This is a linkpost for https://www.ykumar.org/almost-always/

five ways to say "Almost Always" and actually mean it

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1Yudhister Kumar

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I think you mean "The Lebesgue measure of the complement is zero" or something of the sort, not "The Lebesgue measure is non-zero". A set with nonzero measure has to be "fairly big" but it certainly doesn't amount to "almost all".

You can't say "almost all natural numbers are even" unless you have already picked a nonprincipal ultrafilter and it happens to have {even numbers} in it. Since there is no *construction* of a nonprincipal ultrafilter, what one always actually does is to say "Let U be a nonprincipal ultrafilter" and then proceed using just the properties that any nonprincipal ultrafilter has.

There's also "the complement is countable" when you're talking about real numbers, or more generally "the complement is of strictly smaller cardinality than the whole set". (When put like that, of course it's the same as "cofinite" for integers. Speaking of which, your heading for that one has "not finite" but again you mean (and say later on) "cofinite".)

Thanks for the advice! Still learning how to phrase things correctly & effectively.

I wasn't aware that you can't actually explicitly construct a nonprincipal ultrafilter - this is interesting and nonintuitive to me!

One proves the existence of nonprincipal ultrafilters with the axiom of choice. (You can show "every filter is contained in some ultrafilter" pretty directly using Zorn's lemma.) You don't actually need the full strength of AC -- I think "every infinite set has a nonprincipal ultrafilter" is weaker than countable choice, for instance -- but you definitely need *some* choice, and in particular "there are no nonprincipal ultrafilters" is known to be consistent with ZF.

## In English

A boring, colloquial way.

## Not-Finite-ness

A slightly-less-boring mathy way.

There are infinitely many primes. There is exactly one even prime number (2). Infinity minus one is... infinity.

There are infinitely many natural numbers. There are infinitely many natural numbers larger than one-hundred thousand quadrillion quadrillion vigintillion. No practical difference between 10^83 and 1 (other than that one is an upper bound on the number of atoms in the universe).

You formalize this with sets: if you have some (infinite) set, a subset whose complement has a finite cardinality encompasses almost everything in the set. We call this

cofiniteness.## Probability ~One

A way to quantify surety.

I'm not sure what the exact rates are, but I'd bet money on this being true.

It is well known that the number of prime numbers below Nis approximately N/lnN. In the limit as N goes to infinity, the ratio of primes to non-primes numbers goes to approximately zero. So if you choose a random positive integer, I would bet my life savings that it's composite.

An intuitive explanation: if you take all possible combinations of nodes and vertices and let the number of each tend towards infinity, you should expect chaos to triumph over order. (sorry Ramsey). This does depend on a certain definition of symmetry, however, and a clearer statement would be "

almost all graphs have only one automorphism."## The Lebesgue Measure is Not-Zero

A straightforward-yet-strange way for real numbers.

The Lebesgue measure is what you get when you try to generalize length, area, and volume to n-dimensions. It forms the basis for our current understanding of integration, and helps us figure out how big stuff is.

Something with zero volume basically doesn't exist anyway.

Well yes, all the rationals have measure zero. All countable sets have measure zero, and the rationals are countable.

(Measures are nice: they're a neat generalization of physical scales (mass, volume, etc.) to arbitrary mathematical objects. They're particularly useful for dealing with continuous things

~~we love Lebesgue measures~~)Well yes? Of course an arbitrary real number can't be computed to arbitrary precision by a finite algorithm? Computable numbers are

countable,remember? Nevermind thatbasically all the numbers we deal with are computable,this isobvious.(Measure zero stuff basically doesn't exist, even if they're the only things we use on a daily basis)

Well yes, of course! Even though the Cantor set is uncountably infinite, it still has measure zero! It's a weird pseudo-fractal embedding of the real line that somehow manages to lose everything in translation but still keep all the relevant information.

(Idk, the Cantor set is weird)

## It is Contained in a Nonprincipal Ultrafilter

A

filterF on an arbitrary set I is a collection of subsets of I that is closed under set intersections and supersets. (Note that this means that the smallest filter on I is I itself).An

ultrafilteris a filter which, for every A⊆I, contains either A or its complement. Aprincipalultrafilter contains a finite set.A

nonprincipal ultrafilterdoes not.This turns out to be an incredibly powerful mathematical tool, and can be used to generalize the concept of "almost all" to esoteric mathematical objects that might not have well-defined or intuitive properties.

(One of the coolest uses of nonprincipal ultrafilters is in the construction of the hyperreals, post forthcoming).

Let U be a nonprincipal ultrafilter over the natural numbers. It obviously contains no finite sets, but we run into a slight issue when we take the set

E={2,4,6,8,…}and its complement

O={1,3,5,7,…}.By the filter axioms, only one of these can be in U, and one of them

hasto be in U. And thus, we can safely say: