#ultrafilter

[ms2253]

[album] r.n.d. technologies - ultrafilter ep (1996)

[catgirl-lucy]

Ultrafilters & Ultrapowers

Hey! Call me Lucy. I might make an introduction blog later, but I first wanted to make a blog-post about ultrapowers.

Ultrafilters are a concept from set theory, I'll try my best to explain what they are and why they're defined as they are.

First, a quick overview of what we will do: we will extend the real number line by adding new numbers through the use of an ultrapower, these new numbers are called "hyperreals". Roughly, this means that we will have infinite sequences [a₀,a₁,a₂,...] of real numbers representing hyperreal numbers, where similar sequences are regarded as equal. We will also show a surprising theorem: although there are seemingly more hyperreals than reals, hyperreals look the same as real numbers "from within".

If we have a sequence of reals like this: [0,1,1/2,1/3,1/4,...] (I'll call this sequence "ε"), the hyperreal that it represents can be viewed as the "limit" of the sequence. Since a large number of entries of this sequence is smaller than any positive real number r > 0, ε will be smaller than any positive real number r, but since also a large number of entries is larger than 0, ε will be larger than 0. ε is thus an infinitesimal hyperreal number. This is mostly just intuition though, so don't worry if you don't entirely get it.

Two hyperreal numbers x = [x₀,x₁,x₂,...] and y = [y₀,y₁,y₂,...] are equal if x_i = y_i for a large number of indices i. But what does "large" mean in this context?

Well, that's where the ultrafilter comes in. Ultrafilters split a family of sets into sets that are "large" and sets that are "small". In this case, we split sets of natural numbers (numbers 0, 1, 2, 3, etc) into large sets and small sets, so we have an ultrafilter on ℕ, the set of natural numbers. Ultrafilters are identified by the family of large sets: if some set A is in an ultrafilter U, then it is large, and if it's not, then it is small.

We do want our notion of "large sets" and "small sets" to make sense: for example, a hyperreal should always be equal to itself, so we want the whole set of natural numbers, {0, 1, 2, 3, 4, ...} (which is the set of indices for which a sequence is equal to itself), to be large.

Obviously, it would make sense that if a set A is large and B is larger than A, then B is also large. Thus, if A ∈ U is a member of an ultrafilter U ("∈" is the membership symbol), and if B ⊃ A contains everything A contains too ("⊃" is the superset symbol), then B ∈ U is a member of the ultrafilter as well.

We also want hyperreal equality to be transitive, thus if [x₀,x₁,x₂,...] = [y₀,y₁,y₂,...] and [y₀,y₁,y₂,...] = [z₀,z₁,z₂,...], then we want [x₀,x₁,x₂,...] = [z₀,z₁,z₂,...]. If A = {i ∈ ℕ | x_i = y_i} is the set of points at which x and y are equal and B = {i ∈ ℕ | y_i = z_i} is the set of points at which y and z are equal, then C = {i ∈ ℕ | x_i = z_i}, the set of points at which x and z are equal, includes the set A ∩ B = {i ∈ ℕ | x_i = y_i ∧ y_i = z_i}, the set of points at which x is equal to y and y is equal to z. It thus makes sense to have our ultrafilter be closed under intersections: if two sets A and B are large, then the set of points that are both in A and in B, called the "intersection" of A and B (denoted A ∩ B), is a large set as well (and thus also in the ultrafilter).

It would also make sense that, if two hyperreal numbers are nowhere equal, then they aren't equal. So the empty set, {} = ∅, is small.

The five axioms above describe a filter:

A filter F on κ is a family of subsets of κ.

A filter F on κ must contain the whole set κ.

A filter F on κ must be upwards closed, thus for every large set A ∈ F, and every larger set B ⊃ A, B ∈ F is large as well.

A filter F on κ must be downwards directed, thus for every large set A ∈ F and every large set B ∈ F, the intersection of A and B, A ∩ B ∈ F, is large as well.

A filter F on κ may not contain the empty set.

However, these are the axioms of a filter, and not of an ultrafilter. Ultrafilters have one additional axiom.

