# François G. Dorais

less info
reputation
1224
bio website dorais.org location Hanover, NH age member for 3 years, 9 months seen yesterday profile views 535

I like math and a few other things...

# 154 Actions

 Sep1 revised Descriptions of sets and the Axiom of Choice correction Aug31 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. Aug31 answered Descriptions of sets and the Axiom of Choice Aug27 answered Formalizing metamathematics Aug23 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. Aug22 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... Aug22 revised Karatsuba vs. Schönhage-Strassen for multiplication of polynomials small correction Aug22 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. Aug22 answered Can one avoid AC in the proof that in Noetherian rings there is a maximal element for each set? Aug22 revised Karatsuba vs. Schönhage-Strassen for multiplication of polynomials fixed typo Aug22 answered Karatsuba vs. Schönhage-Strassen for multiplication of polynomials Aug9 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... Aug9 revised For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice rewording Aug9 revised For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice correction Aug9 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. Aug9 answered For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice Jul26 awarded Enthusiast Jul21 awarded Yearling Jul19 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. Jul19 comment Uncountable dense subset whose complement is also uncountable and dense It's even simpler to write $(-\infty,0] \setminus \mathbb{Q}$...