Ultraproducts and Compactness (Ultra Series 6)

Previous: Infinitely Large Primes

We’ve seen how to use Łoś’s theorem to prove properties of hyperstructures. We’ve also seen how to use it to prove limitations of first-order logic. Today we’re going to see a major meta-logical application of Łoś’s theorem: proving the compactness theorem! In my opinion, this proof is the most elegant proof of first-order compactness out there. We’ll also see a few other meta-logical results relating to ultraproducts. However, if you’re just interested in learning about more properties of the hypernaturals, then you can skip this post without missing out on anything major!

What is compactness?

A logic is compact if any finitely satisfiable set of sentences is satisfiable. If somebody hands you an infinite set of sentences, and you know how to construct a model to satisfy any finite subset of these sentences, then you can conclude that there must be a model that satisfies all those sentences at once. Compactness is interesting because it allows you to prove surprising results (like that true arithmetic has nonstandard models) from very obvious facts (that any finite set of natural numbers has an upper bound). It also lets you prove the completeness theorem, which says that what you can prove from first-order axioms is exactly what logically follows from those axioms.

Countable Theories

First we’ll prove the compactness theorem for countable theories. I’m including it because it’s conceptually simpler than the proof of the general case. However, it’s not necessary for understanding the general proof, so if you just want to see that you can jump ahead to the next section.

Suppose T is a countably infinite theory that’s finitely satisfiable.
T is countable, so choose an enumeration of T’s sentences: T = {φ1, φ2, φ3, …}.

T is finitely satisfiable, so there’s a model that satisfies φ1. Call this model M1.
There’s also a model M2 that satisfies both φ1 and φ2.
And a model M3 that satisfies φ1, φ2, and φ3.
And so on.

Take the ultraproduct of (M1, M2, M3, …) and call it M.
For every n in N, φn is satisfied by all models from Mk on. (For instance, φ4 is satisfied by M4, M5, M6, and so on.)
This is a cofinite set, so (by Los’s theorem) φn is satisfied by M.

But φn was a totally arbitrary sentence inside T. So M satisfies all of T!

So T is satisfiable. But not only do we get this result, we also have an explicit example of a model that satisfies all of T, which we can start analyzing to prove other properties of.

Compactness for any theory

The proof uses a fact that I haven’t proven yet: If X is a set whose every finite subset has nonempty intersection, then there’s some ultrafilter that extends it. Luckily, it’s easily proven. Start by closing X under supersets (notice that every finite subset of X still has nonempty intersection). Then close it under intersection, and you now have a filter that extends X. (Verify that this really is a filter for bonus points!) And in Part 3 of this series we proved that every filter has an ultrafilter extension. So there’s an ultrafilter U that contains X.

Alright, on to the proof.

Let L be any first-order language and S be any set of L-sentences that is finitely satisfiable. We’re going to build an ultraproduct that satisfies S. This requires three ingredients: an index set I, and a set of L-structures (Mi)i∈I indexed by I, and an ultrafilter U on I. We’ll go through these one at a time.

1. The index set I

I will be the set of all finite subsets of S. This is interestingly different from what we’ve seen before. Previously our indices were always natural numbers. But here our indices are sets of sentences in the language of the structure! This means that we’re going to end up looking at models that look like M{φ,ψ,θ}, where φ,ψ, and θ are sentences in S.

2. The component structures (Mi)i∈I

So, our indices are finite subsets of S. And S is finitely satisfiable. So for each index F (each finite subset of S), there’s a model MF that satisfies everything in F. This is how we get our component structures.

For instance, if φ, ψ, and θ are three sentences in S, then {φ,ψ,θ} is an index, and M{φ,ψ,θ} is any structure that satisfies φ∧ψ∧θ. ∅ is also a finite subset of S, so it’s an index and has a corresponding structure M, which can be any L-structure whatsoever.

To briefly rehash, the important feature of our choice of component structures is that they have the following property: For any finite subset F of S, MF ⊨ φ for every φ in F.

3. The ultrafilter U on I

The choice of ultrafilter here is really important. Our component structures were carefully chosen so that each structure satisfies the finite set of sentences that is its index. This guarantees us that for any sentence φ in S, the set of component structures whose indices contain φ all satisfy φ. This is the set of component structures whose indices are finite supersets of {φ}.

Remember that we’re ultimately aiming to show that each φ is satisfied by the ultraproduct model. By Łoś’s theorem, it suffices to show that the set of component models that satisfy φ is U-large (is in our chosen ultrafilter). Conclusion: we want to make sure that for any φ in S, the set of all finite supersets of {φ} is U-large.

How do we know that there exists any ultrafilter that for every φ contains the set of all finite supersets of {φ}? We apply the finite intersection property! Denote by SF the set of all finite supersets of F. Take any finite set of singletons {{φ1}, {φ2}, {φ3}, …, {φn}}. Is the intersection of S{φ1}, S{φ2}, S{φ3}, …, S{φn} empty? No: the set 1, φ2, φ3, …, φn} is in of all of them! So the set { S{φ} | φ ∈ S } has the finite intersection property, and thus there’s some ultrafilter that extends it. And that’s the ultrafilter we want!

Finalizing things

We’ve now got everything we need to build the structure that satisfies S. Take the ultraproduct Π(MF)F∈I/U. This structure satisfies S. Why?

Consider any φ in S. By Łoś’s theorem, φ is satisfied if a U-large set of the component models satisfy φ. And we know that a U-large set of component models satisfy φ, because φ is satisfied in every model whose index contains φ, i.e. the set of models whose indices are supersets of {φ}. So we’re done!

More implications

The remainder of this post will not be necessary to understand the rest of the series. I want to present a few miscellaneous meta-logical results that relate to ultraproducts that don’t quite fit anywhere else.

First: every finite structure is its own ultrapower. Let’s see why.

Suppose M is a finite structure. Denote by MI/U the ultrapower of M over any arbitrary index set I. First of all, we can easily see that M and MI/U have the same cardinality: if |M| = n for some finite n, then there’s a sentence affirmed by M that says “There are exactly n distinct things”. For instance, if M has exactly two elements, then M ⊨ ∃x ∃y (x≠y ∧ ∀z (z=x ∨ z=y)). Łoś tells us that MI/U also affirms this sentence, so MI/U also has exactly n distinct things.

Furthermore, any finite structure M of size n can be categorically specified up to isomorphism by a single first-order sentence of the form ∃x1∃x2…∃xn (∀x (x = x1 ∨ … ∨ x = xn) ∧ (x1 ≠ x2 ∧ … xn-1 ≠ xn) ∧ ψ(x1, …, xn) ], where ψ specifies the behavior of all relation, function, and constant symbols. Here’s an example for a simple finite structure:

We have four elements, so we use four variables, x1 x2 x3 and x4 to specify the structure.

Since M satisfies this sentence, Łoś tells us that MI/U also affirms this sentence, meaning that MI/U and M are isomorphic. So in other words, every finite structure is its own ultrapower (up to isomorphism). 

What about infinite structures? The situation is the exact opposite! No infinite structure is its own ultrapower.

Let M be an infinite structure and *M be its ultrapower. Add a constant for every object in M. The ultraproduct construction tells you how constants from M translate to constants in *M: for each constant c, if c refers to m in M, then c refers to [m, m, m, …] in *M.

Now we want to construct a new element of *M by writing a sequence that contains infinitely many elements of M, each appearing only finitely many times. Crucially, here we must assume that the index set is the same size as or smaller than M! If we have too many indices, then we aren’t going to be able to refer to each element in our original model only a finite number of times. So for instance, this argument won’t work for the ultrapower ℕ/U.

Now, this new element we’ve constructed is distinct from each constant, because they only agree in finitely many place. The constants covered all of M, but they don’t cover all of *M. So M and *M are not isomorphic!

And some more

These next results relate ultraproducts to compactness and definability.

Fix a first-order language L and consider a class C of L-structures. Say that a set of sentences X is C-satisfiable if there’s a structure in C that satisfies all of X. We say that C is compact if for every set A of L-formulas, finite C-satisfiability implies C-satisfiability. This is a different notion from the compactness discussed in the compactness theorem; the ordinary compactness theorem tells us that for every first-order language L, the class of all L-structures is compact. But it doesn’t tell us anything about other more restricted classes of L-structures.

For instance, let L be the language with an empty signature (no constants, functions or relations). This language is extremely simple: essentially the only contentful sentences you can make are about finite cardinalities. E.g. you can say “there are at least three things” with “∃x ∃y ∃z (x≠y ∧ x≠z ∧ y≠z)” and “there are at most three things” with “∀x ∀y ∀z ∀w (x=y ∨ x=z ∨ x=w ∨ y=z ∨ y=w ∨ z=w)”.

Now let C be the class of all finite L-structures. Is this class compact?

If you said no, good job! For a finite n, let φn be the sentence that says “there are at least n things”, and consider the set of sentences A = {φ1, φ2, φ3, …}. Every finite subset of A is satisfiable by a finite L-structure, but the entire set is clearly not satisfiable by any finite L-structure.

What if we add to C a single countably infinite L-structure M? Then our previous construction no longer works; the entire set A is now satisfied by this new member of C.

Alright, so that’s compactness for classes of structures. The relationship between this and ultraproducts is the following:

Any class of models that’s closed under ultraproducts is also compact.

For any language L and any finite n, consider the class of all L-structures of size n. The ultraproduct of a bunch of size-n L-structures is itself a size-n L-structure, so this class is closed under ultraproducts. And therefore it’s compact!

Now consider our example of the class of all finite structures in the language with empty signature. This was not compact, meaning that it cannot be closed under ultraproducts. This means that the ultraproduct of a bunch of finite structures is not necessarily finite! In particular, suppose that M1 has one element, M2 has two elements, M3 has three elements, and so on. Then the ultraproduct of (M1, M2, M3, …) will be an infinite model. We can also see this by applying Łoś’s theorem to the set of sentences {“There is at least 1 thing.”, “There are at least 2 things.”, “There are at least 3 things.”, …}.

A class C of L-structures is first-order definable if it is the set of all models of some set of L-sentences. First-order definability also relates to ultraproducts, in the following way:

If a class of L-structures is first-order definable, then it’s closed under ultraproducts.

The converse of this is that any class of structures that’s not closed under ultraproducts is not definable.

For any L, the class of finite L-structures is not closed under ultraproducts, so it’s not definable. This tells us that “finiteness” is not first-order definable.

The class of all models of PA is first-order definable, so it’s closed under ultraproducts. This means that even if you take an ultraproduct of infinitely many nonstandard models of PA, you still end up with a model of PA!

Okay, next up we come to the most exciting and powerful property of ultraproducts: countable saturation!

Next: All About Countable Saturation

Infinitely Large Primes (Ultra Series 5)

Previous: Ultraproducts and Łoś’s Theorem

Let’s quickly review what we’ve learned so far.

The hypernaturals, *ℕ, are the ultrapower of ℕ with index set I = ℕ. The first-order language in use here is the language of Peano arithmetic: L = <{0}, {<}, {S,+,⋅}>. Łoś’s theorem from the last post tells us that *ℕ and ℕ are elementarily equivalent. In other words:

For every first-order sentence φ, ℕ ⊨ φ if and only if *ℕ ⊨ φ.

This tells us that *ℕ is not just a model of Peano arithmetic, which is a popular incomplete theory of arithmetic. It’s a model of TRUE arithmetic, the complete set of all first-order truths of arithmetic! Given the insane uncomputability of this theory (If T0 is a Turing machine, and Tn+1 is a Turing machine with an oracle for Tn‘s halting problem, then true arithmetic is not computable by Tn for any finite n), it’s pretty remarkable that we have an explicit example of one of its nonstandard models!

Review over! Now let’s begin with some basic applications of Łoś’s theorem. We can say that a number x is prime in first-order logic: isPrime(x) can be translated as ∀y ∀z (y⋅z = x → (y = 1 ∨ z = 1)). We can also refer to any finite natural number by applying the successor function enough times to zero (e.g. 2 = S(S(0))). Since “2 is a prime” is first-order expressible and ℕ affirms its truth, so must *ℕ.

But remember that in *ℕ, 2 is the equivalence class of the sequence (2,2,2,2,…). Is it really true that for any a, b ∈ *ℕ, if a⋅b = [2,2,2,2,…] then one of a and b is 2 and the other is 1? This is probably not immediately obvious… What about (2,1,2,1,2,1,…)⋅(1,2,1,2,1,2,…)? Certainly this product is (2,2,2,2,…), but neither (2,1,2,1,2,1,…) nor (1,2,1,2,1,2,…) look very much like (1,1,1,1,…) or (2,2,2,2,…).

In fact they are the same. Let’s prove this very generally. Suppose p is any finite prime number, and consider the hypernatural [p, p, p, …], corresponding to the ordinary natural number p. Consider any two hypernaturals a and b such that a⋅b = p. We can choose representative sequences (a0, a1, a2, …) and (b0, b1, b2, …) such that their product is (p, p, p, …). In other words:

(a0⋅b0, a1⋅b1, a2⋅b2, …) = (p, p, p, …)

Since p is prime, we know that for each k, either (ak = p and bk = 1) or (ak = 1 and bk = p). Consider the set of indices A = { k ∈ ℕ | ak = p }. Since each ak is either 1 or p, the complement ℕ\A is equal to { k ∈ ℕ | ak = 1 }. Now, since U is an ultrafilter, either A or ℕ\A is in U. In the first case, [a0, a1, a2, …] = [p, p, p, …] = p. And in the second case, [a0, a1, a2, …] = [1, 1, 1, …] = 1. The same argument applies for [b0, b1, b2, …]. So for any two hypernaturals a and b such that a⋅b = p, both a and b are either 1 or p.

Okay, so we’ve proven that 2 is a prime number. That’s a relief! Let’s do something a bit more substantial now.

Consider the following first-order sentence: ∀x ∃y (y > x ∧ isPrime(y)). This sentence says that there are infinitely many primes, and (as first proven by Euclid) ℕ affirms its truth. So *ℕ ⊨ ∀x ∃y (y > x ∧ isPrime(y)).

Okay, so there are infinitely many hypernatural primes. This doesn’t sound so exciting… after all every ordinary prime is a hypernatural prime too, so didn’t we already know this? Sure, but if we look closely at what exactly the sentence says, it turns out that we get a bit more out of it than just that.

∀x ∃y (y > x ∧ isPrime(y)) says that for every number x, there’s a larger prime number y. Applying this in *ℕ, this means that all hypernatural numbers, even the infinitely large ones, have larger primes. Thus there are nonstandard primes in *ℕ, and in fact infinitely many nonstandard primes.

See if you can think of one for yourself before reading on.




Ok, here’s one: P = [2, 3, 5, 7, 11, 13, 17, …]. This sequence is eventually always larger than the sequence of any standard prime p = [p, p, p, p, …], so P is a nonstandard number. And P is also prime! The argument is very similar to the previous argument: take any two hypernaturals a and b such that a⋅b = k. Consider any representative sequences for a and b whose product is the sequence (2, 3, 5, 7, 11, …), and then consider the set of indices at which 1 appears in a’s sequence. If this set is in the ultrafilter, then a = 1 and b = k. If not, then a = k and b = 1.

Now, a challenge for you: construct an infinite ascending sequence of nonstandard primes starting at P.

And another challenge: construct an infinite descending sequence of nonstandard primes starting at P!

Hint for the first challenge: think about the hypernatural [3, 5, 7, 11, 13, 17, …].

Hint for the second challenge: think about the hypernatural [2, 2, 3, 5, 7, 11, …], or [2, 2, 3, 3, 5, 5, 7, 7, …].

While infinite ascending sequences of numbers can exist in ℕ, an infinite descending sequence of numbers can not, so we’ve identified a concrete difference between *ℕ and ℕ. And by Łoś’s theorem, this difference must not be first-order expressible in the language of PA. In other words, “there are no infinitely descending sequences of numbers” cannot be expressed in first-order.

This illustrates another usage of Łoś’s theorem: rather than starting with a first-order sentence that’s true of ℕ and using it to learn about *ℕ, we can start with some facts about *ℕ to learn about limitations of first-order logic!

Alright, what else? Suppose we have a hypernatural number k that has a sequence (k0, k1, k2, …) such that some formula φ(x) holds of cofinitely many elements of the sequence. Then φ(x) also holds of k. This gives us some hypernatural numbers with awfully strange properties.

Consider the hypernatural E = [1, 2, 4, 8, 16, 32, 64, …]. Now consider the formulas that say “x is divisible by 2”, “x is divisible by 4”, “x is divisible by 8”, “x is divisible by 16”, and so on. Each of these formulas holds true of all but some initial segment of the sequence (1, 2, 4, 8, 16, 32, 64, …). This means that each of these sentences hold true of E. In other words, E is divisible by 2, by 4, by 8, by 16, by 32, and so on forever. In other words, E can be divided by 2 forever!

But how exactly does this work? The sequence (1, 2, 4, 8, 16, 32, 64, …) starts with a 1. So how could we divide it by 2? The key is to remember that hypernaturals are equivalence classes of sequences, not sequences. So any time we run into trouble with a particular sequence, we can jump to an equivalent sequence that solves our problem. To divide (1, 2, 4, 8, 16, 32, 64, …) by 2, we jump to the equivalent sequence (2, 2, 4, 8, 16, 32, 64, …), which is still a representative of E. Now we can divide by 2: E/2 = [1, 1, 2, 4, 8, 16, 32, …]. To divide by 2 again, we jump to the sequence (2, 2, 2, 4, 8, 16, 32, …). Then we get E/4 = [1, 1, 1, 2, 4, 8, 16, …]. We can keep doing this as many times as we like, and thus E is infinitely even! (Notice: another infinitely descending sequence of numbers.)

We can use a similar trick to construct a number that’s divisible by every standard natural number: simply consider M = [0!, 1!, 2!, 3!, 4!, 5!, …]. For any standard natural number n, the set of indices at which the sequence (0!, 1!, 2!, 3!, 4!, 5!, …) can be divided by n is cofinite. So M can be divided by n as well!

This lets us run a kind of fun argument. Remember how Euclid’s original proof of the infinitude of primes went? Imagine that there are finitely many primes. Then we can take their product and add 1, and we’ve constructed a new prime. Well, in *ℕ we can find multiples of infinite sets of numbers! One multiple of all standard prime numbers is [2, 2⋅3, 2⋅3⋅5, 2⋅3⋅5⋅7, 2⋅3⋅5⋅7⋅11, …] = [2, 6, 30, 210, 2310, …]. This is divisible by each standard prime, and is also not twice-divisible by any prime (e.g. you can’t divide it by 4). (Interestingly, it’s not the product of all finite primes; because it’s also divisible by some infinite primes! See if you can think of one such infinite factor.) Now add 1 to get X = [3, 7, 31, 211, 2311, …]. X cannot be divisible by any standard prime, meaning that it’s either divisible by a nonstandard prime, or is itself a nonstandard prime. So just as Euclid’s original proof showed that there can’t be only finitely many primes, this proves that there can’t be only finitely large primes.

Alright, let’s get a little more wild. In ℕ we know that for any number x, there’s a number y that’s divisible by all positive numbers up to and including x (for instance, y = x!). In other words, ℕ ⊨ ∀x ∃y ∀z ((z ≠ 0 ∧ z ≤ x) → (y div by z)), where “y div by z” is shorthand for ∃u (z⋅u = y). In other words, ℕ affirms that for every initial segment S with maximum element x, there’s a number that’s divisible by everything in S (for shorthand we can call this number “a multiple of S”).

Invoke Łoś: anything true in ℕ must also be true in *ℕ. This means that every initial segment of hypernaturals with a maximum element has a multiple! (See if you can do better: can you also prove that every initial segment of hypernaturals with a max has a product? What about initial segments without maximum elements?) For instance, there’s a number that’s divisible by every hypernatural number up to and including the hypernatural K = [0,1,2,3,4,5,…].

What other first-order properties does ℕ have? Here’s one: ℕ is not dense. We can express this by saying ¬(∀x ∀z (x < z → ∃y (x < y ∧ y < z)). We can do better: ℕ affirms that ∀x ¬∃y (x < y ∧ y < S(x)). In other words, for any number x, x and its successor are counterexamples to density (they’re distinct numbers that have nothing between them).

So this must be true of hypernaturals as well! For any hypernatural, say K = [0,1,2,3,4,5,…], there’s no hypernatural between it and its successor K+1 = [1,2,3,4,5,6,…].

However, even though the hypernaturals are not dense, they do contain a dense subset! In fact, we can say something even stronger: no segment of the hypernaturals are dense, but nonetheless *ℕ has a dense subset!

Here’s a rough sketch for how to construct this dense subset. Take any nonstandard hypernatural, say K = [0, 1, 2, 3, 4, 5, …]. For any nonzero finite m, we can construct a hypernatural Km a finite distance from K, such that K’ is divisible by m. Now we can construct n⋅Km, and then (n/m)⋅Km. The set { (n/m)⋅Km | n, m ∈ ℕ } is a subset of the hypernaturals, and it has the order type of the positive rationals!

What other unusual hypernatural numbers can you discover?

Next, we will prove the compactness theorem using ultraproducts. It’s definitely the prettiest proof of compactness out there, so stay tuned!

Ultraproducts and Compactness

Ultraproducts and Łoś’s Theorem (Ultra Series 4)

Previous: Hypernaturals in all their glory

First things first, you’re probably asking yourself… how is Łoś pronounced?? I’m not the most knowledgeable when it comes to Polish pronunciation, but from what I’ve seen the Ł is like a “w”, the o is like the vowel in “thought”, and the ś is like “sh”. So it’s something like “wash”. (I think.)

Ok, on to the math! This is probably going to be the hardest post in this series, so I encourage you to read through it slowly and not give up if you start getting lost. Try to work out some examples for yourself; that’s often the best way to get a grasp on an abstract concept. (In general, my biggest tip for somebody starting to dive into serious mathematical content is to not read it like fiction! In fiction, short sentences can safely be read quickly. But math is a language in which complex ideas can be expressed very compactly. So fight your natural urge to rush through short technical sentences, and don’t feel bad about taking your time in parsing them!)

What the heck is an ultraproduct

Last post we defined an ultrafilter. Now let’s define an ultraproduct.

It turns out we’ve already seen an example of an ultraproduct: the hypernaturals. The hyperreals (denoted *ℝ) are another famous example of an ultraproduct, constructed from ℝ in exactly the same way as we obtained the hypernaturals *ℕ from ℕ. For any structure M whatsoever, there exists a “hyperstructure” *M obtained from M via an ultraproduct. So what exactly is an ultraproduct?

We’ll build up to it in a series of six steps.

(1) Choose an index set I.
(2) Choose a first-order language L and a family of L-structures indexed by I (Mi)i∈I.
(3) Define sequences of elements of these structures.
(4) Define an equivalence relation between these sequences.
(5) Construct the set of equivalence classes under this equivalence relation.
(6) Define the interpretations of the symbols of L in this set.

Let’s get started!

(1) First we select an index set I. This will be the set of indices we use when constructing our sequences of elements. If we want our sequences to be countably infinite, then we choose I = ℕ.

(2) Now, fix some first-order language L = <constants, relations, functions>. Consider any family of L-structures, indexed by I: (Mi)i∈I. For instance, if I = ℕ and L is the language of group theory <{e}, ∅, {⋅}>, then our family of L-structures might be (ℤ1, ℤ2, ℤ3, ℤ4, …), where ℤk is the integers-mod-k (i.e. the cyclic group of size k).

(3) Now we consider sequences of the elements of these structures. For instance, hyperreals are built from sequences of real numbers that look like (a0, a1, a2, a3, …). The set of indices used here is {0, 1, 2, 3, …}, i.e. ℕ.

If our index set isn’t countable, then it’s not possible anymore to visualize sequences like this. For instance, if I = ℝ, then our sequence will have an element for every real number. A more general way to formulate sequences is as maps from the index set I to the elements of the component models. The hyperreal sequence (a0, a1, a2, a3, …) can be thought of as a map a: ℕ → ℝ, where a(n) = an for each n ∈ ℕ.

In general, a sequence will be defined as follows:

f: I → U(Mi)i∈I such that f(i) ∈ Mi for each i ∈ I.

In the ℤk example, a sequence might look like (0, 1, 2, 3, 4, …). Note that 0 ∈ ℤ1, 1 ∈ ℤ2, 2 ∈ ℤ3, and so on. On the other hand (2, 1, 2, 3, 4, …) would not be a valid sequence, because there’s no 2 in ℤ1.

Now, consider the set of all sequences: { f: I → U(Mi)i∈I | f(i) ∈ Mi for each i ∈ I }. This set is called the direct product of (Mi)i∈I and is denoted Π(Mi)i∈I.

(4) We want to construct an equivalence relation on this set. We do so by first defining a free ultrafilter U on I. From the previous post, we know that every infinite set has a free ultrafilter on it, so as long as our index set is infinite, then we’re good to go.

With U in hand, we define the equivalence relation on Π(Mi)i∈I:

Let f and g be sequences (f,g: I → U(Mi)i∈I).
Then f ~ g if and only if { i ∈ I | f(i) = g(i) } ∈ U

You might be getting major deja-vu from the last post. Two sequences are said to be equivalent if the set of places-of-agreement is a member of the chosen ultrafilter. Since every free ultrafilter contains all cofinite sets, any two sequences that agree in all but finitely many places will be equivalent.

(5) Now all the pieces are in place. The ultraproduct of (Mi)i∈I with respect to U is the set of equivalence classes of Π(Mi)i∈I with respect to ~. This is typically written Π(Mi)i∈I/U. For a sequence a: I → U(Mi)i∈I we’ll denote the equivalence class it belongs to as [a].

Now, it’s not necessary for our indexed sequence (Mi)i∈I of L-structures to all be distinct. In fact, all the models can be the same, in which case we have (Mi)i∈I = (M)i∈I, and the ultraproduct Π(M)i∈I/U is called an ultrapower. The ultrapower of M with index set I and ultrafilter U can be written compactly as MI/U.

Some examples: The hypernaturals are the ultrapower ℕ/U = Π(ℕ)i∈ℕ/U where U is any free ultrafilter over ℕ. Similarly, the hyperreals are ℝ/U = Π(ℝ)i∈ℕ/U. The hyperintegers are ℤ/U. And so on.

(6) So far we’ve just defined the ultraproduct as a set (the set of equivalence classes of I-indexed sequences of elements from the models (Mi)i∈I). But we want the ultraproduct to have all the same structure as the models that we used as input. In other words, the ultraproduct of a bunch of L-structures will itself be an L-structure. To make this happen, we need to specify how the relation symbols and function symbols of L work in the ultraproduct model.

Here’s how it works. If f is a unary function symbol in the language L, then we define f on Π(Mi)i∈I/U by applying the function elementwise to the sequences. So:

f: Π(Mi)i∈I/U → Π(Mi)i∈I/U is defined as f([a])(i) = f(a(i)) for every i ∈ I.

