## July 13, 2021

### Large Sets 11

#### Posted by Tom Leinster

Previously: Part 10. Next: Part 12

Measurability is the largest of the “large set” conditions I’m going to talk about in this series. Today I’ll explain how measurability relates to inaccessibility, say a tiny bit about how measurability can arise in analysis problems, and say somewhat more about measurability and codensity monads.

Let’s recall the definition: an uncountable set $X$ is measurable if it admits a nonprincipal ultrafilter that is $X$-complete — in other words, closed under $I$-fold intersections for all $I \lt X$.

We’ll see that if there are any measurable sets at all, they must be enormous. In particular, measurable sets are inaccessible, but there are many many inaccessible sets smaller than the smallest measurable set. In Part 9.5, Mike talked about the various levels of size between inaccessible and measurable: 1-inaccessible, 2-inaccessible, hyperinaccessible, Mahlo, greatly Mahlo, … I recommend it! But here I’m going to stick to measurable versus inaccessible.

Let’s start by showing that every measurable set is inaccessible. (And at the same time, I’ll remind you what inaccessible means.) Let $X$ be a measurable set. To show that $X$ is inaccessible, we have to prove three things:

• $X$ is uncountable. This is an explicit part of the definition of measurability.

• $X$ is regular. In other words, it’s impossible to express $X$ as a union $\bigcup_{i \in I} X_i$ with $I \lt X$ and $X_i \lt X$ for all $i$.

The key here is an observation about nonprincipal $X$-complete ultrafilters. Each member $A$ of such an ultrafilter must have cardinality $\geq X$: for if not, $A$ would be a union of $\lt X$ singletons, and since singletons aren’t in the ultrafilter, neither is $A$.

Since $X$ is measurable, there is some nonprincipal $X$-complete ultrafilter $\mathcal{U}$ on it. If $(X_i)_{i \in I}$ is a family of subsets of $X$ with $I \lt X$ and $X_i \lt X$, then it follows from our observation that no $X_i$ belongs to $\mathcal{U}$. Since $I \lt X$, the union $\bigcup X_i$ is not in $\mathcal{U}$ either, so it definitely can’t be $X$.

• $X$ is a strong limit. In other words, if $Y \lt X$ then $2^Y \lt X$.

This can be proved using a nice lemma: for sets $A$ and $B$, any ultrafilter on $B^A$ closed under both $A$- and $B$-fold intersections must be principal. I leave the proof to you, but it’s quite satisfying — try it! Rot-13ed hint: Sbe rnpu n va N, pbafvqre gur shapgvba sebz O^N gb O tvira ol rinyhngvat ng n.

Against my usual preference, I’ll use contradiction to prove that any measurable set is a strong limit. Let $X$ and $Y$ be sets, suppose that $Y \lt X \leq 2^Y$, and let $\mathcal{U}$ be an $X$-complete ultrafilter on $X$. We’ll show that $\mathcal{U}$ is principal.

For take an injection $i: X \to 2^Y$. Pushing $\mathcal{U}$ forward along $i$ gives an $X$-complete ultrafilter $i_\ast \mathcal{U}$ on $2^Y$. Then $i_\ast \mathcal{U}$ is closed under $Y$-fold intersection, so by the lemma, $i_\ast \mathcal{U}$ is principal. Since $i$ is injective, $\mathcal{U}$ is principal, as required.

We’ve just shown that every measurable set is inaccessible, a result first proved by Stanisław Ulam in 1930. For what happened next historically, I’ll quote from Akihiro Kanamori’s book The Higher Infinite (p.27):

Whether the least measurable cardinal is strictly larger than the least inaccessible cardinal became a focal question, and was settled only thirty years later. For a result that by present-day standards is rather straightforward this may seem like a remarkably long time — even taking into account the convulsive events that took place in Europe. However, this is a typical case of a long-standing mathematical problem suddenly solved in the wake of new techniques.

Kanamori doesn’t say it here, but the answer to the question is yes. So the theorem is:

For every measurable set $X$, there exists an inaccessible set $\lt X$.

In fact, something much stronger is true:

For every measurable set $X$, there are $X$ isomorphism classes of inaccessible sets $\lt X$.

There certainly aren’t more than $X$ of them, for reasons that have nothing to do with largeness axioms: for any set $X$ whatsoever, there are at most $X$ iso classes of sets $\lt X$. So the force of the theorem is that there are as many as $X$ inaccessible sets smaller than any measurable set $X$.

This stronger result has a corollary:

For every measurable set $X$, there are unboundedly many inaccessibles $\lt X$.

That is, for any $Y \lt X$, there is some inaccessible set $Z$ with $Y \lt Z \lt X$. For if not, $Y$ would be an upper bound for the inaccessibles $\lt X$, so there would be at most $Y$ inaccessibles $\lt X$, contradicting the previous theorem.

And this corollary in turn has a corollary:

It is consistent with ETCS + (there are unboundedly many inaccessibles) that there are no measurable sets.

The argument is the same one that’s appeared in many of these posts. Take a model of ETCS with unboundedly many inaccessibles. Call a set in it “small” if it’s $\lt$ every measurable set. Then by the previous corollary, the small sets are a model of ETCS in which there are unboundedly many inaccessible sets but no measurable sets.

The theorem that beneath every measurable set $X$ there are $X$ inaccessibles is by far the hardest result that I’ve mentioned in these posts — as you might guess from that thirty-year gap. The proof is, I think, quite interesting, so I’m going to describe it.

If you’re skim-reading this post, or if you want to get on to the part where I talk about connections with category theory, then you might want to skip ahead to the end of the long bulleted list below.

I’m going to sketch the proof of the theorem, but first I want to highlight a particular feature: it’s one of those mathematical arguments where a thing is proved to exist by showing that the set of such things has positive measure. Such proofs are maybe most familiar from analysis, but the same strategy is behind the probabilistic method in combinatorics.

In the case at hand, “measure” refers to an ultrafilter. We have our measurable set $X$, we have the set

$K(X) = \{\text{iso classes of sets}\ \lt X \},$

and the proof will construct an ultrafilter on $K(X)$ with the property that the set of inaccessible sets $\lt X$ belongs to the ultrafilter. This immediately implies that there’s at least one inaccessible set $\lt X$, as $\emptyset$ does not belong to any ultrafilter. In fact, the ultrafilter has properties implying that every member of it has cardinality $\geq X$, and this gives the full-strength result: there are $X$ iso classes of inaccessible sets $\lt X$.

So we won’t actually construct $X$ inaccessible sets $\lt X$. We’ll just show that there are that many of them.

I’ll now sketch a proof that when $X$ is measurable, there are $X$ isomorphism classes of inaccessible sets $\lt X$. It’s the argument you’ll find in some set theory texts, rewritten in a structural/neutral style. Here we go.

• Choose a nonprincipal $X$-complete ultrafilter $\mathcal{U}$ on $X$, which we can because $X$ is measurable.

• Define $K(X)^X/\mathcal{U}$ to be the set $K(X)^X$ quotiented by the equivalence relation that identifies two functions $X \to K(X)$ when their equalizer belongs to $\mathcal{U}$. (If you know about ultraproducts then you’ll recognize that this is one.) Write $[F]$ for the equivalence class of a function $F: X \to K(X)$.

• Define a relation $\lt$ on $K(X)^X/\mathcal{U}$ by $[F] \lt [G] \ \iff\ \{x \in X: F(x) \lt G(x)\} \in \mathcal{U},$ and check that this is well-defined.

• Show that $\lt$ is a well-order on $K(X)^X/\mathcal{U}$. Ultimately this is because cardinal inequality is a well-order on $K(X)$, but the tricky part of this step is the “well”.

The usual definition of well-order says that every nonempty subset has a least element. But in this step, it’s better (essential?) to use the equivalent condition that there are no countably infinite descending chains $a_0 \gt a_1 \gt \cdots$. The uncountability of $X$ is also needed in this step.

• Each $A \lt X$ gives rise to an element of $K(X)^X/\mathcal{U}$, namely, the equivalence class $[A]$ of the constant function $X \to K(X)$ with value $A$. Call such elements of $K(X)^X/\mathcal{U}$ constant.

• Show that in $K(X)^X/\mathcal{U}$, the set of constant elements is downwards closed, but not every element is constant. One way to construct a nonconstant element is to choose an arbitrary well-order on $X$ and take the equivalence class of the function $x \mapsto \{ y \in X: y \lt x\}$.

• Since $K(X)^X/\mathcal{U}$ is well-ordered and not every element is constant, it has a least nonconstant element $[F]$. Then everything smaller than $[F]$ is constant.

• Push the ultrafilter $\mathcal{U}$ on $X$ forward along $F: X \to K(X)$ to give an ultrafilter $F_\ast \mathcal{U}$ on $K(X)$. Since $\mathcal{U}$ is $X$-complete, so is $F_\ast \mathcal{U}$. Since $[F]$ is not constant, $F_\ast \mathcal{U}$ is not principal.

• Show that:

• the set of iso classes of uncountable sets $\lt X$ belongs to $F_\ast \mathcal{U}$;
• the set of iso classes of strong limits $\lt X$ belongs to $F_\ast \mathcal{U}$;
• the set of iso classes of regular sets $\lt X$ belongs to $F_\ast \mathcal{U}$.

I’m not going to go into the details of these substeps. For the most part, they’re the kind of arguments that you’ll be used to if you’re used to ultraproducts, although there are some not-quite-routine aspects too. In any case, putting together these three results immediately gives:

• the set of iso classes of inaccessible sets $\lt X$ belongs to $F_\ast \mathcal{U}$.
• Since $\mathcal{F}_\ast \mathcal{U}$ is a nonprincipal $X$-complete ultrafilter, every member has cardinality $\geq X$. But it’s an ultrafilter on $K(X)$, which is $\leq X$, so every member has the same cardinality as $X$. In particular, the set of iso classes of inaccessible sets $\lt X$ has cardinality $X$.

That’s it! As I said, it’s much more substantial than any other proof in this series. If you know of a simpler proof, please let me know in the comments.

I promised to say a tiny bit about the connections with analysis. Of course, the concept of measurable set came from analysis historically, but the existence of a measurable set can also be relevant to analysis problems that arise naturally to this day.

Here’s a specific example, from the 1987 book An Introduction to Independence for Analysts by the analyst Garth Dales and the set theorist Hugh Woodin. Recall that the class of Borel sets in a topological space is the $\sigma$-algebra generated by the open sets. That is, it’s the smallest class of subsets containing the open sets and closed under complements and countable unions. Dales and Woodin write (p.viii):

It is not difficult to show that, if $f: \mathbb{R} \to \mathbb{R}$ is a continuous function, then $f(B)$ is Lebesgue measurable for each Borel subset $B$ of $\mathbb{R}$. Now suppose that $f, g: \mathbb{R} \to \mathbb{R}$ are continuous functions. Is $f(\mathbb{R} \setminus g(B))$ Lebesgue measurable for each Borel set $B$? This problem cannot be decided in ZFC. However if there is a measurable cardinal, then these sets are indeed all Lebesgue measurable.

So the measurability of certain subsets of $\mathbb{R}$ can be affected by the existence of a measurable set — which is necessarily much, much, bigger than $\mathbb{R}$.

I’ll finish by talking about how measurability of sets arises in category theory.

Throughout this post and the previous one, I’ve used the correspondence between collections of subsets of $X$ (such as ultrafilters) and functions $2^X \to \{0, 1\}$ (such as measures). But what I haven’t mentioned is the correspondence between measures and integral operators. Given how basic that is to analysis, it’s probably worth thinking about in this context too.

Take a set $X$ and a $\{0, 1\}$-valued probability measure $\mu$ on $X$, defined on every subset of $X$. What kind of function $f: X \to Y$ on $X$ can we integrate against $\mu$? That is, when does the expression

$\int_X f \ d\mu$

define an element of $Y$ in some reasonable way?

Well, if $f$ is constant on some subset of $X$ of measure $1$ then the integral certainly makes sense: it should be the constant value of $f$. (Remember that $\mu$ is a probability measure: $\mu(X) = 1$.) Since $\mu$ is countably additive, this is guaranteed to happen when $Y$ is countable. So, $\mu$ determines an “integration” operator

$\int_X - d\mu : \mathbf{Set}(X, Y) \to Y$

for each countable set $Y$.

Exactly as in analysis, measures and integration operators are in one-to-one correspondence, as long as you set things up right. In this case, we define an integration operator on $X$ to be a family of maps

$\Bigl(\alpha_Y: \mathbf{Set}(X, Y) \to Y\Bigr)_{\text{countable}\ Y}$

natural in countable sets $Y$. Then the following three kinds of structure on $X$ are essentially the same, in the sense of being in bijection with one another:

• integration operators on $X$;

• $\{0, 1\}$-valued probability measures on $X$;

• ultrafilters on $X$ closed under countable intersection.

And now we get to something categorically significant. Write $T(X)$ for the set of integration operators on $X$. Then $T$ is a functor $\mathbf{Set} \to \mathbf{Set}$. In fact, it’s more than just a functor: it’s a monad, the codensity monad of the inclusion

$\mathbf{Ctbl} \hookrightarrow \mathbf{Set}$

of the category of countable sets into the category of sets.

If you like ends, you’ll find it enlightening to write $T(X)$ as one:

$T(X) = \int_{Y \in \mathbf{Ctbl}} Y^{\Set(X, Y)}.$

This is an instance of the general end formula for a codensity monad. Alternatively, the fact that $T$ is this codensity monad follows from the original definition of $T(X)$ as a set of natural transformations.

In summary, the codensity monad $T$ of $\mathbf{Ctbl} \hookrightarrow \mathbf{Set}$ assigns to each set $X$ the set of ultrafilters on $X$ closed under countable intersections. Some elements of $T(X)$ are trivial: there’s the principal ultrafilter $\mathcal{U}_x$ for each $x \in X$. And predictably, the unit $X \to T(X)$ of the monad is the assignment $x \mapsto \mathcal{U}_x$.

But the question is, are there ever any other elements of $T(X)$?

Thinking set-theoretically, the answer is yes if and only if there is some countably measurable set. As we saw last time, is equivalent to the existence of an actually measurable set.

Thinking categorically, the answer is yes if and only if the monad $T$ is not isomorphic to the identity. Now, the codensity monad of a subcategory inclusion $\mathbf{B} \hookrightarrow \mathbf{A}$ is isomorphic to the identity if and only if $\mathbf{B}$ is codense in $\mathbf{A}$ (by definition, if you choose your definitions right). So:

There exists a measurable set if and only if $\mathbf{Ctbl}$ is NOT codense in $\mathbf{Set}$.

This was first shown by Isbell back in 1960, in his paper “Adequate subcategories” (Theorem 2.5), though he proved it directly rather than going via monads.

So, a model of ETCS in which there are no measurable sets is one in which the countable sets play a special role: they are codense in the category of all sets.

A careful examination of the proof shows that the codensity monad of $\mathbf{Ctbl} \hookrightarrow \mathbf{Set}$ is unchanged if we replace $\mathbf{Ctbl}$ by any full subcategory $\mathbf{C}$ containing at least one infinite set. In particular, we could take $\mathbf{C}$ to be the full subcategory consisting of the single set $\mathbb{N}$.

So what?

Well, if we replace $\mathbf{Ctbl}$ by the full subcategory on $\mathbb{N}$ then the description of $T$ in terms of natural transformations (or as an end) becomes something very simple, expressible without categorical language. It tells us that the set of ultrafilters on $X$ closed under countable intersections is in bijection with the set of $End(\mathbb{N})$-equivariant maps

$\mathbb{N}^X \to \mathbb{N}.$

By $End(\mathbb{N})$, I mean the monoid of all functions $\mathbb{N} \to \mathbb{N}$, with composition as the monoid operation. It acts on $\mathbb{N}$ and $\mathbb{N}^X$ in a natural way.

For example, the principal ultrafilter on an element $x \in X$ corresponds to the $x$-projection $\mathbb{N}^X \to \mathbb{N}$.

Putting this together, a set $X$ is countably measurable if and only if the projections are not the only $End(\mathbb{N})$-equivariant maps $\mathbb{N}^X \to \mathbb{N}$. Hence:

There exists a measurable set if and only if there exist a set $X$ and an $End(\mathbb{N})$-equivariant map $\mathbb{N}^X \to \mathbb{N}$ that is not a projection.

All of this codensity stuff can be done with $\mathbf{Ctbl}$ changed to the category $\mathbf{Set}_{\lt Y}$ of sets $\lt Y$, for any fixed $Y$. The codensity monad of $\mathbf{Set}_{\lt Y} \hookrightarrow \mathbf{Set}$ assigns to each set $X$ the set of $Y$-complete ultrafilters on $X$. But that’s for a fixed $Y$. It’s hard to imagine any nontrivial way in which the assignment

$X \mapsto \{ X\text{-complete ultrafilters on}\ X \} = \int_{Y \lt X} Y^{\mathbf{Set}(X, Y)}$

could be functorial, let alone a monad.

That’s it for largeness conditions — I’m not going beyond measurable. It could be fun to carry on… For example, an uncountable set $X$ is strongly compact if every $X$-complete filter on a set can be extended to an $X$-complete ultrafilter, and it’s a nice but not so easy exercise to show that strongly compact implies measurable. (Rot-13ed hint: svefg fubj gung rirel fgebatyl pbzcnpg frg vf erthyne. Guvf vfa’g fb rnfl rvgure.) But enough! I’m stopping here.

#### Next time

Although I’m done with largeness conditions, I do want to say some things about the axiom scheme of replacement. Why? Because it implies several of the largeness conditions in earlier posts, because it’s historically important, because ETCS plus replacement is equivalent to ZFC, because it’s not widely enough known that there are versions of replacement that are entirely natural in structural set theory (it’s not tied to the membership-based approach), and because people seem to like talking about it. That’s for next time.

Posted at July 13, 2021 5:34 PM UTC

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### Re: Large Sets 11

So, in terms of the hierarchy that I described in part 9.5, your proof that measurables are inaccessible actually shows that they are 1-inaccessible, right? I wonder how much harder the proofs are that measurables are 2-inaccessible, hyper-inaccessible, or Mahlo.

Posted by: Mike Shulman on July 14, 2021 4:44 AM | Permalink | Reply to this

### Re: Large Sets 11

[the above] proof that measurables are inaccessible actually shows that they are 1-inaccessible, right?

Yes, right — I thought that, and I think I meant to say it, but I did not, in fact, say it.

I learned this proof from Kuratowski and Mostowski’s book Set Theory, which I think was written relatively soon after the result was first proved, so it might be that the proof hadn’t yet reached its optimal form. I also had to do a lot of work to translate it into structural terms, and the translation might not be optimal either. Perhaps there’s a smoother treatment that with little more effort would prove hyperinaccessbility and Mahloness.

That quote from Kanamori refers to a wider context into which this theorem fits, but I know nothing about it.

Posted by: Tom Leinster on July 14, 2021 11:06 AM | Permalink | Reply to this

### Re: Large Sets 11

To prove that measurable implies Mahlo, I find myself having to build on that ultrapower proof that the smallest measurable set is bigger than the smallest inaccessible set. This is an argument I worked out for myself just now and may be very inefficient. I see that the proof in Jech’s Set Theory (Lemma 10.21 in the 3rd edition) is much shorter. However, it does rely on Theorem 10.20, the proof of which looks quite similar to the argument I found.

I doubt that anyone is following in enough detail to want to read the rest of this comment, but in any case, I’ll sketch the proof I have in mind. Let $X$ be a measurable set. We want to prove that $X$ is Mahlo, which means that whenever $L$ is an an unbounded subset of

$K(X) = \{\text{iso classes of sets} \ \lt X\},$

the set

$L' = \{\text{iso classes of sets} \ Y \lt X \ \text{such that}\ Y = \sup_{Z \in L: Z \lt Y} Z \}$

contains at least one inaccessible.

Let $\mathcal{U}$ be an $X$-complete ultrafilter on $X$. In that long proof in the post, we used the ultrapower $K(X)^X/\mathcal{U}$. We found an element $[F]$ of the ultrapower such that the elements smaller than $F$ are precisely the constants. Then we used the pushforward ultrafilter $F_\ast\mathcal{U}$ on $X$, showing that the set of iso classes of inaccessibles $\lt X$ belongs to $F_\ast \mathcal{U}$.

I claim that $L' \in F_\ast \mathcal{U}$ too. For suppose not. Then for almost all $x \in X$, we have $F(x) \gt \sup_{Z \in L: Z \lt F(x)} Z$. Write $S(x)$ for this sup. Then $S \in K(X)^X$ with $S \lt F$, so $S$ is constant almost everywhere: say $[S] = [A]$, where $A \lt X$. Since $L$ is unbounded, there is some $Z \in L$ with $Z \gt A$. Then $F(x) \leq Z$ for almost all $x \in X$, giving $[F] \leq [Z]$. This contradicts the defining property of $F$, thus proving the claim.

So both $L'$ and the set of iso classes of inaccessibles $\lt X$ belong to the ultrafilter $F_\ast \mathcal{U}$. Hence their intersection is nonempty, which is what had to be proved.

This proof establishes more than stated. It shows that the set of inaccessibles in $L'$ is not only nonempty, but in fact belongs to the nonprincipal $X$-complete ultrafilter on $K(X)$. So, for example, there are $X$ inaccessibles in $L'$. Surely this gets us something, but right now I don’t see what.

Posted by: Tom Leinster on July 17, 2021 10:41 AM | Permalink | Reply to this

### Re: Large Sets 11

That’s a nice proof! I find it more comprehensible than the ones I was able to find online, but I don’t have Jech’s book to compare to. Can you say exactly where the $X$-completeness of $\mathcal{U}$ is used?

My guess is the fact that the set of inaccessibles in $L'$ is not only inhabited but belongs to $F_\ast\mathcal{U}$ is a shadow of the fact that a measurable cardinal is not only Mahlo but also 1-Mahlo, etc. I’m not sure it immediately implies that, but it suggests that similar arguments could be used to establish that.

Posted by: Mike Shulman on July 18, 2021 12:56 PM | Permalink | Reply to this

### Re: Large Sets 11

Can you say exactly where the $X$-completeness of $\mathcal{U}$ is used?

I don’t think it is, as such. But we do use the fact that $\mathcal{U}$ is $\aleph_1$-complete. This is needed to prove that $K(X)^X/\mathcal{U}$ is well-ordered, the point being that an order is well iff it has no countably infinite descending chain. And well-orderedness is needed to construct $[F]$, the least nonconstant element of $K(X)^X/\mathcal{U}$.

(Of course, the full strength of $X$-completeness is used to show that measurable implies inaccessible.)

Posted by: Tom Leinster on July 18, 2021 2:30 PM | Permalink | Reply to this

### Re: Large Sets 11

Personally, I find that this fact:

a model of ETCS in which there are no measurable sets is one in which the countable sets play a special role: they are codense in the category of all sets

makes me feel like measurable sets ought to exist. Why should we expect countable sets to play a special role like that? Moreover, I’ve been told that there’s a strengthening of this fact, namely

There exist unboundedly many measurable sets if and only if $\mathbf{Set}$ does NOT have ANY small codense full subcategory.

Similarly, this makes me feel like there ought to be unboundedly many measurable sets.

Posted by: Mike Shulman on July 14, 2021 5:29 AM | Permalink | Reply to this

### Re: Large Sets 11

I’d be tempted to try to turn that around. Call a set “ridiculously large” if there exists a nonprincipal ultrafilter on it that is closed under countable intersections. Why should we expect a ridiculously large set to exist?

(Why shouldn’t we, you may reply? Well, I don’t know! But for example, it would make $V = L$ impossible.)

Use of words like “special role” or “ridiculously large” may have the effect of swinging one’s feelings in a certain direction.

In some sense, universes without measurable cardinals are “perfect worlds” with perfect dualities, like having the canonical embedding

$X \to Set^{End(\mathbb{N})}(Set(X, \mathbb{N}), \mathbb{N})$

be an isomorphism. In walks a measurable cardinal who ruins everything!

Posted by: Todd Trimble on July 14, 2021 10:13 AM | Permalink | Reply to this

### Re: Large Sets 11

I don’t think that’s fair. There’s something objectively “special” about $\mathbb{N}$ in the absence of a measurable cardinal (i.e. it has a property that no other set has), whereas there’s nothing objectively “ridiculous” (whatever that would even mean) about a countably complete nonprincipal ultrafilter.

Posted by: Mike Shulman on July 14, 2021 2:38 PM | Permalink | Reply to this

### Re: Large Sets 11

Perhaps I misunderstand, but I’m not sure what’s special. Couldn’t you equally well say that measurable cardinals do not exist iff all sets of cardinality up to and including the continuum are codense in the category of sets?

The only thing I understand to be special about $\mathbb{N}$ is that it’s the smallest that has this property.

Posted by: Todd Trimble on July 14, 2021 4:44 PM | Permalink | Reply to this

### Re: Large Sets 11

Yes, I agree, being the smallest with that property is the only thing special about $\mathbb{N}$. I guess one could argue that that’s not really very special: $\mathbb{N}$ is already the smallest infinite set, so really what’s being said is that the boundary between finite and infinite also coincides with the boundary between not-codense and codense. Which, I guess, is not unreasonable. It would be weirder if the boundary of codensity were at some other random place like $\aleph_{\omega+3}$, but we already know that finite and infinite sets differ in a lot of ways, so this could certainly be one of them.

But this does suggest to me that if there are any measurable cardinals, there ought to be unboundedly many of them!

Posted by: Mike Shulman on July 15, 2021 9:24 PM | Permalink | Reply to this

### Re: Large Sets 11

Great, that’s exactly what I was driving at. And I think I do agree at the intuitive level that if there are any measurable cardinals, there “ought” to be unboundedly many of them!

Kanamori has written this wonderful book which sits on my bookshelf collecting dust. I wonder if/when I’ll dive into it.

Posted by: Todd Trimble on July 16, 2021 12:05 AM | Permalink | Reply to this

### Re: Large Sets 11

In some sense, universes without measurable cardinals are “perfect worlds” with perfect dualities

This made me think of how infinite-dimensional vector spaces ruin the perfect duality satisfied by finite-dimensional vector spaces, and how a bit of topology restores it.

This is the story of linearly compact vector spaces, which I’ve told before, but I just want a fragment of it here.

The point is this. For an arbitrary $k$-vector space $X$, the embedding

$X \to Vect(Vect(X, k), k)$

fails to be an isomorphism, and the blame can be laid on the fact that we ignored the natural topology on $Vect(X, k)$. By the “natural topology” I mean the weak${}^\ast$ topology, that is, the product topology $k^X$, treating $k$ as discrete. If we don’t throw this topology away then we do get a perfect duality: the embedding

$X \to TopVect(Vect(X, k), k)$

is an isomorphism for all vector spaces $X$, including the infinite-dimensional ones.

Analogously, is it true that if we give $Set(X, \mathbb{N})$ the weak${}^\ast$/product topology, then we get a perfect duality

$X \to Top^{End(\mathbb{N})}(Set(X, \mathbb{N}), \mathbb{N})$

for all sets $X$? My guess is yes.

In walks a measurable cardinal who ruins everything!

That put into my head a picture like this:

(from a blog post by Vincenzo Dimonte). Though something’s wrong with this picture: measurable cardinals are meant to be large, but this cardinal’s tiny.

Posted by: Tom Leinster on July 14, 2021 11:58 AM | Permalink | Reply to this

### Re: Large Sets 11

Either tiny, or a long, long way away!

Posted by: David Roberts on July 14, 2021 1:24 PM | Permalink | Reply to this

### Re: Large Sets 11

Is there anything to be said here about $Chu(Set,\mathbb{N})$?

Posted by: Mike Shulman on July 14, 2021 3:58 PM | Permalink | Reply to this

### Re: Large Sets 11

Speaking of

This made me think of how infinite-dimensional vector spaces ruin the perfect duality satisfied by finite-dimensional vector spaces, and how a bit of topology restores it,

it may be worth comparing the ‘dual’ control deriving from bornology. Recall Lawvere’s comment from our earlier conversation

Another kind of example concerns bornological spaces. The result always seems to be that the lack of Ulam cardinals is equivalent to the exception-free validity of basic space/quantity dualities.

For some reason, it’s still nagging at me that the discussion we had on condensedness/pyknoticity didn’t yield a nice category-theoretic account of why that is a natural construction.

Recall that Peter Scholze offered us a comparison of bornological and condensed sets here. He had said, as I quoted there,

The goal of this course is to launch a new attack, turning functional analysis into a branch of commutative algebra, and various types of analytic geometry (like manifolds) into algebraic geometry,

rather as Lawvere aimed to do.

Posted by: David Corfield on July 15, 2021 8:45 AM | Permalink | Reply to this

### Re: Large Sets 11

I think this is a great question, David. I’d like to try to outline what I think may be a category-theoretic way to view things, building on that previous discussion. Hopefully Peter Scholze or Dustin Clausen or another expert will see this and be able to say something more substantial.

As in that discussion, for me the principal conceptual aspect of condensed sets are that there is functor $f_{!}$, which we might also denote by $\pi_{\infty}$, from the big étale $(\infty,1)$-topos over a field $k$ to (an $(\infty,1)$-version of) condensed sets, left adjoint to the ‘constant condensed set’ functor $f^{*}$ (I’ll omit to distinguish between condensed sets and their $(\infty,1)$-analogue from now on).

How does this relate to what you are looking for? We first need to take a step back. For any $(\infty,1)$-topos (I might be omitting some hypotheses) and any object $X$ of it, we can construct the ‘shape’ of $X$. In the case of the big étale $(\infty,1)$-topos over a suitable field, this is basically a version of the étale homotopy type of $X$.

This is a very general construction. It is not all that clear what we can do with it in general though, e.g. it is not even all that clear whether shapes of objects in a particular $(\infty,1)$-topos assemble into some nice category. We know from étale homotopy theory that it can be possible to do interesting things with shapes, e.g. one can recover étale cohomology by mapping from the étale homotopy type of a scheme into a suitable notion of Eilenberg-MacLane space. But what kind of framework do we have to do something like this in general?

We could pose a very general and conceptual question: given an $(\infty,1)$-topos $T$, instead of working with shapes, we could ask for an $(\infty,1)$-topos $S$ together with a fully faithful functor $f^{\ast}: S \rightarrow T$ admitting both a right adjoint $f_{\ast}$ and a left adjoint $f_{!}$ (and possibly some more things). This left adjoint $f_{!}$ is then akin to a notion of shape, but the situation is much better. One might be able to make the relationship to shape explicit/part of the requirements (i.e. maybe the shape of $X$ should be equal to the shape of $f^{\ast}f_{!}(X)$ or something like that), but I’ll ignore that for now.

Again, in the case of étale homotopy theory, this is basically exactly the problem that Scholze and Bhatt solved: if one replaces the étale topos with the pro-étale topos, everything works out exactly as in the previous paragraph, taking $S$ to be condensed sets.

We can now come to trying to answer your question. What I think is true, although I’ve not seen this expressed anywhere, is that one can also solve the question I posed in the case of the big topos of topological spaces, and moreover, one can still take $S$ to be condensed sets! In other words, I think there is something like the pro-étale topology on the category of topological spaces which one can use to replace the usual big topos of topological spaces with some pro-étale-like version, and then one has the required functors $f_{!}$, $f^{\ast}$, and $f_{\ast}$.

Now, what I think should be true is that for the latter $f_{!}$, we should have that $f_{!}(\mathbb{R})$, in other words the ‘shape’ of $\mathbb{R}$ as a condensed set, is isomorphic to viewing $\mathbb{R}$ as a condensed set in the usual way that Scholze and Clausen do (emphasising again that I am blurring between 1-topoi and $\infty$-topoi here, in a way that I think is harmless).

For me, if all of this is more or less true, it gives a very conceptual explanation of why one would wish to move to condensed sets. We can have two very different topoi, but the shapes of the objects of these topoi are the same kind of gadgets. Moreover, being a topos, any answer $S$ to the question I posed allows us to immediately develop a theory of abelian groups, etc, internal to it. So we can treat the shapes of topological spaces in the same way as shapes of schemes, and develop the ‘algebraic internalisations’ of both (analogues of topological abelian groups or étale-sheaves of abelian groups on schemes) in the same framework, as just an internalisation of abelian groups in sets. These algebraic internalisations are ‘sensitive’ to the nature of the original topos, unlike the standard definitions of topological abelian group or étale sheaf of abelian groups, where the algebraic and ‘geometric’ notions of isomorphism are not tightly coupled, leading to exactly the kind of technical difficulties which motivated Scholze and Bhatt’s paper, and which motivated Scholze and Clausen’s development of functional analysis in a condensed setting.

I think that the bornological topos is basically an attempt by Lawvere to carry out exactly the picture described here, without quite having language available to fully realise it. In particular, I think that the relationship to shape that I have tried to describe here, and the passage from an object of a topos to its shape qua $f_{!}$, with the idea of developing algebraic geometry with respect to the shape and not the original object, is exactly the kind of thing that Lawvere wished to do.

Posted by: Richard Williamson on July 16, 2021 1:50 AM | Permalink | Reply to this

### Re: Large Sets 11

In the absence of the pieces functor to $Set$, Lawvere himself seems to place the emphasis on comparisons of the discrete and codiscrete functors (adjoints to the constant functor) in Lecture III of Cohesive toposes: combinatorial and infinitesimal cases. But earlier, in Lecture I, he raises the prospect of using a different base topos.

Tom may find interesting from the final page of the lectures:

The study of such examples is always related to double-dualization monads and to the failure of reasonable geometrical theorems in case so-called measurable cardinals are admitted into the category of small sets. This suggests to me that first of all “small” should not be identified with “member of some class”, but should explicitly exclude the measurable cardinals. The category of all small sets is an object we frequently use and if it itself is a measurable cardinal that should not dismay us any more than that the category of all finite sets is not finite.

Posted by: David Corfield on July 18, 2021 10:43 AM | Permalink | Reply to this

### Re: Large Sets 11

I’ve not personally ever yet been convinced of the importance of co-discreteness, etc; most of my intuition relates to the ‘pieces’ functor, in fact I was completely unaware of Lawvere’s work on cohesive toposes when I first discussed this kind of picture with Urs and others over a decade ago.

I’d like to emphasise that whilst I think using a different base topos is a good idea, it is non-trivial to actually realise that picture; for things to work out, one can expect to also have to change the topology in the big topos one is working with, in a non-mechanical way (i.e. one needs some creative input). That is what Bhatt and Scholze did with the pro-étale topology, and what I think can also be done in the case of the big topos of topological spaces.

I’m actually not convinced that there is any completely canonical way to come up with something like the pro-étale topology. It is a bit like with the many topologies used in Voevodsky’s motivic theory: cdh, etc. One can motivate them, but I don’t think there’s anything truly canonical about them (I’m aware of the work of Anel and others on finding reasonably generic ways of constructing Grothendieck topologies, but I don’t see that as giving a canonical characterisation of them). With the pro-étale topology/condensed sets, the choice of topology (jointly epimorphic covers) also seems difficult to characterise in some canonical way; as I wrote in the previous discussion, it is certainly a reasonably choice, but I would expect one to obtain perfectly workable theories with slightly different choices as well.

I’d suggest rather that the way to view things canonically is in answer to the kind of general problem/question that I posed in my previous comment. I doubt that there is any ‘universal’ solution to that problem/question, for the reasons above: I think some creativity is needed. But I could be wrong: maybe if one imposes enough requirements, like compatibility with the usual étale topology in that context, one might be able to obtain something universal.

Something that I have less intuition for as of now is the notion of solid module, etc. I follow Scholze and Clausen’s explanations as to how one might arrive at them, but I do not yet know how to understand the situation more canonically.

Posted by: Richard Williamson on July 22, 2021 1:28 PM | Permalink | Reply to this
Read the post Large Sets 12
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