François G. Dorais
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 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 $(-\infty,0] \setminus \mathbb{Q}$... Jul19 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 $\mathbb{Q}$... Jul19 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.) Jul19 answered Uncountable dense subset whose complement is also uncountable and dense Jul19 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 $\max((|\alpha|^+)^L,\omega_1^L)$. Jul19 comment Silver indiscernibles and constructibility No problem! It was a pleasure! Jul18 revised Silver indiscernibles and constructibility clarification; changed terminology to match the question Jul18 answered Silver indiscernibles and constructibility Jul18 revised How is Kleene's T predicate defined? added wikipedia links Jul18 answered How is Kleene's T predicate defined? Jul17 comment Estimation of factoring time of a $n$-digit number (current state of art) on a desktop @Gerry: I used the formula from Wikipedia, which says that $c = (64/9)^{1/3} + o(1)$. I did ignore the $o(1)$, but all I wanted is to illustrate the fast growth of this function. Jul17 comment Do there exist interesting binary relations satisfying reflexivity and symmetry, but not transitivity? These are well-known graphs with some very interesting open problems. But your post doesn't even mention why anyone should care about these graphs! Jul17 comment Estimation of factoring time of a $n$-digit number (current state of art) on a desktop @TaoLee: The number field sieve is only one step along the path of better and better factoring algorithms. It is conceivable that a clever new algoritm could factor RSA-2048 in a matter of days. Jul16 comment Estimation of factoring time of a $n$-digit number (current state of art) on a desktop @TauLee: You have the formula, just plug in $N$... For RSA-2048, I got $$152373858906444928985207904781622575.47$$ Jul14 answered Order type of uncountable set and order ordering