François G. Dorais
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 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 $(-\infty,0] \setminus \mathbb{Q}$... Jul 19 comment Uncountable dense subset whose complement is also uncountable and dense The only drawback is that you need some Axiom of Choice to get a basis for $\mathbb{R}$ over $\mathbb{Q}$... Jul 19 comment Uncountable dense subset whose complement is also uncountable and dense It doesn't get much simpler than that! (Note that intersecting $(-\infty,0]$ with $\mathbb{R}$ is a bit redundant.) Jul 19 answered Uncountable dense subset whose complement is also uncountable and dense