We can't know how big the real numbers are

Using a classic model theory theorem to show that some things are above our understanding

May 24, 2026

At some point when one is learning mathematics one reaches one of the most important theorems

Theorem (Cantor) #

There is no bijection between the set of natural numbers and the set of real numbers.

Thus, when learning model theory and one reaches the following statement:

Theorem #

There exists a model of set theory (Zermelo-Fraenkel specifically) in which the real numbers (of that model) are in bijection with the natural numbers.

The first instinct is to think that something has gone terribly wrong...

The second instinct is that the word model there has to be doing some heavy lifting for this to make sense.

Model Theory #

Model Theory is a branch of logic that studies how we can realize "concretely" different "universes" for a theory or set of axioms.

Example #

In the theory of groups we need, a "universe" set $G$, a "binary function symbol" for the multiplication $\mu:G \times G \to G$, a "constant symbol" $1 \in G$, a "unary function symbol" for the inversion $\iota:G \to G$, and the group axioms:

Associativity $$\forall x, \forall y, \forall z, \mu(\mu(x,y),z)=\mu(x,\mu(y,z))$$
Neutral element $$\forall x,\mu(x,1) = x = \mu(1,x)$$
Inverses $$\forall x, \mu(x,\iota(x)) = 1 = \mu(\iota(x),x)$$

In this theory, a "model" is simply... a group, namely $G$. What we have done in this process is abstracted a group to a set of symbols $(G,\mu,1,\iota)$ and some essential axioms.

Model theory cares about how different 'models' of a single set of axioms and symbols behave.

Each theory is made up by what is known as a "signature" or "language": a collection of symbols in which to write the axioms.

In the example we gave before there are three symbols: multiplication, the constant and inversion symbols.

We also snuck in another symbol: the equality. This one's special only because we sometimes restrict our interpretations of this symbol to... the equality (duh). But we could have also defined other symbols for other relations not only functions or equality.

Given a language, we can define a (first order) theory given by logical formulas determined by those symbols alongside logical symbols such as $$\vee, \wedge, \neg, \to, \forall, \exists$$ and the parenthesis and commas as auxiliary symbols (to make it easy to read).

First order theories are characterized by only using variables as parameters for the existential ($\exists$) and universal ($\forall$) symbols, higher order theories allow other types of symbols bounded by these.

An interpretation of the language is simply a way to assign the symbols as corresponding functions, constants, relations etc.

A package of a bunch of formulas in a language is what we call a theory.

Finally, an interpretation of a language, in which all formulas of a theory hold -- that is, all the statements 'formalized' by the formulas as symbols translate to actual properties in that interpretation -- is what we call a model of that theory.

Example #

In our example before, we constructed the language of groups and its theory where any model of such theory is actually a group.

We could create the language of rings, of vector spaces, of order relations, etc.

Substructures #

Given a model, we can sometimes find substructures: a piece of the model that is also a model.

Example #

Given a group $G$ any subgroup $H \subseteq G$ will also be a model of the theory of groups, meaning that taking $H$ as an interpretation of the language we get another model of a group that happens to be inside $G$.

It's important to distinguish that $H$ does not need to satisfy all the same formulas as $G$, only the ones we fixed in the theory (the ones that identify them as groups). There might be other formulas that are true in $G$ but not in $H$ and vice versa. For a more concrete example, let $G = Z/2$ be the integers modulo $2$, and its trivial subgroup $H={0}$, then $G$ and $H$ are not isomorphic as groups, so there must be a formula (in the language of groups) that is true in $G$ and not in $H$ and vice versa, right? (Cool exercise to understand what is happening).

Elementary Substructures #

Sometimes a model has substructures that are elementary meaning that the model and the substructure satisfy all the same formulas (not just those in the theory).

Example #

Keeping in line with our group flavored examples, if we take the integers $G=\Z$ as our model of group, then the subgroup given by the even numbers $H = 2\Z$ is also a group that is isomorphic to the whole group, meaning that they must satisfy the same formulas (also cool exercise).

