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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
Jul
19
comment Silver indiscernibles and definable injections
Apostolos has the right idea. Every Silver indiscernible is in fact inaccessible in $L$. There will be a constructible injection $\lambda \to \omega\times\alpha^{<\omega}$ if and only if $\lambda < \max((|\alpha|^+)^L,\omega_1^L)$. If $\alpha < i_\alpha$, then $i_\alpha$ is necessarily much larger than $\max((|\alpha|^+)^L,\omega_1^L)$.
Jul
19
comment Silver indiscernibles and constructibility
No problem! It was a pleasure!