Suppose we have the hyperreal [0,1,0,1,0,1,...]: an alternating sequence of 0's and 1's. Is this equal to 0 = [0,0,0,0,...], or to 1 = [1,1,1,1,...], or is it its own thing? (Note: the 0 in 0 = [0,0,0,0,...] is a hyperreal and the 0's in 0 = [0,0,0,0,...] are real numbers, so they're different (kind of) numbers both called "0"). If it is its own thing, then is it smaller than 1? If it is smaller than 1, then it must be smaller on a large set of indices, meaning it's equal to 0 on a large set of indices, meaning it's equal to 0. If it's not smaller than 1, well, it can't be larger, so it'd only make sense if it's equal to 1, but no axiom about filters says it should! That's why we have this last axiom for ultrafilters, which makes them "decisive": for every set A, it is either large (thus, A ∈ U), or small, meaning that its complement, Ac = {i | i ∉ A}, the set of all points that aren't in A, is large.

And so we have our six axioms of an ultrafilter:

An ultrafilter U on κ is a family of subsets of κ, these subsets are called "large sets".

κ is large.

U is upwards closed.

U is downwards directed.

∅ is not large.

For every set A ⊂ κ, either A ∈ U or Ac ∈ U.

But we're still missing one thing. We can take our ultrafilter U to be the set of all sets of natural numbers that contain 6. ℕ is large, as it contains 6. It is upwards closed: if A contains 6 and B contains everything that A contains and more, then B also contains 6. U is downwards directed: if both A and B contain 6, then the set of all points that are in both A and B still contains 6. The empty set does not contain 6, and every set either does contain 6 or does not contain 6. With this ultrafilter, two hyperreals x and y are equal simply when x₆ and y₆ are equal, so we don't get cool infinitesimals and infinities, and that makes me sad :(

These kinds of boring ultrafilters are called principal ultrafilters. Formally, a principal filter on κ is a filter F on κ for which there is some set X ⊂ κ so that any set A ⊂ κ is large only if it contains everything in X. This filter is often denoted as ↑X. If you want a principal filter U to be an ultrafilter, X needs to be a singleton set, meaning it only contains a single point x. Proving this is left as an exercise for the reader.

Let U be a non-principal ultrafilter on ℕ. This post is getting a bit long, so I won't show why such an ultrafilter exists. Now, we can take the ultrapower of ℝ, the set of real numbers, by U. This ultrapower is often denoted as ℝ^ℕ/U. Members of this ultrapower are (equivalence classes of) functions from ℕ to ℝ, meaning that they send natural numbers/indices to real numbers (the sequence [x₀,x₁,x₂,...] maps the natural number i to the real number x_i). These functions/sequences/equivalence classes are called hyperreal numbers. Two hyperreal numbers, x and y, are equal if {i ∈ ℕ | x(i) = y(i)}, the set of points at which they are equal, is large (i.e. a member of U). We can also define hyperreal comparison and arithmetic operations: x < y if {i | x(i) < y(i)} is large, (x + y)(i) = x(i) + y(i) and (x · y)(i) = x(i) · y(i). Every real number r also has a corresponding hyperreal j(r), which is simply [r,r,r,r,...] (i.e. j(r)(i) = r for all i).

In general, if M is some structure, κ is some set and U is some ultrafilter on κ, then we can take the ultrapower M^κ/U, which is the set of equivalence classes of functions from κ to M, where any relation R in M (for example, "<" in ℝ) is interpreted in M^κ/U as "R(x₁,...,xₙ) if and only if {i ∈ κ | R(x₁(i),...,xₙ(i))} ∈ U is large" and any function f in M (for example, addition in ℝ) is interpreted in M^κ/U as "f(x₁,...,xₙ)(i) = f(x₁(i),...,xₙ(i)) for all i ∈ κ".

A quick note on equivalence classes: in M^κ/U, points aren't actually functions from κ to M, but rather sets of functions from κ to M that are all equal on a large set of values. Given a function f: κ → M, the equivalence classes that f is in is denoted [f]. In this way, if f and g are equal on a large set of values, then [f] and [g] are actually just equal.

The hyperreal [0,1,2,3,4,...], which sends every natural number i to the real number i, is often called ω.

This part of the blog will get a bit more technical, so be warned!

