Voila! We have a suitable definition of “almost all agreement”: if the agreement set I is contained in some arbitrary nonprincipal ultrafilter U.
Isn’t it easier to just say “If the agreement set I has a nonfinite number of elements”? Why the extra complexity?
must contain a set or its complement
Oh I see, so defining it with ultrafilters rules out situations like a=(1,0,1,0,1,0,...) and b=(0,1,0,1,0,1...) where both have infinite zeros and yet their product is zero.
The post is wrong in saying that U contains only cofinite sets. It obviously must contain plenty of sets that are neither finite nor cofinite, because the complements of those sets are also neither finite nor cofinite. Possibly the author intended to type “contains all cofinite sets” instead.
In particular, exactly one of a or b is equivalent to zero in *R.
Which one is equivalent to zero depends upon exactly which non-principal ultrafilter you choose, as there are infinitely many non-principal ultrafilters. Unfortunately (as with many other applications of the Axiom of Choice) there is no finite way to specify which ultrafilter you mean.
The post is wrong in saying that U contains only cofinite sets. It obviously must contain plenty of sets that are neither finite nor cofinite, because the complements of those sets are also neither finite nor cofinite. Possibly the author intended to type “contains all cofinite sets” instead.
Yep, this is correct! I’ve updated the post to reflect this.
E.g. if an ultrafilter contains the set of all even naturals, it won’t contain the set of all odd naturals, neither of which are finite or cofinite.
Of course, this makes all of this rather abstract. It looks to me like for almost any two hyperreals (e.g. a, b as above), the answer to “which of them is larger?” is “It depends on the ultrafilter. Also, I can not tell you if a set is part of any specific ultrafilter. But fear not, for any given ultrafilter, the hyperreals are well-ordered.”
Basically for any usable theorem, one would have to prove that the result is independent of the actual ultrafilter used, which means that numbers such as a and b will probably not feature in them a lot.
I can not fault my analysis 1 professor for opting to stick to the reals (abstract as they are already are) instead.
Isn’t it easier to just say “If the agreement set I has a nonfinite number of elements”? Why the extra complexity?
Oh I see, so defining it with ultrafilters rules out situations like a=(1,0,1,0,1,0,...) and b=(0,1,0,1,0,1...) where both have infinite zeros and yet their product is zero.
The post is wrong in saying that U contains only cofinite sets. It obviously must contain plenty of sets that are neither finite nor cofinite, because the complements of those sets are also neither finite nor cofinite. Possibly the author intended to type “contains all cofinite sets” instead.
In particular, exactly one of a or b is equivalent to zero in *R.
Which one is equivalent to zero depends upon exactly which non-principal ultrafilter you choose, as there are infinitely many non-principal ultrafilters. Unfortunately (as with many other applications of the Axiom of Choice) there is no finite way to specify which ultrafilter you mean.
Yep, this is correct! I’ve updated the post to reflect this.
E.g. if an ultrafilter contains the set of all even naturals, it won’t contain the set of all odd naturals, neither of which are finite or cofinite.
Thanks, this is helpful to point out.
Of course, this makes all of this rather abstract. It looks to me like for almost any two hyperreals (e.g. a, b as above), the answer to “which of them is larger?” is “It depends on the ultrafilter. Also, I can not tell you if a set is part of any specific ultrafilter. But fear not, for any given ultrafilter, the hyperreals are well-ordered.”
Basically for any usable theorem, one would have to prove that the result is independent of the actual ultrafilter used, which means that numbers such as a and b will probably not feature in them a lot.
I can not fault my analysis 1 professor for opting to stick to the reals (abstract as they are already are) instead.