Credit to Joel David Hamkins, who I heard discussing this paradox on an episode of the podcast My Favorite Theorem.

Define a finite game to be any two-player turn-based game such that every possible playthrough ends after finitely many turns. For example, tic-tac-toe is a finite game because every game ends in at most nine turns.. So is chess with the 50-move-rule enforced (if 50 moves are taken without any pawn advances or captures, then the game ends in a draw). In both these examples, there’s an upper bound to how long the game can last, but this is not required. A game of transfinite Nim would count as finite; every game lasts for only finitely many turns, even though there is no upper bound on the number of turns it takes.

Now, consider the game Hypergame. To play Hypergame, Player 1 begins by choosing any finite game G. Then Player 2 plays the first move of G, Player 1 plays the second move of G, and so on until the game is completed. (Since Player 1 chose a finite game, this will always happen after some finite amount of time.)

Is Hypergame a finite game? Yes, we can easily see that it must be. Whatever game Player 1 chooses will be over after n steps, for some finite n. So that playthrough of Hypergame will have taken n+1 steps.

But if Hypergame is a finite game, then it is a valid choice for the first move of Hypergame! So we can now imagine the following playthrough of Hypergame:

A Troubling Playthrough of Hypergame
Player 1: For the finite game that we shall play, I pick Hypergame.
Player 2: Hm, okay. So now I’m playing the first move of Hypergame. So I must now choose any finite game. I’ll choose Hypergame!
Player 1: Alright so I’m again playing the first move of this new game of Hypergame. I’ll choose Hypergame again.
Player 2: And I choose Hypergame again as well.
So on forever…

At no point does either player violate the rules of Hypergame. And yet, we ended up with an infinite playthrough of Hypergame, which we proved was impossible! So we have a contradiction. What is the resolution?

✵✵✵

Here’s one possible resolution, analogous to the resolutions of similar set-theoretic paradoxes.

We can think about a game as a directed rooted tree. The vertices of the tree correspond to game states, and the edges correspond to the allowed moves. The root of the tree corresponds to the starting game state, and Player 1 gets to choose which edge to travel along first. From the new vertex, Player 2 decides the next edge to travel along. And so on. The tree’s leaves correspond to ending states of the game, and each leaf is labelled according to which player won in that ending state.

In this framework, what is a finite game? As I defined it above, a finite game is just any directed rooted tree such that every path starting at the root ends at a leaf after passing through finitely many edges. This corresponds perfectly to the idea that every possible playthrough of the game takes only finitely many turns. Notice that a finite game is not necessarily a finite tree! The game tree of a finite game is only finite if it’s also finitely branching. In other words, for a game to have a finite tree requires not just that every playthrough is finitely long, but that each player always has only finitely many choices on their turn.

For instance, the game tree of Hypergame is not a finite tree, because Player 1 has infinitely many possible finite games to choose from on his first turn. How big exactly is the game tree of Hypergame? We know that we have to have a vertex corresponding to the start of any finite game, so it must be at least as large as the set of all finite games. But how large is this set?

This is where we run into problems. The game tree of a finite game can be arbitrarily large. Consider the game which starts by Player 1 choosing any real number and then immediately losing. The height of the game tree is 1, but its width is the cardinality of the continuum. Similarly for any set X we can find a game whose tree has cardinality |X|. This means that there are finite games of arbitrary cardinalities. But then as a corollary to the nonexistence of a largest cardinality, we know that there is no set of all finite games! And this implies that Hypergame has no game tree! More precisely, there is no set corresponding to the game tree of Hypergame as we defined it.

Couldn’t we instead think about Hypergame as a proper class? Sure! But then when we choose a finite game in our first move, we couldn’t be picking Hypergame, as the property “is a finite game” would only apply to sets and not proper classes. This means that we can’t actually select Hypergame as our first move! And so we avoid the paradoxical conclusion that we can keep picking Hypergame ad infinitum.

I find it quite fascinating that the seemingly innocent notion of a finite game can lead us into paradoxes involving proper classes!

# ZFC as One of Humankind’s Great Inventions

Recently I told a friend that I thought ZFC was one of humankind’s greatest inventions. He pointed out that it was pretty bold to claim this about something that most of mankind has never heard of, which I thought was a fair objection. After thinking for a bit, I reflected that the sense of greatness I meant wasn’t really consequentialist, and thus it was independent of how many people know what ZFC is, or even how many people’s lives are affected in any way by it. Instead I intended greatness in a sort of aesthetic and intellectual sense.

The closest analogy to ZFC outside of math is the idea of a “theory of everything” for physics. If we found a theory of everything for physics, it’d likely have a bunch of important practical consequences, and that’d be part of what makes it a great invention. But it would also be a great invention in an intellectual sense, as a discovery of something fundamental and unifying of many seemingly disparate phenomena we observe. This is what ZFC is like: a mathematical theory of everything. One reason this analogy is imperfect is that due to the incompleteness theorems, we know that there can be no “theory of everything” for mathematics. (Any theory of everything will have at least one thing it can’t prove, namely its own consistency.) So ZFC’s greatness can’t come from being a perfect theory of everything, because we know that it is not. Nonetheless, ZFC serves as a foundation for virtually all known mathematics, and this is what I think is so incredible about it.

What does it mean for something to “serve as a foundation” for math? ZFC is a foundation in (at least) three ways: (1) in terms of its ability to define virtually all mathematical concepts, (2) in terms of its structures being rich enough to contain objects that come from virtually all fields of math, and (3) in terms of being an axiom system that suffices to prove virtually every result in known mathematics.

Syntax

Virtually every mathematical concept you can think of has a definition in the language of ZFC. For example, we have definitions for numbers like “π” and “√2”, sets like ℕ and ℝ, algebraic objects like the group S5 and the ring ℚ[x], geometric objects like Platonic solids and differential manifolds, computational objects like Turing machines and cellular automata, and even logical entities like models of first order theories and proofs within formal systems. What makes this especially impressive is the simplicity of the language: it uses nothing besides the basic symbols of first order logic and one binary relation symbol: ∈. So one thing that ZFC teaches us is that virtually every concept in mathematics can be defined just in terms of the set membership relation, and all mathematics can be understood as exploring the properties of this relation.

Semantics

Models of ZFC are insanely richly structured. You can navigate within them to find sets corresponding to every object that mathematicians study. π has a representative set within any model of ZFC, as does the Monster group or the torus. These representative sets are not always perfect: there are models of ZFC where ℝ is countable, for instance. But within the model, they nonetheless share enough similarities with the original objects that virtually everything you can prove about the original object, remains true of the ZFC-representative.

Proof

Finally, ZFC is a computable set of sentences, and we may inquire about what can be proven from it. Keeping up the ambition of the previous two sections, we might want to claim that all mathematical truths can be proven from ZFC. But due to the limitations of first order logic discovered over the last century, we now know that this goal is not achievable. The set of all first order truths of arithmetic is not computable, and so there must be some such truths that aren’t logical consequences of ZFC. Nonetheless, it is commonly claimed that virtually all mathematical truths can be derived from ZFC using the usual proof system for first order logic.

This is especially remarkable given the simplicity of ZFC. I believe that the intuitive content of each axiom could be explained to a smart middle schooler. Additionally, these axioms are extremely intuitively appealing. the most controversial of them has been choice, which is equivalent to the statement that the Cartesian product of non-empty sets is also non-empty. Second most controversial is probably the axiom of infinity, which just says that there’s an infinite set. The rest are even less hard to accept than these.

Now, the fact that you can prove virtually everything from ZFC doesn’t mean that you should. So don’t interpret me as saying that ZFC is of practical use to the daily work of mathematicians trying to prove things outside of set theory and logic. Again, an analogy to physics: we might discover a theory of everything that we know reproduces all the known phenomena of GR and QM, but find that it’s so hard to prove things that we are practically never better off using this theory to calculate things. Nonetheless, ZFC as a theory of everything teaches us that most of math can be understood as conceptually quite simple: the logical consequences of a fairly simple and computable set of sentences about sets. People make a big deal out of Euclid’s axiomatization of geometry, but this is a small feat relative to the axiomatization of all of mathematics.

Metamath

And not only can ZFC prove virtually everything in ordinary mathematics, but ZFC can prove much of what we know in metamathematics and logic itself. When logicians are studying model theory, or even when set theorists are studying ZFC, they are almost always working with ZFC as their meta-theory, meaning that they are making sure that all of their proofs could ultimately be expanded out as ZFC proofs. So the big results of logic, like the completeness theorem, the compactness theorem, the incompleteness theorems, the Löwenheim-Skolem theorems, are all theorems of ZFC.

The fact that ZFC can even talk about these model theoretic notions means that models of ZFC are able to talk about models of ZFC, which is where things get very meta. One can prove that every model of ZFC – every one of these crazily richly-structured universes containing virtually all of mathematics – contains another such model of ZFC. This follows from the reflection theorem, which again can be proven in ZFC!

Hopefully I have now roused enough interest in you to get you to take a look at some of the actual mathematics. You might be curious to know what exactly this theory is. And you’re in luck, it’s simple enough that I can write the whole theory in just nine lines!

Note that with the exception of the final axiom, Choice, the only symbols I’ve used are logical symbols and ∈. I used shorthand for Choice for the sake of readability, but this could be expanded out just like the others. I’m also using a convention where any free variables are considered to be universally quantified over, which shortens things further.

I’ll close with a one-sentence description for each axiom.

Extensionality: No two distinct sets have all the same elements.
Pairing: For any two sets, there’s a set containing just those two.
Union: The union of any set of sets exists.
Powerset: There is a set of all subsets of any set.
Specification: For any property Φ and any set x, you can form a set out of just those elements of x with that property.
Replacement: For any definable function and any set, the image of that set under the function exists.
Infinity: There’s an infinite set.
Regularity: Every non-empty set has a member that it shares nothing with.
Choice: For any set of nonempty sets, there is a function that picks out one element from each.

The fact that you can prove everything from the infinitude of primes to Fermat’s Last Theorem from just these basic principles, is really quite mind-blowing.

# Forcing and the Independence of CH (Part 2)

Part 1 here

## Part 2: How big is 𝒫(ω)?

Now the pieces are all in place to start applying forcing to prove some big results. Everything that follows assumes the existence of a countable transitive model M of ZFC.

First, a few notes on terminology.

The language of ZFC is very minimalistic. All it has on top of the first-order-logic connectives is a single binary relation symbol ∈. Nonetheless, people will usually talk about theorems of ZFC as if it has a much more elaborate syntax, involving symbols like ∅ and ω and 1 and terms like “bijection” and “ordinal”. For instance, the Continuum Hypothesis can be written as “There’s a bijection between 𝒫(ω) and 1“, which is obviously using much more than just ∈. But these terms are all simply shorthand for complicated phrases in the primitive language of ZFC: for instance a sentence “φ(∅)” involving ∅ could be translated as “∃x (∀y ¬(y ∈ x) ∧ φ(x))”, or in other words “there’s some set x that is empty and φ(x) is true”.

Something interesting goes on when considering symbols like ∅ and ω and 1 that act like proper names. Each name picks out a unique set, but for some of these names, the set that gets picked out differs in different models of ZFC. For example, the interpretation of the name “1” is model-relative; it’s meant to be the first uncountable cardinal, but there are countable models of ZFC in which it’s actually only countably large! If this makes your head spin, it did the same for Skolem: you can find more reading here.

On the other hand, the name “∅” always picks out the same set. In every model of ZFC, ∅ is the unique set that contains nothing at all. When a name’s interpretation isn’t model-relative, it’s called absolute. Examples include “∅” and “1,729”. When a name isn’t absolute, then we need to take care to distinguish between the name itself as a syntactic object, and the set which it refers to in a particular model of ZFC. So we’ll write “1” when referring to the syntactic description of the first uncountable cardinal, and 1M when referring to the actual set that is picked out by this description in the model M.

With that said, let’s talk about a few names that we’ll be using throughout this post:

In M, ω is the conventional name for the set that matches the description “the intersection of all inductive sets”, where inductive sets are those that contain the ordinal 0 and are closed under successor. In any transitive model of ZFC like M, ω is exactly ℕ, the set of natural numbers. Since we’re restricting ourselves to only considering transitive models, we can treat ω as if it’s unambiguous; its interpretation won’t actually vary across the models we’re interested in.

ω1 corresponds to the description “the smallest uncountable ordinal”. This description happens to perfectly coincide with the description “the first uncountable cardinal” in ZFC, which has the name 1. But unlike with ω, different transitive models pick out different sets for ω1 and 1. So in a model M, we’ll write ω1M for the set that M believes to be the smallest uncountable ordinal and 1M for the first uncountable cardinal.

ω2 is the name for the first ordinal for which there’s no injection into ω1 (i.e. the first ordinal that’s larger than ω1). Again, this is not absolute: ω2M depends on M (although again, ω2M is the same as 2M no matter what M is). And in a countable model like those we’ll be working with, ω2M is countable. This will turn out to be very important!

𝒫(ω) is the name for the the set that matches the description “the set of all subsets of ω”. Perhaps predictably at this point, this is not absolute either. So we’ll have to write 𝒫(ω)M when referring to M’s version of 𝒫(ω).

Finally, ZFC can prove that there’s a bijection from 𝒫(ω) to , meaning that this bijection exists in every model. Thus anything we prove about the size of 𝒫(ω) can be carried over to a statement about the size of . Model-theoretically, we can say that in every model M, |𝒫(ω)M| = |M|. It will turn out to be more natural to prove things about 𝒫(ω) than .

Okay, we’re ready to make the continuum hypothesis false! We’ll do this by choosing a partially ordered set (P, ≤) whose extension as a Boolean algebra (B, ≤) has the property that no matter what M-generic filter G in B you choose, M[G] will interpret its existence as proof that |𝒫(ω)| ≥ ℵ2.

Making |𝒫(ω)| ≥ 2

Let P = { f ∈ M | f is a finite partial function from ω2M × ω to {0,1} }. Let’s have some examples of elements of P. The simplest is the empty function ∅. More complicated is the function {((ω+1, 13), 1)}. This function’s domain has just one element, the ordered pair (ω+1, 13), and this element is mapped to 1. The function {((14, 2), 0), ((ω1M•2, 2), 1)} is defined on two elements: (14, 2) and (ω1M•2, 2). And so on.

We’ll order P by reverse inclusion: f ≤ g iff f ⊇ g. Intuitively, f ≤ g if f is a function extension of g: f is defined everywhere that g is and they agree in those places. For instance, say f = {((ω, 12), 0)}, g = {((ω, 12), 0), ((13, 5), 1)}, h = {((ω, 12), 1)}, and p = {((ω, 11), 1)}. Check your understanding by verifying that (1) g ≤ f, (2) f and h are incomparable and have no common lower bound, and (3) f and p are incomparable but do have a common lower bound (what is it?).

The empty function ∅ is a subset of every function in P, meaning that ∅ is bigger than all functions. Thus ∅ is the top element of this partial order. And every f in P is finite, allowing a nice visualization what the partial order (P, ≤) looks like:

Now, G will be an M-generic ultrafilter in the Boolean extension (B, ≤) of (P, ≤). G is the set that we’re ultimately adding onto M, so we want to know some of its properties. In particular, how will we use G to show that |𝒫(ω)| = ℵ2?

What we’re going to do is construct a new set out of G as follows: first take the intersection of P with G. (remember that G is an ultrafilter in B, so it doesn’t only contain elements of P). P⋂G will be some set of finite partial functions from ω2M × ω to {0,1}. We’ll take the union of all these functions, and call the resulting set F. What we’ll prove of F is the following:

(1) F := ⋃(P⋂G) is a total function from ω2M × ω to {0,1}
(2) For every α, β ∈ ω2M, F(α, •) is a distinct function from F(β, •).

A word on the second clause: F takes as input two things: an element of ω2M and an element of ω. If we only give F its first input, then it becomes a function from ω to {0,1}. For α ∈ ω2M, we’ll give the name Fα to the function we get by feeding α to F. Our second clause says that all of these functions are pairwise distinct.

Now, the crucial insight is that each Fα corresponds to some subset of ω, namely {n ω | Fα(n) = 1}. So F defines |ω2M|-many distinct subsets of ω. So in M[G], it comes out as true that |𝒫(ω)M| ≥ |ω2M|. It’s also true in M[G] that |ω2M| = 2M, and so we get that |𝒫(ω)M| ≠ 1M. This is ¬CH!

Well… almost. There’s one final subtlety: |𝒫(ω)M| ≠ 1M is what ¬CH looks like in the model M. For ¬CH to be true in M[G], it must be that that |𝒫(ω)M[G]| ≠ ℵ1M[G]. So what if |𝒫(ω)M| ≠ |𝒫(ω)M[G]|, or 1M ≠ 1M[G]? This would throw a wrench into our proof: it would mean that M[G] believes that M’s version of 𝒫(ω) is bigger than M’s version of 1, but M[G] might not believe that its own version of 𝒫(ω) is bigger than its version of 1. This is the subject of cardinal collapse, which I will not be going into. However, M[G] does in fact believe that 𝒫(ω)M = 𝒫(ω)M[G] and that 1M1M[G].

