20
awarded  Nice Answer
Mar
19
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.
Mar
19
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.)
Mar
19
comment Order homomorphism functions on $\omega_1$
@NoahS: That doesn't satisfy $f \leq h(f)$.
Mar
19
revised Order homomorphism functions on $\omega_1$
added 54 characters in body; edited tags