Orders of Infinity
5 comments
·May 4, 2025btilly
JohnKemeny
I don't think that's the case. They can both not have the property that it is eventually larger than the other.
btilly
No, 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.
LegionMammal978
The 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.
singularity2001
Since we know that these hyper real numbers are well defined we can teach them axiomatically to high school students the way Leibniz used them (and keep the explicit construction via filters to university students just like with a dedekind cut for reals)
Here is the axiomatic approach in Julia and Lean https://github.com/pannous/hyper-lean
I'm not a big fan of using nonstandard analysis for this. We're assuming the existence of arbitrary answers that we cannot ever produce.
For example, which function is eventually larger than the other?
In the ultrafilter, one almost certainly will be larger. In fact the ratio of the two will, asymptotically, approach a specific limit. Which one is larger? What is the ratio? That entirely depends on the ultrafilter.Which means that we can accept the illusionary simplicity of his axiom about every predicate P(N), and it will remain simple right until we try to get a concrete and useful answer out of it.