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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
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Jul
21
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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