### Large Sets 9

#### Posted by Tom Leinster

*Previously: Part 8. Next: Part 9.5*

Today I’ll talk about inaccessibility. A set is said to be “inaccessible” if it cannot be reached or accessed from below using certain operations. We’ve seen this rough idea before — but which operations are the ones in play here, and what makes them especially interesting?

The definition is short and sweet: a set is **inaccessible** if it is
uncountable, a strong limit, and regular. Let’s review what that means:

A set $X$ is a strong limit if $Y \lt X \implies 2^Y \lt X$. That’s the most economical form of the definition, anyway. But an equivalent condition is maybe more illumnating: an uncountable strong limit is a set $X$ such that the sets $\lt X$ are a model of ETCS.

A set $X$ is regular if whenever $(X_i)_{i \in I}$ is a family of sets with $I \lt X$ and $X_i \lt X$ for all $i$, then $\sum_{i \in I} X_i \lt X$. Another way to say this: there’s no map out of $X$ whose codomain and fibres are all smaller than $X$.

So, for a set to be inaccessible means that it’s unreachable from below in two different ways: by the constructions that ETCS provides, and by coproducts of a smaller number of smaller sets.

Some other perspectives on inaccessibility begin to suggest why it’s an important notion:

Every infinite set is either regular or a weak limit. In other words, every successor is regular (as we saw last time). This makes it natural to ask which sets are both regular

*and*a weak limit. The uncountable such sets are called**weakly inaccessible**. Accessibility itself is the corresponding notion with “strong limit” in place of “weak limit”.(What I’m calling “inaccessible” used to be called “strongly inaccessible”, but I believe I’m following the dominant modern usage by dropping the “strongly”.)

We saw last time that regularity is a natural condition. Now regularity involves sums (coproducts) of sets. What if we change them to products? That is, call a set $X$

**product-regular**if whenever $(X_i)_{i \in I}$ is a family of sets with $I \lt X$ and $X_i \lt X$ for all $i$, then $\prod X_i \lt X$. Which sets are product-regular? Certainly $\mathbb{N}$ is, but what else?In fact, for uncountable sets, $\text{product-regular} \iff \text{inaccessible}.$ This is another hint that inaccessibility is important.

Can I give you an example of an inaccessible set? No! Inaccessible sets — if they exist — are larger than anything we’ve contemplated before, and beyond the realm where we can just “write one down”.

Let’s dig into that claim.

The largest kinds of sets we’ve considered so
far are the beth fixed
points. As
I’ll explain, every inaccessible set *is* a beth fixed point… but most
beth fixed points aren’t inaccessible. (They’re “accessible”, I guess, but
do people really say that?)

The proof that inaccessible sets are beth fixed points is really nice, so I’ll show it to you in full. It rests on a fact I mentioned in Part 7:

A set $X$ is a beth fixed point if and only if the sets $\lt X$ are a model of ETCS + (all beths exist).

We’ll show that every inaccessible set $X$ satisfies this equivalent condition. The sets $\lt X$ are certainly a model of ETCS, since $X$ is a strong limit. What we have to show, then, is that for every well-ordered set $W$ whose underlying set $U(W)$ is $\lt X$, the beth $\beth_W$ exists and is $\lt X$. We do this by induction on $W$:

If $W$ is empty then $\beth_W$ exists (it’s $\beth_0 = \mathbb{N}$) and is $\lt X$

*since $X$ is uncountable*.If $W$ is a successor, say $W = V^+$, then by inductive hypothesis, $\beth_V$ exists and is $\lt X$. Now $\beth_W$ exists; it’s $2^{\beth_V}$, which is $\lt X$

*since $X$ is a strong limit*.If $W$ is a nonempty limit then by inductive hypothesis, $\beth_V$ exists and is $\lt X$ for all $V \prec W$. Now $\beth_W = \sup_{V \prec W} \beth_V = \sup_{w \in W} \beth_{↡ w}$ where $↡ w = \{ w' \in W : w' \lt w \}$. This is a supremum of sets $\lt X$ indexed by $U(W) \lt X$, and is therefore $\lt X$

*since $X$ is regular*.

What I like about this proof is that the three cases of the transfinite induction naturally use the three parts of the definition of inaccessibility: uncountability, the strong limit property and regularity.

So: every inaccessible set is a beth fixed point. But inaccessibility is a much stronger condition. The smallest beth fixed point (if it exists) is not inaccessible. Nor is the second-smallest, nor the third-smallest. In fact:

For any inaccessible set $X$, there are unboundedly many beth fixed points $\lt X$.

This means that for any set $Y \lt X$, there is some beth fixed point $B$ with $Y \leq B \lt X$.

We can even *construct* such a $B$, in some sense of “construct”:

Take our starting set $Y$, and put $B_0 = Y$.

Form an initial well order $I(B_0)$ with underlying set $B_0$. Now $X$ is a beth fixed point, meaning that $\beth_{I(X)}$ exists and is $\cong X$. Since $B_0 \lt X$, it follows that $\beth_{I(B_0)}$ exists and is $\lt X$. Put $B_1 = \beth_{I(B_0)}$. Then from our starting set $B_0 \lt X$, we’ve constructed a new set $B_1 \lt X$.

Repeat this process to get a sequence of sets $Y = B_0 \leq B_1 \leq B_2 \leq \cdots,$ which are all $\lt X$.

This sequence has an upper bound, $X$, so it has a

*least*upper bound, $B$, satisfying $Y \leq B \leq X$. Better still, $B$ is*strictly*smaller than $X$: for $X$ is uncountable and regular, so it can’t be the supremum of the sequence $(B_n)_{n \in \mathbb{N}}$ of strictly smaller sets. So $Y \leq B \lt X$. And finally, $B$ is a beth fixed point, as mentioned in Part 7.

This result gives us another little independence theorem:

It is consistent with ETCS + (there are unboundedly many beth fixed points) that there are no inaccessible sets.

The proof is the same old argument we keep on seeing. Take a model of ETCS with unboundedly many beth fixed points. Call a set “small” if it is $\lt$ every inaccessible set in the model. Then the previous result implies that the small sets form a model of ETCS with unboundedly many beth fixed points, in which there are no inaccessible sets.

Perhaps I should state that proof a *little* more carefully. If the
original model contains no inaccessible sets, we’re done. If it does
contain an inaccessible set, there’s a smallest one, $X$. Then by the
previous result, the sets $\lt X$ (the “small sets”) are a
model of ETCS + (there are unboundedly many beth fixed points) + (there are
no inaccessible sets).

So if there are any inaccessible sets at all, they’re among the beth fixed points, but the smallest inaccessible set is bigger than the smallest beth fixed point.

#### Next time

In the next of my posts we’ll look at measurable sets. Measurability is the largest large set axiom I’ll talk about in this series, and it’s a nice one: it connects to both measurability in the sense of measure theory and codensity monads in category theory.

*Added later*: but first, Mike will talk about the size levels in between inaccessibility and measurability.

## Re: Large Sets 9

Now seems like a good moment to cite Mike Shulman’s Set theory for category theory, both for this post in particular and the series of posts in general.

I didn’t want to talk about the connection between inaccessibility and Grothendieck universes, but if you’re interested, you can find that topic discussed in section 8 of Mike’s paper.