What if f is a binary function symbol? Then:

f: (Π(Mi)i∈I/U)2 → Π(Mi)i∈I/U is defined as f([a], [b])(i) = f(a(i), b(i)) for every i ∈ I.

This generalizes in the obvious way to trinary function symbols, quaternary function symbols, and so on.

What about relations? Suppose R is a unary relation symbol in the language L. We need to define R on the ultraproduct Π(Mi)i∈I/U, and we do it as follows:

Π(Mi)i∈I/U ⊨ R([a]) if and only if { i ∈ I | Mi ⊨ R(a(i)) } ∈ U.

In other words, Π(Mi)i∈I/U affirms R([a]) if and only if the set of indices i such that Mi affirms R(a(i)) is in the ultrafilter. For example, if R holds for cofinitely many members of the sequence a, then R holds of [a].

If R is binary, we define it as follows:

Π(Mi)i∈I/U ⊨ R([a], [b]) if and only if { i ∈ I | Mi ⊨ R(a(i), b(i)) } ∈ U.

And again, this generalizes in the obvious way.

This fully defines the ultraproduct Π(Mi)i∈I/U as an L-structure! (If you’re thinking ‘what about constant symbols?’, remember that constants are just 0-ary functions)

Say that again, slower

That was really abstract, so let’s go through it again with the (hopefully now-familiar) example of the hypernaturals.

We start by defining an index set I. We choose I = ℕ.

Now define the language we’ll use. This will be the standard language of Peano arithmetic: one constant symbol (0), one relation symbol (<), and three function symbols (S, +, ×).

The family of structures in this language that we’ll consider (Mi)i∈I will just be a single structure repeated: for each i, Mi will be the L-structure ℕ (the natural numbers with 0, <, +, and × defined on it). So our family of structures is just (ℕ)i∈ℕ = (ℕ, ℕ, ℕ, ℕ, …).

Our sequences are functions from I to U(Mi)i∈I such that for each i∈I, f(i) ∈ Mi. For the hypernaturals, I and U(ℕ)i∈ℕ are both just ℕ, so our sequences are functions from ℕ to ℕ. We can represent the function f: ℕ → ℕ in the familiar way: (f(0), f(1), f(2), f(3), …).

The set of all sequences is the set of all functions from ℕ to ℕ. This is the direct product Π(ℕ)i∈ℕ = ℕ.

Now, we take any ultrafilter on I = ℕ. Call it U. We use U to define the equivalence relation on the direct product ℕ:

(a(0), a(1), a(2), …) ~ (b(0), b(1), b(2), …) if and only if { i ∈ ℕ | a(i) = b(i) } ∈ U

And taking equivalence classes of this relation, we’ve recovered our original definition of the hypernatural numbers! ℕ/U = *ℕ. Now we finish up by defining all functions and relations on *ℕ.

Functions are defined pointwise:

0 = [0, 0, 0, 0, …]
S[a(0), a(1), a(2), …] = [Sa(0), Sa(1), Sa(2), …]
[a(0), a(1), a(2), …] + [b(0), b(1), b(2), …] = [a(0) + b(0), a(1) + b(1), a(2) + b(2), …]
[a(0), a(1), a(2), …] ⋅ [b(0), b(1), b(2), …] = [a(0) ⋅ b(0), a(1) ⋅ b(1), a(2) ⋅ b(2), …]

We just have one relation symbol <, and relations are defined according to the ultrafilter:

[a(0), a(1), a(2), …] < [b(0), b(1), b(2), …] iff { i ∈ ℕ | ℕ ⊨ (a(i) < b(i)) } ∈ U

And we’re done!

Łoś’s theorem

Now we’re ready to prove Łoś’s theorem in its full generality. First, let’s state the result:

Fix any index set I, any language L, and any family of L-structures (Mi)i∈I. Choose a free ultrafilter U on I and construct the ultraproduct structure Π(Mi)i∈I/U. Łoś’s theorem says:

For every L-sentence φ, Π(Mi)i∈I/U ⊨ φ if and only if { i ∈ I | Mi ⊨ φ } ∈ U.

A special case of this is where our ultraproduct is an ultrapower of M, in which case it reduces to:

For every L-sentence φ, MI/U ⊨ φ if and only if M ⊨ φ

In other words, any ultrapower of M is elementary equivalent to M!

The proof is by induction on the set of all L-formulas.

Base case: φ is atomic

Atomic sentences are either of the form R(t1, …, tn) or (t1 = t2) for an n-ary relation symbol R and terms t1, … tn.

Suppose φ is R(t1, …, tn). This case is easy: it was literally the way we defined the interpretation of relation symbols in the ultraproduct model that Π(Mi)i∈I/U ⊨ R(t1, …, tn) if and only if {i ∈ I | Mi ⊨ R(t1, …, tn)} ∈ U.

Suppose φ is (t1 = t2). t1 and t2 are terms, so the ultraproduct model (Π(Mi)i∈I/U) interprets them as I-sequences, i.e. functions from I to U(Mi)i∈I such that t1(i) and t2(i) are both in Mi. We’ll write the denotations of t1 and t2 as [t1] and [t2]. Now, [t1] = [t2] iff { i ∈ I | t1 = t2 } ∈ U iff { i ∈ I | Mi ⊨ (t1 = t2) } ∈ U, which is what we want.

Inductive step: φ is ¬ψ, ψ∧θ, or ∃x ψ

Assume that Los’s theorem holds for ψ and θ. Now we must show that it holds for φ

Suppose φ is ¬ψ.

Then Π(Mi)i∈I/U ⊨ φ
iff Π(Mi)i∈I/U ⊭ ψ
iff { i ∈ I | Mi ⊨ ψ } ∉ U (by the inductive hypothesis)
iff { i ∈ I | Mi ⊨ ψ }c ∈ U (by the ultra property of U)
iff { i ∈ I | Mi ⊭ ψ } ∈ U

iff { i ∈ I | Mi ⊨ ¬ψ } ∈ U
iff { i ∈ I | Mi ⊨ φ } ∈ U

Suppose φ is ψ∧θ.

Then Π(Mi)i∈I/U ⊨ φ
iff Π(Mi)i∈I/U ⊨ ψ and Π(Mi)i∈I/U ⊨ θ
iff { i ∈ I | Mi ⊨ ψ } ∈ U and { i ∈ I | Mi ⊨ θ } ∈ U (by the inductive hypothesis)
iff { i ∈ I | Mi ⊨ ψ } ⋂ { i ∈ I | Mi ⊨ θ } ∈ U (by closure-under- of U)
iff { i ∈ I | Mi ⊨ ψ and Mi ⊨ θ } ∈ U
iff { i ∈ I | Mi ⊨ ψ∧θ } ∈ U
iff { i ∈ I | Mi ⊨ φ } ∈ U

Suppose φ is ∃x ψ.

Then Π(Mi)i∈I/U ⊨ φ
iff Π(Mi)i∈I/U ⊨ ∃x ψ
iff Π(Mi)i∈I/U ⊨ ψ(a) for some [a] ∈ Π(Mi)i∈I/U
iff { i ∈ I | Mi ⊨ ψ(a(i)) } ∈ U (by the inductive hypothesis)
iff { i ∈ I | Mi ⊨ ∃x ψ(x) } ∈ U

And that completes the proof! We don’t need to consider ∨, →, ↔, or ∀, because these can all be defined in terms of ¬, ∧, and ∃.

Now we know that the first-order properties of ultraproducts are tied closely to those of their component structures. The ultraproduct of any collection of two-element structures is itself a two-element structure. Same with the ultraproduct of any collection of structures, cofinitely many of which are two-element structures!

The ultraproduct of any collection of PA models is itself a PA model. The ultraproduct of any collection of groups is itself a group. But the ultraproduct of all finite groups need not itself be finite, because “I am finite” isn’t first-order expressible.

And in particular, an ultrapower of a structure M perfectly mimics ALL of the first-order properties of M!

Łoś’s theorem is an incredibly powerful tool we can wield to illuminate the strange structure of the hypernatural numbers. We’re now positioned to discover nonstandard prime numbers, infinitely even numbers, numbers that are divisible by every standard natural number, and infinitely large prime gaps. All of this (and more) in the next post!

Next: Weird nonstandard numbers

Hypernaturals in all their glory (Ultra Series 3)

Previous: Hypernaturals simplified

What is an ultrafilter? (with pretty pictures)

To define an ultrafilter we need to first define a filter. Here’s a pretty good initial intuition for what a filter is: a filter on a set X is a criterion for deciding which subsets of X are “large”. In other words, a filter provides us one way of conceptualizing the idea of large and small subsets, and it allows us to do so in a way that gives us more resolution than the cardinality approach (namely, assess size of sets just in terms of their cardinality). For example, in a countably infinite set X, the cofinite subsets of X (those that contain all but finitely many elements of X) have the same cardinality as the subsets of X that are infinite but not cofinite. But there’s some intuitive sense in which a set that contains all but finitely many things is larger than a set that leaves out infinitely many things. Filters allow us to capture this distinction.

Alright, so given a set X, a filter F on X is a collection of subsets of X (i.e. it’s a subset of 𝒫(X)) that satisfies the following four conditions:

(i) X ∈ F … “X is large”
(ii) ∅ ∉ F … “the empty set is not large”
(iii) If A ⊆ B and A ∈ F, then B ∈ F … “supersets of large sets are large”
(iv) If A ∈ F and B ∈ F, then A ⋂ B ∈ F … “intersections of large sets are large”

In other words, a filter on X is a set of subsets of X that contains X, doesn’t contain the empty set, and is closed under supersets and intersection. Note that a filter is also closed under union, because of (iii) (the union of A and B is a superset of A).

An ultrafilter is a filter with one more constraint, namely that for any subset of X, either that subset or its complement is in the filter.

(v) For any A ⊆ X, either A ∈ F or (X\A) ∈ F … “a set is either large, or if not, then its complement is large”

There’s a nice way to visualize filters and ultrafilters that uses the Hasse diagram of the power set of X. For a concrete example, let X = {a, b}. We can draw the power-set of X as follows:

We draw an arrow from A to B when A is a subset of B. Now, what are the possible filters on X? There are three, see if you can find them all before reading on.

Only two of these are ultrafilters. Which two?

Remember that for an ultrafilter U, every subset or its complement is in U. So an ultrafilter always contains half of all subsets. This gives an easy way to rule out the first one.

Another example: let X = {a, b, c}. Then the power-set of X looks like:

Note that we’ve left out some arrows, like the arrow from {a} to {a,b,c}. This is okay, because transitivity of the subset relation makes this arrow redundant. Anyway, what are some filters on X? Here are three of them:

Only one of these is an ultrafilter! You should be able to identify it pretty easily. See if you can pick out the other four filters, and identify which of them are ultrafilters (there should be two). And another exercise: why is the following not a filter?

Does it have any extension that’s an ultrafilter?

One thing to notice is that in all of these examples, when something is in the filter then everything it points to is also in the filter. This corresponds to ultrafilters being closed under supersets. Also, for any two things in the filter, their meet (their greatest lower bound; the highest set on the diagram that points to both of them) is also is the filter. This corresponds to closure under intersections.

Imagine that there is a stream flowing up the Hasse diagram through all the various paths represented by arrows. Choose any point on the diagram and imagine dripping green dye into the water at that point. The green color filters up through the diagram until it reaches the top. And everything that’s colored green is in the filter! This captures the idea that filters are closed under superset, but what about intersection? If X is finite, this corresponds to the dye all coming from a single source, rather than it being dripped in at multiple distinct points. The infinite case is a little trickier, as we’ll see shortly.

One other important thing to notice is that whenever we had an ultrafilter, it always contained a singleton. An ultrafilter that contains a singleton is called a principal ultrafilter, and an ultrafilter that doesn’t contain any singletons is called a free ultrafilter. So far we haven’t seen any free ultrafilters, and in fact as long as X is finite, any ultrafilter on X will be principal. (Prove this!) But the situation changes when X is an infinite set.

The Hasse diagram for an infinite set is a bit harder to visualize, since now we have uncountably many subsets. But let’s try anyway! What does the Hasse diagram of ℕ look like? Well, we know that ∅ is at the bottom and ℕ is at the top, so let’s start there.

Next we can draw all the singleton sets. ∅ points at all of these, so we’re not going to bother drawing each individual arrow.

Next we have all the pair sets, and then the triples. Each singleton points at infinitely many pairs, and each pair points at infinitely many triples.

And so on through all finite cardinalities.

Now what? We’ve only exhausted all the finite sets. We can now start from the top with the cofinite sets, those that are missing only finitely many things. First we have the sets that contain all but a single natural number:

Then the sets containing all but a pair of naturals, and so on through all the cofinite sets.

But we’re not done yet. We haven’t exhausted all of the subsets of ℕ; for instance the set of even numbers is neither finite nor cofinite. In fact, there are only countably many finite and cofinite sets, but there are uncountably many subsets of ℕ, so there must be a thick intermediate section of infinite sets that are not cofinite (i.e. infinite sets with infinite complements).

A sanity check that this diagram makes sense: start with a finite set and then add elements until you have a cofinite set. Between the finite set and the cofinite set there’s always an intermediate set that’s infinite but not cofinite. This matches with our image: any path from the finite to the cofinite passes through the middle section.

Now, what would a filter on the naturals look like on this diagram? If our filter is principal, then we can still roughly sketch it the same way as before:

How about an ultrafilter? Depends on whether it’s principal or free. Any principal ultrafilter must look like the third image above; it must start at the “finite” section and filter upwards (remember that principal means that it contains a singleton).

Any principal ultrafilter on ℕ can be written as { A ⊆ ℕ | n ∈ A } for some n ∈ ℕ.

What about free ultrafilters? A free ultrafilter contains no singletons. This implies that it contains no finite set. See if you can come up with a proof, and only then read on to see mine.

Suppose that U is a free ultrafilter on X and contains some finite set F. U is free, so it contains no singletons. So for every a ∈ F, the singleton {a} ∉ U. By ultra, X\{a} ∈ U. By closure-under-finite-intersection, the intersection of {X\{a} | a ∈ F} is in U. So X\F ∈ U. But now we have F ∈ U and X\F ∈ U, and their intersection is ∅. So ∅ ∈ U, contradicting filter.

So a free ultrafilter must contain no finite sets, meaning that it contains all the cofinite sets. Since it’s ultra, it also contains “half” of all the intermediate sets. So visually it’ll look something like:

That’s what a free ultrafilter on the naturals would look like if such a thing existed. But how do we know that any such object actually does exist? This is not so trivial, and in fact the proof of existence uses the axiom of choice. Here’s a short proof using Zorn’s Lemma (which is equivalent to choice in ZF).

Let F be any filter on X. Consider the set Ω of all filters on X that extend F. (Ω, ⊆) is a partially ordered set, and for any nonempty chain of filters C ⊆ Ω, the union of C is itself a filter on X. (Prove this!) The union of C is also an upper bound on C, meaning that every nonempty chain of filters has an upper bound. Now we apply Zorn’s Lemma to conclude that there’s a maximal filter U in Ω. Maximality of U means that U is not a subset of V for any V ∈ Ω.

Almost done! U is maximal, but is it an ultrafilter? Suppose not. Then there’s some A in X such that A ∉ U and (X\A) ∉ U. Simply extend U by adding in A and all supersets and intersections. This is a filter that extends F and contains U, contradicting maximality. So U is an ultrafilter on X!

Now, F was a totally arbitrary filter. So we’ve shown that every filter on X has an ultrafilter extension. Now let X be infinite and take the filter on X consisting of all cofinite subsets of X (this is called the Fréchet filter). Any ultrafilter extension of the Fréchet filter also contains all cofinite subsets of X, and thus contains no singletons. So it’s free! Thus any infinite set has a free ultrafilter.

Hypernatural numbers

Still with me? Good! Then you’re ready for the full definition of the hypernatural numbers, using ultrafilters. Take any free ultrafilter U on ℕ. U contains all cofinite sets and no finite sets, and is also decisive on all the intermediate sets. If you remember from the last post, this makes U a perfect fit for our desired “decisiveness criterion”.

Now consider the set of all countable sequences of natural numbers. Define the equivalence relation ~ on this set as follows:

(a1, a2, a3, …) ~ (b1, b2, b3, …) iff { k ∈ ℕ | ak = bk } ∈ U

Note the resemblance to our definition last post:

(a1, a2, a3, …) ~ (b1, b2, b3, …) iff { k ∈ ℕ | ak = bk } is cofinite

This previous definition corresponded to using the Fréchet filter for our criterion. But since it was not an ultrafilter, it didn’t suffice. Now, with an ultrafilter in hand, we get decisiveness!

Addition and multiplication on the hypernaturals is defined very easily:

[a1, a2, a3, …] + [b1, b2, b3, …] = [a1+b1, a2+b2, a3+b3, …]
[a1, a2, a3, …] ⋅ [b1, b2, b3, …] = [a1⋅b1, a2⋅b2, a3⋅b3, …]

Let’s now define < on the hypernaturals.

(a1, a2, a3, …) < (b1, b2, b3, …) if { k ∈ ℕ | ak < bk } ∈ U

The proof of transitivity in the previous post still works here. Now let’s prove that < is a total order.

Consider the following three sets:

X = { k ∈ ℕ | ak < bk }
Y = { k ∈ ℕ | ak > bk }
Z = { k ∈ ℕ | ak = bk }

The intersection of any pair of these sets is empty, meaning that at most one of them is in U. Could none of them be in U? Suppose X, Y, and Z are not in U. Then ℕ\X and ℕ\Y are in U. So (ℕ\X) ⋃ (ℕ\Y) is in U as well. But (ℕ\X) ⋃ (ℕ\Y) = Z! So Z is in U, contradicting our assumption.

So exactly one of these three sets is in U, meaning that a < b or b < a or a = b. This proves that using an ultrafilter really has fixed the problem we ran into previously. This problem was that the hypernaturals were quite different from the naturals in undesirable ways (like < not being a total order). The natural question to ask now is “Just how similar are the hypernaturals to the naturals?”

The answer is remarkable. It turns out that there are no first-order expressible differences between the naturals and the hypernaturals! Any first-order sentence that holds true of the natural numbers also holds true of the hypernatural numbers! This result is actually just one special case of an incredibly general result called Łoś’s theorem. And in the next post we are going to prove it!

Next up: Łoś’s theorem and ultraproducts!

Hypernaturals Simplified (Ultra Series 2)

Previous: Introduction

What is a hypernatural number? It is a collection of infinitely long sequences of natural numbers. More precisely, it is an equivalence class of these infinite sequences.

An equivalence class under what equivalence relation? This is a little tricky to describe.

I’ll start with a slight lie to simplify the story. When we see the trouble that results from our simple definition, I will reveal the true nature of the equivalence relation that gives the hypernaturals. In the process you’ll see how the notion of an ultrafilter naturally arises.

So, hypernaturals are all about infinite sequences of natural numbers. Some examples of these:


We’ll define an equivalence relation ~ between sequences as follows:

Let x and y be infinite sequences of natural numbers.
Then x ~ y iff x and y agree in all but finitely many places.

For example, (0,1,2,3,4,5,6,…) ~ (19,1,2,3,4,5,6,…), because these two sequences only disagree at one spot (the zero index).

(1,1,2,2,4,4,…) and (1,2,4,8,…) are not equivalent, because these sequences disagree at infinitely many indices (every index besides the zeroth index).

Same with (0,1,2,3,4,5,6,…) and (1,2,3,4,5,6,7,…); even though they look similar, these sequences disagree everywhere.

(2,4,6,8,10,12,14,…) and (2,0,6,0,10,0,14,…) are not equivalent, because these sequences disagree at infinitely many indices (every odd index).

One can easily check that ~ is an equivalence relation, and thus it partitions the set of sequences of naturals into equivalence classes. We’ll denote the equivalence class of the sequence (a1, a2, a3, …) as [a1, a2, a3, …]. These equivalence classes are (our first stab at the definition of) the hypernaturals!

For instance, the equivalence class of (0,0,0,0,0,…) contains (1,4,2,0,0,0,0,0,…), as well as (0,2,4,19,0,0,0,0,…), and every other sequence that eventually agrees with (0,0,0,0,…) forever. So all of these correspond to the same hypernatural number: [0,0,0,0,…]. This object is our first hypernatural number! It is in fact the hypernatural number that corresponds exactly to the ordinary natural number 0. In other words 0 = [0,0,0,0,…].

[1,1,1,1,…] is a distinct equivalence class from [0,0,0,0,…]. After all, the sequences (0,0,0,0,…) and (1,1,1,1,…) disagree everywhere. You might guess that [1,1,1,1,…] is the hypernatural analogue to the natural number 1, and you’d be right!

For any standard natural number N, the corresponding hypernatural number is [N,N,N,N,N,…], the equivalence class of the sequence consisting entirely of Ns.

Now consider the hypernatural [0,1,2,3,4,5,6,…]. Let’s call it K. Does K = N for any standard natural number N? In other words, is (0,1,2,3,4,5,6,…) ~ (N,N,N,N,N,…) true for any finite N? No! Whatever N you choose, it will only agree with (0,1,2,3,4,5,6,…) at one location. We need cofinite agreement, and here we have merely finite agreement. Not good enough! This means that K is our first nonstandard natural number!

How does K relate to the standard naturals in terms of order? We haven’t talked about how to define < on the hypernaturals yet, but it’s much the same as our definition of =.

[a1, a2, a3, …] = [b1, b2, b3, …]
{ k∈ℕ | ak = bk } is cofinite

[a1, a2, a3, …] < [b1, b2, b3, …]
{ k∈ℕ | ak < bk } is cofinite

Exercise: Verify that this is in fact a well-defined relation. Every equivalence class has many different possible representatives; why does the choice of representatives not matter for the purpose of describing order?

Now we can see that K > N for every standard N. Look at (0,1,2,3,4,5,…) and (N,N,N,N,…). The elements of the first sequence are only less than the elements of the second sequence at the first N indices. Then the elements of K are greater than the elements of N forever. So elements of K’s representative sequence are greater than elements of N’s representative sequence in a cofinite set of indices. Thus, K > N for every standard N. So K is an infinitely large number!

Here’s another one: K’ = [0,2,4,6,8,…]. You can see that K’ > K, because the elements of K’ are greater than those of K at all but one index (the first one). So we have another, bigger, infinite number.

Addition and multiplication are defined elementwise, so

K + K
= [0,1,2,3,4,…] + [0,1,2,3,4,…]
= [0+0, 1+1, 2+2, 3+3, 4+4, …]
= [0,2,4,6,8,…]
= K’

= [0,2,4,6,8,…]
= [2⋅0, 2⋅1, 2⋅2, 2⋅3, 2⋅4, …]
= 2⋅[0,1,2,3,4,…]
= 2⋅K

Predictably, we get many many infinities. In fact, there are continuum many nonstandard hypernatural numbers!

Proof: we construct an injection f from ℝ to *ℕ. If x is a real number, then f(x) := [floor(x), floor(10x), floor(100x), floor(1000x), …]. For example, f(35.23957…) = [35,352,3523,35239,352395, …]. For any two distinct reals x and y, the sequences x and y will eventually disagree forever. So each real is mapped to a distinct hypernatural, meaning that there are no more reals than hypernaturals. At the same time, there are no more hypernaturals than reals, because there are only continuum many countable sequences of natural numbers. So |*ℕ| = |ℝ|.

It turns out that every nonstandard hypernatural number is also larger than every standard natural number. We’ll see why in a bit, but it’ll take a bit of subtlety that I’ve yet to introduce.

Now, > is transitive in ℕ. Is it also transitive in *ℕ? Yes! Suppose A > B and B > C. Choose any representative sequences (a1, a2, a3, …), (b1, b2, b3, …), and (c1, c2, c3, …) for A, B, and C. Then X = { k∈ℕ | ak > bk } and Y = { k∈ℕ | bk > ck } are both cofinite. The intersection of cofinite sets is also cofinite, meaning that X⋂Y = { k∈ℕ | ak > bk and bk > ck } = { k∈ℕ | ak > ck } is cofinite. So A > C!

It’s a good sign that > is transitive. But unfortunately, the story I’ve told you thus far starts to break down here. The greater-than relation is a total order on the natural numbers. For any naturals a and b, exactly one of the following is true: a = b, a > b, b > a. But this is not true of the hypernaturals!

Consider the two hypernatural numbers n = [0,1,0,1,0,1,…] and m = [1,0,1,0,1,0,…]. Are n and m equal? Clearly not; they disagree everywhere. So n ≠ m.

Is n > m? No. The set of indices where n’s sequence is greater than m’s sequence is {1, 3, 5, 7, …}, which is not cofinite.

So is m > n? No! The set of indices where m’s sequence is greater than n’s sequence is {0, 2, 4, 6, …}, which is also not cofinite!

So as we’ve currently defined the hypernatural numbers, the > relation is not a total relation on them. This might be fine for some purposes, but we’ll be interested in defining the hypernaturals to mirror the properties of the naturals as closely as possible. So we’ll have to tweak our definition of the hypernaturals. The tweak will occur way back at the start where we defined our equivalence relation on sequences of naturals.

Recall: we said that two sequences (a1, a2, a3, …) and (b1, b2, b3, …), are equivalent if they agree in all but finitely many places. Said another way: a ~ b if { k∈ℕ | ak = bk } is cofinite. We defined > similarly: a > b if the agreement set for > is cofinite.

The problem with this definition was that it wasn’t definitive enough. There are cases where the agreement set is neither cofinite nor finite. (Like in our previous example, where the agreement set was the evens.) In such cases, our system gives us no direction as to whether a > b or b > a. We need a criterion that still deals with all the cofinite and finite cases appropriately, but also gives us a definitive answer in every other case. In other words, for ANY possible set X of indices of agreement, either X or X’s complement must be considered “large enough” to decide in its favor.

For example, maybe we say that if { k∈ℕ | ak = bk } = {0,2,4,6,8,…}, then a > b. Now our criterion for whether a > b is: the set of indices for which ak = bk is either cofinite OR it’s the evens. This implies that [1,0,1,0,1,0,…] > [0,1,0,1,0,1,…].

Once we’ve made this choice, consistency forces us to also accept other sets besides the evens as decisive. For instance, now compare (0,0,1,0,1,0,…) and (0,1,0,1,0,1,…). The set of indices where the first is greater than the second is {2,4,6,8,…}. But notice that the first differs only cofinitely from (1,0,1,0,…), meaning that [0,0,1,0,1,0,…] = [1,0,1,0,1,0,…]. The conclusion is that [0,0,1,0,1,0,…] > [0,1,0,1,0,1,…], which says that the set of indices {2,4,6,8,…} must also be decisive. And in general, once we’ve accepted the evens as a decisive set of indices, we must also accept the evens minus any finite set.

