### Large Sets 8

#### Posted by Tom Leinster

*Previously: Part 7. Next: Part 9*

If you’ve been wanting to follow this series but haven’t had time to keep up, now’s a good moment to hop back on board — I won’t assume much of what’s gone before.

Back in the mists of time, when I took a first undergraduate course on axiomatic set theory, I was exhilarated by the extraordinary world of infinite sets I saw opening up before me. In that world, addition is the same as multiplication! Which is the same as maximum! That is,

$X + Y \cong X \times Y \cong max(X, Y)$

for all infinite $X$ and $Y$. It seemed unthinkably exotic.

I then heard this part of cardinal arithmetic called “trivial” for exactly the reasons just stated. Although that description is technically correct, it poured a bucket of cold water over my enthusiasm in a way that only mathematicians can.

So with apologies to my past self, I give you the informal title of this post: the nontrivial part of cardinal arithmetic.

Once we’ve seen that binary addition and multiplication are “trivial”,
what’s left to do in cardinal arithmetic? Well, there’s addition and
multiplication of *infinitely* many sets, and there’s exponentiation.
Some of the theory of those operations is trivial too, in the sense of
following immediately from the isomorphisms

$X + Y \cong X \times Y \cong max(X, Y)$

for infinite $X$ and $Y$. Let’s get that trivial stuff out of the way first.

Addition, even infinitary addition, is more or less the same as supremum. An easy little argument shows that for any family $(X_i)_{i \in I}$ of nonempty sets,

$\sum_{i \in I} X_i \cong \max(I, \sup_{i \in I} X_i)$

— unless, of course, it’s a *finite* family of *finite* sets. (The
sum of finitely many natural numbers is not the same as its sup!)

Exponentiation is a special case of infinitary multiplication: $Y^X = \prod_{x \in X} Y$. Still, it’s an important enough case that it’s well worth thinking about separately.

In many cases, exponentiation is trivial too. It’s trivial when the expression

$Y^X$

is top-heavy — when $X$ is bigger than $Y$. It’s even trivial when $X$ is just slightly smaller than $Y$. Indeed,

$Y^X = 2^X \quad \text{ if } \quad 2^X \geq Y,$

assuming that $X$ and $Y$ are infinite. The proof is a pushover:

$2^X \leq Y^X \leq (2^X)^X \cong 2^{X \times X} \cong 2^X.$

For example, $\mathbb{N}^X \cong \mathbb{R}^X \cong 2^X$ for all infinite sets $X$.

There’s another case where exponentiation is trivial: when the base is the power set of something. Another one-line proof shows that if $Y$ is a power set then

$Y^X \cong \max(Y, 2^X).$

Here I’ve assumed again that $X$ and $Y$ are infinite. If $X$ or $Y$ is finite then exponentiation is trivial too.

So the only occasions when $Y^X$ is *not* trivial to compute are when $Y$
is not a power set and $2^X \lt Y$ — the expression $Y^X$ is
“bottom-heavy”. It’s certainly the case that

$Y \leq Y^X \leq 2^Y.$

But even assuming the generalized continuum hypothesis, this still leaves two options for $Y^X$: is it $Y$ or $2^Y$?

Now we’re moving across the boundary into what I’m calling the “nontrivial” part of cardinal arithmetic. (Of course, any professional would regard everything I’m saying here as trivial.) And it’s time to talk about the wonderful König’s theorem.

König’s theorem states that for any families of sets $(X_i)_{i \in I}$ and $(Y_i)_{i \in I}$ whatsoever,

$X_i \lt Y_i \ \text{for all}\ i \quad \implies \quad \sum_{i \in I} X_i \lt \prod_{i \in I} Y_i.$

Let’s spend a while contemplating this result, before we start to think about the proof. Some observations:

There are many variant statements that can be obtained by taking König’s theorem and replacing one or both $\lt$ signs by $\leq$, or by changing the sum to a product, or vice versa. All of those variants are either trivially true or false. (For example, changing both $\lt$s to $\leq$s gives a false statement.) Among these variants, König’s theorem is unique in being both nontrivial and true.

If we assume the generalized continuum hypothesis then König’s theorem does become trivial, at least when $X_i$ and $Y_i$ are infinite. For then $2^{X_i} \leq Y_i$ for every $i$, and taking the product of these inequalities over all $i$ gives $2^{\sum X_i} \leq \prod Y_i$, hence the result.

When every $X_i$ is empty, König states that a product of nonempty sets is nonempty. This is equivalent to the axiom of choice (in the presence of the other ETCS axioms). So, we’re going to have to use the axiom of choice in the proof.