Alright, so now all we need to do to show that M[G] believes ¬CH is to prove (1) that F is a total function from ω2M × ω to {0,1}, and (2) that for every α, β ∈ ω2M, Fα is distinct from Fβ. We do this in three steps:

1. F is a function.
2. F is total.
3. For every α, β ∈ ω2M, Fα is distinct from Fβ.

Alright so how do we know that F is a function? Remember that F = ⋃(P⋂G). What if some of the partial functions in P⋂G are incompatible with each other? In this case, their union cannot be a function. So to prove step 1 we need to prove that all the functions in P⋂G are compatible. This follows easily from two facts: that G is an ultrafilter in B and P is dense in B. The argument: G is an ultrafilter, so for any f, g ∈ P there’s some element f∧g ∈ B that’s below both f and g. All we know about f∧g is that it’s an element of B, but we have no guarantee that it’s also a member of P, our set of partial functions. In other words, we can’t say for sure that f∧g is actually a partial function from ω2M × ω to {0,1}. But now we use the fact that P is dense in B! By the definition of density, since f∧g is an element of B, P must contain some h ≤ f∧g. Now by transitivity, h ≤ f and h ≤ g. So f and g have a common function extension, meaning they must be compatible functions! Pretty magical right?

So F is a function. But how do we know that it’s total? We prove this by looking closely at the dense subsets of P. In particular, for any α ∈ ω2M and any n ∈ ω, define Dα,n := {f ∈ P | (α, n) ∈ dom(f)}. This is a dense subset of P. Why? Well, regardless of what α and n are, any function f in P is either already defined on (α, n), in which case we’re done, or it’s not, in which case f has a function extension with (α, n) in its domain. So from any element of P, you can follow the order downwards until you find a function with (α, n) in its domain. Since Dα,n is dense in P and G is M-generic, G must have some element in common with Dα,n. Thus G contains an element f of P that has (α, n) in its domain. This f is one of the functions that we union over to get F, so F must have (α, n) in its domain as well! And since α and n were totally arbitrary, F must be total.

Finally, why are the “Fα”s pairwise distinct? Again we construct dense subsets for our purposes: for any α, β ∈ ω2M, define Dα,β := {f ∈ P | fα ≠ fβ}. This is clearly dense (you can always extend any f in P to make fα and fβ disagree somewhere). So G contains an element f of P for which fα ≠ fβ. And thus it must also be true of F that Fα ≠ Fβ!

And we’re done! We’ve shown that once we add G to M, we can construct a new set F = ⋃(P⋂G) such that F encodes |ω2M|-many distinct subsets of ω, and thus that M[G] ⊨ |𝒫(ω)| ≥ ℵ2.

Making |𝒫(ω)| ≥ ℵ420

What’s great is that this argument barely relied on the “2” in ω2M. We could just as easily have started with P = {f ∈ M | f is a finite partial function from ω420M × ω to {0,1}}, constructed an M-generic filter G in the Boolean extension of P, then defined F to be ⋃(P⋂G).

G is still an ultrafilter in B and P is still dense in B, so F is still a function.

Dα,n := {f ∈ P | (α, n) ∈ dom(f)} is still a dense subset of P for any α ∈ ω420M and any n ∈ ω, so F is still total.

And Dα,β := {f ∈ P | fα ≠ fβ} is still a dense subset of P for any α, β ∈ ω420M, so all the “Fα”s are pairwise distinct.

And now F encodes |ω420M|-many distinct subsets of ω! The final step, which I’m going to skip over again, is showing that M[G] believes that |ω420M| = |ω420M[G]|, i.e. that cardinal collapse doesn’t occur.

And there we have it: this choice of P gives us a new model M[G] of ZFC that believes that |𝒫(ω)| ≥ ℵ420! Note how easy and quick this argument is now that we’ve gone through the argument for how to make M[G] believe |𝒫(ω)| ≥ ℵ2. This is a great thing about forcing: once you really understand one application, other applications become immensely simpler and easier to understand.

Making |𝒫(ω)| = ℵ1

Okay, we’ve made CH false. Now let’s make it true! This time our choice for (P, ≤) will be the following:

P = {f ∈ M | f is an M-countable partial function from ω1M to 𝒫(ω)M}. Once more we order by reverse inclusion: f ≤ g iff f ⊇ g.

A crucial thing to notice is that I’ve merely required the functions in P to be M-countable, rather than countable. For a set to be M-countable is for there to exist an injection in M from set to ω. Any M-countable set is countable, but some countable sets may not be M-countable (like M itself!). In particular, we know that ω1M is actually a countable set, meaning that if we required true countability instead of just M-countability, then P would include some total functions! But M has no injection from ω1M to ω, so no function in P can be total. This will be important in a moment!

We get our M-generic ultrafilter G and define F := ⋂(P⋂G). Now we want to show that F is a total and surjective function from ω1M to 𝒫(ω). As before, we proceed in three steps:

1. F is a function
2. F is total
3. F is surjective

Step 1: G is an ultrafilter and P is dense in G, so the same argument works to show that all elements of P⋂G are compatible: (1) ultrafilter implies that any f, g in P⋂G have a least upper bound f∧g in B, (2) density of P in G implies that f∧g has a lower bound h in P, so (3) f and g have a common function extension . Thus F is a function.

Step 2: Dα := {f ∈ P | α ∈ dom(f)} is dense in P for any α ∈ ω1M, so F is total.

Step 3: DA := {f ∈ P | A ∈ image(f)} is dense in P for any A 𝒫(ω)M. Why? No f in P is total, so any f in P can be extended by adding one more point (α, A) where α ∉ dom(f). Thus A ∈ image(F) for any A 𝒫(ω), so F is surjective.

So F is a total surjective function from ω1M to 𝒫(ω)M, meaning that in M[G] it’s true that |𝒫(ω)M| ≤ ℵ1M. And since ZFC proves that |𝒫(ω)| > ℵ0, it follows that M[G] ⊨ |𝒫(ω)| = ℵ1.

You might have noticed that each of these arguments could just as easily have been made if we had started out with defining P as the finite partial functions in M from ω1M to 𝒫(ω)M. This is true! If we had done this, then we still would have ended up proving that F was a total surjective function from ω1M to 𝒫(ω)M, and therefore that |𝒫(ω)M| = ℵ1M. However, this is where cardinal collapse rears its ugly head: the final step is to prove that |𝒫(ω)M| = |𝒫(ω)M[G]| and that 1M = 1M[G]. This requires that P be the countable partial functions rather than just the finite ones.

And with that final caveat, we see that M[G] believes that |𝒫(ω)| = ℵ1.

Making |𝒫(ω)| ≤ ℵ314

Let P = {f ∈ M | f is an M-countable partial function from ω314M to 𝒫(ω)} and order by reverse inclusion: f ≤ g iff f ⊇ g.

⋂(P⋂G) is again a total and surjective function from ω314M to 𝒫(ω), following the exact same arguments as in the previous section. Cardinal collapse doesn’t occur here either, so M[G] believes that |𝒫(ω)| ≤ ℵ314.

✯✯✯

We’ve proven the independence of the Continuum Hypothesis from ZFC! We’ve done more: we’ve also shown how to construct models of ZFC in which we have lots of control over the size of 𝒫(ω). Want a model of ZFC in which |𝒫(ω)| is exactly 1729? Just force with the right P and you’re good to go! There’s something a bit unsatisfactory about all this though, which is that in each application of forcing we’ve had to skim over some details about cardinal collapse that were absolutely essential to the proof going through. I hope to go into these details in a future post to close these last remaining gaps.

# Forcing and the Independence of CH (Part 1)

## Part 1: What is Forcing?

Forcing is a set-theoretic technique developed by Paul Cohen in the 1960s to prove the independence of the Continuum Hypothesis from ZFC. He won a Fields Medal as a result, and to this day it’s the only Fields Medal to be awarded for a work in logic. It’s safe to say that forcing is one of the most powerful techniques that set theorists have at their disposal as they explore the set theoretic multiverse.

Forcing has an intimidating reputation, which is somewhat deserved. The technique involves many moving parts and it takes some time to see how they all fit together. I found these three resources to be excellent for learning the basics of the technique:

Each of these sources takes a slightly different approach, and I think that they complement each other quite well. I’ve produced a high-level outline of the argument for why forcing works, as well as how to use forcing to prove the independence of the CH. I don’t feel quite competent to present all of forcing from the ground up, so this outline is not meant to be self-contained. In particular, I’m gonna assume you’re familiar with the basics of first order logic and ZFC and are comfortable with Boolean algebras. I’m also going to skim over the details of cardinal collapse, which are still fairly opaque to me.

Forcing in 5 lines!

1. Fix a countable transitive model M ⊨ ZFC and a Boolean algebra (B, ≤) ∈ M.
2. Define MB := { f ∈ M | f: MB ⇀ B } ⊆ M.
3. Choose an M-generic ultrafilter G ⊆ B.
4. For every n in MB, define valG(n) := { valG(m) | (m, b) ∈ n for some b in G }.
5. Define M[G] := { valG(n) | n ∈ MB }. This is a countable transitive model of ZFC containing all of M as well as G!

We start with a countable transitive model M of ZFC, and then construct a larger countable transitive model M[G] of ZFC that contains a new set G. Let me say a few words on each step.

