# Computing truth values of sentences of arithmetic, or: Math is hard

Previously I talked about the arithmetic hierarchy for sets, and how it relates to the decidability of sets. There’s also a parallel notion of the arithmetic hierarchy for sentences of Peano arithmetic, and it relates to the difficulty of deciding the truth value of those sentences.

Truth value here and everywhere else in this post refers to truth value in the standard model of arithmetic. Truth value in the sense of “being true in all models of PA” is a much simpler matter; PA is recursively axiomatizable and first order logic is sound and complete, so any sentence that’s true in all models of PA can be eventually proven by a program that enumerates all the theorems of PA. So if a sentence is true in all models of PA, then there’s an algorithm that will tell you that in a finite amount of time (though it will run forever on an input that’s false in some models).

Not so for truth in the standard model! As we’ll see, whether a sentence evaluates to true in the standard model of arithmetic turns out to be much more difficult to determine in general. Only for the simplest sentences can you decide their truth value using an ordinary Turing machine. And the set of all sentences is in some sense infinitely uncomputable (you’ll see in a bit in what sense exactly this is).

What we’ll discuss is a way to convert sentences of Peano arithmetic to computer programs. Before diving into that, though, one note of caution is necessary: the arithmetic hierarchy for sentences is sometimes talked about purely syntactically (just by looking at the sentence as a string of symbols) and other times is talked about semantically (by looking at logically equivalent sentences). Here I will be primarily interested in the entirely-syntactic version of the arithmetic hierarchy. If you’ve only been introduced to the semantic version of the hierarchy, what you see here might differ a bit from what you recognize.

Let’s begin!

The simplest types of sentences have no quantifiers at all. For instance…

0 = 0
2 ⋅ 2 < 7
(2 + 2 = 4) → (2 ⋅ 2 = 4)

Each of these sentences can be translated into a program quite easily, since +, ⋅, =, and < are computable. We can translate the → in the third sentence by converting it into a conjunction:

``````## (2 + 2 = 4) → (2 ⋅ 2 = 4)
not(2 + 2 == 4 and not 2 * 2 == 4)``````

Slightly less simple-looking are sentences with bounded quantifiers:

∀x < 10 (x + 0 = x)
∃x < 100 (x + x = x)
∀x < 5 ∃y < 7 (x > 1 → x⋅y = 12)
∃x < 5 ∀y < x ∀z < y (y⋅z ≠ x)

In each of these examples, the bounded quantifier could in principle be expanded out, leaving us with a finite quantifier-free sentence. This should suggest to us that adding bounded quantifiers doesn’t actually increase the computational difficulty.

We can translate sentences with bounded quantifiers into programs by converting each bounded quantifier to a for loop. The translation slightly differently depending on whether the quantifier is universal or existential:

``````def Aupto(n, phi):
for x in range(n):
if not phi(x):
return False
return True``````
``````def Elessthan(n, phi):
for x in range(n):
if phi(x):
return True
return False``````

Note that the second input needs to be a function; reflecting that it’s a sentence with free variables. Now we can quite easily translate each of the examples, using lambda notation to more conveniently define the necessary functions

``````## ∀x<10 (x + 0 = x)
Aupto(10, lambda x: x + 0 == x)

## ∃x<100 (x + x = x)
Elessthan(100, lambda x: x + x == x)

## ∀x<5 ∃y<7 ((x > 1) → (x*y = 12))
Aupto(5, lambda x: Elessthan(7, lambda y: not (x > 1 and x * y != 12)))

## ∃x<5 ∀y<x ∀z<y (y⋅z ≠ x)
Elessthan(5, lambda x: Aupto(x, lambda y: Aupto(y, lambda z: y * z != x)))``````

Each of these programs, when run, determines whether or not the sentence is true. Hopefully it’s clear how we can translate any sentence with bounded quantifiers into a program of this form. And when we run the program, it will determine the truth value of the sentence in a finite amount of time.

So far, we’ve only talked about the simplest kinds of sentences, with no unbounded quantifiers. There are two names that both refer to this class: Π0 and Σ0. So now you know how to write a program that determines the truth value of any Σ00 sentence!

We now move up a level in the hierarchy, by adding unbounded quantifiers. These quantifiers must all appear out front and be the same type of quantifier (all universal or all existential).

Σ1 sentences: ∃x1 ∃x2 … ∃xk Phi(x1, x2, …, xk), where Phi is Π0.
Π1 sentences: ∀x1 ∀x2 … ∀xk Phi(x1, x2, …, xk), where Phi is Σ0.

Some examples of Σ1 sentences:

∃x ∃y (x⋅x = y)
∃x (x⋅x = 5)
∃x ∀y < x (x+y > x⋅y)

And some Π1 sentences:

∀x (x + 0 = x)
∀x ∀y (x + y < 10)
∀x ∃y < 10 (y⋅y + y = x)

We can translate unbounded quantifiers as while loops:

``````def A(phi):
x = 0
while True:
if not phi(x):
return False
x += 1

def E(phi):
x = 0
while True:
if phi(x):
return True
x += 1``````

There’s a radical change here from the bounded case, which is that these functions are no longer guaranteed to terminate. A(Φ) never returns True, and E(Φ) never returns False. This reflects the nature of unbounded quantifiers. An unbounded universal quantifier is claiming something to be true of all numbers, and thus there are infinitely many cases to be checked. Of course, the moment you find a case that fails, you can return False. But if the universally quantified statement is true of all numbers, then the function will have to keep searching through the numbers forever, hoping to find a counterexample. With an unbounded existential quantifier, all one needs to do is find a single example where the statement is true and then return True. But if there is no such example (i.e. if the statement is always false), then the program will have to search forever.

I encourage you to think about these functions for a few minutes until you’re satisfied that not only do they capture the unbounded universal and existential quantifiers, but that there’s no better way to define them.

Now we can quite easily translate our example sentences as programs:

``````## ∃x ∃y (x⋅x = y)
E(lambda x: E(lambda y: x * x == y))

## ∃x (x⋅x = 5)
E(lambda x: x * x == 5)

## ∃x ∀y < x (x+y > x⋅y)
E(lambda x: Aupto(x, lambda y: x + y > x * y))

## ∀x (x + 0 = x)
A(lambda x: x + 0 == x)

## ∀x ∀y (x + y < 10)
A(lambda x: A(lambda y: x + y < 10))

## ∀x ∃y < 10 (y⋅y + y = x)
A(lambda x: Elessthan(10, y * y + y == x))``````

The first is a true Σ1 sentence, so it terminates and returns True. The second is a false Σ1 sentence, so it runs forever. See if you can figure out if the third ever halts, and then run the program for yourself to see!

The fourth is a true Π1 sentence, which means that it will never halt (it will keep looking for a counterexample and failing to find one forever). The fifth is a false Π1 sentence, so it does halt at the first moment it finds a value of x and y whose sum is 10. And figure out the sixth for yourself!

The next level of the hierarchy involves alternating quantifiers.

Σ2 sentences: ∃x1 ∃x2 … ∃xk Φ(x1, x2, …, xk), where Φ is Π1.
Π2 sentences: ∀x1 ∀x2 … ∀xk Φ(x1, x2, …, xk), where Φ is Σ1.

So now we’re allowed sentences with a block of one type of unbounded quantifier followed by a block of the other type of unbounded quantifier, and ending with a Σ0 sentence. You might guess that the Python functions we’ve defined already are strong enough to handle this case (and indeed, all higher levels of the hierarchy), and you’re right. At least, partially. Try running some examples of Σ2 or Π2 sentences and see what happens. For example:

``````## ∀x ∃y (x > y)
A(lambda x: E(lambda y: x > y))``````

It runs forever! If we were to look into the structure of this program, we’d see that A(Φ) only halts if it finds a counterexample to Φ, and E(Φ) only halts if it finds an example of Φ. In other words A(E(Φ)) only halts if A finds out that E(Φ) is false; but E(Φ) never halts if it’s false! The two programs’ goals are diametrically opposed, and as such, brought together like this they never halt on any input.

The same goes for a sentence like ∃x ∀y (x > y): for this program to halt, it would require that ∀y (x > y) is found to be true for some value of x, But ∀y (x > y) will never be found true, because universally quantified sentences can only be found false! This has nothing to do with the (x > y) being quantified over, it’s entirely about the structure of the quantifiers.

No Turing machine can decide the truth values of Σ2 and Π2 sentences. There’s a caveat here, related to the semantic version of the arithmetic hierarchy. It’s often possible to take a Π2 sentence like ∀x ∃y (y + y = x) and convert it to a logically equivalent but Π1 sentence like ∀x ∃y<x (y + y = x). This translation works, because y + y = x is only going to be true if y is less than or equal to x. Now we have a false Π1 sentence rather than a false Π2 sentence, and as such we can find a counterexample and halt.

We can talk about a sentence’s essential level on the arithmetic hierarchy, which is the lowest level of the logically equivalent sentence. It’s important to note here that “logically equivalent sentence” is a cross-model notion: A and B are logically equivalent if and only if they have the same truth values in every model of PA, not just the standard model. The soundness and completeness of first order logic, and the recursive nature of the axioms of PA, tells us that the set of sentences that are logically equivalent to a given sentence of PA is recursively enumerable. So we can generate these sentences by searching for PA proofs of equivalence and keeping track of the lowest level of the arithmetic hierarchy attained so far.

Even when we do this, we will still find sentences that have no logical equivalents below Σ2 or Π2. These sentences are essentially uncomputable; not just uncomputable in virtue of their form, but truly uncomputable in all of their logical equivalents. However, while they are uncomputable, they would become computable if we had a stronger Turing machine. Let’s take another look at the last example:

``````## ∀x ∃y (x > y)
A(lambda x: E(lambda y: x > y))``````

Recall that the problem was that A(E(Φ)) only halts if E(Φ) returns False, and E(Φ) can only return True. But if we had a TM equipped with an oracle for the truth value of E(Φ) sentences, then maybe we could evaluate A(E(Φ))!