In the beginning of this blog-post, I mentioned that hyperreals look the same as real numbers. I'll make this statement more formal:

For any formula φ that can be built up in the following way:

φ ≡ "x = y" for expressions x and y (expressions are variables and "a + b" and "a · b" for other expressions a and b)

φ ≡ "x < y" for expressions x and y

φ ≡ "ψ ∧ ξ" (ψ and ξ are both true) for formulas ψ and ξ

φ ≡ "ψ ∨ ξ" (ψ or ξ is true (or both)) for formulas ψ and ξ

φ ≡ "¬ψ" (ψ is not true) for an formula ψ

φ ≡ "∃x ψ(x)" (there exists a value for x for which ψ is true) for a variable x and an formula ψ

φ ≡ "∀x ψ(x)" (for all values of x, ψ is true) for a variable x and an formula ψ

We have that ℝ ⊧ φ (φ is true when evaluating equality, comparison and expressions from within ℝ, where variables can have real number values) if and only if ℝ^ℕ/U ⊧ φ (φ is true when evaluating equality, comparison and expressions from within ℝ^ℕ/U, where variables can have hyperreal number values).

In other words: ℝ and ℝ^ℕ/U are elementary equivalent.

So, how will we prove this? Well, we will use induction: "if something being true for all m < n implies it being true for n itself, then it must be true for all n (where m and n are natural numbers)". Specifically, we will use induction on the length of formulas: we will show that, if the above statement holds for all formulas ψ shorter than φ, then it must also hold for φ.

However, we won't use the exact statement above. Instead, we will use the following:

Given a formula φ(...) and hyperreal numbers x₁,...,xₖ, ℝ^ℕ/U ⊧ φ(x₁,...,xₖ) if and only if {i | ℝ ⊧ φ(x₁(i),...,xₖ(i))} is large.

Now, why does this imply the original statement? Well, when k = 0, {i | ℝ ⊧ φ} can only be ∅ or ℕ. It being ∅ is equivalent to φ being false in ℝ and, if the statement is true, also equivalent to φ being false in ℝ^ℕ/U. And it being ℕ is equivalent to φ being true in ℝ and, again, if the statement is true, it is also equivalent to φ being true in ℝ^ℕ/U. We thus have that φ being true in ℝ is equivalent to φ being true in ℝ^ℕ/U.

Note: M ⊧ φ simply means that the formula φ is true when interpreted in M.

Now, why do we need this stronger statement? Well, it makes induction a lot easier: given that this statement holds for all ψ shorter than φ, it's easier to prove it also holds for φ.

Now, we can actually do the induction.

First, if φ ≡ "x = y", then we need to show that (1) ℝ^ℕ/U ⊧ φ(x,y) iff (2) {i | ℝ ⊧ φ(x(i),y(i))} is large. This follows immediately from the definition of equality in ℝ^ℕ/U, the same holds for "<".

Now, if φ(x₁,...,xₖ) ≡ "ψ(x₁,...,xₖ) ∧ ξ(x₁,...,xₖ)", we have that {i | ℝ ⊧ φ(x₁(i),...,xₖ(i))} = {i | ℝ ⊧ ψ(x₁(i),...,xₖ(i)) ∧ ℝ ⊧ ξ(x₁(i),...,xₖ(i))} = {i | ℝ ⊧ ψ(x₁(i),...,xₖ(i))} ∩ {i | ℝ ⊧ ξ(x₁(i),...,xₖ(i))}. Since {i | ℝ ⊧ ψ(x₁(i),...,xₖ(i))} is large iff ψ(x₁,...,xₖ) is true in ℝ^ℕ/U, and {i | ℝ ⊧ ξ(x₁(i),...,xₖ(i))} iff ξ(x₁,...,xₖ) is true in ℝ^ℕ/U, and U is closed under intersections, we have that {i | ℝ ⊧ φ(x₁(i),...,xₖ(i))} is large iff φ holds in ℝ^ℕ/U. A similar argument works for ∨.

If φ(x₁,...,xₖ) ≡ "¬ψ(x₁,...,xₖ)", then we can just use the ultraness of the ultrafilter.

