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I don't think that's the case. They can both not have the property that it is eventually larger than the other.
by JohnKemeny 1y ago
I don't think that's the case. They can both not have the property that it is eventually larger than the other.
- LegionMammal978 1y agoThe axioms demand that either one function is eventually dominated by the other, or both functions are of the same order. But which of these is the case will strongly depend on which subsequence you look at.
- btilly 1y agoYou may have missed the same subtlety that I did. Because pi is irrational, the functions are different at all integers. Therefore, in the total order, these two functions cannot have the same order. That still doesn't resolve which one is larger though.
- LegionMammal978 1y agoWell, as presented in Tao's post, the set Ω can be either the natural numbers or the real numbers. So I'm assuming the "subsequence" is a (perhaps uncountable?) set of real parameters, in the latter case.
- btilly 1y agoAh. Good point.
- btilly 1y agoNo, it is the case. Look for the comment in the article, after passing to a subsequence if necessary. The ultrafilter produces the necessary subsequence for any question that you ask, and will do so in such a way as to produce logically consistent answers for any combination of questions that you choose. That is why the ultrafilter axiom is a weak version of choice. Take the set of possible yes/no questions that we can ask as predicates, such that each answer shows up infinitely often. The ultrafilter results in an arbitrary yet consistent set of choices of yes/no for each predicate.
- JohnKemeny 1y agoOkay, yes, I see. But then it seems that O doesn't obey some very natural standard schools, and then what is it good for? O is a total order, but functions aren't in any way a total order, so what's the point?
- btilly 1y agoAnd now you see what I don't like about it!