The three criterion we’ve seen as desirable for what sets of indices will count as decisive are (1) includes all cofinite sets, (2) for any set X, either X or X’s complement is decisive, and (3) consistency. These requirements turn out to correspond perfectly to a type of mathematical object called a free ultra-filter!

In the next post, we will define ultrafilters and finalize our definition of the hypernatural numbers!

Ultrafilters, Ultraproducts, and Hypernaturals 1: Introduction

This is the series I wish would have existed a month ago when I first started learning about ultrafilters and ultraproducts. First of all, what’s the motivation for learning about these ultra-objects? I imagine that if you’re here, you probably already have some degree of interest in ultrafilters, ultraproducts, or nonstandard models of arithmetic. But I’ll see if I can bolster that interest a little bit.

The most exciting application to me personally is that ultraproducts give you a recipe for constructing new mathematical structures out of familiar ones. Out of some model M one can construct a a new model *M, which typically has a much richer and stranger structure, but is nonetheless elementarily equivalent to M (this is called Łoś’s theorem). Elementary equivalence means that all the expressive resources of first-order logic are insufficient to tell M and *M apart: they agree on all first-order sentences.

For example, the ultraproduct of ℝ (the real numbers) is *ℝ, the hyperreal numbers. The hyperreals contain an enormous supply of infinitesimal quantities clustered around every real, as well as infinitely large quantities. And all the usual operations on ℝ, like addition and multiplication, are already nicely defined in *ℝ, meaning that these infinitesimals and infinities have a well-defined algebraic structure. And Łoś’s theorem tells us that ℝ and *ℝ are elementary equivalent: any first-order sentence true of the reals is also true of the hyperreals.

Another example: the ultraproduct of ℕ (the natural numbers) is *ℕ, the hypernaturals. The hypernaturals don’t contain any infinitesimals, but they do contain an uncountable infinity of infinite numbers. And since *N and N are elementarily equivalent, *ℕ is a model of true arithmetic! This is super exciting to me. True arithmetic is this unimaginably complicated set of sentences; there’s no Turing machine that decides which sentences reside in true arithmetic, nor is there a Turing machine with a halting oracle that does the job, nor even a Turing machine with a halting oracle for Turing machines with halting oracles! The computational complexity of true arithmetic is the limit of this sequence of progressively harder halting problems: you can only decide the set if you have an oracle for the halting problem, plus an oracle for the halting problem for TMs with oracles for the halting problem, plus an oracle for the halting problem for TMs with oracles for the halting problem for TMs with oracles for the halting problem, and so on. This level of complexity is known as 0(ω).

So true arithmetic is this impossibly uncomputable set of sentences. We know trivially that ℕ is a model of true arithmetic, because that’s how true arithmetic is defined { φ | ℕ ⊨ φ }. And we also know that true arithmetic has nonstandard models, because it’s a first-order theory and no first order theory can categorically define ℕ (I think the easiest way to see why this is true is to apply the Löwenheim-Skolem theorem, but – self-promotion alert – there’s also this very nice and simple proof from first principles). And ultraproducts allow us to construct an explicit example of one of these nonstandard models! You can actually write down some of these nonstandard numbers (they look like infinite sequences of natural numbers) and discover their strange properties. For example, (2, 3, 5, 7, 11, …) is a nonstandard prime number, (1, 2, 4, 8, 16, …) is infinitely even, and (0!, 1!, 2!, 3!, 4!, …) is a multiple of every standard natural number. We’ll dive into all of this soon enough.

Ultrafilters and ultraproducts also have applications outside of logic. A fairly basic result about ultrafilters (that every ultrafilter over a finite set contains a singleton) is equivalent to Arrow’s impossibility theorem in voting theory (that any voting system with unanimity, independence of irrelevant alternatives (IIR), and finitely many voters contains a dictator). And the existence of a singleton-free ultrafilter over an infinite set (a free ultrafilter) shows that with infinitely many voters, there’s a non-dictatorial voting system with unanimity and IIR. There’s a pretty good description of those results here, but finish reading my post first!

Nick Bostrom in an old paper titled Infinite Ethics describes how to apply ultrafilters to resolve some of the issues that arise when trying to apply utilitarian ethics in a universe with infinitely many experiencers. Suppose that the current universe contains an infinite number of experiencers, each of whom is having an experience with a valence of +1. By pressing a button, you can immediately transition to a universe where everybody is now experiencing a valence of +2. Clearly pressing the button is a utilitarian good. But 1+1+1+… = ∞ = 2+2+2+…. So it looks like a standard account of utility as real numbers says that the utility of the +2 world is the same as the utility of the +1 world (both just ∞). But if we treat utilities as hyperreal numbers instead of ordinary real numbers, then we can do the comparison and get the right answer. It’s worth noting that this approach doesn’t fix all the problems raised by an infinite universe; for instance, if pressing the button only increases a finite number of experiencers from valence +1 to valence +2, then the net utility of the universe is unchanged, even with hyperreal utilities. (This corresponds to a property that we’ll see soon, which is that a free ultrafilter contains no finite sets.)

Okay, introduction over. Hopefully you’re now excitedly asking yourself the question “So what are these ultrafilters and ultraproducts anyway??” The story begins in the next post!

Hypernaturals Simplified

Even Numbers are Tautologies

To each atomic proposition P assign a natural number p. If that natural number is even, then its corresponding proposition is true. If the number is odd then the proposition is false.

If you have two numbers n and m and you multiply them, the result is even so long as either n or m is even. So multiplication is like or; P∨Q corresponds to pq.

Negation is like adding 1: even becomes odd and odd becomes even. So ¬P corresponds to p+1.

Consider addition: p+q is even if p and q are both even or both odd. So p+q is like the biconditional ↔.

Other connectives can be formed out of these. Take P∧Q: P∧Q is equivalent to ¬(¬P∨¬Q), which is (p+1)(q+1) + 1. So P∧Q corresponds to pq+p+q+2. P→Q is ¬P∨Q, which is (p+1)q.

The logical constants ⊤ and ⊥ correspond to numerical constants: ⊤ can be assigned 0 and ⊥ assigned 1, for instance.

Tautologies translate to algebraic expressions which are always even. For instance, P→P translates to (p+1)p, which is always even. P→(Q→P) translates to (p+1)(q+1)p, which is also always even. P∨¬P translates to p(p+1), always even.

Contradictions translate to algebraic expressions which are always odd. P∧¬P translates to (p+1)(p+2) + 1, which is always odd. And so on.

Inference rules can be understood as saying: if all the premises are even, then the conclusion will be even. Take conjunction elimination: P∧Q ⊨ P. This says that if (p+1)(q+1) + 1 is even then p is even, which ends up being right if you work it out. Modus ponens: P, P→Q ⊨ Q. This says that if p and (p+1)q are both even, then q is even. Again works out!

You can also work the other way: For any number p, p(p+1) is even. Translating into propositional logic, this says that P∨¬P is a tautology. We’ve proved the law of the excluded middle! It’s interesting to note that earlier we saw that P→P translates to (p+1)p. So in some sense the law of the excluded middle is just self-implication but with the two products reversed!

Notes on motivation and self-improvement

I’ve recently been thinking about something that strikes me as a remarkable fact about human psychology. Nearly everybody knows of multiple things that consistently bring them significant happiness, and despite this they don’t do these things, or do them much less frequently than they would like.

For example, virtually every time that I meditate I feel better off at the end of it. There’s usually long-term effects too, like that the rest of my day is improved, and sometimes multiple days in the future are improved. There’s even some longer-term benefits; my sense is that consistent meditation allows me to think more clearly and get insight about the structure of my own mind. I’ve meditated hundreds of times and have literally never had a negative experience. The most I can say is I’ve had some neutral experiences, but the vast majority have been positive experiences, and many of them have been extremely positive experiences. And yet, I don’t regularly meditate! If you picked a random moment in the last ten years of my life and asked me “Did you meditate today? What about this week? Or this month?”, the answer to all three would more often than not be “no”.

And it’s not that I just sometimes forget about meditation. It’s actually worse than this! I feel an active resistance towards meditating whenever I think of doing it. One reason I’m saying this is that I used to be very into psychological egoism. The idea is that you can boil down all of our motivations to finding personal happiness. The claim is that if you interrogate your own motivations and keep asking “well why do you want that?”, you’ll eventually find a basic seeking after positive valence experiences at the end of the chain. But when you actually start to introspect on your own mind, it feels like there are these very clear cases, cases that aren’t even very rare, of your mind rejecting happy experiences. One could say “Well really this is a short-term versus long-term type of thing and you’re really rejecting the long-term happiness in exchange for some short term gratification”. But it’s not even that! For me, the benefits I get from meditation happen basically immediately. If you were to try to model me as a simple happiness maximizer, even one that prefers short-term gratification, you would end up making really bad predictions about how I live my life.