Step 1
1. Fix a countable transitive model M ⊨ ZFC and a Boolean algebra (B, ≤) ∈ M.

There’s not much to say about Step 1. I suppose it’s worth noting that we’re assuming the existence of a countable transitive model of ZFC, meaning that the results of forcing are all conditioned not just on the consistency of ZFC, but on the existence of transitive models of ZFC, which is significantly stronger. (The existence of countable transitive models is no stronger than the existence of transitive models, by the downwards Lowenheim-Skolem theorem.)

Moving on!

Step 2
2. Define MB := { f ∈ M | f: MB ⇀ B }

MB is sometimes called a Boolean-valued model of ZFC. The elements of MB are called “B-names” or “B-valued sets” or just “B-sets” (as I’ll call them). These B-sets are partial functions in M that go from MB to B. “Huh? Hold on,” I hear you saying, “Isn’t that a circular definition?”

It might look circular, but it’s not; it’s inductive. First of all, regardless of what MB is, the empty set counts as a partial function from MB to B: all its inputs are in MB, all its outputs are in B, and no input gets sent to multiple outputs! ∅ is also in M, so ∅ is an element of MB. But now that we know that ∅ is in MB, we can also construct the function {(∅, b)} for each b in B. This satisfies the criterion of “being a partial function from MB to B”, and it is in M because B is in M. So for any b in B, {(∅, b)} is in MB. And now we can construct more partial functions with these new elements of MB, like {(∅, b), ({(∅, b’)}, b’’)} and so on.

If you’re still not convinced that the definition of MB is sensible, we can define it without any apparent circularity, making the induction explicit:

• V0 =
• Vα+1 = { f ∈ M | f is a partial function from Vα to B } for any ordinal α in M.
• Vλ = ⋃{Vα | α ∈ λ} for any limit ordinal λ in M.
• Finally, MB = ⋃{Vα | α ∈ M}.

Okay, so now we know the definition of MB. But what is it? How do we intuitively think about the elements of MB? One way is to think of them as “fuzzy sets”, sets that contain each other to varying degrees. In this way of seeing things, the Boolean algebra B is our extension of the trivial Boolean algebra {True, False} to a more complicated Boolean algebra that allows intermediate truth values as well. So for instance, of the B-set {(∅, b)} we can say “it contains ∅ to degree b, and contains nothing else”. Of the B-set {(∅, b), ({(∅, b’)}, b’’)} we may say “it contains ∅ to degree b, and contains [the set that contains ∅ to degree b’] to degree b’’, and nothing else.” And so on.

Every Boolean algebra has a top element ⊤ corresponding to “definite truth” and a bottom element ⊥ corresponding to “definite falsity”, so our B-sets don’t have to be entirely fuzzy. For instance the B-set {(∅, ⊤)} definitely contains ∅ and definitely doesn’t contain anything else. It can be thought of as intuitively similar to the set {∅} in M. And the B-set {(∅, ⊥)} definitely doesn’t contain ∅, or anything else for that matter. This brings up an interesting question: how is the B-set {(∅, ⊥)} any different from the B-set ∅? They have the same intuitive properties: both contain nothing (or said differently, contain each thing to degree ⊥). Well that’s right! You can think of {(∅, ⊥)} and ∅ as two different names for the same idea. They are technically distinct as names, but in Steps 3 and 4 when we collapse MB back down to an ordinary model of ZFC, the distinction between them will vanish. (The details of how exactly we assign B-values to sentences about MB are interesting in their own right, and might deserve their own post.)

So intuitively what we’ve done is take our starting model M and produce a new structure that looks a lot like M in many ways except that it’s highly ambiguous: in addition to all the old sharp sets we have all these new fuzzy sets that are unsure about their members. In the next two steps we collapse all this ambiguity back down, erasing all of the fuzziness. It turns out that there are many ways to collapse the ambiguity (one for each ultrafilter in B in fact!), but we’ll want to choose our ultrafilter very carefully so that we (1) don’t end up back where we started, (2) end up with a model of ZFC, and (3) end up with an interesting model of ZFC.

Step 3
3. Choose an M-generic ultrafilter G ⊆ B.

There’s some terminology here that you might be unfamiliar with.

Firstly, what’s an ultrafilter? It’s a nonempty proper subset of B that’s closed upwards, closed under intersection, and contains one of b or ¬b for each b in B. I talked about ultrafilters here in the context of power sets, but it generalizes easily to Boolean algebras (and there are some pretty pictures to develop your intuitions a bit).

Okay, so that’s what ultrafilters are. What is it to be a generic ultrafilter? A generic ultrafilter is one that intersects every dense subset of B. What’s a dense subset of B? A dense subset of a Boolean algebra (B, ≤) is any subset of B\{⊥} that lower-bounds all of B\{⊥}. In other words, D is a dense subset of B if for every element of B\{⊥} there’s a lower element of D. Intuitively, D is like a foundation that the rest of B\{⊥} rests upon (where “upwards” corresponds to “greater than”). No matter where you are in B\{⊥}, you can follow the order downwards and eventually find yourself at an element of D. And a generic ultrafilter is an ultrafilter that shares something in common with each of these foundational subsets.Why have we excluded the bottom element of B, ⊥? If we hadn’t done so then the dense subsets would just be any and all sets containing ⊥! Then any generic ultrafilter would have to contain ⊥, but this is inconsistent with the definition of an ultrafilter!

Finally, what’s an M-generic ultrafilter? Recalling that a generic ultrafilter has to intersect every dense subset of B, an M-generic ultrafilter only needs to intersect every dense subset of B that’s in M.

I hear a possible objection! “But you already told us that B is in M! And since ZFC has the power set axiom, don’t we know that M must also contain all of B’s subsets? If so then M-genericity is no different from genericity!” The subtlety here is that the power set axiom guarantees us that M contains a set 𝒫(B) that contains all the subsets of B that are in M. Nowhere in ZFC is there a guarantee that every subset of B exists, and in fact such a guarantee is not possible in any first-order theory whatsoever! (Can you see how this follows from the downward Lowenheim-Skolem theorem?) The power set axiom doesn’t create any subsets of B, it merely collects together all the subsets of B that already exist in M.

Now, why do we care about G being M-generic? A few steps down the line we’ll see why M-genericity is such an important property, but in all honesty my intuition is murky here as to how to motivate it a priori. For now, take it on faith that M-genericity ends up being exactly the right property to require of G for our purposes. We’ll end up making extensive use of the fact that G intersects every dense subset of B in M.

Very keen readers might be wondering: how do we know that an M-generic ultrafilter even exists? This is where the countability of M saves us: since M is countable, it only contains countably many dense subsets of B: (D0, D1, D2, …). We construct G as follows: first we choose any b0 ∈ D0. Then for any n ∈ ℕ, since Dn+1 is dense in B, we can find a dn+1 ∈ Dn+1 such that dn+1 ≤ dn. Now we have an infinite descending chain of elements from dense subsets. Define G to be the upwards closure of {dn | n ∈ ℕ}, i.e. G = {b ∈ B | b ≥ dn for some n ∈ ℕ}. This is M-generic by construction, and it’s an ultrafilter: for any b ∈ B, either {b’ ∈ B | b’ ≤ b} or {b’ ∈ B | b’ ≤ ¬b} is dense, implying that G contains either b or ¬b. (Convince yourself that it’s also closed upwards and under ∧.)

Really important note: this proof of G’s existence relies on the countability of M. But from M’s perspective, it isn’t countable; i.e. M doesn’t believe that there’s a bijection that puts each set in correspondence with an element of omega. So M can’t prove the existence of G, and in fact G is a set that doesn’t exist within M at all.

Step 4
4. For every n in MB, define valG(n) := {valG(m) | (m, b) ∈ n for some b in G}.

Ok, next we use G to define a “valuation function” valG(•). This function takes the B-sets created in Step 2 and “collapses” them into ordinary sets. It’s defined inductively: the G-valuation of a B-set n is defined in terms of the G-valuations of the elements of n. (As a side note, the fact that this definition works relies on the transitivity of our starting model! Can you see why?)

Let’s work out some simple examples. Start with n = ∅. Then valG(∅) = {valG(m) | (m, b) ∈ ∅ for some b in G} = ∅, because there is no (m, b) in ∅. Thus G evaluates ∅ as ∅. So far so good! Now for a slightly trickier one: n = {(∅, b)}, where b is some element of B.

valG( {(∅,b)} ) = {valG(m) | (m, b) ∈ {(∅, b)} for some b in G} = {valG(m) | m = ∅ & b ∈ G}

There are two cases: either b is in G or b is not in G. In the first case, {(∅, b)} is evaluated to be {∅}. In the second, {(∅,b)} is evaluated to be {}. What’s the intuition here? G is a subset of B, and we can think of it as a criterion for determining which Boolean values will be collapsed to True. And since G is an ultrafilter, every Boolean value will either be collapsed to True (if b ∈ G) or to False (if ¬b ∈ G). Recall that we thought of {(∅, b)} as a B-set that “contained ∅ to degree b”. Our G-evaluation of this set removed all fuzziness: if b was included in G then we evaluate {(∅, b)} as actually containing ∅. And if not, then we said that it didn’t. Either way, all fuzziness has been removed and we’ve ended up with an ordinary set!

The same happens for every B-set. When we feed it into the function valG(•), we collapse all its fuzziness and end up with an ordinary set, whose members are determined by our choice of B.

Step 5
5. Define M[G] := {valG(n) | n ∈ MB}. M[G] is a countable transitive model of ZFC such that M ⊆ M[G] and G ∈ M[G].

Now we simply evaluate all the B-sets and collect them together, and give the result a name: M[G]. This part is easy: the hard work has already been done in defining MB and valG. What remains is to show that this new set is a countable transitive model of ZFC containing all of M as well as G. Let’s do this now.

First of all, why is M[G] countable? Well, M is countable and MB is a subset of M, so MB is countable. And valG is a surjective function from MB into M[G], so |M[G]| ≤ |MB|.

Second, why is M[G] transitive? This follows immediately from its definition: the elements of M[G] are G-valuations, and elements of G-valuations are themselves G-valuations. So elements of elements of M[G] are themselves elements of M[G]. That’s transitivity!

Third, why is M[G] a model of ZFC? Er, we’ll come back to this one.

Fourth, why is M ⊆ M[G]? To show this, we’ll take any arbitrary set x in M and show that it exists in M[G]. To do this, we first define a canonical name for x: a B-set Nx that acts as the surrogate of x in MB. Define Nx to be {(Ny, ⊤) | y ∈ x}. (Make sure that Nx is actually in MB!) What’s the G-valuation of Nx? It’s just x! Here’s a proof by ∈-induction:

First, N = {(Ny,⊤) | y ∈ ∅} = ∅.
So valG(N) = valG(∅) = {valG(m) | (m, b) ∈ ∅ for some b in G} = ∅.

Now, assume that valG(Ny) = y for every y ∈ x.
Then valG(Nx) = {valG(Ny) | (Ny, b) ∈ Nx for some b in G} = {y | (Ny, ⊤) ∈ Nx} = {y | y ∈ x} = x.

Thus by ∈-induction, for all x, valG(Nx) = x.

Finally, why is G ∈ M[G]? We prove this by finding a B-set that G-valuates to G itself. Define A := {(Nb, b) | b ∈ B}. Then valG(A) = { valG(m) | (m, b) ∈ A for some b in G } = {valG(Nb) | b ∈ G} = {b | b ∈ G} = G.

And there it is! We’re done!

Ha ha, tried to pull a fast one on you. We still have one crucial step remaining: proving that M[G] is actually a model of ZFC! As far as I know, there’s no concise way to do this… you have to actually go through each axiom of ZFC and verify that M[G] satisfies it. I’m not going to do that here, but I will provide proofs of four axioms to give you a flavor.

Infinity: M contains a set I satisfying the axiom of infinity and M ⊆ M[G]. So I ∈ M[G], and it still satisfies infinity (still contains ∅ and is closed under successor).

Pairing: Let x and y be any two sets in M[G]. Then for some n1 and n2 in MB, x = valG(n1) and y = valG(n2). Now define n := {(n1, ⊤), (n2, ⊤)} ∈ MB. Then valG(n) = {x, y}.

Union: Let x be any element of M[G]. Since M[G] is transitive, every element y of an element of x is in M[G]. So for each such y there’s a B-set ny such that valG(ny) = y. Define n := {(ny,⊤) | y ∈ z for some z ∈ x}. Then valG(n) = {valG(ny) | y ∈ z for some z ∈ x} = {y | y ∈ z for some z ∈ x} = ⋃x.

Comprehension: Let φ(y) be any first-order formula in the language of ZFC, and x ∈ M[G]. For every element y of x, let ny be any B-set that G-valuates to y. Define n := {(ny,⊤) | y ∈ x & φ(y)}. Then valG(n) = {y ∈ x | φ(y)}.

And I leave you to fill in the rest of the axioms of ZFC for yourself. Much of it looks very similar to the proofs of pairing, union, and comprehension: make the natural choice for a B-set which G-valuates to the particular set whose existence you’re trying to prove. And THEN you’re done!

✯✯✯

Okay! If you’ve made it this far, take a breather and congratulate yourself. You now understand how to adjoin a new set to an existing model of ZFC so long as this new set is an M-generic ultrafilter in a Boolean algebra B ∈ M). And this process works equally well no matter what Boolean algebra you pick!

In fact, the choice of the Boolean algebra in step 1 is the key degree of freedom we have in this whole process. B isn’t the set that we end up adjoining to M, and in fact B is always chosen to be an element of our starting model. But the fact that G is always chosen to be an M-generic filter in B means that the structure of B greatly influences what properties G has.

You might notice that we also technically have another degree of freedom in this process: namely step 3 where I said “choose an M-generic filter G in B”. While this is technically true, in practice much of forcing is about finding really clever choices of B so that any M-generic filter in B has whatever interesting property we’re looking for.

There’s another subtlety about the choice of B, namely that in applications of forcing one generally starts with a partially ordered set (P, ≤) and then extends it to a Boolean algebra B. A really cool result with an even cooler topological proof is that this extension is always possible. For any choice of P one can construct a complete Boolean algebra B such that P is order-isomorphic to a dense subset of B. (In a sentence, we turn P into a topological space whose open sets are the downwards-closed subsets and then let B be the set of regular open sets ordered by inclusion. More on this below.)

What’s great about this is that since P is dense in B, all dense subsets of P are also dense subsets of B. This means that we can analyze properties of our M-generic ultrafilter G in B solely by looking at the dense subsets of P! This means that for many purposes you can simply ignore the Boolean algebra B and let it do its work in the background, while focusing on the partially ordered set (P, ≤) you picked out. The advantage of this should be obvious: there are many more partially ordered sets than there are Boolean algebras, so we have much more freedom to creatively choose P.

Let’s summarize everything by going over the outline we started with and filling in a few details:

Forcing in slightly more than 5 lines

1. Fix a countable transitive model M ⊨ ZFC and a partial order (P, ≤) ∈ M.
2. Extend P to a Boolean algebra (B, ≤) ∈ M.
1. Define T := {A ⊆ P | A is closed downwards}.
2. Define B := {S ∈ T | Int(Cl(S)) = S}.
3. (B, ⊆) is a Boolean algebra: ¬A = Int(Ac), A∧B = A⋂B, A∨B = Int(Cl(A∪B)).
4. P is order-isomorphic to a dense subset of B
1. For p ∈ P, define S(p) := {q ∈ P | q ≤ p} ∈ B.
2. S is an order-isomorphism from P to B and {S(p) | p ∈ P} is dense in B.
3. Define MB := { f ∈ M | f: MB ⇀ B }.
4. Choose an M-generic ultrafilter G ⊆ B.
1. Since M is countable, we can enumerate the dense subsets of B in M: (D0, D1, D2, …)
2. Choose any b0 ∈ D0.
3. For any n ∈ ℕ, since Dn+1 is dense in B, we can find a dn+1 ∈ Dn+1 s.t. dn+1 ≤ dn.
4. G := { b ∈ B | b ≥ dn for some n ∈ ℕ } is an M-generic ultrafilter in B.
5. For every n in MB, define valG(n) := { valG(m) | (m, b) ∈ n for some b in G }.
6. Define M[G] := { valG(n) | n ∈ MB }.
1. M[G] is a countable transitive model of ZFC.
2. M ⊆ M[G]
1. For any x ∈ M, define Nx := {(Ny,⊤) | y ∈ x} ∈ MB.
2. Then valG(Nx) = x, so x ∈ M[G].
3. G ∈ M[G]
1. Define A := {(Nb, b) | b ∈ B} ∈ MB.
2. valG(A) = G, so G ∈ M[G].

We are now ready to apply forcing to prove the independence of CH from ZFC! In Part 2 you will learn exactly what choice of P makes the cardinality of the continuum ≥ ℵ2 (thus making CH false) and what choice of P makes the cardinality of the continuum exactly ℵ1 (thus making CH true). In fact, I’ll do you one better: in Part 2 you’ll learn what to make P in order to make the cardinality of the continuum larger than ℵ69, ℵ420, or virtually any other cardinality you like!

End of Part 1. Part 2 here.

# The Ultra Series: Guide

Assumed background knowledge: basic set theory lingo (∅, singleton, subset, power set, cardinality), what is first order logic (structures, universes, and interpretations), what are ℕ and ℝ, what’s the difference between countable and uncountable infinities, and what “continuum many” means.

1 Introduction
Here I give a high-level description of what an ultraproduct is, and provide a few examples. Skippable if you want to jump straight to the math!