When $X_i = 1$ and $Y_i = 2$ for all $i$, König’s theorem becomes Cantor’s theorem: $I \lt 2^I$ for all sets $I$. So, we can expect the proof to be a generalization of the diagonal argument.

As for the proof, I won’t spoil it! I *promise* you that if you haven’t
seen it before and you figure it out for yourself, it will brighten your
day. It’s the proof of *strict* inequality that’s the really nice part.

When I first learned König’s theorem, I got a bit overexcited. I thought that such a general statement would be sure to have a wealth of applications. But as far as I know, there’s basically only one, and it’s this:

Let $X$ and $I$ be infinite sets. Suppose that $X$ can be expressed as a coproduct of $I$ sets that are all $\lt X$. Then $X \lt X^I$.

Indeed, if $X \cong \sum_{i \in I} X_i$ with $X_i \lt X$ for each $i$, then König’s theorem with $Y_i = X$ for all $i$ gives the result.

For example, suppose we have a model of ETCS in which $\aleph_\omega$ exists. Then

$\aleph_\omega \cong \sup_{n \in \mathbb{N}} \aleph_n \cong \sum_{n \in \mathbb{N}} \aleph_n,$

so

$\aleph_\omega \lt \aleph_\omega^\mathbb{N}.$

The right-hand side here is a *very* bottom-heavy exponential:
$\aleph_\omega$ is vastly bigger than $\mathbb{N}$.

At this point I want to talk about some bad intuition that I carried around with me for many years. I’ll explain it using the example of two particular sets, $\mathbb{R} = 2^\mathbb{N}$ and $\mathbb{N}$, but other examples would work too.

$\mathbb{R} + \mathbb{N} \cong \mathbb{R}$, and my mental picture is that $\mathbb{R}$ is bigger than $\mathbb{N}$ and therefore “absorbs” it: adjoining tiny little $\mathbb{N}$ to great big $\mathbb{R}$ doesn’t make it any bigger.

*That*intuitive idea is fine. (It’s crude, because $X + Y \cong X$ even if $X$ and $Y$ have the*same*cardinality. But that doesn’t make it wrong.)$\mathbb{R} \times \mathbb{N} \cong \mathbb{R}$, and my mental picture for this is similar: since $\mathbb{R}$ is bigger than $\mathbb{N}$, taking $\mathbb{N}$ copies of it doesn’t make it any bigger.

*That*intuitive idea is fine too.$\mathbb{R}^\mathbb{N} \cong \mathbb{R}$, and my mental picture for this used to be similar: since $\mathbb{R}$ is bigger than $\mathbb{N}$, taking mere

*sequences*of reals doesn’t give you a larger set than taking reals themselves.*This intuition is completely wrong!*

It’s wrong because there are sets $X$ much bigger than $\mathbb{N}$ such that $X^\mathbb{N} \gt X$. Or at least, such sets exist as long as we add to ETCS a very mild extra axiom: the existence of an uncountable weak limit. Then $\aleph_\omega$ exists, and as we just saw,

$\aleph_\omega^\mathbb{N} \gt \aleph_\omega.$

That’s why it’s nonsense to think that $\mathbb{R}^\mathbb{N} \cong \mathbb{R}$ because $\mathbb{R}$ is bigger than, or “absorbs”, $\mathbb{N}$.

End of digression.

For results like the “$X \lt X^I$” one, it’s useful to think in terms of
cofinality. By definition, the **cofinality** $cf(X)$ of an infinite set
$X$ is the smallest set $I$ with the following property:

there is some family of sets $(X_i)_{i \in I}$ such that $X \cong \sum X_i$ and $X_i \lt X$ for all $i$.

The word “cofinality” comes from an equivalent order-theoretic formulation which I won’t talk about.

Certainly the set $I = X$ has the stated property, since $X \cong \sum_{x \in X} 1$. So $cf(X)$ is well defined and $cf(X) \leq X$. Some examples:

What’s the cofinality of $\mathbb{N}$? Since a finite coproduct of finite sets is finite, any set $I$ with the property above must be infinite. So $cf(\mathbb{N})$ is infinite. But $cf(\mathbb{N}) \leq \mathbb{N}$, so $cf(\mathbb{N}) = \mathbb{N}$.

How about $\aleph_1$, the smallest set $\gt \aleph_0 = \mathbb{N}$? If $(X_i)_{i \in I}$ is a family with $I \lt \aleph_1$ and $X_i \lt \aleph_1$, then $I \leq \aleph_0$ and $X_i \leq \aleph_0$. It follows that $\sum_{i \in I} X_i \leq \aleph_0 \times \aleph_0 \cong \aleph_0 \lt \aleph_1.$ So $\aleph_1$ cannot be expressed as a coproduct of $\lt \aleph_1$ sets that are all $\lt \aleph_1$. That is, $cf(\aleph_1) = \aleph_1$.

