# The lattice of sets of natural numbers is rich

$\newcommand\of{\subseteq}\newcommand\N{\mathbb{N}}\newcommand\R{\mathbb{R}}\def\<#1>{\left\langle#1\right\rangle}\newcommand\Z{\mathbb{Z}}\newcommand\Q{\mathbb{Q}}\newcommand\ltomega{{{<}\omega}}\newcommand\unaryminus{-}\newcommand\intersect{\cap}\newcommand\union{\cup}\renewcommand\emptyset{\varnothing}$Let us explore the vast and densely populated expanses of the lattice of all sets of natural numbers—the power set lattice $\<P(\N),\of>$.

We find in this lattice every possible set of natural numbers $A\of\N$, and we consider them with respect to the subset relation $A\of B$. This is a lattice order because the union of two sets $A\union B$ is their least upper bound with respect to inclusion and the intersection $A\intersect B$ is their greatest lower bound. Indeed, it is a distributive lattice in light of $A\intersect(B\union C)=(A\intersect B)\union(A\intersect C)$, but furthermore, as a power set algebra it is a Boolean algebra and every Boolean algebra is a distributive lattice.

The empty set $\emptyset$ is the global least element at the bottom of the lattice, of course, and the whole set $\N$ is the greatest element, at the top. The singletons $\set{n}$ are atoms, one step up from $\emptyset$, while their complements $\N\setminus\set{n}$ are coatoms, one step down from $\N$. One generally conceives of the finite sets as clustering near the Earthly bottom of the lattice, finitely close to $\emptyset$, whereas the cofinite sets soar above, finitely close to the heavenly top. In the vast central regions between, in contrast, we find the infinite-coinfinite sets, including many deeply interesting sets such as the prime numbers, the squares, the square-free numbers, the composites, and uncountably many more.

Question. Which familiar orders can we find as suborders in the power set lattice $P(\N)$ of all sets of natural numbers?

We can easily find a copy of the natural-number order $\<\N,\leq>$ in the power set lattice $P(\N)$, simply by considering the chain of initial segments:
$$\emptyset\of\set{0}\of\set{0,1}\of\set{0,1,2}\of\cdots$$
This sequence climbs like a ladder up the left side of the main figure. Curious readers will notice that although every one of these sets is finite and hence conceived as clinging to Earth, as we had said, nevertheless there is no upper bound for this sequence in the lattice other than the global maximum $\N$—any particular natural number is eventually subsumed. Thus one climbs to the heavens on this ladder, which stretches to the very top of the lattice, while remaining close to Earth all the while on every rung.

The complements of those sets similarly form a infinite descending sequence:
$$\cdots\ \of\ \N\setminus\set{0,1,2}\ \of\ \N\setminus\set{0,1}\ \of\ \N\setminus\set{0}\ \of\ \N$$
This chain of cofinite sets dangles like a knotted rope from the heavens down the right side of the main figure. Every knot on this rope is a cofinite set, which we conceived as very near the top, and yet the rope dangles temptingly down to Earth, for there is no lower bound except $\emptyset$.

Can we find a copy of the discrete integer order $\<\Z,\leq>$ in the power set lattice?

Yes, we can. An example is already visible in the red chain at the right side of main lattice diagram, a sky crane hanging in space. Standing at the set of odd numbers, we can ascend in discrete steps by adding additional even numbers one at a time; or we can descend similarly by removing successive odd numbers. The result is a discrete chain isomorphic to the integer order $\Z$, and it is unbounded except by the extreme points $\emptyset$ and $\N$.

Can we find a dense order, say, a copy of the rational line $\<\Q,\leq>$?

What? You want to find a dense order in the lattice of sets of natural numbers? That might seem impossible, if one thinks of the lattice as having a fundamentally discrete nature. The nearest neighbors of any set, after all, are obtained by adding or removing exactly one element—doing so makes a discrete step up or down with no strictly intermediate set. Does the essentially discrete nature of the lattice suggest that we cannot embed the dense order $\<\Q,\leq>$?

No! The surprising fact is that indeed we can embed the rational line into the power set lattice $P(\N)$. Indeed, much more is true—it turns out that we can embed a copy of any countable order whatsoever into the lattice $P(\N)$. Every countable order that occurs anywhere at all occurs in the lattice of sets of natural numbers.

Theorem. The power set lattice on the natural numbers $P(\N)$ is universal for all countable orders. That is, every order relation on a countable domain is isomorphic to a suborder of $\<P(\N),\of>$.

Proof. Let us prove first that every order relation $\<L,\preccurlyeq>$ is isomorphic to the subset relation $\of$ on a collection of sets. Namely, for each $q\in L$ let $q{\downarrow}$ be the down set of $q$, the set of predecessors of $q$:
$$q{\downarrow}=\set{p\in L\mid p\preccurlyeq q}\of L$$
This is a subset of $L$, and what I claim is that
$$p\preccurlyeq q\quad\text{ if and only if }\quad p{\downarrow}\of q{\downarrow}.$$
This is almost immediate. For the forward implication, if $p\preccurlyeq q$, then any order element below $p$ is also below $q$, and so $p{\downarrow}\of q{\downarrow}$. Conversely, if $p{\downarrow}\of q{\downarrow}$, then since $p\in p{\downarrow}$, we must have $p\in q{\downarrow}$ and hence $p\preccurlyeq q$ as desired. So the map $p\mapsto p{\downarrow}$ is an isomorphism of the original order $\<L,\preccurlyeq>$ with the family of all downsets $\set{p{\downarrow}\mid p\in L}$ using the subset relation.

To complete the proof, we simply observe that the power set lattice $P(L)$ of a countably infinite set $L$ is isomorphic to $P(\N)$, and for finite $L$ it is isomorphic to the power set lattice below a finite set of natural numbers. This is because any bijection of $L$ with a set of natural numbers induces a corresponding isomorphism of the respective subsets. So we may find a copy of $L$ inside $P(\N)$, as desired. $\Box$

One may make the proof more concrete by combining the two steps, like this: if $\<L,\preccurlyeq>$ is any countable order relation, then enumerate the domain set $L=\set{p_0,p_1,p_2,\ldots}$, and for each $p\in L$ let $A_p=\set{n\in\N\mid p_n\preccurlyeq p}\of\N$. This is like the down set of $p$, except we take the indices of the elements instead of the elements themselves. It follows that $p\preccurlyeq q$ if and only if $A_p\of A_q$, and so the mapping $p\mapsto A_p$ is an order isomorphism of $\<L,\preccurlyeq>$ with the collection of $A_p$ sets under $\of$, as desired.

Since the set of rational numbers $\Q$ is countable, it follows as an instance of the universality theorem above that we may find a copy of the rational line $\<\Q,\leq>$ inside the power set lattice $\<P(\N),\of>$. Can you find an elegant explicit presentation of rational line in the powerset lattice? In light of the characterization of $\Q$ as a countable endless dense linear order, it will suffice to find any chain in $P(\N)$ that forms a countable endless dense linear order.

But wait, there is more! We can actually find a copy of the real line $\<\R,\leq>$ in the power set lattice on the naturals! Wait…really? How is that possible? The reals are uncountable, and so this would be an uncountable chain the lattice of sets of natural numbers, but every subset $A\of\N$ is countable, and there are only countably many elements to add to $A$ as you go up. How could there be an uncountable chain in the lattice?

Let me explain.

Theorem. The power set lattice on the natural numbers $\<P(\N),\of>$ has an uncountable chain, and indeed, it has a copy of the real continuum $\<\R,\leq>$.

Proof. We have already observed that there must be a copy of the rational line $\<\Q,\leq>$ in the power set lattice $P(\N)$. In other words, there is a map $q\mapsto A_q\of\N$ such that
$$p<q\quad\text{ if and only if }\quad A_p\of A_q$$
for any rational numbers $p,q\in\Q$.

We may now find our desired copy of the real continuum simply by completing this embedding. Namely, for any real number $r\in\R$, let
$$A_r=\bigcup_{\substack{q\leq r\\ q\in\Q}} A_q.$$
Since the sets $A_q$ for rational numbers $q$ form a chain, it follows easily that $r\leq s$ implies $A_r\of A_s$. And if $r<s$, then $A_r$ will be strictly contained in some $A_q$ for a rational number $q$ strictly between $r<q<s$, and consequently strictly contained in $A_s$. From this, it follows that $r\leq s$ if and only if $A_r\of A_s$, and so we have found a copy of the real number order in the power set lattice, as desired. $\Box$

Let us now round out the surprising events of this section by proving another shocking fact—we can also find an uncountable antichain in the power set lattice $P(\N)$.

Theorem. The power set lattice on the natural numbers $\<P(\N),\of>$ has an uncountable antichain, a collection of continuum many pairwise incomparable sets.

Proof. Consider the tree of finite binary sequences $2^{\ltomega}$. This is the set of all finite binary sequences, ordered by the relation $s\leq t$ that holds when $t$ extends $s$ by possibly adding additional binary digits on the end. This is a countable order relation, and so by universality theorem above we may find a copy of it in the power set lattice $P(\N)$. But let us be a little more specific. We shall label each node $s$ of the tree $2^{\ltomega}$ with a distinct natural number. For any infinite binary path $x\in 2^\omega$ through the tree, let $A_x$ be the set of numbers that appear along the path. Every $A_x$ is infinite, but if $x\neq y$, then $x$ and $y$ will eventually branch away from each other, and consequently $A_x$ and $A_y$ will have only finitely many numbers in common. In particular, neither $A_x$ nor $A_y$ will be a subset of the other, and so we have found an antichain in the powerset lattice. The number of paths through the binary tree is the number of infinite binary sequences, which is the same as the cardinality of the continuum of real numbers, as desired. $\Box$

So the lattice $P(\N)$ offers a rich complexity. But is there also homogeneity to be found here? Place yourself at one of the sets in the central region of the lattice, an infinite coinfinite set. There are various new elements for you to add to make a larger set—you can add just this element, or just that one, or any of infinitely many new atoms for you to add to make a larger set. And you could add any combination of those atoms. Above any coinfinite point in the lattice, therefore, one finds a perfect copy of the whole lattice again. Looking upward from any coinfinite set, it always looks the same.

Theorem. The lattice of all sets of natural numbers is homogenous in several respects.

1. The lattice of sets above any given coinfinite set $A\of\N$ is isomorphic to the whole power set lattice $P(\N)$.
2. The lattice of sets below any given infinite set $B\of\N$ is isomorphic to the whole power set lattice $P(\N)$.
3. For any two infinite coinfinite sets $A,B\of \N$, there is an automorphism of the lattice $P(\N)$ carrying $A$ to $B$.

Proof. Let us prove statement (2) first. Consider any infinite set $B\of\N$ and the sublattice in $P(\N)$ below it, which is precisely the power set lattice $P(B)$ of all subsets of $B$. Since $B$ is an infinite set of natural numbers, there is a bijection $\pi:B\to\N$. We may therefore transform subsets of $B$ to subsets of $\N$ by applying the bijection. Namely, for any $X\of B$ define $\bar\pi(X)=\pi[B]=\set{\pi(b)\mid b\in X}$. Since $\pi$ is injective, we have $X\of Y$ if and only if $\bar\pi(X)\of\bar\pi(Y)$, and the map $\bar\pi$ maps onto $P(\N)$ precisely because $\pi$ is surjective, so every $Z\of\N$ is $Z=\bar\pi(X)$, where $X=\pi^{\unaryminus 1}Z=\set{b\in B\mid \pi(b)\in Z}$. So $\bar\pi$ is an isomorphism of the lattice $P(B)$ with $P(\N)$, as desired.

For statement (1), consider any cofinite set $A\of\N$ and let $A{\uparrow}$ be the part of the power set lattice above $A$, meaning $A\uparrow=\set{X\of\N\mid A\of X}$. Since the complement $B=\N\setminus A$ is infinite, there is a bijection of it with the whole set of natural numbers $\pi:B\to\N$. I claim that the lattice $A{\uparrow}$ is isomorphic with the lattice $P(B)$, which by statement (1) we know is isomorphic to $P(\N)$. To see this, we simply cast $A$ out of any set $X$ containing $A$. What remains is a subset of $B$, and every subset of $B$ will be realized. More specifically, if $A\of X$ we define $\tau(X)=X\setminus A=X\intersect B$, and observe that $X\of Y$ if and only if $\tau(X)\of\tau(Y)$, for the sets already containing $A$. And if $Z\of B$, then $Z=\tau(A\union Z)$. So $\tau$ is an isomorphism of the lattice $A{\uparrow}$ with $P(B)$, which by statement (1) is isomorphic to the whole lattice $P(\N)$, as desired.

Finally, let us fix two infinite coinfinite sets $A,B\of \N$. In particular, $A$ and $B$ are in bijection, as are the complements $\N\setminus A$ and $\N\setminus B$. By combining these two bijections, we get a bijection $\pi:\N\to\N$ of $\N$ with itself that carries $A$ exactly to $B$, that is, for which $\pi[A]=B$. (The reader will have an exercise in my book to give a careful account of this construction.) Because $\pi$ is a bijection of $\N$ with itself, it induces a corresponding automorphism of the lattice, defined by $\bar\pi(X)=\pi[X]$ for $X\of\N$. This is an isomorphism of $P(\N)$ with itself, and $\bar\pi(A)=\pi[A]=B$, as desired. $\Box$

Statement (3) of the theorem shows that any two infinite coinfinite sets, being automorphic images of one another, play the same structural role in the lattice $P(\N)$—they will have all the same lattice-theoretic properties. In light of this theorem, therefore, one should not look for structure theorems concerning particular sets in the lattice. Rather, the structure theory of $P(\N)$ is about the structure as a whole, as with the theorems I have proved above.

This is a selection from my book in progress Topics in Logic. It appears in chapter 3, which is on relational logic, and which includes a section on orders with a subsection specifically on the lattice of all sets of natural numbers, because I find it fascinating.

# The modal logic of arithmetic potentialism and the universal algorithm

[bibtex key=”Hamkins:The-modal-logic-of-arithmetic-potentialism”]

Abstract. Natural potentialist systems arise from the models of arithmetic when they are considered under their various natural extension concepts, such as end-extensions, arbitrary extension, $\Sigma_n$-elementary extensions, conservative extensions and more. For these potentialist systems, I prove, a propositional modal assertion is valid in a model of arithmetic, with respect to assertions in the language of arithmetic with parameters, exactly when it is an assertion of S4. Meanwhile, with respect to sentences, the validities of a model are always between S4 and S5, and these bounds are sharp in that both endpoints are realized. The models validating exactly S5 are the models of the arithmetic maximality principle, which asserts that every possibly necessary statement is already true, and these models are equivalently characterized as those satisfying a maximal $\Sigma_1$ theory. The main proof makes fundamental use of the universal algorithm, of which this article provides a self-contained account.

In this article, I consider the models of arithmetic under various natural extension concepts, including end-extensions, arbitrary extensions, $\Sigma_n$-elementary extensions, conservative extensions and more. Each extension concept gives rise to an arithmetic potentialist system, a Kripke model of possible arithmetic worlds, and the main goal is to discover the modal validities of these systems.

For most of the extension concepts, a modal assertion is valid with respect to assertions in the language of arithmetic, allowing parameters, exactly when it is an assertion of the modal theory S4. For sentences, however, the modal validities form a theory between S4 and S5, with both endpoints being realized. A model of arithmetic validates S5 with respect to sentences just in case it is a model of the arithmetic maximality principle, and these models are equivalently characterized as those realizing a maximal $\Sigma_1$ theory.

The main argument relies fundamentally on the universal algorithm, the theorem due to Woodin that there is a Turing machine program that can enumerate any finite sequence in the right model of arithmetic, and furthermore this model can be end-extended so as to realize any further extension of that sequence available in the model. In the paper, I give a self-contained account of this theorem using my simplified proof.

The paper concludes with philosophical remarks on the nature of potentialism, including a discussion of how the linear inevitability form of potentialism is actually much closer to actualism than the more radical forms of potentialism, which exhibit branching possibility. I also propose to view the philosphy of ultrafinitism in modal terms as a form of potentialism, pushing the issue of branching possibility in ultrafinitism to the surface.

# The universal algorithm: a new simple proof of Woodin’s theorem

This is the third in a series of posts I’ve made recently concerning what I call the universal algorithm, which is a program that can in principle compute any function, if only you should run it in the right universe. Earlier, I had presented a few elementary proofs of this surprising theorem: see Every function can be computable! and A program that accepts exactly any desired finite set, in the right universe.

$\newcommand\PA{\text{PA}}$Those arguments established the universal algorithm, but they fell short of proving Woodin’s interesting strengthening of the theorem, which explains how the universal algorithm can be extended from any arithmetic universe to a larger one, in such a way so as to extend the given enumerated sequence in any desired manner. Woodin emphasized how his theorem raises various philosophical issues about the absoluteness or rather the non-absoluteness of finiteness, which I find extremely interesting.

Woodin’s proof, however, is a little more involved than the simple arguments I provided for the universal algorithm alone. Please see the paper Blanck, R., and Enayat, A. Marginalia on a theorem of Woodin, The Journal of Symbolic Logic, 82(1), 359-374, 2017. doi:10.1017/jsl.2016.8 for a further discussion of Woodin’s argument and related results.

What I’ve recently discovered, however, is that in fact one can prove Woodin’s stronger version of the theorem using only the method of the elementary argument. This variation also allows one to drop the countability requirement on the models, as was done by Blanck and Enayat. My thinking on this argument was greatly influenced by a comment of Vadim Kosoy on my original post.

It will be convenient to adopt an enumeration model of Turing computability, by which we view a Turing machine program as providing a means to computably enumerate a list of numbers. We start the program running, and it generates a list of numbers, possibly finite, possibly infinite, possibly empty, possibly with repetition. This way of using Turing machines is fully Turing equivalent to the usual way, if one simply imagines enumerating input/output pairs so as to code any given computable partial function.

Theorem.(Woodin) There is a Turing machine program $e$ with the following properties.

1. $\PA$ proves that $e$ enumerates a finite sequence of numbers.
2. For any finite sequence $s$, there is a model $M$ of $\PA$ in which program $e$ enumerates exactly $s$.
3. For any model $M$ in which $e$ enumerates a (possibly nonstandard) sequence $s$ and any $t\in M$ extending $s$, there is an end-extension $N$ of $M$ in which $e$ enumerates exactly $t$.

It is statement (3) that makes this theorem stronger than merely the universal algorithm that I mentioned in my earlier posts and which I find particularly to invite philosophical speculation on the provisional nature of finiteness. After all, if in one universe the program $e$ enumerates a finite sequence $s$, then for any $t$ extending $s$ — we might imagine having painted some new pattern $t$ on top of $s$ — there is a taller universe in which $e$ enumerates exactly $t$. So we need only wait long enough (into the next universe), and then our program $e$ will enumerate exactly the sequence $t$ we had desired.

Proof. This is the new elementary proof.  Let’s begin by recalling the earlier proof of the universal algorithm, for statements (1) and (2) only. Namely, let $e$ be the program that undertakes a systematic exhaustive search through all proofs from $\PA$ for a proof of a statement of the form, “program $e$ does not enumerate exactly the sequence $s$,” where $s$ is an explicitly listed finite sequence of numbers. Upon finding such a proof (the first such proof found), it proceeds to enumerate exactly the numbers appearing in $s$.  Thus, at bottom, the program $e$ is a petulant child: it searches for a proof that it shouldn’t behave in a certain way, and then proceeds at once to behave in exactly the forbidden manner.

(The reader may notice an apparent circularity in the definition of program $e$, since we referred to $e$ when defining $e$. But this is no problem at all, and it is a standard technique in computability theory to use the Kleene recursion theorem to show that this kind of definition is completely fine. Namely, we really define a program $f(e)$ that performs that task, asking about $e$, and then by the recursion theorem, there is a program $e$ such that $e$ and $f(e)$ compute the same function, provably so. And so for this fixed-point program $e$, it is searching for proofs about itself.)

It is clear that the program $e$ will enumerate a finite list of numbers only, since either it never finds the sought-after proof, in which case it enumerates nothing, and the empty sequence is finite, or else it does find a proof, in which case it enumerates exactly the finitely many numbers explicitly appearing in the statement that was proved. So $\PA$ proves that in any case $e$ enumerates a finite list. Further, if $\PA$ is consistent, then you will not be able to refute any particular finite sequence being enumerated by $e$, because if you could, then (for the smallest such instance) the program $e$ would in fact enumerate exactly those numbers, and this would be provable, contradicting $\text{Con}(\PA)$. Precisely because you cannot refute that statement, it follows that the theory $\PA$ plus the assertion that $e$ enumerates exactly $s$ is consistent, for any particular $s$. So there is a model $M$ of $\PA$ in which program $e$ enumerates exactly $s$. This establishes statements (1) and (2) for this program.

Let me now modify the program in order to achieve the key third property. Note that the program described above definitely does not have property (3), since once a nonempty sequence $s$ is enumerated, then the program is essentially finished, and so running it in a taller universe $N$ will not affect the sequence it enumerates. To achieve (3), therefore, we modify the program by allowing it to add more to the sequence.

Specfically, for the new modified version of the program $e$, we start as before by searching for a proof in $\PA$ that the list enumerated by $e$ is not exactly some explicitly listed finite sequence $s$. When this proof is found, then $e$ immediately enumerates the numbers appearing in $s$. Next, it inspects the proof that it had found. Since the proof used only finitely many $\PA$ axioms, it is therefore a proof from a certain fragment $\PA_n$, the $\Sigma_n$ fragment of $\PA$. Now, the algorithm $e$ continues by searching for a proof in a strictly smaller fragment that program $e$ does not enumerate exactly some explicitly listed sequence $t$ properly extending the sequence of numbers already enumerated. When such a proof is found, it then immediately enumerates (the rest of) those numbers. And now simply iterate this, looking for new proofs in still-smaller fragments of $\PA$ that a still-longer extension is not the sequence enumerated by $e$.

Succinctly: the program $e$ searches for a proof, in a strictly smaller fragment of $\PA$ each time, that $e$ does not enumerate exactly a certain explicitly listed sequence $s$ extending whatever has been already enumerated so far, and when found, it enumerates those new elements, and repeats.

We can still prove in $\PA$ that $e$ enumerates a finite sequence, since the fragment of $\PA$ that is used each time is going down, and $\PA$ proves that this can happen only finitely often. So statement (1) holds.

Again, you cannot refute that any particular finite sequence $s$ is the sequence enumerated by $e$, since if you could do this, then in the standard model, the program would eventually find such a proof, and then perhaps another and another, until ultimately, it would find some last proof that $e$ does not enumerate exactly some finite sequence $t$, at which time the program will have enumerated exactly $t$ and never afterward add to it. So that proof would have proved a false statement. This is a contradiction, since that proof is standard.

So again, precisely because you cannot refute these statements, it follows that it is consistent with $\PA$ that program $e$ enumerates exactly $s$, for any particular finite $s$. So statement (2) holds.

Finally, for statement (3), suppose that $M$ is a model of $\PA$ in which $e$ enumerates exactly some finite sequence $s$. If $s$ is the empty sequence, then $M$ thinks that there is no proof in $\PA$ that $e$ does not enumerate exactly $t$, for any particular $t$. And so it thinks the theory $\PA+$ “$e$ enumerates exactly $t$” is consistent. So in $M$ we may build the Henkin model $N$ of this theory, which is an end-extension of $M$ in which $e$ enumerates exactly $t$, as desired.

If alternatively $s$ was nonempty, then it was enumerated by $e$ in $M$ because in $M$ there was ultimately a proof in some fragment $\PA_n$ that it should not do so, but it never found a corresponding proof about an extension of $s$ in any strictly smaller fragment of $\PA$.  So $M$ has a proof from $\PA_n$ that $e$ does not enumerate exactly $s$, even though it did.

Notice that $n$ must be nonstandard, because $M$ has a definable $\Sigma_k$-truth predicate for every standard $k$, and using this predicate, $M$ can see that every $\PA_k$-provable statement must be true.

Precisely because the model $M$ lacked the proofs from the strictly smaller fragment $\PA_{n-1}$, it follows that for any particular finite $t$ extending $s$ in $M$, the model thinks that the theory $T=\PA_{n-1}+$ “$e$ enumerates exactly $t$” is consistent. Since $n$ is nonstandard, this theory includes all the actual $\PA$ axioms. In $M$ we can build the Henkin model $N$ of this theory, which will be an end-extension of $M$ in which $\PA$ holds and program $e$ enumerates exactly $t$, as desired for statement (3). QED

Corollary. Let $e$ be the universal algorithm program $e$ of the theorem. Then

1. For any infinite sequence $S:\mathbb{N}\to\mathbb{N}$, there is a model $M$ of $\PA$ in which program $e$ enumerates a (nonstandard finite) sequence starting with $S$.
2. If $M$ is any model of $\PA$ in which program $e$ enumerates some (possibly nonstandard) finite sequence $s$, and $S$ is any $M$-definable infinite sequence extending $s$, then there is an end-extension of $M$ in which $e$ enumerates a sequence starting with $S$.

Proof. (1) Fix $S:\mathbb{N}\to\mathbb{N}$. By a simple compactness argument, there is a model $M$ of true arithmetic in which the sequence $S$ is the standard part of some coded nonstandard finite sequence $t$. By the main theorem, there is some end-extension $M^+$ of $M$ in which $e$ enumerates $t$, which extends $S$, as desired.

(2) If $e$ enumerates $s$ in $M$, a model of $\PA$, and $S$ is an $M$-infinite sequence definable in $M$, then by a compactness argument inside $M$, we can build a model $M’$ inside $M$ in which $S$ is coded by an element, and then apply the main theorem to find a further end-extension $M^+$ in which $e$ enumerates that element, and hence which enumerates an extension of $S$. QED

# Jacob Davis, PhD 2016, Carnegie Mellon University

Jacob Davis successfully defended his dissertation, “Universal Graphs at $\aleph_{\omega_1+1}$ and Set-theoretic Geology,” at Carnegie Mellon University on April 29, 2016, under the supervision of James Cummings. I was on the dissertation committee (participating via Google Hangouts), along with Ernest Schimmerling and Clinton Conley.

The thesis consisted of two main parts. In the first half, starting from a model of ZFC with a supercompact cardinal, Jacob constructed a model in which $2^{\aleph_{\omega_1}} = 2^{\aleph_{\omega_1+1}} = \aleph_{\omega_1+3}$ and in which there is a jointly universal family of size $\aleph_{\omega_1+2}$ of graphs on $\aleph_{\omega_1+1}$.  The same technique works with any uncountable cardinal in place of $\omega_1$.  In the second half, Jacob proved a variety of results in the area of set-theoretic geology, including several instances of the downward directed grounds hypothesis, including an analysis of the chain condition of the resulting ground models.

# Universal structures, GC MathFest, February 2014

This will be a talk for the CUNY Graduate Center MathFest, held on the afternoon of Februrary 4, 2014, intended for graduate-school-bound undergraduate students, including prospective students for the CUNY Graduate Center, giving them a chance to meet graduate students and faculty at the CUNY Graduate Center and see the kind of mathematics that is done here.

In this 30 minute talk, I’ll introduce the concept of a universal structure, with various examples, including the countable random graph, the surreal number line and the hypnagogic digraph.

MathFest Program/schedule

# Universal structures: the countable random graph, the surreal numbers and the hypnagogic digraph, Swarthmore College, October 2013

I’ll be speaking for the Swarthmore College Mathematics and Statistics Colloquium on October 8th, 2013.

Abstract.  I’ll be giving an introduction to universal structures in mathematics, where a structure $\mathcal{M}$ is universal for a class of structures, if every structure in that class arises as (isomorphic to) a substructure of $\mathcal{M}$.  For example, Cantor proved that the rational line $\mathbb{Q}$ is universal for all countable linear orders.  Is a corresponding fact true of the real line for linear orders of that size? Are there countably universal partial orders? Is there a countably universal graph? directed graph? acyclic digraph?  Is there a countably universal group? We’ll answer all these questions and more, with an account of the countable random graph, generalizations to the random graded digraphs, Fraïssé limits, the role of saturation, the surreal numbers and the hypnagogic digraph.  The talk will conclude with some very recent work on universality amongst the models of set theory.

Poster

# Universality, saturation and the surreal number line, Shanghai, June 2013

This will be a short lecture series given at the conclusion of the graduate logic class in the Mathematical Logic group at Fudan University in Shanghai, June 13, 18 (or 20), 2013.

I will present an elementary introduction to the theory of universal orders and relations and saturated structures.  We’ll start with the classical fact, proved by Cantor, that the rational line is the universal countable linear order.  But what about universal partial orders, universal graphs and other mathematical structures?  Is there a computable universal partial order?  What is the countable random graph? Which orders embed into  the power set of the natural numbers under the subset relation $\langle P(\mathbb{N}),\subset\rangle$? Proceeding to larger and larger universal orders, we’ll eventually arrive at the surreal numbers and the hypnagogic digraph.