Let’s think about that for a minute more. What would an oracle for the truth value of Σ1 sentences be like? One thing that would work is if we could run E(Φ) “to infinity” and see if it ever finds an example, and if not, then return False. So perhaps an infinite-time Turing machine would do the trick. Another way would be if we could simply ask whether E(Φ) ever halts! If it does, then ∃y (x > y) must be true, and if not, then it must be false.

So a halting oracle suffices to decide the truth values of Σ1 sentences! Same for Π1 sentences: we just ask if A(Φ) ever halts and return False if so, and True otherwise.

If we run the above program on a Turing machine equipped with a halting oracle, what will we get? Now we can evaluate the inner existential quantifier for any given value of x. So in particular, for x = 0, we will find that Ey (x > y) is false. We’ve found a counterexample, so our program will terminate and return False.

On the other hand, if our sentence was true, then we would be faced with the familiar feature of universal quantifiers: we’d run forever looking for a counterexample and never find one. So to determine that this sentence is true, we’d need an oracle for the halting problem for this new more powerful Turing machine!

Here’s a summary of what we have so far:

TM = Ordinary Turing Machine
TM2 = TM + oracle for TM
TM3 = TM + oracle for TM2

The table shows what type of machine suffices to decide the truth value of a sentence, depending on where on the arithmetic hierarchy the sentence falls and whether the sentence is true or false.

We’re now ready to generalize. In general, Σn sentences start with a block of existential quantifiers, and then alternate between blocks of existential and universal quantifiers n – 1 times before ending in a Σ0 sentence. Πn sentences start with a block of universal quantifiers, alternates quantifiers n – 1 times, and then ends in a Σ0 sentence. And as you move up the arithmetic hierarchy, it requires more and more powerful halting oracles to decide whether sentences are true:

If we define Σω to be the union of all the Σ classes in the hierarchy, and Πω the union of the Π classes, then deciding the truth value of Σω ⋃ Πω (the set of all arithmetic sentences) would require a TMω – a Turing machine with an oracle for TM, TM2, TM3, and so on. Thus the theory of true arithmetic (the set of all first-order sentences that are true of ℕ), is not only undecidable, it’s undecidable with a TM2, TM3, and TMn for every n ∈ ℕ. At every level of the arithmetic hierarchy, we get new sentences that are essentially on that level (not just sentences that are superficially on that level in light of their syntactic form, but sentences which, in their simplest possible logically equivalent form, lie on that level).

This gives some sense of just how hard math is. Just understanding the first-order truths of arithmetic requires an infinity of halting oracles, each more powerful than the last. And that says nothing about the second-order truths of arithmetic! That would require even stronger Turing machines than TMω – Turing machines that have halting oracles for TMω, and then TMs with oracles for that, and so on to unimaginable heights (just how high we must go is not currently known).

# The Rise of Anti-Set Theory

This report details the discovery and proliferation of a radical new form of set-theory that has come to be known across math departments across the world as Anti-Set Theory. The earliest appearance of Anti-Set Theory is shrouded in mystery, but my best efforts have traced it to the work of two little-known logicians named Narl Cowman and Bishi Ranger. The germinal notion that would eventually grow into Anti-Set Theory was the idea of a theory of sets obtained by taking all of our familiar intuitions about sets, and require their exact opposites to hold. Any property that would hold of all sets, would of necessity not hold of any the objects in this theory. An attractive idea by any standard, and one self-evidently worthy of pursuit. And thus Anti-Set Theory was born.

The development of Anti-Set Theory progressed in three phases: the Era of Simple Negation (which was plagued with unattractively permissive axioms and a proliferation of models), the Age of Negation after the Universal Quantifier (where the previous problems were resolved, but new problems of consistency and paradox arose), and finally the Modern Age of Axiomatic Anti-Set Theory (where there arose the split between “revolutionary” and “traditionalist” Anti-Settists, as they came to be known). Each of these phases will be explained in more depth in what follows.

The Era of Simple Negation

The first formulations of Anti-Set Theory suffered for practical reasons that were obvious in retrospect, but took embarrassingly long to remedy. The early anti-settists naively took their project to be to simply take the axioms of existing set theories and negate each of them, then to collect them into a new axiomatic theory and explore the consequences. While the idea to build off of existing set theories was a worthy one, there was a serious problem with this approach. Namely, the axioms of existing set theories were designed to be quite strong and restrictive, which made their negations overly weak and permissive. Many axioms had the form “All sets have this property”, the negation of which becomes “At least one set doesn’t have this property”, not “NO sets have this property.” The latter was the ambition of the original anti-settists, but it would take them until the Age of Negation After the Universal Quantifier about a half hour later to realize the crucial error they were making.

For instance, in the axioms of ZF (the orthodox axiomatization of set theory until Anti-Set Theory took the world by storm), there is an axiom known as “the Axiom of Pairing.” This axiom says, in plain English, that for any two sets x and y, there exists a set containing just x and y and nothing else (which we typically write as {x, y}). In the language of first-order logic, it said “∀x∀y∃z∀w (w∈z ↔ (w=x ∨ w=y))” (or in a more readable short-hand, “∀x∀y∃z (z = {x, y})”). Anti-Settists saw clearly the obvious benefit of having a set theory with the opposite property, namely that for any two sets x and y, the pair set {x, y} must NOT exist. But their original formulation of the Axiom of Anti-Pairing used simple negation: “¬∀x∀y∃z (z = {x, y})”, which is equivalent to “∃x∃y∀z (z ≠ {x, y})”, or “there are two sets x and y such that no set contains just x and y.” This was obviously too weak for what was desired.

The same problem cropped up for anti-union, anti-comprehension, anti-foundation, and virtually every other axiom of ZF. A new idea was needed. And soon enough, it came: we should place our negations AFTER the initial block of universal quantifiers, not before. The brilliance of this breakthrough is hard to overestimate. Fields Medals were awarded and backs were patted. This brings us to the second phase of the history of Anti-Set Theory.

The Age of Negation After the Universal Quantifier

Now that the Anti-Settists knew how they were going to proceed, they got busy axiomatizing their new theory. Every set has a power set? No more, now no set has a power set. The union of each set exists? Nope! Unions are banned. Infinite sets? Absolutely not.

The first few axioms to be proposed were anti-pairing, anti-union, anti-powerset, and anti-foundation and they took the following form:

Anti-Pairing: ∀x∀y¬∃z∀w (w ∈ z ↔ (w = x ∨ w = y))
Anti-Union: ∀x¬∃y∀z (z ∈ y ↔ ∃w (w ∈ x ∧ z ∈ w))
Anti-Powerset: ∀x∀y∃z (z ⊆ y ∧ z ∉ x)
Anti-Foundation: ∀x∀y (y ∈ x → ∃z (z ∈ x ∧ z ∈ y))
Anti-Infinity: ∀x¬∃y∀z (z ∈ x → ∀w (w = z⋃{z} → w ∈ y))

But soon they began running into trouble. The first hint that something was amiss appeared with the Axiom of Anti-Comprehension. The Axiom of Comprehension says that for any property Φ definable as a sentence of first-order ZF, and for any set X, there exists the subset of X consisting of those elements that satisfy Φ. Anti-Comprehension could thus be written:

Anti-Comprehension: ∀x∀y (y ≠ {z∈x | φ(z)})

Perhaps you can see the problem. What if φ is a tautology? Or in other words, what if φ(z) is a property satisfied by ANY and EVERY set, like, say, z=z? Then our axiom tells us that for any set x, the subset consisting of those elements that are equal to themselves cannot exist. But that’s just x itself! In other words, for any set x, x does not exist. The first Anti-Settists had unwittingly destroyed the entire universe of sets!

Another problem arose with Anti-Replacement. Recall that Replacement says that for any function F(x) definable in a sentence of first-order ZF, and for any set X, the image of X under F exists. So Anti-Replacement states that this image does not exist. But what if F is the identity function? Then Anti-Replacement tells us that the image of X under the identity map (namely, X itself) does not exist. And again, we’ve proved that no sets exist.

Quite simply, the problem the Anti-Settists were running into was that while their original axioms were too weak, their new axioms were too strong! So strong that they DESTROYED THE UNIVERSE. That’s strong.

The solution? Anti-Comprehension and Anti-Replacement were discarded. But there was another problem in the Axiom of Anti-Extensionality. Extensionality says that any two sets are the same if they share all the same elements. So Anti-Extensionality says that if two sets share all the same elements, then they are not the same. But if we compare any set X with itself using this standard, we find that X cannot equal X! This was even worse than before, because it violates the first-order tautology ∀x (x = x). Not only did this second wave of Anti-Settists destroy the universe, they also broke the rules of logic!

So Anti-Extensionality had to go. That much everybody agreed on. And the removal of this axiom settled the last of the paradoxes of Anti-Set Theory. But no sooner had the dust settled than a new controversy arose as to the nature of Extensionality…

The Great Schism: The Modern Age of Anti-Set Theory

So far, the Anti-Settists had compiled the following list of axioms:

Anti-Pairing: ∀x∀y¬∃z∀w (w ∈ z ↔ (w=x ∨ w=y))
Anti-Union: ∀x¬∃y∀z (z∈y ↔ ∃w (w∈x ∧ z∈w))
Anti-Powerset: ∀x∀y∃z (z⊆y ∧ z∉x)
Anti-Foundation: ∀x∀y (y∈x → ∃z (z∈x ∧ z∈y))
Anti-Infinity: ¬∃x∀y (y ∈ x → ∀z (z = y⋃{y} → z ∈ x))

You might have noticed that the Axiom of Anti-Infinity looks a little strange. The Axiom of Infinity tells us that there’s a set X that contains all empty sets, and such that for any set Y contained in X, if Y has a successor then that successor is also in X. So the Axiom of Anti-Infinity will say that there is no such set. We’ll prove shortly that in fact, the other axioms entail that there are no empty sets, so it’s vacuously true of every set that it “contains all empty sets.” Thus the only restriction placed on our universe by the Axiom of Anti-Infinity is that there’s no set that contains the successors of all its members with successors.

For a while, everybody agreed on this list of axioms and all was well. But after a couple of hours, new rumbles arose about Extensionality. A new contingent of Anti-Settists began arguing in favor of including the Axiom of Extensionality. It’s hard to describe exactly what was going in the minds of these individuals, who appeared to be turning their backs on everything that Anti-Set Theory was all about by accepting an ORDINARY axiom alongside all of their beautiful Anti-Axioms. Some of them expressed a concern that they had gone too far by turning their back on extensionality, and worried that Anti-Set Theory was so far removed from our intuitions that it no longer deserved to be called a theory of sets. Others pointed out that Anti-Set Theory as it was currently formulated ruled out certain models that they felt deserved to belong to the pantheon of Anti-Set Universes. The rest of the Anti-Settists called them crazy, but they persisted. The fighting reached fever pitch one afternoon in a meeting of the leading Anti-Settists, where one individual who shan’t be named accused another of “selling out” to Traditional Set Theory, and a fist-fight nearly broke out. This led to the Great Schism.

Anti-Settists fractured into two contingents that became known as the traditionalists, who advocated an extensional Anti-Set Theory, and the revolutionaries, who wanted an intensional Anti-Set Theory where two sets could share all the same elements but still be different. In recent history the fighting has cooled off, probably because both sides noticed that there actually didn’t seem to be all that huge of a difference between extensional and intensional anti-set theory. The primary realization was that the other axioms banned any objects based off of the elements they contained (so that anti-pairing, for instance, really says that there can be no set with the same elements as the pair of x and y, not just that the pair of x and y doesn’t exist).

Extensional Anti-Set Theory
Extensionality: ∀x∀y (∀z (z ∈ x ↔ z ∈ y) → x = y)
Anti-Pairing: ∀x∀y¬∃z∀w (w ∈ z ↔ (w = x ∨ w = y))
Anti-Union: ∀x¬∃y∀z (z ∈ y ↔ ∃w (w ∈ x ∧ z ∈ w))
Anti-Powerset: ∀x∀y∃z (z⊆y ∧ z∉x)
Anti-Foundation: ∀x∀y (y ∈ x → ∃z (z ∈ x ∧ z ∈ y))

Intensional Anti-Set Theory
Anti-Pairing: ∀x∀y¬∃z∀w (w ∈ z ↔ (w = x ∨ w = y))
Anti-Union: ∀x¬∃y∀z (z ∈ y ↔ ∃w (w ∈ x ∧ z ∈ w))
Anti-Powerset: ∀x∀y∃z (z⊆y ∧ z∉x)
Anti-Foundation: ∀x∀y (y ∈ x → ∃z (z ∈ x ∧ z ∈ y))

That covers the history of Anti-Set Theory up to modern times. Let’s now take a look at some of the peculiar details of the theory.

Theorem 1: No anti-sets contain exactly one element.

Proof: Suppose that there was an anti-set X such that X contained Y and nothing else. Then the sentence “Aw (w ∈ X ↔ (w = Y ∨ w = Y))” would be true. But this would be a violation of anti-pairing, as there can be no set whose elements are the same as the pair {Y, Y}. So no such anti-set can exist.

Theorem 2: No anti-sets contain exactly two elements.

Proof: Suppose there was an anti-set X such that X contained Y, Z, and nothing else. Then the sentence “Aw (w ∈ X ↔ (w = Y ∨ w = Z))” would be true. But this would be a violation of anti-pairing. So no such anti-set can exist.

Theorem 3: No anti-sets contain an empty anti-set.

Proof: Suppose that some set X contained a set Y, where Y is empty. By Anti-Foundation, every element of X must share an element with X. So Y must share some element with X. But Y contains nothing, so it can’t share any element with X. Contradiction. So no such anti-set exists.

Theorem 4: No anti-set can be its own union.

Proof: Follows trivially from the axiom of Anti-Union: UX doesn’t exist for any x, so if X = UX, then X doesn’t exist. (This rules out anti-sets with the same structure as limit ordinals.)

Theorem 5: No empty anti-sets exist.

Proof: Suppose there exists an empty anti-set X. Then the union of X is also an empty anti-set. The Axiom of Anti-Union says that there can be no anti-set with the same elements as the union of any anti-set. So there can be no empty anti-set. Contradiction, so no empty anti-set exists.

These theorems tell us a lot about the structure of anti-sets. In particular, every anti-set contains at least three elements, and each of those elements in turn contains at least three elements, and so on forever. So we only have infinite descending membership-chains of anti-sets.

Also, we’ve ruled out anti-sets with zero, one, and two elements, so you might think that no three-element anti-sets can exist either. But the same argument we used for one- and two-element anti-sets doesn’t work any longer, since pairing never produces three-element sets. In fact, the first model of Anti-Set Theory discovered contained exactly five anti-sets, each of which had three elements. This model is an intensional model, as it’s crucial that two of the sets (C and X) contain the same elements. We’ll call this the Primordial Anti-Set Model.

Universe: P, A, B, C, X
P = {A, B, C}
A = {B, C, X}
B = {A, C, X}
C = {A, B, X}
X = {A, B, X}

Here’s two attempts to visualize this model:

Beautiful! Let’s check each of the axioms to convince ourselves that it really is a valid model.

Anti-Union: U(P) = A ⋃ B ⋃ C = {A, B, C, X}. Similarly, you can show that U(A), U(B), U(C), and U(X) all contain A, B, C, and X. And {A, B, C, X} is not an anti-set in the universe, so no violations of Anti-Union occur.

Anti-Pairing: This one’s easy, since each of our sets contains three elements, and no three-element set can be formed by pairing.

Anti-Powerset: Note that the axiom of power set only says “there’s a set containing all the existing sets that are subsets of X, for each X”, but it doesn’t bring into existence any new subsets. So for instance, in this model, the power set of P is just {P}, as P is the only subset of P that exists. But {P} doesn’t exist! Similarly, for each element, its power set is just the set of itself, which doesn’t exist by anti-pairing.

Anti-Foundation: Does any set contain an element that it doesn’t have anything in common with? You can just verify this by inspection.

Anti-Infinity: (EDIT: this model and the following one actually violate Anti-Infinity! Realized this after originally posting. A future post will provide more details.)

Now, this was a model only of Intensional Anti-Set Theory, not Extensional Anti-Set Theory. Here’s a way to modify it to get a model that works in both theories:

Universe: T, A, B, C
T = {A, B, C}
A = {T, B, C}
B = {A, T, C}
C = {A, B, T}

Again, convince yourself that this universe satisfies the axioms.

Using a similar construction, we can show that there are models of anti-set theory with exactly N sets, for every N ≥ 4. Our universe: T, A1, A2, …, AN-1. T = {An | n ∈ {1,2,…,N-1}, Ak = {T} ⋃ {Aj | j ∈ {1,2,…,N-1} \ {k}}.

Many lines of research remain open in the field of Anti-Set Theory. Can Anti-Sets serve as a new foundation for mathematics? Are there models of Anti-Set Theory that contain structures isomorphic to the natural numbers? Is the theory still consistent if we include an Axiom of Anti-Choice? (Adding Anti-Choice rules out the models we’ve discussed so far. Perhaps only infinite universes are allowed with Anti-Choice.) Other proposed additions to the axioms include the simple negation of extensionality, and an “Axiom of Undefinability”, which says that no set exists whose elements satisfy a definable property. How would these additions affect the universe of sets?

I want to close by noting that set theory sometimes gets a bad rap among mathematicians and the wider community for being too abstract or not “useful enough.” We hope that the advent of Anti-Set Theory will make it plain to the public that there is a future world in which set theory can be of GREAT practical use to everybody. The possible applications of Anti-Set Theory are too numerous to count, so numerous in fact that I regard as unnecessary naming any specific examples.

# The Transfinite Factorial

Chapter 1: When Circularity is not Paradoxical

Circular definitions can be quite bad. Take the definition “a is the natural number that is equal to a”, or worse, “a is the natural number that is NOT equal to a”. Or, for an even more perverse one, “a+1, where a is the smallest number that can be defined in fewer than 50 words.” But just as not all self-reference is paradoxical, similarly not all circularity is vicious. Circular definitions form a crucial component of every area of math, where they go under the more polite term “recursion.”

The key component of a benign circular definition is that it avoids infinite regress. So “a=a” is unhelpful, because to evaluate a you must first know what a evaluates to. But what about “an = n ⋅ an-1, and a0 = 1″? Now to know a500, you only need to know a499, which in turn only requires you know a498, which requires… and so on, each step moving a little closer to the base case “a0 = 1″, at which point you can travel all the way back up the chain to discover that a500 = 500!. (That’s 500 factorial, not an excitable 500.)

Douglas Hofstadter has a parable about the “porpuquine”, a rare cousin of the porcupine that grows smaller porpuquines on its back rather than quills. The market value of such a beast is determined by the total number of porpuquine noses you can find on it. And if you’re worried about an infinite regress, don’t fear! There’s a smallest porpuquine, which is entirely bald.

Recursive definitions are extremely nice when we want to define a function whose outputs are related to each other in a more simple way than they are to the inputs. Some examples (in which S stands for the successor function, which takes 0 to 1 to 2 to …):

∀x (x + 0 = x)
∀x∀y (x + S(y) = S(x+y))

Multiplication
∀x (x ⋅ 0 = 0)
∀x∀y (x ⋅ S(y) = x⋅y + x)

Notice the similarity between these two definitions. In each case we define what happens when we add 0 to a number, and then we define what happens when we add a successor to a number. This covers all the natural numbers! Let’s show this in action by proving that 9+3 = 12.

9 + 3
9 + S(2)
S(9 + 2)
S(9 + S(1))
S(S(9 + 1))
S(S(9 + S(0)))
S(S(S(9 + 0)))
S(S(S(9)))
S(S(10))
S(11)
12

The colorful step is where we hit our base case (evaluating x+0), at which point we need merely to pass our answers up the levels to get back to the solution of our original problem.

Another: Let’s show that 3⋅2 = 6.

3⋅2
3⋅S(1)
3⋅1 + 3
3⋅S(0) + 3
(3⋅0 + 3) + 3
(0 + 3) + 3
(0 + S(2)) + 3
S(0 + 2) + 3
S(0 + S(1)) + 3
S(S(0 + 1)) + 3
S(S(0 + S(0))) + 3
S(S(S(0 + 0))) + 3
S(S(S(0))) + 3
S(S(1)) + 3
S(2) + 3
3 + 3
3 + S(2)
S(3 + 2)
S(3 + S(1))
S(S(3 + 1))
S(S(3 + S(0)))
S(S(S(3 + 0)))
S(S(S(3)))
S(S(4))
S(5)
6

Each green line is where we use a “base case rule” – the first is the base case for multiplication, the second and third for addition. I like that you can see the complexity growing in the length of the line until you hit the base case, at which point the lines start shortening until we next need to apply the recursive step.

Chapter 2: Infinite Recursion

Now what happens if we want to add x + y but y is neither 0 nor a successor. “Huh? Isn’t every non-zero number a successor of some number?” I hear you ask. Not infinity! Or more precisely, not ω. Our recursive definitional scheme is all well and good for finite numbers, but it fails when we enter the realm of transfinite ordinals.

No problem! There’s an easy patch. The problem is that we don’t know how to add or multiply x and y when y is a limit ordinal. So we just add on an extra rule to cover this case!

α + 0 = α
α + S(β) = S(α + β)
α + λ = U{α + β | β ∈ λ}

Multiplication
α ⋅ 0 = 0
α ⋅ S(β) = α⋅β + α
α + λ = U{α ⋅ β | β ∈ λ}

If you’re scratching your head right now at the notion of “the union of a set of numbers” or “a number being an element of another number”, go read this for a thorough summary. In short, in ZFC we encode each number as the set of all smaller numbers (e.g. 3 = {0, 1, 2}), and ω (the first infinite number) is defined to be the set of all natural numbers (ω = {0, 1, 2, 3, …}). So 3 ⋃ 5 = {0,1,2} ⋃ {0,1,2,3,4} = {0,1,2,3,4} = 5. In general, the union of a set of numbers is just the maximum number among them (or if there is no maximum number, then it’s the supremum of all the numbers).

Outside the context of ZFC, we can equally write α + λ = lim (α + β) as β → λ. We can also write α + λ = sup {α + β | β < λ}. And for those more familiar with ZFC, we don’t need to take the union over all β ∈ λ, we can make do with any cofinal subset (we’ll do this to simplify calculations greatly later in the post).

Now we’ve extended + and × into the infinite, and what we get is a little unusual. The first major sign that things have changed is that addition and multiplication are no longer commutative.

1 + ω = ⋃{1 + n: n ∈ ω} = U{1, 2, 3, 4, …} = ω
ω + 1 = ω + S(0) = S(ω + 0) = S(ω)

ω certainly doesn’t equal S(ω), so ω + 1 ≠ 1 + ω.

Distributivity also fails! For instance (ω + 3) ⋅ 2 = ω⋅2 + 3, not ω⋅2 + 6. But we don’t lose every nice algebraic feature. Associativity still holds for both addition and multiplication! I.e. α + (β + γ) = (α + β) + γ and α ⋅ (β ⋅ γ) = (α ⋅ β) ⋅ γ. This result is really important.

Earlier we showed that 1 + ω = ω. One can easily show that in general, n + ω = ω for any finite n, and the same holds for α + β is “infinitely smaller” than β. This notion of “infinitely smaller” is a little subtle; ω is not infinitely smaller than ω⋅2, but it is infinitely smaller than ω2. (It should be noted that everything up to here has been standard textbook stuff, but from here on I will be introducing original notions that I haven’t seen elsewhere. So treat what follows this with skepticism.)

A useful definition for us will be α << β (read as “α is infinitely less than β”) if α + β = β. We’ll also say that α and β are “comparable” if neither is infinitely smaller than the other: α ~ β if ¬(α << β) and ¬(β << α). Some conjectures I have about the comparability relation:

Conjecture 1: Commutativity holds between comparable ordinals.

Conjecture 2: α ~ β if and only if there’s some finite n such that α < β⋅n and β < α⋅n.

I’m actually fairly confident that Conjecture 1 is false, and would love to see a counterexample.

Here are some more useful identities for transfinite addition and multiplication:

Theorem 1: For finite n and limit ordinal λ, α⋅(λ + n) = α⋅λ + α⋅n

Proof sketch: α⋅(λ + n) = α⋅(λ + (n-1)) + α = α⋅(λ + (n-2)) + α + α = … = α⋅(λ + 0) + α + α + … + α = α⋅λ + α⋅n.

Theorem 2: For finite n and β << α, (α + β)⋅n = α⋅n + β.

Proof sketch: (α + β)⋅n = (α + β)⋅(n-1) + (α + β) = (α + β)⋅(n-2) + (α + β) + (α + β) = … = (α + β)⋅0 + (α + β) + … + (α + β) = (α + β) + (α + β) + … + (α + β) + (α + β) = α + (β + α) + (β + … + α) + β = α + α + α + … + α + β = α⋅n + β.

In case it’s not clear what’s going with this proof, the idea is that we expand (α + β)⋅n out to (α + β) + … + (α + β) (n times). Then we shift parentheses to enclose all the middle βs, making them look like (β + α). But then by assumption, since β << α, the β vanishes in each of these middle terms, leaving just the final one: α + α + α + … + α + β = α⋅n + β.

Infinite exponentiation, tetration and so on can be defined in the exact same way as addition and multiplication: first cover the 0 case, then the successor case, and finally the limit case. And the limit case is defined by just taking the union of all smaller cases.

Chapter 3: Transfinite Factorials

Now, what’s the first function you think of when you think of recursively defined functions? That’s right, the factorial! I mentioned it at the beginning of this post, long ago: n! = n⋅(n-1)! and 0! = 1. So let’s extend this definition to obtain a transfinite factorial!

Factorial
0! = 1
S(α)! = α! ⋅ S(α)
λ! = U{β! | β ∈ λ}

Now we can ask ω! is.

Theorem: ω! = ω

Proof: ω! = U{n! | n ∈ ω} = U{1, 2, 6, 24, 120, …} = ω (since {1, 2, 6, 24, 120, …} is a cofinal subset of ω).

Next, (ω+1)! = ω! ⋅ (ω + 1) = ω ⋅ (ω + 1) = ω⋅ω + ω⋅1 = ω2 + ω. Note that to prove this, we’ve used Theorem 1 from before.

And we can keep going from there.

Theorem: (ω + n)! = ωn+1 + ωn⋅n + ωn-1⋅(n – 1) + … + ω2⋅2 + ω.

Try to prove this for yourself!

Now that we know what (ω + n)! is for each finite n, we are able to calculate what (ω⋅2)! is (using the fact that {ω + n | n ∈ ω} is a cofinal subset of ω⋅2). I’ll write its value, as well as the next few interesting ordinals:

(ω⋅2)! = U{ω!, (ω+1)!, (ω+2)!, …} = ωω
(ω⋅3)! = ωω⋅2
(ω⋅n)! = ωω⋅(n-1)

Using the same trick as before, this enables us to solve (ω2)!

2)! = U{ω!, (ω⋅2)!, (ω⋅3)!, …} = U{ω, ωω, ωω⋅2, ωω⋅3, …} = ωω2

From here, things suddenly take on a remarkable regularity (for a while at least). Looking at limit ordinals above ω2, they appear to have the following structure, at least up until we arrive at the first ordinal where standard notation fails us: ωωω = ε0.

Conjecture: For λ a limit ordinal such that ω2 ≤ λ < ε0, λ! = ωλ

If this conjecture is right, then (ωω)! = ωωω, and (ωωω)! = ωωωω, and so on. We can also write this using tetration notation by saying that (nω)! = n+1ω.

But something different happens once we get to ε0!

Conjecture: ε0! = U{(nω)! | n ∈ ω} = U{ω, 2ω, 3ω, 4ω, …} = ε0

If this is right, then this is pretty interesting behavior! The fixed points of the factorial function are 1, ω, and ε0, at the very least. What happens after ε0? The factorial function “loops” in a certain sense! Just as (ω+1)! was ω2 + ω, so (ε0+1)! ends up being ε02 + ε0. And (ε0⋅2)! = ε0ε0, paralleling (ω⋅2)! = ωω. And so on.

Conjecture: εn is a fixed point of the factorial function for all n.

There are plenty more questions to be answered. Is ω1 another fixed point of the factorial function? What exactly does the list of all the fixed points look like? Are there any cardinals that aren’t equal to their own factorial? What exactly is the general pattern for factorials of successor ordinals? How does α! compare to αα in general?

# Finiteness can’t be captured in a sound, complete, finitary proof system

Consider the sentence “This blog has finitely many posts.” Do you understand what this sentence means? Then no set of rules (even infinite, even uncomputable!) can model your reasoning. This claim may sound shocking, but it can be justified on solid metamathematical grounds.

Another example: the sentence “There are finitely many planets in the universe.” You don’t have to think it’s true, you just have to think you understand what it means. What’s the common theme? It’s the notion of there being ‘finitely many’ of some class of objects. Let’s imagine building a language that has the expressive resources of first-order logic (which are quite modest), plus an additional quantifier F, whose semantics are given by the following rule: Fx φ(x) is satisfied by a model iff there are only finitely many objects in that model that satisfy φ(x).

It turns out that any language consisting of first order logic plus the quantifier F can’t be axiomatized in any sound, complete, and finitary proof system. Notice that effective enumerability of the rules of the proof system is not a requirement here! So long as the language is strong enough to express the semantics of {∧, ¬, ∀, F, variables xn, and relations Rn}, no set of sentences and sentence-manipulation rules in that language will suffice to capture these semantics.

Here’s a proof: consider the first-order theory of Peano arithmetic. This theory has nonstandard models (as any theory of arithmetic must have in a logic that is compact). All of these nonstandard models have the following feature: that there are numbers that are larger than infinitely many numbers. Think about what this means: this is a common feature of all nonstandards, so if we could write a sentence to express this feature then we could rule them all out! This is where the quantifier F steps in. With F, we can write the following simple sentence and add it to PA as an axiom:

∀x Fy (y < x)

In English: every number has finitely many numbers less than it. And with that, we’ve ruled out all nonstandard models! So now we have a theory that is categorical for ℕ. And that’s a big deal, metamathematically speaking!

Why? Well, as I’ve talked about in a previous post, you can prove some pretty strong limitative results about any logic that can produce a theory that’s categorical for ℕ. In particular, if we can produce such a theory then its logic cannot be compact. Quick proof: suppose a set of sentences Σ models ℕ. Add to Σ a constant c and the axioms “c ≠ 0”, “c ≠ 1”, “c ≠ 2”, and so on, and call this new set Σ’. Every finite subset of Σ’ models ℕ. So by compactness, Σ’ has a model. But this model is nonstandard – it contains numbers that aren’t natural numbers. And since Σ is a subset of Σ’, any model of Σ’ is also a model of Σ.

So compactness implies that no theory is categorical for ℕ. But compactness follows from the following three properties: a sound and complete proof system (Σ ⊢ α if and only if Σ ⊨ α), and that all proofs are only finitely long (try expressing this property without F!). Quick proof: If a set of sentences is finitely satisfied, then every finite subset of it has a model (by definition), so no finite subset of it can be refuted (by soundness), so the entire set can’t be refuted (by finite proofs), so the entire set is satisfied (by completeness).

So soundness + completeness + finiteness ⇒ compactness ⇒ the existence of nonstandard models of arithmetic in any theory that models ℕ. Which means that the semantics of F cannot be captured in any sound, complete, and finite proof system!

Take your pick: either you don’t really understand the semantics of the “finitely many” quantifier F, or no set of rules (not even requiring this set to be finite or computable) can fully capture your reasoning in finite-length proofs.

More information about related extensions of first-order logic and their limitations can be found here. The result I describe here is a rephrasing of results discussed there.

# The Mindblowing Goodstein Sequences

Goodstein sequences are amazingly strange and unintuitive. I want to tell you all their unusual properties up-front, but I also don’t want to spoil the surprise. So let me start by defining the Goodstein sequence of a number.

We start by defining the hereditary base-n representation of a number m. We first write m as a sum of powers of n (e.g. for m = 266 and n = 2, we write 266 = 28 + 23 + 21). Now we write each exponent as a sum of powers of n. (So now we have 266 = 223 + 22+1 + 21). And we continue this until all numbers in the representation are ≤ n. For 266, our representation stabilizes at 266 = 222+1 + 22+1 + 21.

Now we define Gn(m) as follows: if m = 0, then Gn(m) = 0. Otherwise, Gn(m) is the number produced by replacing every n in the hereditary base-n representation of m by n+1 and subtracting 1. So G2(266) = G2(222+1 + 22+1 + 21) = 333+1 + 33+1 + 31 – 1 ≈ 4.4 × 1038.

Finally, we define the Goodstein sequence for a number n as the following:

a0 = n
a1 = G2(a0)
a2 = G3(a1)
a3 = G4(a2)
And so on.

Let’s take a small starting number and look at its Goodstein sequence. Suppose we start with 4.

a0 = 4 = 22
a1 = 33 – 1 = 2⋅32 + 2⋅3 + 2 = 26
a2 = 2⋅42 + 2⋅4 + 1 = 41
a3 = 2⋅52 + 2⋅5 = 60
a4 = 2⋅62 + 2⋅6 – 1 = 2⋅62 + 6 + 5 = 83
a5 = 2⋅72 + 7 + 4 = 109

Now, notice that the sequence always seems to increase. And this makes sense! After all, each step we take, we increase the base of each exponent by 1 (a big increase in the value) and only subtract 1. Here’s a plot of the first 100 values of the Goodstein sequence of 4:

And for further verification, here are the first million values of the Goodstein sequence of 4:

Looks like exponential growth to me! Which, again, is exactly what you’d expect. In addition, the higher the starting value, the more impressive the exponential growth we see. Here we have the first 100 values of the Goodstein sequence of 6:

Notice how after only 100 steps, we’re already at a value of 5 × 1010! This type of growth only gets more preposterous for larger starting values. Let’s look at the Goodstein sequence starting with 19.

a0 = 19
a1 ≈ 7 × 1012
a2 ≈ 1.3 × 10154
a3 ≈ 1.8 × 102184
And so on.

We can’t even plot a sequence like this, as it’ll just look like two perpendicular lines:

Alright, so here’s a conjecture that hopefully you’re convinced of: Goodstein sequences are always growing. In fact, let’s conjecture something even weaker: Goodstein sequences never terminate (i.e. never go to zero).

We’re finally ready for our first big reveal! Both of these conjectures are wrong! Not only do some Goodstein sequences terminate, it turns out that EVERY Goodstein sequence eventually terminates!

This is pretty mind-blowing. What about all those graphs I showed? Well, the sequences do increase for a very very long time, but they eventually turn around. How long is eventually? The Goodstein sequence of 4 takes 3 × 2402,653,210 – 1 steps to turn around!! This number is too large for me to be able to actually show you a plot of the sequence turning around. But we do know that once the Goodstein sequence stops increasing, it actually stays fixed for 3 × 2402,653,209 steps before finally beginning its descent to zero. Heuristically, it would look something like this:

(Why does it stay constant, incidentally? The reason is that when we get to a number whose hereditary base-B notation just looks like B + n for some n < B, we increase B by 1 and decrease n by 1 each step. This doesn’t change the sequence’s value! So it stays constant until we get to n = 0. At that point, we decrease by 1 until we descend all the way to zero.)

So, how do we prove this? We do it using infinite ordinals. (Now you see why this topic caught my eye recently!) The plan is to associate with Goodstein sequence a “matching sequence” of ordinals. The way to do this is pretty simple: for a number written in hereditary base-n notation, just replace every n in that representation with ω. Now you have an infinite ordinal!

So, for example, here’s the first few numbers in the Goodstein sequence of 4, along with their translation to ordinals:

4 = 22 becomes ωω
26 = 33 – 1 = 2⋅32 + 2⋅3 + 2 becomes 2ω2 + 2ω + 2
41 = 2⋅42 + 2⋅4 + 1 becomes 2ω2 + 2ω + 1
60 = 2⋅52 + 2⋅5 becomes 2ω2 + 2ω
83 = 2⋅62 + 2⋅6 – 1 = 2⋅62 + 6 + 5 becomes 2ω2 + ω + 5
109 = 2⋅72 + 7 + 4 becomes 2ω2 + ω + 4

Examine this sequence of ordinals for a couple of minutes. You might notice something interesting: at each step, the ordinal is decreasing! This turns out to always hold true. Let’s see why.

Notice that there are two cases: either the ordinal is a successor ordinal (like ω2 + 4), or it’s a limit ordinal (like ω2 + ω). If it’s a successor ordinal, then the next ordinal in the sequence is just its predecessor. (So ω2 + 4 becomes ω2 + 3). This is obviously a smaller ordinal.

What if it’s a limit ordinal? Then the smallest limit ordinal in its expanded representation drops to a smaller limit ordinal and some finite number is added. So for instance, in the ordinal ω2 + 2ω, 2ω will drop to ω and some finite number will be added on depending on which step in the sequence you’re at. So we’ll end up with ω2 + ω + n, for some finite n. This ordinal is smaller than the original one, because we’ve made an infinitely large jump downwards (from one limit ordinal to a lower limit ordinal), and only increased by a finite amount.

So in either case, we go to a smaller ordinal. And now the key realization is that every decreasing sequence of ordinals MUST terminate! (This is one of my favorite facts about ordinals, by the way. Even though we’re looking at arbitrarily large infinite ordinals, it’s still the case that there is no decreasing sequence that goes on forever. Whenever we hit a limit ordinal, we have to make an infinitely large jump downwards to get to the next element in the sequence, as the ordinal has no predecessor.)

So we’ve paired every Goodstein sequence with a decreasing sequence of ordinals. And every decreasing sequence of ordinals eventually terminates. So the Goodstein sequence must terminate as well! (One easy way to see this is to notice that every value in a Goodstein sequence is ≤ the ordinal assigned to it. Another way is to notice that the only number that could possibly be assigned to the ordinal 0 is 0 itself. So when the decreasing sequence of ordinals hits 0, as it must eventually, so must the Goodstein sequence) And that’s our proof!

That Goodstein sequences always terminate is the big flashy result for this post. But there’s much more that’s cool about them. For instance, the proof that we just used did not rely on the fact that we just increased the base by 1 at each step. In fact, even if we update the base by an arbitrarily large function at each stage, the sequences still must terminate!

Even cooler, the proof that all Goodstein sequences terminate cannot be formalized in Peano arithmetic! And in fact, no proof can be found of this theorem using PA. Laurie Kirby and Jeff Paris proved in 1982 that the statement that Goodstein sequences always terminate could be reduced to a theorem of Gentzen from which the consistency of PA could be deduced. So if PA could prove that Goodstein sequences always terminate, then it could prove its own consistency! Gödel’s second incompleteness theorem tells us that this cannot be, and thus assuming that PA really is consistent, it cannot prove that that these sequences all terminate.

This was historically the third result to be found independent of PA, and it was the first that was purely number-theoretic in character (as opposed to meta-mathematical, like the Gödel sentences). It gives us a very salient way of expressing the limitations of Peano arithmetic: simply write a program that computes the Goodstein sequence of any given input, terminating only when it reaches 0 and outputting the length of the sequence. (Code Golf Stack Exchange shows that this can be done using just 77 characters of Haskell! And here’s a nice way to do it in Python). Now, it happens to be true that this program will always terminate for any natural number input (if run on sufficiently powerful hardware). But Peano arithmetic cannot prove this!

As a final fun suprise, let’s define G(n) to be the length of the Goodstein sequence starting at n. As we’ve already seen, this sequence gets really big really quickly. But exactly how quickly does it grow? Turns out that it grows much much faster than other commonly known fast-growing computable functions. For instance, G(n) grows much faster than Ackermann’s function, and G(12) is already greater than Graham’s number!

# Wacky Non-Standard Models of PA

Peano arithmetic is unable to pin down the natural numbers, despite its countable infinity of axioms. In fact, assuming its consistency Peano arithmetic has models of every cardinality, meaning that as far as PA is aware, there might be uncountably many real numbers. (If PA is not consistent then it has no models.) I want to take a look at these non-standard models of PA, especially the countable ones. A natural question is, how many countable non-standard models are there alongside the standard models?

Assuming the consistency of PA (which I will leave out from now on, as it’s assumed in all that follows), there are continuum many non-isomorphic countable models. That’s a lot! That means that there’s a distinct countable non-standard model of arithmetic for every real number. This is our first result: the number of nonstandard models of PA of cardinality ℵ0 is 20. Interestingly, this result generalizes! For every infinite cardinal κ, there are 2κ non-isomorphic nonstandard models of cardinality κ. That’s a lot of nonstandard models! In fact, since any model of cardinality κ involves a specification of some number of constants, binary relations over κ and functions from κ to κ, we know that the maximum number of models of cardinality κ is 2κ. So in this sense, Peano arithmetic has as many nonstandard models of each cardinality as it can have!

Let’s take a closer look at the countable non-standard models. It turns out that we can say a lot about their order type. Namely, all countable non standard models of PA have order type ω + (ω*+ω)·η. What does this notation mean? Let me break down each of the order types involved in that formula:

• ω: order type of the naturals
• ω*: order type of the negative integers
• η: order type of the rationals

So ω*+ω is the order type of the integer line, and (ω*+ω)·η is the order type of a structure that resembles an integer line for each rational number. Thus, every countable non-standard model of PA looks like a copy of natural numbers followed by as many copies of integers as there are rational numbers. The order on this structure is lexicographic: two nonstandards on the same integer line are judged according to their position on the integer line, and two nonstandards on different integer lines are judged according to the position of these integer lines on the rational line. It’s not the easiest thing to visualize, but here’s my attempt:

So if all countable non-standard models have the same order type, then where do they differ? It turns out that they differ in the details of how addition and multiplication work on the non-standards. (After all, a model of PA is defined by the size of its universe and its interpretation of ≤, +, and ×. Same order type means same ≤, so what’s left to vary is + and ×.) In each of the models, + and × work exactly like normal on the naturals. And + and × must operate on the non-standards in such a way as to maintain the truth of all the axioms of PA. So, for instance, since PA can prove that 2x = x + x, the same must be true for nonstandard x. And so on. But even given these restrictions, there are still uncountably many ways to define + and × on the non-standards.

What’s more, we have a theorem known as the Tennenbaum Theorem, which tells us that it’s impossible to give recursive definitions to + and × in non-standard models. Said more simply, addition and multiplication are uncomputable in every non-standard model of arithmetic!

One thing I remain unsure of is how the nonstandard models of Peano arithmetic compare to the structure of ω in nonstandard models of ZFC. We know that there must be models of ZFC where ω is nonstandard, by the compactness theorem (define ZFC* to be ZFC with an extra constant c, with the extra axioms “c ∈ ω”, “c ≠ 0”, “c ≠ 1”, “c ≠ 2”, and so on. ZFC* has a model by compactness, and this model is also a model of ZFC by monotonicity.) But is ZFC “better” at ruling out nonstandard models of the naturals than PA is? Or is there a nonstandard ω for every nonstandard model of Peano arithmetic?

# Constructing the Church-Kleene Ordinal

I want to share a way to construct the order type of the Church-Kleene ordinal ω1CK. As far as I’m aware, this construction is original to me, and I think it’s much simpler and more elegant than others that I’ve found on the internet. Please let me know if you see any issues with it!

So, we start with the notion of a computable ordinal. A computable ordinal is a computable well-order on the natural numbers. In other words, it’s a well-order < on ℕ such that there exists a Turing machine that, given any two natural numbers n and m, outputs whether n < m according to that order.

There’s an injection from computable well-orders on ℕ to Turing machines, so there are countably many computable well-orders (at most one for each Turing machine). This means that there’s a list of all such computable well-orders. Let’s label the items on this list in order: {<0, <1, <2, <3, …}. Each <n stands for a Turing machine which takes in two natural number inputs and returns true or false based on the nth computable well-ordering.

Now, we want to construct a set with order type ω1CK. We’re going to have it be an order on the natural numbers, so what we have to work with from the start is the set {0, 1, 2, 3, 4, 5, 6, …}.

We start by splitting up this set into an infinite collection of sets. First, we take every other number, starting with 0: {0, 2, 4, 6, 8, …}. That’ll be our first set. We’ve left behind {1, 3, 5, 7, 9, …}, so next we take every other number from this set starting with 1: {1, 5, 9, 13, …}. Now we’ve left behind {3, 7, 11, 15, …}. Our next set, predictably, will be every other number from this set starting with 3. This is {3, 11, 19, 27, …}. And we keep on like this forever, until we’ve exhausted every natural number.

This gives us the following set of sets:

{{0, 2, 4, 6, …}, {1, 5, 9, 13, …}, {3, 11, 19, 27, …}, {7, 23, 39, 47, …}, …}

Now, we define our order to make everything in the first set less than everything in the second set, and everything in the second set less than everything in the third set, and so on. So far so good, but this only gives us order type ω2. Next we define the order WITHIN each set.

To do this, we’ll use our enumeration of computable orders. In particular, within the nth set, the ith number will be less than the jth number exactly in the case that i <n j. You can think about this as just treating each set “as if” it’s really {0, 1, 2, 3, …}, then ordering it according to <n, and then relabeling the numbers back to what they started as.

This fully defines our order. Now, what’s the order type of this set? Well, it can’t be any particular computable order type, because every computable order type can be found as a sub-order of it! So it must be some uncomputable order type. And by construction, we know it must be exactly the order type that contains all and only computable order types as suborders. But this is just the order type of the Church-Kleene ordinal!

So there we have it! We’ve constructed ω1CK. Now, remember that the Church-Kleene ordinal is uncomputable. But I just described an apparently well-defined process for creating an ordered set with its order! This means that some part of the process I described must be uncomputable. Challenge question for you: Which part is uncomputable, and can you think of a proof of its uncomputability that is independent of the uncomputability of ω1CK?

# What ordinals can be embedded in ℚ and ℝ?

Last time we talked a little bit about some properties of the order type of ℚ. I want to go into more detail about these properties, and actually prove them to you. The proofs are nice and succinct, and ultimately rest heavily on the density of ℚ.

Every Countable Ordinal Can Be Embedded Into ℚ

Take any countable well-ordered set (X, ≺). Its order type corresponds to some countable ordinal. Since X is countable, we can enumerate all of its elements (the order in which we enumerate the elements might not line up with the well-order ≺). Let’s give this enumeration a name: (x1, x2, x3, …).

Now we’ll inductively define an order-preserving bijection from X into ℚ. We’ll call this function f. First, let f(x1) be any rational number. Now, assume that we’ve already defined f(x1) through f(xn-1) in such a way as to preserve the original order ≺. All we need to do to complete the proof is to assign to f(xn) a rational number such that the ≺ is still preserved.

Here’s how to do that. Split up the elements of X that we’ve already constructed maps for as follows: A = {xi | xi ≺ xn} and B = {xi | xi > xn}. In other words, A is the subset of {x1, x2, …, xn-1} consisting of elements less than x_n and B is the subset consisting of elements greater than xn. Every element of B is strictly larger than every element of A. So we can use the density of the rationals to find some rational number q in between A and B! We define f(xn) to be this rational q. This way of defining f(xn) preserves the usual order, because by construction, f(xn) < f(xi) for any i less than n exactly in the case that xn < xi.

By induction, then, we’ve guaranteed that f maps X to ℚ in such a way as to preserve the original order! And all we assumed about X was that it was countable and well-ordered. This means that any countable and well-ordered set can be found within ℚ!

No Uncountable Ordinals Can Be Embedded Into ℝ

In a well-ordered set X, every non-maximal element of X has an immediate successor (i.e. a least element that’s greater than it.) Proof: Take any non-maximal x ∈ X. Consider the subset of X consisting of all elements greater than x: {y ∈ X | x < y}. This set is not empty because α is not maximal. Any non-empty subset of a well-ordered set has a least element, so this subset has a least element. I.e, there’s a least element greater than x. Call this element S(x), for “the successor of x”,

Now, take any well-ordered subset X ⊆ ℝ (with the usual order). Since it’s well-ordered, every element has an immediate successor (by the previous paragraph). We will construct a bijection that maps X to ℚ, using the fact that ℚ is dense in ℝ (i.e. that there’s a rational between any two reals). Call this function f. To each element x ∈ X, f(x) will be any rational such that x < f(x) < S(x). This maps every non-maximal element of X to a rational number. To complete this, just map the maximal element of X to any rational of your choice. There we go, we’ve constructed a bijection from X to ℚ!

The implication of this is that every well-ordered subset of the reals is only countably large. In other words, even though ℝ is uncountably large, we can’t embed uncountable ordinals inside it! The set of ordinals we can embed within ℝ is exactly the set of ordinals we can embed within ℚ! (This set or ordinals is exactly ω1: the set of all countable ordinals).

Final Note

Notice that the previous proof relied on the fact that between any two reals you can find a rational. So this same proof would NOT go through for the hyper-reals! There’s no rational number (or real number, at that!) in between 1 and 1+ϵ. And in fact, you CAN embed ω1 into the hyperreals! This is especially interesting because the hyperreals have the same cardinality as the reals! So the embeddability of ω1 here is really a consequence of the order type of the hyperreals being much larger than the reals. And if we want to take a step towards even crazier extensions of ℝ, EVERY SINGLE ordinal can be embedded within the surreal numbers!

# The order type of the rational numbers

Take any set and place an order on it. If the order is a well-order (i.e. if every subset has a least element), then the ordered set has some particular order type. It’s isomorphic to one particular ordinal. But the notion of order type can be extended beyond well-ordered sets. Any two ordered sets are said to have the same order type if they are order isomorphic: if there’s a map f from one set to the other such that both f and f-1 preserve the ordering of elements.

One structure that has a very interesting order type is the rational numbers. After all, the rationals are a countable set, but every rational number resembles a “limit ordinal” in the sense that it has no immediate predecessor.

One question that we can ask to get some insight about the order type of the rational numbers is: what ordinals can be found within the rationals? That is, take some well-ordered subset of the rationals. Look at the order type of this subset. This order type corresponds to some ordinal. And different choices of well-ordered subsets of the rationals give us different ordinals! So which ordinals can we “find” within the rationals in this sense?

First off, every finite ordinal can obviously be found. To find the ordinal n, just take the subset {0, 1, 2, …, n-1}. We can also find ω! You can just take the subset {0, 1, 2, 3, …}. What about ω+1? Try for yourself: can you construct a subset of the rationals with the order type of ω+1?

There’s no unique way to do it, but one easy way is to take the subset {1/2, 3/4, 7/8, …, 1}. This set has a countable infinity of elements, one for each natural number, after all of which comes a single element: exactly the order type of ω+1!

If we want a subset of the rationals with order type ω+2, we can use the same trick: {1/2, 3/4, 7/8, …, 1, 2}. And clearly this extends for any ω+n. But how about ω+ω? Can you construct a subset of the rationals with this order type? (Do it yourself before reading on!)

Here’s one way: {1/2, 3/4, 7/8, …, 1+1/2, 1+3/4, 1+7/8, …}. It should be easy to see how to extend this trick to get ω⋅3, ω⋅4, and indeed ω⋅n for any finite n. You can even naturally extend this to get to ω⋅ω. But then what? Are we finished?

If you guessed no, you’re right! We can find subsets of the rationals with order type ω3, and ωn for any finite n, and ωω. (Try it!) And we can keep going beyond that as well. So how high can you go?

Turns out that EVERY countable ordinal can be embedded into the rationals, under the usual order! So in some sense the order type of the rationals is as complicated as possible while still being countable. Also, remember that the set of countable ordinals is uncountably large So this means that the rationals are a countable ordered set that has all of the uncountably many countable ordinals embedded within it! Isn’t that great?

(One thing that makes this seem less insane is that when we’re looking at what ordinals can be embedded in the rationals, we’re searching through different subsets of the rationals. And even though the rationals are countable, the set of all subsets of the rationals is uncountably large.)

One more interesting thing: the set of all subsets of the rationals has cardinality ℶ1. But the set of all ordinals that can be embedded into the rationals is ω1, which has cardinality ℵ1. So if we start with the set of all subsets of the rationals, then strip away all but those that are well-ordered, and then choose just a single representative for each order type, we get a set of cardinality ℵ1. And there is no guarantee that this final set has the same cardinality as the original set, because this is the continuum hypothesis! This is a way to think about how to “construct” a real life set with cardinality ℵ1: look at the set of well-ordered subsets of the rationals, and split it into order-type equivalence classes.

# Can you compute all the countable ordinals?

One can live without knowing the ordinals, to be sure, but not as well.

Quote from this paper

What is an ordinal number? If you read my previous post, you might be convinced that an ordinal number is a particular type of set that contains all the ordinal numbers less than it. In particular, the ordinal 0 is the empty set, the ordinal 1 is the set {0}, the ordinal 2 is the set {0,1}, and so on. This is a fine way to think about ordinals, but there’s a much deeper view of them that I want to present in this post.

The fundamental concept is that of order types. Order types are to order-preserving bijections as cardinalities are to bijections. If you’ve seen pictures like these before, then they’re good to have in mind when you think about what order types are: Let’s get into more detail about order types.

Take any two sets. If there is a bijection between them, then we say that they have the same cardinality. So we can think of each cardinal number as a class of sets, all in bijective correspondence with one another.

Take any two well-ordered sets. (Recall, a set is well-ordered if each of its subsets has a least element.) If there is an order-preserving bijection between them, then we say that the two sets have the same order type. So we can think of each ordinal number as a class of sets, all of the same order type.

For instance, the cardinal number “2” is the class of all sets with two elements (as all such sets are in bijective correspondence with one another). And the ordinal number “2” is the class of all well-ordered two-element sets. In particular, Von Neumann’s ordinal 2 = {0, 1} = {∅, {∅}} has this order type.

The finite cardinalities and finite order types line up nicely: any two well-ordered sets with the same finite cardinality are guaranteed to have the same order type. It’s in the realm of the infinite where order types and cardinalities come apart.

The cardinal number ℵ0 is the class of all sets that can be put into bijective correspondence with the set of natural numbers ω. This includes the set of even numbers and the set of rational numbers. So how many order types are there of sets with cardinality ℵ0?

Consider the following two ordered sets: X = {0 < 1 < 2 < 3 < …} and Y = {1 < 2 < 3 < … < 0}. They both have cardinality ℵ0, but do they have the same order types? If so, then there must be a bijection f: X → Y such that f is order-preserving (for all a ∈ X and b ∈ X, if a < b then f(a) < f(b)). But what element of X is mapped to 0 in Y? In the set Y, 0 is larger than every element besides itself, but there is no element of X that has this property! So no, X and Y do not have the same order types.

X is the order type of ω (a countably infinite list of elements with no elements that have infinitely many predecessors), and Y is the order type of ω+1 (a countably infinite list of elements with a single element coming after all of them). This tells us that there are at least two different order types within the cardinality ℵ0.

Can we construct an order on the set of natural numbers that has the same order type as ω+2? Well, all we need is for there to be two elements that are larger than all the rest. So we can write something like {2 < 3 < 4 < … < 0 < 1}. This generalizes easily to ω+n, for any finite n: {n+1 < n+2 < … < 0 < 1 < … < n} has the same order type as ω+n.

But now what about ω+ω, i.e. ω⋅2? To get a set of naturals with the same order type, we have to use a new trick. We need two countably infinite sequences of numbers that we can place beside one another. An easy way to accomplish this is by using the evens and odds: {0 < 2 < 4 < … < 1 < 3 < 5 < …}.

We can get an order on the naturals with the same order type as ω⋅2 + 1 by just choosing a single natural number to place after all the others. For instance: {2 < 4 < 6 < … < 1 < 3 < 5 < … < 0}. It should be easy to see how to get ω⋅2 + n for any finite n: just do {n < n+2 < n+4 < … < n+1 < n+3 < n+5 < … < 0 < 1 < 2 < … < n-1} And now how about ω⋅3?

(Try it out for yourself before reading on!)

To get ω⋅3, we need to place three countably infinite sequences of naturals side by side. For instance: {0 < 3 < 6 < … < 1 < 4 < 7 < … < 2 < 5 < 8 < …}.

For ω⋅4, we can use a similar trick: {0 < 4 < 8 < … < 1 < 5 < 9 < … < 2 < 6 < 10 < … < 3 < 7 < 11 < …}. Again, we can quite easily generalize this to ω⋅n for any finite n: {0 < n < 2n < … < 1 < n+1 < 2n+1 < … < 2 < n+2 < 2n+2 < … < … < n-1 < 2n-1 < 3n-1 < …}. But how about ω⋅ω? How do we deal with ω2?

This is a fun exercise to try for yourself at home. Construct an order on the set of natural numbers that has the same order type as ω2. Then try to generalize the trick to get ω3, ω4, and ωn for any finite n.

Next construct an order on the naturals with the same order type as ωω! Can you do it for ω^ω^ω? And ω^ω^ω^…? One thing that should become clear is that the larger an ordinal you’re dealing with, the harder it becomes to construct an order on the naturals with the same order type. The natural question is: is it always possible to do this type of construction, or do we at some point run out of clever tricks to use to get to higher and higher countable ordinals?

Well, if an ordinal α is countable, then it is a well-ordered set with the same cardinality as the naturals. So there always exists some order that can be placed on the naturals to mimic the order type of α. But is this order always computable?

We call an order on the naturals computable if there’s some Turing machine which takes as input two naturals x and y and outputs whether x < y according to this order. We call an order type computable if there’s some computable order on the naturals with that order type. The standard order (0 < 1 < 2 < 3 < …) is computable, and it’s easy to see how to compute all the other orders we’ve discussed so far. But are ALL the countable order types computable?

The wonderful and strange answer is no. There are countable order types that are uncomputable! There must exist orders on the naturals with such order types, but these orders are so immensely complicated and strange that they cannot be defined by ANY Turing machine! Here’s a quick and easy proof of this. (The next three paragraphs are the coolest and most important part of this whole post, so pay attention!)

Turn your mind back to Von Neumann’s construction of the ordinals, according to which each ordinal was exactly the set of all smaller ordinals. More technically, a set α is a Von Neumann ordinal iff α is well ordered with respect to ∈ AND every element of α is also a subset of α. Consider now the set of all countable ordinals. It can be seen that this set is itself an ordinal. Can it be a countable ordinal? No, because if it were then it would have to be an element of itself! (This is not allowed in ZFC by the axiom of regularity.) This means that there are uncountably many countable ordinals.

Ok, so far so good. Now we just observe that there are only countably many Turing machines. So there are only countably many Turing machines that compute orders on the naturals. And therefore there are only countably many computable ordinals.

Putting this together, we see that there are uncountably many uncomputable countable order types (say that sentence out loud). After all, there are uncountably many countable ordinals, and only countably many computable ordinals!

And if there’s only countably many computable ordinals, then there’s some countable Von Neumann ordinal consisting of the set of all computable ordinals! This set couldn’t possibly be computable, because then it would be contained within itself. So it’s the smallest uncomputable ordinal! This set is called the Church-Kleene ordinal. Its order type is literally so convoluted and complex that no Turing machine can compute it!