2 Hypernaturals Simplified
Here you get a first glimpse of the hypernaturals. It’s a fuzzy glimpse from afar, and our first attempt to define them is overly simplified and imperfect. Nonetheless, we get some good intuitions for how hypernatural numbers are structured, before eventually confronting the problem at the core of the definition.

3 Hypernaturals in all their glory
We draw some pretty pictures and introduce the concept of an ultrafilter. The concept is put to work immediately, allowing us to give a full definition of the hypernaturals with no simplifications. The issues with the previous definition have now been patched, and the hypernaturals are a well-defined structure ripe to be explored.

4 Ultraproducts and Łoś’s theorem
We describe how to pronounce “Łoś”, define what an ultraproduct is, and see how the hypernaturals are actually just the ultraproduct of the naturals. And then we prove Łoś’s theorem!

5 Infinitely Large Primes
With the newfound power of Łoś’s theorem at our hands, we return to the realm of the hypernaturals and start exploring its structure. We describe some infinitely large prime numbers, and prove that there are infinitely many of them. We find more strange infinitely large hypernatural numbers in our exploration: numbers that can be divided by 2 ad infinitum, numbers that are divisible by every finite number, and more. We learn that there’s a subset of the hypernaturals that is arranged just like the positive rational numbers, but that the hypernaturals are not dense.

6 Ultraproducts and Compactness
We zoom out from the hypernaturals, and show that ultraproducts can be used to give the prettiest proof of the compactness theorem for first order logic. We prove it first for countable theories, and then for all theories. We then get a little wild and discuss some meta-logical results involving ultraproducts, definability, and compactness.

We now describe the most powerful property of ultraproducts: countable saturation. And then we prove it! With our new tool, we dive back into the hypernaturals to learn more about their structure. We show that for any countable set of hypernaturals, there’s a hypernaturals that’s divisible by them all, and see that this entails the existence of uncountably many hypernatural primes. We prove that the hypernaturals have uncountable cofinality and coinitiality. And from this we see that no two hypernaturals are countably infinitely far apart; all distances are finite or uncountable! We wrap up with a quick proof that ultraproducts are always either finite or uncountable, and a mind-blowing result that relates ultraproducts to the continuum hypothesis.

7.5 Shorter Proof of Countable Saturation
I give a significantly shorter and conceptually simpler proof of countable saturation than the previous post. Then I wax philosophical for a few minutes about constructivism in the context of ultraproduct-related proofs.

# No more than half of anti-sets are big

Background on anti-set theory here.

Suppose that somebody tells you that they have in their hands a model of anti-set theory containing 50 anti-sets. The axiom of anti-infinity gives us some nice immediate restrictions on how large these anti-sets can be. As a quick reminder, the axiom of anti-infinity tells us that every anti-set X must contain another anti-set Y that has a successor that’s not in X. The successor of X is defined to be X ⋃ {X} (an anti-set containing everything in X plus X itself). One big difference between successors in ZFC and successors in anti-set-theory is that anti-sets aren’t always guaranteed to have successors. Another is that anti-sets can be their own successors (this happens if the anti-set contains itself).

Ok, so anti-infinity immediately tells us that there can be no universal set (i.e. a set containing the entire universe). Since every set X must contain a set that has a successor outside X, no set can contain everything, or else it would contain all the successors!

So there are no anti-sets of size 50. How about anti-sets of size 49? It’s recently been proved that there can only be at most 25 sets of size 49. And in general, in a universe of size N no more than half of the anti-sets can be size N-1. Let’s prove it!

First, a naming convention: in a universe of size N, let’s call any sets of size N-1 big sets, and any other sets small sets. Big sets are as big as an anti-set can be. To satisfy anti-infinity, any set X must contain an element Y that has a successor that’s not in X. We’ll call Y the anti-infinity-satisfier for X.

Now, some observations. Suppose that a big set X has a successor. This successor is either size N-1 (if X contains itself) or size N (if not). But no set is size N! So any big set that has a successor, must be its own successor. But this means that big sets cannot be anti-infinity-satisfiers. If a big set X was an anti-infinity-satisfier for some other set Y, then its successor could not be in Y. But its successor is X, and X is in Y!

What this means is that every set must contain a small set as its anti-infinity-satisfier. Now, consider any big set X. X must contain some small set z that serves as an anti-infinity-satisfier for it. But for z to serve as an anti-infinity-satisfier, z must have a successor that’s not in X. What this means is that X must be missing z’s successor.

In other words, every big set must be missing at least one small set’s successor. But remember, big sets are size N-1. This means that they contain every set in the universe besides one. So now we know that the “missing set” that characterizes each big set is always a successor of a small set!

Let’s say that our universe contains k small sets. Then there are at most k successors of small sets. Any big set must be the set of all sets besides one of these successors. Thus by extensionality, there are at most k big sets. Therefore there can never be more big sets than small sets!

And that’s the result: In any finite universe, at most half of the anti-sets are big.

This resolves a conjecture that I posted earlier as an open question: in an N-element model, are there always at least three small sets? The answer is yes. If we had fewer than three small sets, then we’d have to also have fewer than three big sets. This means that our universe is at most size four (two small sets and two big sets). But no four-element models exist! QED.

# Epsilon-induction and the cumulative hierarchy

The axiom of foundation says that every non-empty set must contain an element that it is disjoint with. One implication of this is that no set can contain itself. (Why? Suppose ∃x (x ∈ x). Then you can pair x with itself to form {x}. {x} is non-empty, so it must contain an element that it’s disjoint with. But its only element is x, and that element is not disjoint with {x}. In particular, they both have as an element x.)

Another implication of foundation is that there can be no infinite descending sequence of sets. A sequence of sets is a function f whose domain is ω. For each k ∈ ω we write f(k) as fk. Suppose that f is a function with the property that for each n ∈ ω, fn+1 ∈ fn. Then by the axiom of replacement, there would exist the set {f0, f1, f2, …}. Each element of this set contains the following element, so none of the elements are disjoint with the entire set. So the sequence we started with could not exist.

This allows us to prove a pretty powerful and surprising result about set theory. The result: For any sentence Φ, if Φ holds of at least one set, then there must exist a set X such that Φ(X) holds, but Φ doesn’t hold for any element of X.

Formally:

(∃A Φ(A)) → ∃X (Φ(X) ∧ ∀Y ∈ X (¬Φ(Y))).

Suppose this was false. Then Φ would be satisfied by at least one set, and every set that satisfied Φ would contain an element that satisfied Φ. We can use this to build an infinite descending sequence of sets. Take any set that satisfies Φ and call it X0. Since X0 satisfies Φ, it must contain at least one element that satisfies Φ. Define X1 to be any such element. X1 satisfies Φ as well, so it must contain at least one element that satisfies Φ, which we’ll call X2. And so on forever. Since no set can contain itself, Xn+1 is always distinct from Xn. We can use recursion and then replacement to construct the set {X0, X1, X2, …}, which is an infinite descending sequence of sets. Contradiction!

This peculiar principle turns out to be useful to prove some big results, and the proofs are always a little funny and enjoyable to me. I’m not aware of any name for it, so I’ll call it the “far-from-the-tree” principle: for any property that holds of at least one set, there’s a set that has that property, none of whose elements have the property. I’ll now show two big results that are proven by this principle.

Everything’s in the Cumulative Hierarchy

The cumulative hierarchy is constructed recursively as follows:

V0 = ∅
Vα+1 = 𝒫(Vα), for each ordinal α
Vλ = U{Vβ : β < λ}, for each limit ordinal λ

V0 = ∅
V1 = {∅}
V2 = {∅, {∅}}
V3 = {∅, {∅}, {{∅}}, {∅, {∅}}}

One can prove that Vα is transitive for each ordinal α (in other words, that every element of Vα is also a subset of Vα). Also, due to the nature of the power-set operation, for any two ordinals α < β, Vα ⊆ Vβ.

Now, the reason the cumulative hierarchy is interesting is that we can prove that every set is on some level of the hierarchy. We do so with the far-from-the-tree principle:

Suppose that there was a set X that wasn’t on the hierarchy. This means that ∀α, X ∉ Vα. We use the far-from-tree principle with the property of “not being on the hierarchy” to conclude that there must be a set Y such that Y is not on the hierarchy, but every element of Y is on the hierarchy. So for each element y ∈ Y there’s some ordinal αy such that y ∈ Vαy. Take the highest such ordinal, and now we have a level of the hierarchy that every element of Y is on. Now just go up by one level! Since every element of Y was in the previous level, the set consisting of each of those elements (i.e. Y) is in the power set of that level. So Y actually is on the hierarchy! Contradiction.

This shows that any set you choose shows up on some level of the cumulative hierarchy. Incidentally, the rank of a set is defined as the first level that the set appears on, and one can prove that for each ordinal α, rank(α) = α.