If φ ≡ "∃y ψ(y,x₁,...,xₖ)", then {i | ℝ ⊧ φ(x₁(i),...,xₖ(i))} = {i | ℝ ⊧ ∃y ψ(y,x₁(i),...,xₖ(i))} = {i | ∃y ∈ ℝ. ℝ ⊧ ψ(y,x₁(i),...,xₖ(i))} = ∪_{y ∈ ℝ} {i | ℝ ⊧ ψ(y,x₁(i),...,xₖ(i))}. We have that the set {i | ℝ ⊧ ψ(y,x₁(i),...,xₖ(i))} for y ∈ ℝ is large iff ℝ^ℕ/U ⊧ ψ(j(y),x₁,...,xₖ). If this set is large for some y ∈ ℝ, and thus if ℝ^ℕ/U ⊧ φ(x₁,...,xₖ), then ∪_{y ∈ ℝ} {i | ℝ ⊧ ψ(y,x₁(i),...,xₖ(i))} is larger than that set, so it is large as well. For the converse direction, if ∪_{y ∈ ℝ} {i | ℝ ⊧ ψ(y,x₁(i),...,xₖ(i))} is large, then we can create a hyperreal z where ψ ⊧ ψ(z(i),x₁(i),...,xₖ(i)) for all i for which ℝ ⊧ ∃y ψ(y,x₁(i),...,xₖ(i)), and we have ℝ^ℕ/U ⊧ ψ(z,x₁(i),...,xₖ(i)), and thus ℝ^ℕ/U ⊧ φ(x₁(i),...,xₖ(i)). Again, a similar argument works for ∀.

(Sorry if you couldn't follow along, I'm not good at explaining these things in an intuitive way.)

This result can be extended to show that M^κ/U is elementary equivalent to M for every structure M, every set κ and every ultrafilter U on κ.

Now, this result might be surprising, as we have a new number ω in ℝ^ℕ/U. Surely, there is a formula that states the existence of this number, right?

Well, it turns out, such a formula does not exist! You can try something like "there is no natural number n so that 1+...+1 w/ n 1's is greater than ω", but ω+1 is a natural number in the hyperreals, so such a natural number does exist. Similarly, any formula you can come up with, as long as it is created using the rules above (using conjunction, disjunction, negation, qauntification, etc), cannot state the existence of an infinite number ω.

But if ℝ^ℕ/U and ℝ are seemingly indistinguishable, might there already be an undetectable infinite real number in ℝ? Well, maybe~ :3 But it's undetectable anyways, so you don't have to worry about it.

Before I end this blog-post, I want to give some more intuition on what filters & ultrafilters actually are. To me, ultrafilters, and filters in general, are like "limits of sets". The principal filter ↑X has X as limit, while non-principal filters and ultrafilters have limits that aren't really sets, but look like ones. For example, you might have the set of prime numbers in your filter, and then the limit of that filter will be a "set" in which all numbers are prime numbers. And if your ultrafilter is non-principal (so for every n, there is a set A ∈ U in the filter that does not contain n), then the limit of that ultrafilter will be a "set" in which all numbers don't actually exist. In the case of filters, this "set" can be any "set" (though it still isn't really a set), but in the case of ultrafilters, this limit looks like a singleton set (i.e. it only has one "element": ω).

I don't know if my intuition of filters and ultrafilters will help anyone, tho, but I think it's cool!

That's all I had to say.

Bye!~ Have a nice day.

[bauliya]

hey what the FUCK

[brightwaterfilters]

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[flameshadowconjuring]

I really like ultrafilters and other choice principles. Did you know that the ultrafilter lemma is equivalent to Tychonov's theorem when restricted to Hausdorff spaces? Also, the absolute coolest use of ultrafilters is the construction of really weird first-order structures, like the hyperreals. Though sometimes not so weird. I did once want to construct an algebraically closed field from an ultraproduct of the F_p spaces and ended up constructing a space isomorphic to C.

rb this with your favorite math concepts/books/videos... things u enjoy and that make you excited! (or reply but i want to hear about it and if you rb it then i hear more cool stuff from more people)

my favorite books are the grapes of math and things to make and do in the fourth dimension. i'm also reallyyyy wanting to read number freak and godel, escher, bach. concepts i love are chaos theory, non-euclidean geometry, and dimensions beyond 3rd!

