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bio website dorais.org
location Hanover, NH
age
visits member for 3 years, 9 months
seen yesterday

I like math and a few other things...


Sep
1
revised Descriptions of sets and the Axiom of Choice
correction
Aug
31
comment Descriptions of sets and the Axiom of Choice
Well, $OD$ is transitive whenever $V = OD$ since $V$ is always transitive, so $V = OD$ and $V = HOD$ mean exactly the same thing.
Aug
31
answered Descriptions of sets and the Axiom of Choice
Aug
27
answered Formalizing metamathematics
Aug
23
comment Can one avoid AC in the proof that in Noetherian rings there is a maximal element for each set?
Good point, Arturo. Note that Hodges separates the three definitions in the paper cited above.
Aug
22
comment Karatsuba vs. Schönhage-Strassen for multiplication of polynomials
This answer assumes that the coefficients of the polynomials are integers. I thought that was part of the question, but I now see that it wasn't...
Aug
22
revised Karatsuba vs. Schönhage-Strassen for multiplication of polynomials
small correction
Aug
22
comment Can one avoid AC in the proof that in Noetherian rings there is a maximal element for each set?
Yes, see W. Hodges, Six impossible rings, J. Algebra 31 (1974), 218-244.
Aug
22
answered Can one avoid AC in the proof that in Noetherian rings there is a maximal element for each set?
Aug
22
revised Karatsuba vs. Schönhage-Strassen for multiplication of polynomials
fixed typo
Aug
22
answered Karatsuba vs. Schönhage-Strassen for multiplication of polynomials
Aug
9
comment For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice
@Asaf: I'm not used to the quality standards of this site, if you (or anyone else) want to flesh out my answer, please go right ahead...
Aug
9
revised For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice
rewording
Aug
9
revised For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice
correction
Aug
9
comment For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice
Yes, @Asaf. The argument is essentially the same as yours, but it avoids the use of ordinals as the OP requested.
Aug
9
answered For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice
Jul
26
awarded  Enthusiast
Jul
21
awarded  Yearling
Jul
19
comment Uncountable dense subset whose complement is also uncountable and dense
@tomcuchta: Yes, because a Bernstein set must meet every closed interval that are not singletons.
Jul
19
comment Uncountable dense subset whose complement is also uncountable and dense
It's even simpler to write $(-\infty,0] \setminus \mathbb{Q}$...