∈-Induction

∈-induction is a really strong principle of induction, because it is induction over all sets, not just the natural numbers or the ordinals. The principle of ∈-induction is the following:

Suppose that for every set X, if Φ(x) holds for every x ∈ X, then Φ(X) holds.
Then Φ holds of every set.

Suppose that this principle was false. Then it’s true that for every set X, if Φ(x) holds for every x ∈ X, then Φ(X) holds, but false that Φ holds of every set. So there’s some set Y such that ¬Φ(Y). We use the far-from-the-tree principle with the property “¬Φ”. This tells us that there’s some set Z such that ¬Φ(Z), but Φ(z) for every z in Z. But we already said that if Φ(z) for every z in Z, then Φ(Z)! Contradiction.

I hope you like these applications as much as I do. They feel to me like very strong and surprising results, especially given how simply they can be proved.

# Nonstandard integers, rationals, and reals

In discussions of first-order logic and set theory, one often hears about nonstandard natural numbers and their unusual properties. But rarely do you hear about nonstandard integers, nonstandard rationals, or nonstandard reals. This post will be devoted to the ways in which these other number systems pick up unintended interpretations in first-order logic, and in particular first-order ZFC.

Nonstandard ℕ

Nonstandard interpretations of the natural numbers are the most well-known, so I won’t spend much time on them. In ZFC, the set of natural numbers is called ω, and is defined to be the intersection of all inductive sets. An inductive set is one that includes zero and is closed under successorship (if n is in the set, then so is S(n)).

This definition feels so right when you think about it for a little while that it’s hard to see how it could go wrong. How could there be anything OTHER than the standard natural numbers in the intersection of all inductive sets?

Well, one can prove from the compactness of first order logic that there must be such nonstandard numbers, and the proof is so simple and pretty that it’s probably already appeared on this blog six or seven times. Add to ZFC a constant K and the axioms “K ∈ ω”, “K > 0”, “K > 1”, “K > 2”, and so on for all the standard naturals (this is a decidable set of sentences). Every finite subset of this new theory has a model, so by compactness the theory as a whole also has a model. This model is nonstandard by construction, and is also a model of ZFC by monotonicity (removing axioms never removes models).

So how do nonstandard naturals appear in the intersection of all inductive sets? The key term to focus on is the “all” in “all inductive sets”. While ZFC guarantees the existence of an inductive set (by the axiom of infinity) and every definable subset of this set (by comprehension/separation), it does not actually guarantee the existence of ALL subsets of this set. (This is actually a general feature of first-order theories, that they are able to guarantee the existence of all definable subsets of a set, but not all subsets.) There are models where the inductive set given to us by the axiom of infinity is enormous (uncountably large), and all of its subsets contain (in addition to the standard naturals) infinitely descending membership chains of sets, each of which contains every standard natural. In these models, omega is not just the standard naturals, but also includes these infinite elements, these numbers with infinitely many predecessors. And the proof from compactness shows us that we can’t eliminate these nonstandard natural numbers.

Nonstandard ℤ

The usual way of defining the integers in ZFC is as equivalence classes of pairs of natural numbers. In particular, we define the equivalence relation ~ on ω×ω as follows: (a, b) ~ (c, d) if and only if a + d = b + c. Two ordered pairs of naturals are equivalent under this relation so long as their difference is the same. So (1, 0) is in the same class as (2, 1), which is in the same class as (15, 14).

The idea here is that the integer represented by the equivalence class of (a, b) is supposed to be what we think of as a – b. So the integer 2 is the equivalence class {(2, 0), (3, 1), (4, 2), (5, 3), …}. And the integer -2 is the equivalence class {(0, 2), (1, 3), (2, 4), (3, 5), …}. For shorthand, we’ll write the equivalence class of (a, b) as [a, b]. Thus for each positive integer n, we can write n = [n, 0], and for each negative integer n, n = [0, n].

Addition and multiplication can be defined on the integers in terms of addition and multiplication on ω.

[a, b] + [c, d] = [a + c, b + d]
[a, b] ⋅ [c, d] = [a⋅c + b⋅d, a⋅d + b⋅c]

Check to make sure that this makes sense in light of our interpretation of [a, b] as a – b. We can also define new operations on the integers that didn’t apply to the naturals, like negation.

-[a, b] = [b, a]

And we can define an order on the integers as follows:

[a, b] < [c, d] if and only if a + d < b + c

Check to make sure that this makes sense.

Now, the integers are built out of ω, so it makes sense that the nonstandard interpretations of ω carry over to nonstandard interpretations of ℤ. But there are in fact two ways in which integers can be nonstandard.

(1) The natural number 0 refers to the same object (the empty set) in every model of ZFC. But the set of natural numbers ω refers to different objects in different models, as we saw a moment ago. In this sense, ω is not categorically defined, while 0 is. But since integers are infinite sets of pairs of elements of ω, individual integers aren’t categorically defined either!

For instance, the integer 0 is the set {(n, m) ∈ ω×ω | n = m}. The number of elements in this set is the same as the number of elements of ω. So the integer 0 has as many elements as ω! This means that in a nonstandard model of ω that has nonstandard numbers like K, 0 will also contain nonstandard ordered pairs like (K, K) and (K+1, K+1).

(2) Consider the nonstandard element of ω that we’re calling K, which is larger than every standard natural number. There’s an integer [K, 0], which is the equivalence class of the ordered pair (K, 0). [K, 0] > [1, 0], because K + 0 > 0 + 1. And [K, 0] > [55, 0], because K + 0 > 0 + 55. And so on for every finite integer. Just like ω, there are models in which ℤ has integers larger than all the standard integers.

But unlike ω, in nonstandard models ℤ also has nonstandard integers less than all standard integers. Consider -[K, 0] = [0, K]. Again, you can prove that this integer is less than the integer 0, the integer -55, and so on for every standard integer you can think of.

From now on I’ll write the integer [n, 0] as n and the integer [0, n] as -n.

Nonstandard ℚ

Just as the integers were equivalence classes of pairs of naturals, the rationals are equivalence classes of pairs of integers. We define a new equivalence relation (which I’ll use the same symbol for) as follows:

For integers a, b, c, and d, (a, b) ~ (c, d) if and only if a⋅d = b⋅c.

This equates any two pairs of integers that have the same ratio. So (1, 2) ~ (2, 4) ~ (15, 30), and (-33, 17) ~ (66, -34) ~ (-99, 51). The equivalence classes of this relationship are the rational numbers. Like before, we’ll write [a, b] to represent the equivalence class that (a, b) belongs to. You can think of [a, b] as representing the rational number a/b.

Defining addition, multiplication, subtraction, and division on the rationals is pretty easy:

[a, b] + [c, d] = [a⋅d + b⋅c, b⋅d]
[a, b] ⋅ [c, d] = [a⋅c, b⋅d]
[a, b] – [c, d] = [a⋅d – b⋅c, b⋅d]
[a, b] / [c, d] = [a⋅d, b⋅c]

And the order is similarly easy to define in terms of the order on integers:

[a, b] < [c, d] if and only if a⋅d < b⋅c

Once we’ve moved to ℤ, there’s an explosion of nonstandard elements. For instance, we still have nonstandard rationals larger than every ordinary rational (like [K, 1]). But now we also have rationals like [1, K]. This rational is smaller than every ordinary positive rational [1, 10], [1, 100], [1, 1000], and so on. And it is greater than the rational number 0! So in other words we now have infinitesimal rational numbers!

What’s more, we can construct a rational like [K+1, K]. This guy is infinitesimally larger than the rational number 1. So is [K+1, K⋅2], but this one is even closer to 1 than the previous. And we also have [K+1, K2], which is even closer!

The nonstandard rational number line is crowded with these infinitesimal and infinite elements. The ordinary rationals are usually thought to be dense, but we’ve just seen that there are nonstandard rationals that are “too close together” to fit any standard rationals in. However, between any two nonstandard rationals K and K’, there’s another nonstandard rational (K + K’) / 2.

If you’ve heard of the hyperreals, you might be wondering if any of the nonstandard rationals have the same structure. The simple reason why they don’t is that the hyperreals are complete – there are no gaps – whereas the rationals are not. For instance, ZFC can prove that there is no rational [a, b] such that [a, b] ⋅ [a, b] = 2, but there is a hyperreal with that property. To get completion of the number line, we have to move on to the next step, the reals.

Nonstandard ℝ

The construction of the reals from the rationals is slightly different from the previous constructions. Instead of considering ordered pairs of rationals, we’ll consider sets of rationals. In particular, we’ll look at nonempty sets of rationals that are closed downwards (if n is in the set and m < n, then m is in the set also), and have no greatest element, but don’t include all rationals.

Each real number is one of these subsets, and ℝ is the set of all such subsets. So for instance, the real number √2 is the set {x ∈ ℚ : x2 < 2}. The order on the reals is simple to describe:

x ≤ y if and only if x ⊆ y

x + y = {r + s : (r ∈ x) ∧ (s ∈ y)}

Multiplication of reals has a messier definition, but it’s nothing too crazy. And importantly, ZFC can prove that ℝ has the following feature: any non-empty subset of ℝ which is bounded above has a least upper bound.

Now, in nonstandard models, there are real numbers like K, 1/K, 1 + 1/K2 and so on. But now there’s also real numbers like √K, 2K, log(K), and any other function you can define on the reals, applied to infinite reals and infinitesimals. So in one sense, the nonstandard models have many more reals than the “ordinary reals” we learn about in calculus.

But the way we constructed the reals as subsets of the rationals opened up a new type of nonstandard phenomena, in which there end up being many fewer reals that we expect there to be. Remember that we identified reals with subsets of the rationals, and that earlier we said that there is in general no way to guarantee the existence of all subsets of an infinite set. The same applies here; when we say that ℝ = {A ⊆ ℚ | A is a Dedekind cut}, we are actually only guaranteeing the existence of those reals that can be definable as Dedekind cuts. So for instance, ZFC can prove the existence of √2 as a real number, because ZFC can define the Dedekind cut {x ∈ ℚ : x2 < 2}. Same with most real numbers you’re probably familiar with, like π and e. But considering that there are only countably many definitions in ZFC, and uncountably many reals, there must be uncountably many reals that are undefinable! These undefinable reals cannot be proven to exist, and thus there are models in which they don’t exist.

In fact, the set ℝ is the first of the sets we’ve discussed that not only has nonstandard interpretations in which it’s too large, but also nonstandard interpretations in which it’s too small. There are models of ZFC where ℝ has only countably many items! There’s a subtlety here, which is that ZFC can prove that |ℝ| > |ω|. How is this possible if there are models where ℝ is countable?

It’s possible because these models, which are missing many real numbers, are ALSO missing lots of functions, in particular the functions that would put ℝ and ω in bijective correspondence! So even though ℝ is “in reality” countable in these models, the model itself doesn’t know that, because it’s missing the functions that prove its countability.

# The Anti-Set Program

In ZFC set theory, we specify a collection of sentences within a first-order language to count as our axioms. The models of this collection of sentences are the set-theoretic universes (and many of these models are “unintended” – pesky perversions in which the set of naturals ω is uncountably large, as one example – but we’ll put this aside for this post). Most of the axioms act as constraints that all sets must follow. For instance, the axiom of pairing says that “For any sets x and y, there must exist another set containing as elements just x and y and nothing else”. This is an axiom that begins with universal quantification over the sets of the universe, and then states some requirement that must hold of all these sets.

The anti-set program is what you get when you take each of these restriction axioms, and negate the restriction it imposes on all sets. So, for instance, the axiom of anti-pairing says that for any sets x and y, there must NOT exist any set {x, y}. Contrast this with the simple negation of pairing, which would tell us only that there exist two sets x and y such that their pair doesn’t exist. The anti-axiom is much stronger than the negated axiom, in that it requires NO pairs to exist.

Not all axioms begin with universal quantifiers, in particular the axiom of infinity, which simply asserts that a set exists that satisfies a certain property. To form the axiom of anti-infinity, we simply negate the original axiom (so that no sets with that property exist).

As it turns out, the anti-set program, if applied to ALL the axioms of ZFC, ends in disaster and paradox. In particular, a contradiction can be derived from anti-comprehension, from anti-replacement, and from anti-extensionality. We don’t handle these cases the same way. Anti-comprehension and anti-replacement are simply discarded, being too difficult to patch. By contrast, anti-extensionality is replaced by ordinary extensionality. What’s up with that? The philosophical justification is simply that extensionality, being the most a priori of the bunch, is needed to justify us calling the objects in our universe sets at all.

There’s one last consideration we must address, which regards the axiom of anti-choice. I am currently uncertain as to whether adding this axiom makes the theory inconsistent. One thing that is currently known about the axiom of anti-choice is that with its addition, out go all finite models (there’s a really pretty proof of this that I won’t include here). In the rest of this post, I will be excluding anti-choice from the axioms and only exploring models of anti-ZF.

With that background behind us, let me list the axioms of anti-ZF.

Anti-Pairing
∀x∀y∀z∃w (w ∈ z ∧ w ≠ x ∧ w ≠ y)
No set is the pair of two others.

Anti-Union
∀x∀y∃z (z ∈ y ∧ ¬∃w (z ∈ w ∧ w ∈ x))
No set is the union of another.

Anti-Powerset
∀x∀y∃z (z ∈ y ∧ ∃w (w ∈ z ∧ w ∉ x))
No set contains all existing subsets of another.

Anti-Foundation
∀x∀y (y ∈ x → ∃z (z ∈ x ∧ z ∈ y))
Every set’s members have at least one element in common with it.

Anti-Infinity
∀x∃y ((y is empty ∧ y ∉ x) ∨ (y ∈ x ∧ ∃z (z = S(y) ∧ z ∉ x)))
Every set either doesn’t contain all empty sets, or has an element whose successor is outside the set.

If the axioms of ZF are considered to be a maximally nice setting for mathematics, then perhaps the axioms of anti-ZF can be considered to be maximally bad for mathematics.

We need to now address some issues involving the axiom of anti-infinity. First, the abbreviations used: “y is empty” is shorthand for “∀z (z ∉ y)” and “z = S(y)” is shorthand for “∀w (w ∈ z ↔ (w ∈ y ∨ w = y))” (i.e. z = y ⋃ {y}).

Second, it turns out that from the other axioms we can prove that there are no empty sets. So the first part of the disjunction is always false, meaning that the second part must always be true. Thus we can simplify the axiom of anti-infinity to the following statement, which is logically equivalent in the context of the other axioms:

Anti-Infinity
∀x∃y (y ∈ x ∧ ∃z (z = S(y) ∧ z ∉ x))
Every set has an element whose successor is outside the set.

Let’s now prove some elementary consequences of these axioms.

Theorems

>> There are no empty sets.
Anti-Union ⊢ ¬∃x∀y (y ∉ x)
Suppose ∃x∀y (y ∉ x). Call this set ∅. So ∀y (y ∉ ∅) Then ⋃∅ = ∅. But this implies that the union of ∅ exists. This contradicts anti-union.

>> There are no one-element sets.
Anti-Pairing ⊢ ¬∃x∃y∀z (z ∈ x ↔ z = y)
Suppose that ∃x∃y∀z (z ∈ x ↔ z = y). Then ∃x∃y∀z (z ∈ x ↔ (z = y ∨ z = y)). But this is a violation of anti-pairing, as then x would be the pair of y and y. Contradiction.

>> There are no two-element sets.
Anti-Pairing ⊢ ¬∃x∃y∃z∀w (w ∈ x ↔ (w = y ∨ w = z))
Suppose that ∃x∃y∃z∀w (w ∈ x ↔ (w = y ∨ w = z)). Then w is the pair of y and z, so anti-pairing is violated. Contradiction.

>> No models of anti-ZF have just N-element sets (for any finite N).
Suppose that a model of anti-ZF had only N-element sets. Take any of these sets and call it X. By anti-infinity, X must contain an element with a successor that is outside the set. Call this element Y and its successor S(Y). Y cannot be its own successor, as then its successor would be inside the set. This means that Y ∉ Y. Also, Y is an N-element set by assumption. But since Y ≠ S(Y), S(Y) must contain all the elements of Y in addition to Y itself. So S(Y) contains N+1 elements. Contradiction.

>> There is no set of all sets.
Suppose that there is such a set, and call it X. By Anti-Infinity, X must contain an element Y with a successor that’s outside of X. But no set is outside of X, by assumption! Contradiction.

>> No N-element model of anti-ZF can have N-1 sets that each contain N-1 elements.
Suppose that this were true. Consider any of these sets that contain N-1 elements, and call it X. By the same argument made two above, this set must contain an element whose successor contains N elements. But there are only N sets in the model, so this is a set of all sets! But we already know that no such set can exist. Contradiction.

>> Every finite set of disjoint sets must be its own choice function.
By a choice function for a set X, I mean a set that contains exactly one element in common with each element of X. Suppose that X is a finite set of disjoint sets. Let’s give the elements of X names: A1, A2, A3, …, AN. By anti-foundation, each An must contain at least one element of X. Since the Ans are disjoint, these elements cannot be the same. Also, since there are only N elements of X, no An can contain more than one element of X. So each element of X contains exactly one element of X. Thus X is a choice function for X!

The next results come from a program I wrote that finds models of anti-ZF of a given size.

>> There are no models of size one, two, three or four.

>> There are exactly two non-isomorphic models of size five.

Here are pictures of the two:

Now, some conjectures! I’m pretty sure of each of these, especially Conjecture 2, but haven’t been able to prove them.

Conjecture 1: There’s always a set that contains itself.

Conjecture 2: There can be no sets of disjoint sets.

Conjecture 3: In an N-element model, there are never less than three sets with fewer than N-1 elements.