[ceausescue]

taking an ultrafilter over the oppression poset and using it to construct non-standard dnis

[imperialtommy]

I was reading the wikipedia article on measurable cardinals (https://en.wikipedia.org/wiki/Measurable_cardinal) just now, and the following things made me crack up:

large cardinal

singletons

ultrafilter

Scott's trick

"...it follows from ZFC that every measurable cardinal is inaccessible (and is ineffable, Ramsey, etc.)

huge cardinal

"...real valued measurable cardinals are weakly inaccessible (they are in fact weakly Mahlo)

set theory is deeply unserious

[theglowsociety]

PIÑA COLADA RICE PUDDING 🍍

Ingredients

1 L (4 cups) Joyya by Neilson 2% Ultrafiltered Milk

125 ml (½ cup) Arborio rice or any other short grain rice

75 ml (⅓ cup) sugar

10 ml (2 tsp) cornstarch

180 ml (¾ cup) coconut milk

5 ml (1 tsp) vanilla extract

250 ml (1 cup) fresh pineapple, sliced

Directions

In a large pot, gently bring the Joyya ultrafiltered milk and rice to a boil.

Cover and simmer over low heat for approximately 30 minutes, stirring occasionally, until tender.

Meanwhile, mix the sugar and cornstarch in a bowl. Add the coconut milk and mix well with a whisk.

Pour into the cooked rice mixture and bring to a boil while continuing to stir.

Simmer for 1 minute. Add the vanilla and pineapple.

Transfer mixture to a bowl and cover with plastic wrap directly on the surface of the rice pudding.

Let cool, then refrigerate until completely cooled.

Stir gently before serving.

[noneofthisisreal]

What is Fromager D’Affinois?

Fromager d’Affinois is a soft white mould cheese from the Auvergne region of France. It is produced on a vast commercial scale and is often mistakenly referred to as Brie. As a matter of fact, D’Affinois has very little in common with Brie or Camembert other than the white mould around it.

Indeed, D’Affinois is a double cream soft cheese that is made by adding cream to ultrafiltered milk during production. As a result, the cheese is very high in fat (around 60%) and develops a rich and creamy texture.

[flameshadowconjuring]

I am really tired of all the propaganda which says that the axiom of choice is this necessary evil, because it is really useful in proofs but the choice functions are not constructive. Because yes, the axiom itself gives you no scheme for constructing an explicit choice function. That much is true. But I would argue that the axiom's negation is actually way more non-constructive!

The axiom of constructibility (also known as V=L) implies the axiom of choice, so the negation of the axiom of choice implies the existence of sets which are not ordinal definable, and thus cannot be constructed by exclusively appealing to the axioms of ZF (which, with the exception of extensionality and foundation, are all axioms which exclusively describe closure properties of the set-theoretic universe).

In fact, the proof of V=L implying choice is constructive! If V=L holds, we can explicitly define a well-ordering of the entire set-theoretic universe! Not only that, but the existence of a transitive model of ZF implies a transitive model of ZF+V=L which is point-definable, i.e. every set in that model is uniquely characterized by a first-order formula of set theory. Every set being definable strikes me as very constructive.

Sure, things like Banach-Tarski are pretty weird results that require the axiom of choice or a similar axiom. But the only real issue here is that we ascribe our geometric intuition to a partition of a sphere that is impossible to do in real life! Not only can we not divide real-life objects infinitely often, but even if we could, we physically could not select points the way that Banach-Tarski requires.

I think it is interesting to think about set theories where choice does not hold or may not hold. Certainly, a set theory where all subsets of R are measurable is appealing, if only for convenience. And it is legitimately fun to prove equivalences between the axiom of choice and various other propositions, such as the total ordering of the cardinals or Tychonov's theorem, or to reason about the relative strengths of different weaker variants like dependent choice, or my personal favorite, the ultrafilter lemma. It gives you more insight into the beauty of choice. But, in as much as you can 'believe' in an axiom, I believe that the axiom of choice is true in the platonic ideal of mathematics, and I will not stand for slander against it.

[shoggothgf]

you too can add ultrafiltered milk and stevia to instant coffee for something that's unpleasantly off tasting in 3 different dimensions but otherwise has a strong resemblance to coffee

[brightwaterfilters]

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[commercialroplants]

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