There's a really nice PBS Infinite Series video on Youtube How Big are all Infinities Combined, exploring the sizes of infinite sets. (If you haven't watched it, this would be a great time. I'll wait.) In this video they make heavy use of Cantor's theorem, which says

Given any set \(S\), its power set \(\mathcal{P}(S)\) is bigger than \(S\)where \(\mathcal{P}(S)\) is the set of all subsets of \(S\).

The amazing thing about this is that it is true even when \(S\) is infinite, though we have to work a little to say what we mean by *bigger than* here: what we mean is that there is no injective function \(\mathcal{P}(S) \to S\), though there is an obvious injective function \(i:S \to \mathcal{P}(S)\), given by \(i(x)=\{x\}\) for each \(x \in S\).

It's a non-trivial fact (the Cantor-Schröder-Bernstein theorem) that given two sets \(A\) and \(B\), if there is an injection from \(A\) to \(B\) and an injection from \(B\) to \(A\), then there is a bijection from \(A\) to \(B\); so it makes sense to say that \(A\) is smaller than (or equal to) \(B\) in size if there is an injection from \(A\) to \(B\), but none from \(B\) to \(A\).

This immediately tells us that there is a whole tower of infinite sets that we can make; starting with the natural numbers \(\mathbb{N}\), then we can think about \(\mathcal{P}(\mathbb{N})\), \(\mathcal{P}(\mathcal{P}(\mathbb{N})\), and so on, each of them strictly larger than its predecessor.

It's not very hard to prove that the second of these, \(\mathcal{P}(\mathbb{N})\), is the same size as \(\mathbb{R}\). It is an interesting question whether there is any infinite size between that of \(\mathbb{N}\) and \(\mathbb{R}\), but one which I won't consider here.

In the video (you have watched it now, right?) the question is how big all the infinities combined is.

But there's even an issue with what it means for one infinite set to be bigger than another, and that's what I want to explore here.

The problem is, to a large extent, knowing just what sets and functions are.

It's easy enough to decide what a function is if you know what a set is.

If \(A\) and \(B\) are sets, we think informally as a function \(A \to B\) as a rule that assigns to each element of \(A\) an element of \(B\). We can formalise this by saying that a function \(A \to B\) is a subset of \(A \times B\) (the set of all ordered pairs \((a,b)\) where \(a \in A\) and \(b \in B\)) which has the property that each element of \(A\) appears in exactly one of the ordered pairs.

But what's a set?

It would be really nice if we could just say that a set was a (somehow) well-defined collection of objects. But trying to do without some kind of restriction on how we define the sets tends to lead to real problems, the most famous of which is Russell's paradox:

Let \(X\) be the set of all sets which are not elements of themselves. Then if \(X \in X\), by the definition of \(X\), \(X \not \in X\). But conversely, if \(X \not \in X\) then again by the definition of \(X\), \(X \in X\).This is a genuine paradox in the world of naive set theory, so we need to be a bit more careful about what counts as a description of the elements comprising a set if we want to avoid problems like this.

The (ok, a) solution is to develop axiomatic set theory, and one particularly popular version of that is ZFC (which stands for Zermelo-Fraenkel plus the Axiom of Choice). I won't explain just how ZFC works here, but in essence it is a set of rules that tell us how we can build new sets out of existing ones, and a couple of particular sets that we insist on (so that we can build others out of them). One of the sets we insist on is the set of von Neumann integers, so \(\mathbb{N}\) is definitely in in there.

There's always the question of whether our axioms are consistent. Most mathematicians are happy that the ZFC axioms are in fact consistent, partly because we can describe a collection of objects (which we call sets) that seem to satisfy all the axioms. The study of particular collections of objects and operations on them which satisfy particular sets of axioms is called **model theory**, and it throws up some surprising results.

We'll call a collection of sets that satisfy the axioms of ZFC a **universe**.

There is a fundamental theorem of model theory, the **downward Löwenheim-Skolem theorem** that tells us that there is a countable universe (yes, the *entire* universe is countable) that satisfies the axioms of ZFC.

But there's clearly something fishy about this. Cantor's theorem also follows from these axioms of set theory, and that tells us that the power set of \(\mathbb{N}\) is uncountable.

How does an uncountable set fit into a countable universe? This is **Skolem's paradox**.

And this is where the issue of just what is a set rears up in all its glory and horror.

Let's review the situation.

The entire universe is countable, so in this model of set theory there are only countably many subsets of \(\mathbb{N}\). (This model clearly fails to include most of the subsets that we might think ought to be there.) Since the universe is countable and \(\mathcal{P}(\mathbb{N})\) is countable, there is a bijection between \(\mathbb{N}\) and \(\mathcal{P}(\mathbb{N})\).

But Cantor has proved that there can be no such bijection.

But remember what a function \(A \to B\) is: it is a subset of \(A \times B\) satisfying a certain constraint, namely that every element of \(A\) occurs in exactly on of the ordered pairs.

We know that the \(\mathcal{P}(\mathbb{N})\) in this universe is countable, so there is a collection of ordered pairs \((n,X)\), one for each natural number \(n\) and such that every \(X \subseteq \mathbb{N}\) in this universe appears in the collection.

The escape clause is the one that tells us that this collection is not—cannot be—one of the sets in the universe.

Looking at the universe from outside, we see a function—what we might call an **external** function—giving a bijection between \(\mathbb{N}\) and this universe's model of \(\mathcal{P}(\mathbb{N})\).

But if we live inside the universe, we can't see this: it doesn't exist in the universe. There is no **internal** function which matches the natural numbers with the sets of natural numbers.

You may well have an objection to raise at this point: the real numbers are a complete ordered field, and it's also a theorem that this defines them up to isomorphism. And we know that the real numbers (at least in the mathematical universe we're accustomed to) really are uncountable. So how do they fit into this countable universe?

The answer to this is unavoidably a little technical.

The completeness axiom for the real numbers says something about all subsets of \(\mathbb{R}\); this is a second order axiom. The axioms of ZFC aren't powerful enough for this type of operation. They are first order, which (roughly speaking) means that they let you say something about all elements of a set, but not all subsets of it.

If you use the first order version of the axioms for a complete ordered field, there is more than one structure satisfying the axioms. Some are countable (as must be the case in the countable universe of ZFC guaranteed by the Löwenheim-Skolem Theorem), and some are larger (indeed there are models of all infinite cardinalities).

So all the above is true if we use the power of first order axioms to specify what a set is. If we allow second order axioms, then the game changes. Then why don't we use some kind of second order theory, and avoid this kind of nonsense?

It's very tempting to do this. Unfortunately, there are problems with second order theories too. Though it solves some of the problems, for example tying down the real numbers much more specifically, the notion of proof in a second order theory is not as well behaved as in first order theory. Consequently, logicians who work on the foundations of mathematics explore the strengths and weaknesses of first and second order theories, while 'working mathematicians' tend not to think about either too much, but use first or second order axioms as they are more convenient.

### Acknowledgement

This post was sparked off by the Infinite Series video*How Big are all Infinities Combined*. Gabe Perez-Giz (@fizziksgabe) also bears some responsibility, since when I made a brief comment on twitter about this he asked if I had a blog post about it. Well, now I do.