Not always a model admits substructures, much less elementary substructures.

If one thinks for a bit, for a model to admit an elementary substructure it should behave much like infinity: a part is 'equivalent' to the whole in some way. But it is not evidently true that an infinite model admits elementary substructures.

Löwenheim-Skolem theorem #

Given infinite model $M$ then for any infinite cardinality $\kappa$:

if $\kappa \leq |M|$, there are elementary submodels $N \subseteq M$ such that $|N|=\kappa$.
if $\kappa \geq |M|$, there are models $N'$ with $|N'|=\kappa$ such that $M$ is an elementary substructure of $N'$.

This is incredible! If we have an infinite model, then there must be models (equivalent even!) on any infinite cardinality. (Cool exercise to show that the infinity part is required).

If you need the details Wikipedia's proof is ok. The TL;DR is

for the first part, use the axiom of choice to "fill" a subset $A \subseteq M$ with the required elements of $M$ so that every formula is either true (adding elements of M) or false.
for the second part we can create a new language using $M$ "as symbols" and any number (uncountable) of new symbols different from those of $M$. Using compactness it's easy to show that this new theory also has a model (using the first part one can find a model of the required cardinality).

Example #

Returning to our integer group example $G=\Z$ this means that there are groups $N'$ that are 'the same' as the integers but with any infinite cardinality! A group like $N'$ won't be isomorphic to $\Z$ (because they would need to be in bijection, meaning they would have to have the same cardinality), but from the perspective of group theory they are indistinguishable, they satisfy exactly the same properties (of group theory!).

Peano arithmetic and 'weird naturals' #

There are even more extreme examples. Usually when we work with the arithmetic of natural numbers we essentially use the Peano axioms. A very small set of (first order) formulas that seem to condense our understanding of the natural numbers: we start at 1 (or 0 for some), we know how to add 1 and we know it satisfies the principle of induction.

What the Löwenheim-Skolem theorem tells us is that there are other versions of the natural numbers (with other cardinalities) that also satisfy the same Peano axioms! This is like a supercharged version of Gödel's incompleteness theorem: we can't possibly characterize the natural numbers, there will always be other weird uncountable versions of them!

Skolem's Paradox #

There's a final example I'd like to share here today; one related to what I mentioned at the beginning. The theory of sets can be formulated in a first order language; there are actually multiple ways to do it but the most accepted one is the Zermelo-Fraenkel (or ZF for short).

As a direct corollary from the Löwenheim-Skolem theorem is the following: we have a model of ZF - the one we 'use' to make math every day; that model is infinite (we have the set of natural numbers inside). Therefore, there must be a model of ZF that is countable (the smallest possible infinity). Furthermore, the real numbers $\R$ can be defined in the model we usually use, meaning that it can be defined in that model as well, mutatis mutandis.

In this model $\R$ must be countable, since the entire model is countable... but Cantor's theorem is also true in this model so...

what is happening?

This apparent contradiction is known as Skolem's Paradox.

Cantor's theorem only says: "a bijection from $\N$ to $\R$ does not exist"; if one understands a function as a set, the statement can be reinterpreted as: "a specific set (with some required properties) does not exist in this model". From that reinterpretation, one can gather is that it's possible that there is a bijection from the real numbers (in that model) to the natural numbers, it's just that the bijection is not a set inside that model! In other words, the concept of cardinality in that model works the same as in our usual model, but Cantor's theorem does not apply to sets "across models"!

How large are the real numbers? #

One thing that we can take from Skolem's paradox is that we actually don't know how large the real numbers are. We don't really know in which model of ZF we 'live'; we know that in all ZF models, we can't find a bijection from the real numbers to the natural numbers insde the model, but the fact that we can't find it doesn't mean that the bijection doesn't exist somewhere else.