So, there’s a whole set of strategies that I just know will effectively bring me happiness. And I can implement these strategies at virtually any time. And yet I don’t do them! Or at least, I do them much less frequently than I could be. And I’m willing to bet that the same is true of you. I want you to think about the following: What are three concrete actions you can take that aren’t very costly, and that pretty much always make you feel happier, healthier, or better in any way. Just think of these three things, maybe even write them down. And then ask yourself: “Well, how often do I actually do these things? Do I do them at a rate that I think makes sense, or do I do them less than that?”

I’ve already given the example of meditation for myself. Another example is exercise. Exercise is a less good example for the psychological egoism point I was making, because there is a good case to be made that it’s a trade off between present suffering and future happiness. But regardless, I know that when I exercise, I feel good for the rest of my day. I have more energy, I’m typically more focused, and I feel better about myself. Long-run consistent exercise makes you look better, which raises self-esteem and makes you more likely to attract social interactions that you might desire, like a romantic partner. And due to the halo effect, you’re probably able to more easily get friends. (Even if you think it’s a bad thing that our society prizes attractiveness so much, it still is a true fact about society. Just because something shouldn’t be the case doesn’t mean you should act as if it isn’t the case!)

Another one for me is taking a walk in the morning. If instead of going on the computer, the first thing I do is just get up and take a walk outside, it consistently makes the rest of my day feel better. I don’t think that this one has longer term effects beyond that single day, but it’s still something I could do which takes maybe 30 minutes and isn’t physically exhausting or anything. And I have a lot of resistance to doing it!

One more example I think I can give is writing. If in the course of a day I write an essay or make progress on an essay, I feel a lot better. This is one which is actually situational, because it’s not the case that at every moment I have an idea for a good essay I can write. If I just sort of sit down and start writing without any idea of something I want to write about, that typically isn’t going to have the same effect. But nonetheless, it’s often the case that there are topics that I want to write about and yet I don’t. So that’s another interesting phenomenon.

So, exercise, meditation, walking in the morning, and writing, these are all concrete things I can put into action literally today. And in fact, I have been doing these for last few weeks and I hope that continues; I’m trying to build them as actual habits rather than just some temporary phase I’m in. But taking an outside view I can see that throughout most of my life I have not had a really consistent practice of any of these things.

Ultimately I just think it’s a good thing to take a moment out of your day and reflect on the things you can do to help yourself, and that you KNOW will help yourself. And possibly just thinking about these things will be motivating to some degree to actually do them! Put a special focus on the fact that you really could just start doing them right now; there’s nothing stopping you. Of course, if one of your choices was something like “every time I vacation in Hawaii it consistently raises my happiness for a period of time”, well… that’s true but not very actionable. On the other hand, things like exercising or meditation can be done in almost any environment! You don’t have to go to the gym, you could just search “high intensity interval training” on Youtube, do ten to fifteen minutes of exercise and you will almost certainly feel better for the rest of your day. So ask yourself the question “why am I not doing that?” And then do it!

First Order Truth Trees

The analytic tableaux (also called truth tree) style of proof system is really nice. The tree structure tends to resemble the actual process of mathematical reasoning better than the Hilbert-style linear proof systems, and I find that the formal proofs are much easier to follow. For first-order logic (without equality), the system involves four new inference rules to deal with quantifiers. There are a few different logically equivalent choices for these rules, but here is my favorite:

Universal instantiation ∀
∀x φ(x)

(where t is any closed term – any term without free variables)

Existential instantiation ∃
∃x φ(x)

(where a is any new constant)

Negated universal ¬∀
¬∀x φ

∃x ¬φ

Negated existential ¬∃
¬∃x φ

∀x ¬φ

The instantiation rules are actually quite interesting. Suppose that you’re working in the language of ZFC. ZFC has no constant or function symbols, only a single binary relation symbol: ∈, interpreted as “is an element of”. This seems to raise a problem for using the instantiation rules. For universal instantiation, we can only apply it to closed terms. But a language with no constants has no closed terms! And looking at existential instantiation, there’s a similar problem: we’re supposed to derive φ(a), where a is a new constant, but our language has no constants!

So what’s up? Does the standard first-order analytical tableaux approach not work for simple languages like ZFC? Insofar as ZFC is a shining beacon of what first-order logic can accomplish, this is a little disconcerting.

Actually, the instantiation rules work exactly as stated for ZFC. The subtlety in their interpretation is that when running proofs, you are allowed to (and in fact, often must) expand the language. So when you do existential instantiation, you don’t choose a constant symbol that already exists inside your language. Instead, you add a totally new constant symbol to your language, and then declare that φ holds of this symbol. Similarly, you can do universal instantiation in a language with no closed terms, simply by expanding the language to include some constants.

Now, if we’re expanding the language in the course of the proof, then who’s to say that in the end our proof will still be valid for our original language? Well, suppose we’re working in some minimal language L that has no constants. Let T be our background theory (the set of non-tautologies that we’re taking as assumptions). When we do existential instantiation, we always create a new constant. Let’s call it a. We then expand the language into L’ = L ⋃ {a}, and we expand our assumptions to T’ = T ⋃ {φ(a)}. When we do universal instantiation, we either use an existing closed term or create a new one. In the second case, we create a new constant b and form a new language L’ = L ⋃ {b}. We also expand our assumptions to a new set T’ = T ⋃ {φ(b)}.

The important thing to note is that we haven’t done anything that invalidates any models of T! If T originally contained a sentence of the form ∀x φ(x), then adding c and declaring φ(c) doesn’t conflict with this. And if T originally contained a sentence of the form ∃x φ(x), then in every model of T at least one thing satisfies φ. So when we add a new constant c, that constant can just refer back to any of these φ-satisfiers.

You might think: “But hold on! How can we be sure that it’s safe to just add new constants? Couldn’t we expand the domain too much, in a way that’s inconsistent with our original theory?” The answer to this is that the domain doesn’t have to expand to accommodate new constants! These constants can refer to existing elements of the domain. For instance, suppose T contains six existential statements and a sentence that says there are exactly five objects. Each time we run existential instantiation on one of the six existential sentences, we create a new constant. So we’ll get six new constants. But these constants can refer to the same value! And since our theory already says that there are five objects, the models of our expanded theory will also contain exactly five objects, meaning that in every model of our original theory, the new constants will refer to elements of the domain that already exist. No domain expansion!

Ok, but how do we know that there isn’t some very complicated sentence ψ that was originally true in every model of T, but becomes false in some models of T’? To fully prove that this never happens, we could do induction over all sentences in the language, showing that any sentence that is entailed by T is also entailed by T’. But considering the details of the expansion process, and trying to come up with ways that the expansion might fail, is a good way to get an intuition for why this proof system works.

I’ll close with a few examples of using analytic tableaux to prove basic results in PA and ZFC. Think of this as a proof of concept! (Each of these proofs also uses some rules regarding equality, which are obvious enough despite that I haven’t explicitly defined them).

First, we show that PA proves ∀x (0 + x = x). This is nontrivial, despite that ∀x (x + 0 = x) is an axiom!

Next, a proof from ZFC that the empty set exists and is unique:

Now a proof from PA that addition is commutative. This one is shockingly complicated for such a simple concept, and requires three uses of induction, but this goes to show how basic the axioms of PA really are! If you look closely, you’ll also notice that this proof replicates the proof of ∀x (0 + x = x) from above!

Final example, here’s a PA proof that no number is its own successor. This rules out nonstandard models of arithmetic with one nonstandard number!

One nonstandard is worth infinitely many standards

Suppose that M is a nonstandard model of true arithmetic (the set of all first-order sentences in the language of PA that are true in the standard model of arithmetic, ℕ). Now, take any sentence φ(x) with one free variable. Suppose that there’s some nonstandard number k in M such that φ holds of k. Since k is larger than every standard natural, the following infinite set of sentences are all true:

∃x (x > 0 ∧ φ(x))
∃x (x > 1 ∧ φ(x))
∃x (x > 2 ∧ φ(x))

∃x (x > 1000000 ∧ φ(x))

Since these sentences are true in M and M is a model of true arithmetic, these sentences must also be true in the standard model ℕ. So it must be true for every standard natural that there’s a larger standard natural that satisfies φ. In other words, you can guarantee that there are infinitely many standard naturals that satisfy a property φ just by finding a single nonstandard number k that satisfies φ in a model of true arithmetic!

Furthermore, since in ℕ it is true that every standard natural has a larger standard natural satisfying φ, the sentence ∀x ∃y (y > x ∧ φ(y)) is true in ℕ. So this sentence must be true in every model of true arithmetic, including M! This means that just by finding a single nonstandard satisfying φ, you can immediately be sure that there are infinitely many standard numbers AND infinitely many nonstandard numbers (in every nonstandard model of TA!) satisfying φ. This is pretty dramatic!

As an example, consider the twin prime conjecture. We can construct the predicate isPrime(x) in first-order PA with the formula ∀y (∃z (y⋅z = x) → (y=1 ∨ y=x)). Then the predicate isTwinPrime(x) is just: isPrime(x) ∧ isPrime(x+2). Now the twin prime conjecture just says that ∀x ∃y (y > x ∧ isTwinPrime(y)), which is exactly the form we saw in the last paragraph! So to prove the twin prime conjecture, it suffices to demonstrate a single nonstandard twin prime in a model of true arithmetic.