Generally, the same argument shows that if $X$ is a successor (that is, there’s some set $Y$ such that $X$ is the smallest set $\gt Y$), then $cf(X) = X$.

Assuming that $\aleph_\omega$ exists, we have $\aleph_\omega = \sup_{n \in \mathbb{N}} \aleph_n = \sum_{n \in \mathbb{N}} \aleph_n$ with $\aleph_n \lt \aleph_\omega$ for each $n$. Hence $\cf(\aleph_\omega) = \mathbb{N} = \aleph_0$.

As these examples show, many sets $X$ have the property that $cf(X) \cong
X$. Such sets are called **regular**. So

$\mathbb{N} = \aleph_0, \aleph_1, \aleph_2, \ldots$

are all regular, as is any successor. An infinite set that is not regular
is called **singular**. Thus, a set $X$ is singular if and only if it can
be expressed as a coproduct of $\lt X$ sets that are each $\lt X$.

Since the smallest example of a singular set is $\aleph_\omega$, and since it is consistent with ETCS that $\aleph_\omega$ does not exist,

It is consistent with ETCS that every infinite set is regular.

(I’m only defining cofinality, and regular vs. singular, for *infinite*
sets. I don’t know whether there’s a sensible way to set up the definitions
for finite sets.)

AsideRegularity comes up quite naturally in category theory. For instance, take the result that finite limits commute with filtered colimits in $\mathbf{Set}$. The definitions of finite limit and filtered colimit both refer (at least implicitly) to the concept of a set being finite, and finite means $\lt \mathbb{N}$. One might seek to generalize the commutativity result by replacing $\mathbb{N}$ by some larger set $K$. But if one tries to do this, one quickly finds oneself needing the assumption that $K$ is regular. There’s a nice explanation of this point in Francis Borceux’sHandbook of Categorical Algebra, Volume 1, Section 6.4.

The cofinality of a set is always regular:

$cf(cf(X)) \cong cf(X)$

for every infinite set $X$. (I won’t give the proof.) So $cf$ is an operation converting arbitrary sets into regular sets.

This is another moment at which a category theorist might get overexcited. Perhaps $cf$ feels like some kind of reflector of sets into regular sets, and perhaps the statements

$cf(X) \leq X, \quad cf(cf(X)) \cong cf(X)$

suggest there’s a monad around. Not so fast! *Cofinality is not
order-preserving:*

$\aleph_1 \lt \aleph_\omega \quad \text{ but } \quad cf(\aleph_1) = \aleph_1 \gt \aleph_0 = cf(\aleph_\omega).$

The subtle behaviour of the cf operation has driven a lot of research in set theory. For example, the extraordinarily prolific set theorist Saharon Shelah found new results in cardinal arithmetic using his PCF theory, where PCF stands for “potential (or possible) cofinalities”.

At the very least, the cf notation is highly convenient. Recall this result from earlier:

Let $X$ and $I$ be infinite sets. Suppose that $X$ can be expressed as a coproduct of $I$ sets that are all $\lt X$. Then $X \lt X^I$.

We can rephrase it more compactly like this:

$X \lt X^{cf(X)}$ for all infinite $X$.

This result, a corollary of König’s theorem, has an important corollary itself. Let $X$ be an infinite set and $Y$ a set with at least two elements. Replacing $X$ by $Y^X$ in the last result gives

$Y^X \lt (Y^X)^{cf(Y^X)} \cong Y^{max(X, cf(Y^X))}$

and so

$X \lt cf(Y^X).$

In other words, you can’t partition $Y^X$ into $X$ smaller pieces.

In particular, this is true when $Y = 2$, giving

$X \lt cf(2^X)$

for all infinite sets $X$. Since $cf(S) \leq S$ for all $S$, this is a sharpening of Cantor’s result that $X \lt 2^X$.

A nice application of the fact that $X \lt cf(2^X)$ is to show that $\aleph_\omega$ is not the power set of anything in the world. (Allen Knutson mentioned this back in Part 4, and Todd Trimble gave a proof.) For if $\aleph_\omega \cong 2^X$ then

$X \lt cf(2^X) = cf(\aleph_\omega) = \aleph_0,$

so $X$ is finite. But $\aleph_\omega \cong 2^X$, so $\aleph_\omega$ is finite, a contradiction.

#### Next time

Next time, we’ll climb another rung up the ladder of largeness, where we’ll
find the *inaccessible* sets.

## Re: Large Sets 8

Do you mean “if $X$ xor $Y$ is finite”? (-: