# A model of set theory with a definable copy of the complex field in which the two roots of -1 are set-theoretically indiscernible


$$(a+bi)+(c+di)=(a+c)+(b+d)i$$ $$(a+bi)\cdot(c+di)=(ac-bd)+(ad+bc)i.$$

The complex field thus provides a system of numbers giving sense to expressions like $\sqrt{–1}$, while obeying the familiar algebraic rules of a field. Hamilton had presented this conception of complex numbers as pairs of real numbers to the Royal Irish Academy in 1833.

One may easily observe in the complex numbers, however, that $–i$ is also a square root of $–1$, because

$$(–i)\cdot(–i)=(–1)^2\cdot i^2=i^2=-1.$$

Thus, both $i$ and $–i$ have the property of being square roots of $–1$, and indeed, these are the only square roots of $–1$ in the complex field.

A small conundrum may arise when one realizes that $–i$ therefore also fulfills what might have been taken as the “defining” property of the ideal element $i$, namely, that it squares to $–1$. So this property doesn’t actually define $i$, in light of the fact that there is another distinct object $–i$ that also has this property. Can we tell $i$ and $–i$ apart?

Not in the complex field, no, we cannot. The basic fact is that $i$ and $–i$ are indiscernible as complex numbers with respect to the algebraic structure of $\C$—any property that $i$ has in the structure $\langle\C,+,\cdot,0,1\rangle$ will also hold of $–i$. One way to see this is to observe that complex conjugation, the map $$a+bi\quad\mapsto\quad a-bi$$ is an automorphism of the complex number field, an isomorphism of the structure with itself. And since this automorphism swaps $i$ with $–i$, it follows that any statement true of $i$ in the complex numbers, expressible in the language of fields, will also hold of $–i$.

In fact, the complex number field $\C$ has an extremely rich automorphism group, and every irrational complex number is indiscernible from various doppelgängers. There is an automorphism of $\C$ that swaps $\sqrt{2}$ and $–\sqrt{2}$, for example, and another that permutes the cube roots of $5$, mapping the real root $\sqrt[3]{5}$ with the two nonreal roots. So these numbers can have no property not shared by their various automorphic images. The general fact is that every complex number, except the rational numbers, is moved by some automorphism of $\C$. One can begin to see this by noticing that there are two ways to embed the algebraic field extensions $\newcommand\Q{\mathbb{Q}}\Q(\sqrt{2})$ into $\C$, and both embeddings extend fully to automorphisms of $\C$.

Because there is an automorphism of $\C$ swapping $\sqrt{2}$ and $–\sqrt{2}$, it means that these two numbers are also indiscernible as complex numbers, just like $i$ and $–i$ were—any property that $\sqrt{2}$ holds in the complex numbers is also held by $–\sqrt{2}$. But wait a minute, how can that be? After all, $\sqrt{2}$ is positive and $–\sqrt{2}$ is negative, and isn’t this a property that separates them? Well, yes, in the real numbers $\R$ this is a separating property, and since the order is definable from the algebraic structure of the real field (positive numbers are exactly the nonzero squares), it is a real algebraic property that distinguishes $\sqrt{2}$ from $–\sqrt{2}$, as only the former has itself a square root in $\R$. But this definition does not work in $\C$, since both have square roots there, and more generally, the surprise is that the real numbers $\R$ are not definable as a subfield in the complex field $\C$—there is no property expressible in the language of fields that picks out exactly the real numbers. There are $2^{2^{\aleph_0}}$ many distinct ways to embed $\R$ as a subfield of $\C$, and none of them is definable in $\C$.

The conclusion is that if we regard the complex numbers with the field structure only, $\langle\C,+,\cdot,0,1\rangle$, then we cannot refer unambiguously to $i$ or $–i$, to $\sqrt{2}$ or $–\sqrt{2}$, or indeed to any irrational complex number. Every irrational number is moved by some automorphism of the complex field. The irrational algebraic numbers can be permuted in their finite sets of indiscernible roots of their irreducible polynomial, and any two transcendental complex numbers (transcendental over $\Q$) are automorphic. For example, there is an automorphism of $\C$ moving $e+2i$ to $1+\sqrt{\pi}i$.

Finding a path out of that chaos, mathematicians like to conceive of $\C$ as a field extension of $\R$, in effect fixing the copy of $\R$ in $\C$. It is as though we are working in the structure $\langle\C,+,\cdot,0,1,\R\rangle$, where we have augmented the complex field structure with a predicate picking out the real numbers. So this isn’t just a field, but a field with an identified subfield. In this structure, $\sqrt{2}$ and $\sqrt[3]{5}$ and so on are definable, since one has identified the real numbers and within that subfield the order on the reals is definable, and so we can define every real algebraic number using this order. With the predicate for $\R$ picking out the reals, the structure has only the one nontrivial automorphism, complex conjugation, and to my way of thinking, this is the reason that the indiscernibility issue is usually considered more prominently with $i$ and $–i$.

The indiscernibility of $i$ and $–i$ in the complex field has been written on at length in the philosophical literature, since it seems to refute a certain philosophical account of structuralism that might otherwise have seemed appealing. Namely, the relevant view is a version of abstract structuralism, the view that what mathematical objects are is the structural role that they play in a mathematical system. On this view the natural number $2$ simply is the role that $2$ plays in Dedekind arithmetic, the role of being the successor of the successor of zero (Dedekind arithmetic is the categorical second-order axiomatization of $\langle\newcommand\N{\mathbb{N}}\N,0,S\rangle$). The view is that what mathematical structure is is the structural roles that objects play in any instance of the structure. The structural role is exactly what is preserved by isomorphism, and so it would seem to be an invariant for the isomorphism orbits of an indidvidual with respect to a structure.

The problem with this version of abstract structuralism is that it seems to be refuted by the example of $i$ and $–i$ in the complex field. Precisely because these numbers are automorphic, they would seem each to play exactly the same role in the complex field—the two numbers are isomorphic copies of one another via complex conjugation. Thus, they are distinct numbers, but play the same structural role, and so we cannot seem to identify the abstract number with the structural roles. This problem occurs, of course, in any mathematical structure that is not rigid.

The numbers $i$ and $–i$ are indiscernible in the field structure of $\C$, but of course we can distinguish them in contexts with additional structure. For example, if we use the Hamilton presentation of the complex numbers as pairs of real numbers, representing $a+bi$ with the pair $(a,b)$, then the number $i$ has coordinates $(0,1)$ and $–i$ has coordinates $(0,-1)$. The complex field equipped with this coordinate structure, perhaps given by the real and imaginary parts operators—let us call it the complex plane, as opposed to the complex field—is a rigid structure in which $i$ and $–i$ are discernible and indeed definable.

Finally, this brings me to the main point of this blog post. What I would like to do is to prove that it is relatively consistent with ZFC that we can definably construct a copy of the complex numbers $\C$ in such a way that not only are $i$ and $–i$ indiscernible in the field structure, but actually the particular set-theoretic objects $i$ and $–i$ are indiscernible in the set-theoretic background in which the construction is undertaken.

Goal. A definable copy of the complex field in which the two square roots of $–1$ are indiscernible not only in the field structure, but also in the set-theoretic background in which the construction of the field takes place.

These two aims are in tension, for we want the particular copy $\C$ to be definable (as a particular set-theoretic object, not just defined up to isomorphism), but the individual square roots of $–1$ to be set-theoretically indiscernible.

The goal is not always possible. For example, some models of ZFC are pointwise definable, meaning that every individual set is definable in them by some distinguishing set-theoretic property. More generally, if the V=HOD axiom holds, then there is a definable global well order of the set-theoretic universe, and with any such order we could define a linear order on $\{i,–i\}$ in any definable copy of $\C$, which would allow us to define each of the roots. For these reasons, in some models of ZFC, it is not possible to achieve the goal, and the most we can hope for a consistency result.

But indeed, the consistency goal is achievable.

Theorem. If ZFC is consistent, then there is a model of ZFC that has a definable complete ordered field $\R$ with a definable algebraic closure $\C$, such that the two square roots of $–1$ in $\C$ are set-theoretically indiscernible, even with ordinal parameters.

Proof. The proof makes use of what are known as Grozek-Laver pairs, definable pair sets having no ordinal-definable element. See M. Groszek & R. Laver, Finite Groups of OD-conjugates, Periodica Mathematica Hungarica, v. 18, pp. 87–97 (1987), for a very general version of this. This theorem also appears at theorem 4.6 in my paper Ehrenfeucht’s lemma in set theory, joint with Gunter Fuchs, Victoria Gitman, and myself. The arguments provide a model of set theory with a definable pair set $A=\{i,j\}$, such that neither element $i$ nor $j$ is definable from ordinal parameters. The pair set is definable, but neither element is definable.

To undertake the construction, we start with one of the standard definable constructions of the real field $\R$. For example, we could use Dedekind cuts in $\Q$, where $\Q$ is constructed explicitly as the quotient field of the integer ring $\mathbb{Z}$ in some canonical definable manner, and where the integers are definably constructed from a definable copy of the natural numbers $\mathbb{N}$, such as the finite von Neumann ordinals. So we have a definable complete ordered field, the real field $\R$.

Given this and the set $A$, we follow a suggestion of Timothy Gowers in the discussion of this problem on Twitter. Namely, we use the elements of $A$ as variables to form the polynomial ring $\R[A]$, meaning $\R[i,j]$, where $i$ and $j$ are the two elements of $A$. It is not necessary to distinguish the elements of $A$ to form this ring of polynomials, since we take all finite polynomial expressions using real coefficients and elements of $A$ raised to a power. (In particular, although I have referred to the elements as $i$ and $j$, there is to be no suggestion that I am somehow saying $i$ is the “real” $i$; I am not, for I could have called them $j$,$i$ or $j$,$k$ or $a$,$a’$, and so on.) Then we quotient by the ideal $(i^2+1,i+j)$, which is defined symmetrically in the elements of $A$, since it is the same ideal as $(j^2+1,j+i)$. Let $\C$ be the quotient $\C=\R[i,j]/(i^2+1,i+j)$, which will make both $i$ and $j$ the two square roots of $–1$, and so by the fundamental theorem of algebra this is a copy of the complex numbers.

Since $\R$ and $A$ were definable, and we didn’t need ever to choose a particular element of $A$ in the construction to define the polynomial ring or the ideal, this copy of $\C$ is definable without parameters. But since $i$ and $j$ are set-theoretically indiscernible in the model of set theory in which we are undertaking the construction, it follows that their equivalence classes in the quotient are also indiscernible. And so we have a definable copy of the complex field $\C$, extending a definable copy of $\R$, in which the two square roots of $–1$ are indiscernible not just in the field structure, but fully in the set-theoretic background in which the fields were constructed. $\Box$

In particular, in this model of set theory, there will be absolutely no way to distinguish the two roots by any further definable structure, whether using second-order or higher-order definitions of the field $\C$ or using any definable set-theoretic property whatsoever.

The analysis suggests a natural further inquiry. Namely,

Question. Is there a model of set theory with a definable copy of the complex field $\C$, such that the hierarchy of relative definability and indiscernibility in $\C$ matches the set-theoretic relative definability and indiscernibility of the objects?

That is, we would want to mimic the phenomenon of $i$ and $–i$ in the above construction with all complex numbers, so that $\sqrt{2}$ and $–\sqrt{2}$ were also indiscernible, not just in this copy of $\C$ but also in the set-theoretic background, and $\sqrt[4]{2}$ was set-theoretically indiscernible from the other new fourth-root of $2$, but can set-theoretically define both $\sqrt{2}$ and $–\sqrt{2}$. In other words, I want the set-theoretic definability hierarchy to match the complex-number-theoretic definability hierarchy. I may post this question on MathOverflow, when I formulate a version of it with which I am satisfied. I believe it will be answered by iterated Sacks forcing in a manner similar to that used in many papers by Marcia Groszek, and in particular, in my paper with her, The Implicitly constructible universe.

# Pointwise definable and Leibnizian models of arithmetic and set theory, realized in end extensions of a given model, Notre Dame Logic Seminar, October 2022

This will be a talk for the Notre Dame logic seminar, 11 October 2022, 2pm in Hales-Healey Hall.

Abstract.  I shall present very new results on pointwise definable and Leibnizian end-extensions of models of arithmetic and set theory. Using the universal algorithm, I shall present a new flexible method showing that every countable model of PA admits a pointwise definable $\Sigma_n$-elementary end-extension. Also, any model of PA of size at most continuum admits an extension that is Leibnizian, meaning that any two distinct points are separated by some expressible property. Similar results hold in set theory, where one can also achieve V=L in the extension, or indeed any suitable theory holding in an inner model of the original model.

# Every countable model of arithmetic or set theory has a pointwise definable end extension

• J. D. Hamkins, “Every countable model of arithmetic or set theory has a pointwise definable end extension,” mathematics arXiv, 2022.
[Bibtex]
@ARTICLE{Hamkins:Every-countable-model-of-arithmetic-or-set-theory-has-a-pointwise-definable-end-extension,
author = {Joel David Hamkins},
title = {Every countable model of arithmetic or set theory has a pointwise definable end extension},
journal = {mathematics arXiv},
year = {2022},
volume = {},
number = {},
pages = {},
month = {},
note = {manuscript under review},
abstract = {},
keywords = {under-review},
source = {},
doi = {10.48550/ARXIV.2209.12578},
eprint = {2209.12578},
archivePrefix={arXiv},
primaryClass={math.LO},
url = {http://jdh.hamkins.org/pointwise-definable-end-extensions},
}

arXiv:2209.12578

Abstract. According to the math tea argument, there must be real numbers that we cannot describe or define, because there are uncountably many real numbers, but only countably many definitions. And yet, the existence of pointwise definable models of set theory, in which every individual is definable without parameters, challenges this conclusion. In this article, I introduce a flexible new method for constructing pointwise definable models of arithmetic and set theory, showing furthermore that every countable model of Zermelo-Fraenkel ZF set theory and of Peano arithmetic PA has a pointwise-definable end extension. In the arithmetic case, I use the universal algorithm and its $\Sigma_n$ generalizations to build a progressively elementary tower making any desired individual $a_n$ definable at each stage $n$, while preserving these definitions through to the limit model, which can thus be arranged to be pointwise definable. A similar method works in set theory, and one can moreover achieve $V=L$ in the extension or indeed any other suitable theory holding in an inner model of the original model, thereby fulfilling the resurrection phenomenon. For example, every countable model of ZF with an inner model with a measurable cardinal has an end extension to a pointwise-definable model of $\text{ZFC}+V=L[\mu]$.

# Pointwise definable end-extensions of the universe, Sophia 2022, Salzburg

This will be an online talk for the Salzburg Conference for Young Analytical Philosophy, the SOPhiA 2022 Salzburgiense Concilium Omnibus Philosophis Analyticis, with a special workshop session Reflecting on ten years of the set-theoretic multiverse. The workshop will meet Thursday 8 September 2022 4:00pm – 7:30pm.

The name of the workshop (“Reflecting on ten years…”), I was amazed to learn, refers to the period since my 2012 paper, The set-theoretic multiverse, in the Review of Symbolic Logic, in which I had first introduced my arguments and views concerning set-theoretic pluralism. I am deeply honored by this workshop highlighting my work in this way and focussing on the developments growing out of it.

In this talk, I shall engage in that discussion by presenting some very new work connecting several topics that have been prominent in discussions of the set-theoretic multiverse, namely, set-theoretic potentialism and pointwise definability.

Abstract. Using the universal algorithm and its generalizations, I shall present new work on the possibility of end-extending any given countable model of arithmetic or set theory to a pointwise definable model, one in which every object is definable without parameters. Every countable model of Peano arithmetic, for example, admits an end-extension to a pointwise definable model. And similarly, every countable model of ZF set theory admits an end-extension to a pointwise definable model of ZFC+V=L, as well as to pointwise definable models of other sufficient theories, accommodating large cardinals. I shall discuss the philosophical significance of these results in the philosophy of set theory with a view to potentialism and the set-theoretic multiverse.

# Quantifer elimination

A theory admits quantifier-elimination when every assertion is logically equivalent over the theory to a quantifier-free assertion. This is quite a remarkable property when it occurs, because it reveals a severe limitation on the range of concepts that can be expressed in the theory—a quantifier-free assertion, after all, is able to express only combinations of the immediate atomic facts at hand. As a result, we are generally able to prove quantifier-elimination results for a theory only when we already have a profound understanding of it and its models, and the quantifier-elimination result itself usually leads quickly to classification of the definable objects, sets, and relations in the theory and its models. In this way, quantifier-elimination results often showcase our mastery over a particular theory and its models. So let us present a few quantifier-elimination results, exhibiting our expertise over some natural theories.

## Endless dense linear orders


Consider any two rational numbers $x,y$ in the structure $\<\Q,<>$. What can one say about them? Well, we can certainly make the atomic assertions that $x$ to a Boolean combination of these assertions.

Theorem. The theory of the rational order $\<\Q,<>$ admits elimination of quantifiers—every assertion $\varphi(x,\ldots)$ is logically equivalent in the rational order to a quantifier-free assertion.

Proof. To see this, observe simply by Cantor’s categoricity theorem for countable dense linear orders that any pair $x<y$ in $\Q$ is automorphic to any other such pair $x'<y’$, and similarly for pairs with $x=y$ or $y<x$. Consequently, $\varphi(x,y)$ either holds of all pairs with $x<y$ or of none of them, of all pairs with $x=y$ or none, and of all pairs with $y<x$ or none. The assertion $\varphi(x,y)$ is therefore equivalent to the disjunction of the three atomic relations for which it is realized, including $\top$ as the disjunction of all three atomic possibilities and $\bottom$ as the empty disjunction.

More generally, a similar observation applies to assertions $\varphi(x_1,\ldots,x_n)$ with more free variables. By Cantor’s theorem, every $n$-tuple of points in $\Q$ is automorphic with any other such $n$-tuple of points having the same atomic order relations. Therefore any assertion holding of one such $n$-tuple holds of all $n$-tuples with that same atomic type, and consequently every assertion $\varphi(x_1,\ldots,x_n)$ is logically equivalent in $\<\Q,<>$ to a disjunction of those combinations of atomic relations amongst the variables $x_1,\ldots,x_n$ for which it holds. In particular, every assertion is equivalent in $\<\Q,<>$ to a quantifier-free assertion. In short, the theory of this model $\Th(\<\Q,<>)$ admits elimination of quantifiers. $\Box$

What about other endless dense linear orders? The argument we have given so far is about the theory of this particular model $\<\Q,<>$. In fact, the theory of the rational order is exactly the theory of endless dense linear orders, because this theory is complete, which one can see as an immediate consequence of the categoricity result of Cantor’s theorem and the downward Löwenheim-Skolem theorem. In my book, I have not yet proved the Löwenheim-Skolem theorem at this stage, however, and so let me give a direct proof of quantifier-elimination in the theory of endless dense linear orders, from which we can also derive the completeness of this theory.

Theorem. In the theory of endless dense linear orders, every statement is logically equivalent to a quantifier-free statement.

Proof. To clarify, the quantifier-free statement will have the same free variables as the original assertion, provided we allow $\bottom$ and $\top$ as logical constants. We argue by induction on formulas. The claim is of course already true for the atomic formulas, and it is clearly preserved under Boolean connectives. So it suffices inductively to eliminate the quantifier from $\exists x\, \varphi(x,\ldots)$, where $\varphi$ is itself quantifier-free. We can place $\varphi$ in disjunctive normal form, a disjunction of conjunction clauses, where each conjunction clause is a conjunction of literals, that is, atomic or negated atomic assertions. Since the atomic assertions $x<y$, $x=y$ and $y<x$ are mutually exclusive and exhaustive, the negation of any one of them is equivalent to the disjunction of the other two. Thus we may eliminate any need for negation. By redistributing conjunction over disjunction as much as possible, we reduce to the case of $\exists x\,\varphi$, where $\varphi$ is in disjunctive normal form without any negation. The existential quantifier distributes over disjunction, and so we reduce to the case $\varphi$ is a conjunction of atomic assertions. We may eliminate any instance of $x=x$ or $y=y$, since these impose no requirement. We may assume that the variable $x$ occurs in each conjunct, since otherwise that conjunct commutes outside the quantifier. If $x=y$ occurs in $\varphi$ for some variable $y$ not identical to $x$, then the existential claim is equivalent to $\varphi(y,\ldots)$, that is, by replacing every instance of $x$ with $y$, and we have thus eliminated the quantifier. If $x<x$ occurs as one of the conjuncts, this is not satisfiable and so the assertion is equivalent to $\bottom$. Thus we have reduced to the case where $\varphi$ is a conjunction of assertions of the form $x<y_i$ and $z_j<x$. If only one type of these occurs, then the assertion $\exists x\,\varphi$ is outright provable in the theory by the endless assumption and thus equivalent to $\top$. Otherwise, both types $x<y_i$ and $z_j<x$ occur, and in this case the existence of an $x$ obeying this conjunction of assertions is equivalent over the theory of endless dense linear orders to the quantifier-free conjunction $\bigwedge_{i,j}z_j<y_i$, since there will be an $x$ between them in this case and only in this case. Thus, we have eliminated the quantifier $\exists x$, and so by induction every formula is equivalent over this theory to a quantifier-free formula. $\Box$

Corollary. The theory of endless dense linear orders is complete.

Proof. If $\sigma$ is any sentence in this theory, then by theorem above, it is logically equivalent to a Boolean combination of quantifier-free assertions with the same variables. Since $\sigma$ is a sentence and there are no quantifier-free atomic sentences except $\bottom$ and $\top$, it follows that $\sigma$ is equivalent over the theory to a Boolean combination of $\bottom$ or $\top$. All such sentences are equivalent either to $\bottom$ or $\top$, and thus either $\sigma$ is entailed by the theory or $\neg\sigma$ is, and so the theory is complete. $\Box$

Corollary. In any endless dense linear order, the definable sets (allowing parameters) are precisely the finite unions of intervals.

Proof. By intervals we mean a generalized concept allowing either open or closed endpoints, as well as rays, in any of the forms:
$$(a,b)\qquad [a,b]\qquad [a,b)\qquad (a,b]\qquad (a,\infty)\qquad [a,\infty)\qquad (\unaryminus\infty,b)\qquad (\unaryminus\infty,b]$$
Of course any such interval is definable, since $(a,b)$ is defined by $(a<x)\wedge(x<b)$, taking the endpoints $a$ and $b$ as parameters, and $(-\infty,b]$ is defined by $(x<b)\vee (x=b)$, and so on. Thus, finite unions of intervals are also definable by taking a disjunction.

Conversely, any putative definition $\varphi(x,y_1,\ldots,y_n)$ is equivalent to a Boolean combination of atomic assertions concerning $x$ and the parameters $y_i$. Thus, whenever it is true for some $x$ between, above, or below the parameters $y_i$, it will be true of all $x$ in that same interval, and so the set that is defined will be a finite union of intervals having the parameters $y_i$ as endpoints, with the intervals being open or closed depending on whether the parameters themselves satisfy the formula or not. $\Box$

## Theory of successor

Let us next consider the theory of a successor function, as realized for example in the Dedekind model, $\<\N,S,0>$, where $S$ is the successor
function $Sn=n+1$. The theory has the following three axioms:
$$\forall x\, (Sx\neq 0)$$

$$\forall x,y\, (Sx=Sy\implies x=y)$$

$$\forall x\, \bigl(x\neq 0\implies \exists y\,(Sy=x)\bigr).$$
In the Dedekind model, every individual is definable, since $x=n$ just in case $x=SS\cdots S0$, where we have $n$ iterative applications of $S$. So this is a pointwise definable model, and hence also Leibnizian. Note the interplay between the $n$ of the object theory and $n$ of the metatheory in the claim that every individual is definable.

What definable subsets of the Dedekind model can we think of? Of course, we can define any particular finite set, since the numbers are definable as individuals. For example, we can define the set ${1,5,8}$ by saying, “either $x$ has the defining property of $1$ or it has the defining property of $5$ or it has the defining property of $8$.” Thus any finite set is definable, and by negating such a formula, we see also that any cofinite set—the complement of a finite set—is definable. Are there any other definable sets? For example, can we define the set of even numbers? How could we prove that we cannot? The Dedekind structure has no automorphisms, since all the individuals are definable, and so we cannot expect to use automorphism to show that the even numbers are not definable as a set. We need a deeper understanding of definability and truth in this structure.

Theorem. The theory of a successor function admits elimination of quantifiers—every assertion is equivalent in this theory to a quantifier-free assertion.

Proof. By induction on formulas. The claim is already true for atomic assertions, since they have no quantifiers, and quantifier-free assertions are clearly closed under the Boolean connectives. So it suffices by induction to eliminate the quantifier from assertions of the form $\exists x\, \varphi(x,\ldots)$, where $\varphi$ is quantifier free. We may place $\varphi$ in disjunctive normal form, and since the quantifier distributes over disjunction, we reduce to the case that $\varphi$ is a conjunction of atomic and negated atomic assertions. We may assume that $x$ appears in each atomic conjunct, since otherwise we may bring that conjunct outside the quantifier. We may furthermore assume that $x$ appears on only one side of each atomic clause, since otherwise the statement is either trivially true as with $SSx=SSx$ or $Sx\neq SSx$, or trivially false as with $Sx=SSx$. Consider for example:
$$\exists x\,\bigl[(SSSx=y)\wedge (SSy=SSSz)\wedge (SSSSSx=SSSw)\wedge{}$$
$$\hskip1in{}\wedge (Sx\neq SSSSw)\wedge (SSSSy\neq SSSSSz)\bigr]$$
We can remove duplicated $S$s occurring on both sides of an equation. If $x=S^ky$ appears, we can get rid of $x$ and replace all occurrences with $S^ky$. If $S^nx=y$ appears, can add $S$’s everywhere and then replace any occurrence of $S^nx$ with $y$. If only inequalities appear, then the statement is simply true.

For example, since the third clause in the formula above is equivalent to $SSx=w$, we may use that to omit any need to refer to $x$, and the formula overall is equivalent to
$$(Sw=y)\wedge (y=Sz)\wedge (w\neq SSSSSw)\wedge (y\neq Sz),$$ which has no quantifiers.
Since the method is completely general, we have proved that the theory of successor admits elimination of quantifiers. $\Box$

It follows that the definable sets in the Dedekind model $\<\N,S,0>$, using only the first-order language of this structure, are precisely the finite and cofinite sets.

Corollary. The definable sets in $\<\N,S,0>$ are precisely the finite and cofinite sets

Proof. This is because an atomic formula defines a finite set, and the collection of finite or cofinite sets is closed under negation and Boolean combinations. Since every formula is equivalent to a quantifier-free formula, it follows that every formula is a Boolean combination of atomic formulas, and hence defines a finite or cofinite set. $\Box$

In particular, the concepts of being even or being odd are not definable from the successor operation in $\<\N,S,0>$, since the set of even numbers is neither finite nor cofinite.

Corollary. The theory of a successor function is complete—it is the theory of the standard model $\<\N,S,0>$.

Proof. If $\sigma$ is a sentence in the language of successor, then by the quantifier-elimination theorem it is equivalent to a quantifier-free assertion in the language with the successor function $S$ and constant symbol $0$. But the only quantifier-free sentences in this language are Boolean combinations of equations of the form $S^n0=S^k0$. Since all such equations are settled by the theory, the sentence itself is settled by the theory, and so the theory is complete. $\Box$

We saw that the three axioms displayed on the previous page were true in the Dedekind model $\<\N,S,0>$. Are there any other models of these axioms? Yes, there are. For example, we can add another $\Z$-chain of successors on the side, as with $\N+\Z$ or $\N\sqcup\Z$, although we shall see that the order is not definable. What are the definable elements in the enlarged structure? Still $0$ and all its finite successors are definable as before. But no elements of the $\Z$-chains can be definable, because we may perform an automorphism of the structure that translates elements within the $\Z$-chain by a fixed amount.

Let me prove next that the theory implies the induction axiom schema.

Corollary. The theory of successor (the three axioms) implies the induction axiom schema in the language of successor, that is, the following assertion for any assertion $\varphi(x)$:
$$\left[\varphi(0)\wedge\bigl(\forall x\,\bigl(\varphi(x)\implies\varphi(Sx)\bigr)\right]\implies\forall x\,\varphi(x)$$

Proof. Consider the set defined by $\varphi(x)$. By the earlier corollary, it must be eventually periodic in the standard model $\<\N,S,0>$. But by the induction assumption stated in the theorem, it must hold of every number in the standard model. So the standard model thinks that $\forall x\,\varphi(x)$. But the theory of the standard model is the theory of successor, which is complete. So the theory of successor entails that $\varphi$ is universal, as desired. $\Box$

In other words, in the trivial theory of successor–the three axioms—we get the corresponding induction axiom for free.

## Presburger arithmetic

Presburger arithmetic is the theory of addition on the natural numbers, that is, the theory of the structure $\<\N,+,0,1>$. The numbers $0$ and $1$ are actually definable here from addition alone, since $0$ is the unique additive identity, and $1$ is the only number $u$ that is not expressible as a sum $x+y$ with both $x\neq u$ and $y\neq u$. So we may view this model if desired as a definitional expansion of $\<\N,+>$, with addition only. The number $2$ is similarly definable as $1+1$, and indeed any number $n$ is definable as $1+\cdots+1$, with $n$ summands, and so this is a pointwise definable model and hence also Leibnizian.

What are the definable subsets? We can define the even numbers, of course, since $x$ is even if and only if $\exists y\,(y+y=x)$. We can similarly define congruence modulo $2$ by $x\equiv_2 y\iff \exists z\,\bigl[(z+z+x=y)\vee (z+z+y=x)\bigr]$. More generally, we can express the relation of congruence modulo $n$ for any fixed $n$ as follows:
$$x\equiv_n y\quad\text{ if and only if }\exists z\,\bigl[(\overbrace{z+\cdots+z}^n+x=y)\vee(\overbrace{z+\cdots+z}^n+y=x)\bigr].$$
What I claim is that this exhausts what is expressible.

Theorem. Presburger arithmetic in the definitional expansion with all congruence relations, that is, the theory of the structure
$$\<\N,+,0,1,\equiv_2,\equiv_3,\equiv_4,\ldots>$$
admits elimination of quantifiers. In particular, every assertion in the language of $\<\N,+,0,1>$ is equivalent to a quantifer-free assertion in the language with the congruence relations.

Proof. We consider Presburger arithmetic in the language with addition $+$, with all the congruence relations $\equiv_n$ for every $n\geq 2$, and the constants $0$ and $1$. We prove quantifier-elimination in this language by induction on formulas. As before the claim already holds for atomic assertions and is preserved by Boolean connectives. So it suffices to eliminate the quantifier from assertions of the form $\exists x\,\varphi(x,\ldots)$, where $\varphi$ is quantifier-free. By placing $\varphi$ into disjunctive normal form and distributing the quantifier over the disjunction, we may assume that $\varphi$ is a conjunction of atomic and negated atomic assertions. Note that negated congruences are equivalent to a disjunction of positive congruences, such as in the case:
$$x\not\equiv_4 y\quad\text{ if and only if }\quad (x+1\equiv_4y)\vee(x+1+1\equiv_4y)\vee (x+1+1+1\equiv_4 y).$$
We may therefore assume there are no negated congruences in $\varphi$. By canceling like terms on each side of an equation or congruence, we may assume that $x$ occurs on only one side. We may assume that $x$ occurs nontrivially in every conjunct of $\varphi$, since otherwise this conjunct commutes outside the quantifier. Since subtraction modulo $n$ is the same as adding $n-1$ times, we may also assume that all congruences occurring in $\varphi$ have the form $kx\equiv_n t$, where $kx$ denotes the syntactic expression $x+\cdots+x$ occurring in the formula, with $k$ summands, and $t$ is a term not involving the variable $x$. Thus, $\varphi$ is a conjunction of expressions each having the form $kx\equiv_n t$, $ax+r=s$, or $bx+u\neq v$, where $ax$ and $bx$ similarly denote the iterated sums $x+\cdots+x$ and $r,s,u,v$ are terms not involving $x$.

If indeed there is a conjunct of the equality form $ax+r=s$ occurring in $\varphi$, then we may omit the quantifier as follows. Namely, in order to fulfill the existence assertion, we know that $x$ will have to solve $ax+r=s$, and so in particular $r\equiv_a s$, which ensures the existence of such an $x$, but also in this case any inequality $bx+u\neq v$ can be equivalently expressed as $abx+au\neq av$, which since $ax+r=s$ is equivalent to $bs+au\neq av+br$, and this does does not involve $x$; similarly, any congruence $kx\equiv_n t$ is equivalent to $akx\equiv_{an}at$, which is equivalent to $s\equiv_{an} r+at$, which again does not involve $x$. Thus, when there is an equality involving $x$ present in $\varphi$, then we can use that fact to express the whole formula in an equivalent manner not involving $x$.

So we have reduced to the case $\exists x\,\varphi$, where $\varphi$ is a conjunction of inequalities $bs+u\neq v$ and congruences $kx\equiv_n t$. We can now ignore the inequalities, since if the congruence system has a solution, then it will have infinitely many solutions, and so there will be an $x$ solving any finitely given inequalities. So we may assume that $\varphi$ is simply a list of congruences of the form $kx\equiv_n t$, and the assertion is that this system of congruences has a solution. But there are only finitely many congruences mentioned, and so by working modulo the least common multiple of the bases that occur, there are only finitely many possible values for $x$ to be checked. And so we can simply replace $\varphi$ with a disjunction over these finitely many values $i$, replacing $x$ in each conjunction with $1+\cdots+1$, using $i$ copies of $1$, for each $i$ up to the least common multiples of the bases that arise in the congruences appearing in $\varphi$. If there is an $x$ solving the system, then one of these values of $i$ will work, and conversely.

So we have ultimately succeeded in expressing $\exists x\,\varphi$ in a quantifier-free manner, and so by induction every assertion in Presburger arithmetic is equivalent to a quantifier-free assertion in the language allowing addition, congruences, and the constants $0$ and $1$. $\Box$

Corollary. The definable sets in $\<\N,+,0,1>$ are exactly the eventually periodic sets.

Proof. Every periodic set is definable, since one can specify the set up to the period $p$, and then express the invariance modulo $p$. Any finite deviation from a definable set also is definable, since every individual number is definable. So every eventually period set is definable. Conversely, every definable set is defined by a quantifier-free assertion in the language of $\<\N,+,0,1,\equiv_2,\equiv_3,\equiv_4,\ldots>$. We may place the definition in disjunctive normal form, and again replace negated congruences with a disjunction of positive congruences. For large enough values of $x$, the equalities and inequalities appearing in the definition become irrelevant, and so the definition eventually agrees with a finite union of solutions of congruence systems. Every such system is periodic with a period at most the least common multiple of the bases of the congruences appearing in it. And so every definable set is eventually periodic, as desired. $\Box$

Corollary. Multiplication is not definable in $\<\N,+,0,1>$. Indeed, the squaring operation is not definable, and neither is the divisibility relation $p\divides q$.

Proof. If we could define multiplication, or even the squaring operation, then we would be able to define the set of perfect squares, but this is not eventually periodic. Similarly, if we could define the divides relation $p\divides q$, then we could define the set of prime numbers, which is not eventually periodic. $\Box$

## Real-closed field

Let us lastly consider the ordered real field $\<\R,+,\cdot,0,1,<>$. I want to mention (without proof) that a deep theorem of Tarski shows that this structure admits elimination of quantifiers: every assertion is equivalent in this structure to a quantifier-free assertion. In fact all that is need is that this is a real-closed field, an ordered field in which every odd-degree polynomial has a root and every positive number has a square root.

We can begin to gain insight to this fact by reaching into the depths of our high-school education. Presented with an equation $ax^2+bx+c=0$ in the integers, we know by the quadratic formula that the solution is $x=\left(-b\pm\sqrt{b^2-4ac}\right)/2a$, and in particular, there is a solution in the real numbers if and only if $b^2-4ac\geq 0$, since otherwise a negative discriminant means the solution is a complex number. In other words,
$$\exists x\,(ax^2+bx+c=0)\quad\text{ if and only if }\quad b^2-4ac\geq 0.$$
The key point is that this an elimination of quantifiers result, since we have eliminated the quantifier $\exists x$.

Tarski’s theorem proves more generally that every assertion in the language of ordered fields is equivalent in real-closed fields to a quantifier-free assertion. Furthermore, there is a computable procedure to find the quantifier-free equivalent, as well as a computable procedure to determine the truth of any quantifier-free assertion in the theory of real-closed fields.

What I find incredible is that it follows from this that there is a computable procedure to determine the truth of any first-order assertion of Cartesian plane geometry, since all such assertions are expressible in the language of $\<\R,+,\cdot,0,1,<>$. Amazing! I view this as an incredible culmination of two thousand years of mathematical investigation: we now have an algorithm to determine by rote procedure the truth of any statement in Cartesian geometry. Meanwhile, a counterpoint is that the decision procedure, unfortunately, is not computationally feasible, however, since it takes more than exponential time, and it is a topic of research to investigate the computational running time of the best algorithms.

# Definability and the Math Tea argument: must there be numbers we cannot describe or define? University of Warsaw, 22 January 2021

This will be a talk for a new mathematical logic seminar at the University of Warsaw in the Department of Hhilosophy, entitled Epistemic and Semantic Commitments of Foundational Theories, devoted to formal truth theories and implicit commitments of foundational theories as well as their conceptual surroundings.

My talk will be held 22 January 2021, 8 pm CET (7 pm UK), online via Zoom https://us02web.zoom.us/j/83366049995.

Abstract. According to the math tea argument, perhaps heard at a good afternoon tea, there must be some real numbers that we can neither describe nor define, since there are uncountably many real numbers, but only countably many definitions. Is it correct? In this talk, I shall discuss the phenomenon of pointwise definable structures in mathematics, structures in which every object has a property that only it exhibits. A mathematical structure is Leibnizian, in contrast, if any pair of distinct objects in it exhibit different properties. Is there a Leibnizian structure with no definable elements? We shall discuss many interesting elementary examples, eventually working up to the proof that every countable model of set theory has a pointwise definable extension, in which every mathematical object is definable.

Pointwise definable models of set theory

• J. D. Hamkins, D. Linetsky, and J. Reitz, “Pointwise definable models of set theory,” Journal of Symbolic Logic, vol. 78, iss. 1, p. 139–156, 2013.
[Bibtex]
@article {HamkinsLinetskyReitz2013:PointwiseDefinableModelsOfSetTheory,
AUTHOR = {Hamkins, Joel David and Linetsky, David and Reitz, Jonas},
TITLE = {Pointwise definable models of set theory},
JOURNAL = {Journal of Symbolic Logic},
FJOURNAL = {Journal of Symbolic Logic},
VOLUME = {78},
YEAR = {2013},
NUMBER = {1},
PAGES = {139--156},
ISSN = {0022-4812},
MRCLASS = {03E55},
MRNUMBER = {3087066},
MRREVIEWER = {Bernhard A. König},
DOI = {10.2178/jsl.7801090},
URL = {http://jdh.hamkins.org/pointwisedefinablemodelsofsettheory/},
eprint = "1105.4597",
archivePrefix = {arXiv},
primaryClass = {math.LO},
}

# Categorical cardinals, CUNY Set Theory Seminar, June 2020

This will be an online talk for the CUNY Set Theory Seminar, Friday 26 June 2020, 2 pm EST = 7 pm UK time. Contact Victoria Gitman for Zoom access.

Abstract: Zermelo famously characterized the models of second-order Zermelo-Fraenkel set theory $\text{ZFC}_2$ in his 1930 quasi-categoricity result asserting that the models of $\text{ZFC}_2$ are precisely those isomorphic to a rank-initial segment $V_\kappa$ of the cumulative set-theoretic universe $V$ cut off at an inaccessible cardinal $\kappa$. I shall discuss the extent to which Zermelo’s quasi-categoricity analysis can rise fully to the level of categoricity, in light of the observation that many of the $V_\kappa$ universes are categorically characterized by their sentences or theories. For example, if $\kappa$ is the smallest inaccessible cardinal, then up to isomorphism $V_\kappa$ is the unique model of $\text{ZFC}_2$ plus the sentence “there are no inaccessible cardinals.” This cardinal $\kappa$ is therefore an instance of what we call a first-order sententially categorical cardinal. Similarly, many of the other inaccessible universes satisfy categorical extensions of $\text{ZFC}_2$ by a sentence or theory, either in first or second order. I shall thus introduce and investigate the categorical cardinals, a new kind of large cardinal. This is joint work with Robin Solberg (Oxford).

# The $\Sigma_1$-definable universal finite sequence

• J. D. Hamkins and K. J. Williams, “The $\Sigma_1$-definable universal finite sequence,” Journal of Symbolic Logic, 2021.
[Bibtex]
@ARTICLE{HamkinsWilliams2021:The-universal-finite-sequence,
author = {Joel David Hamkins and Kameryn J. Williams},
title = {The $\Sigma_1$-definable universal finite sequence},
journal = {Journal of Symbolic Logic},
year = {2021},
volume = {},
number = {},
pages = {},
month = {},
note = {},
abstract = {},
keywords = {},
eprint = {1909.09100},
archivePrefix = {arXiv},
primaryClass = {math.LO},
source = {},
doi = {10.1017/jsl.2020.59},
}

Abstract. We introduce the $\Sigma_1$-definable universal finite sequence and prove that it exhibits the universal extension property amongst the countable models of set theory under end-extension. That is, (i) the sequence is $\Sigma_1$-definable and provably finite; (ii) the sequence is empty in transitive models; and (iii) if $M$ is a countable model of set theory in which the sequence is $s$ and $t$ is any finite extension of $s$ in this model, then there is an end extension of $M$ to a model in which the sequence is $t$. Our proof method grows out of a new infinitary-logic-free proof of the Barwise extension theorem, by which any countable model of set theory is end-extended to a model of $V=L$ or indeed any theory true in a suitable submodel of the original model. The main theorem settles the modal logic of end-extensional potentialism, showing that the potentialist validities of the models of set theory under end-extensions are exactly the assertions of S4. Finally, we introduce the end-extensional maximality principle, which asserts that every possibly necessary sentence is already true, and show that every countable model extends to a model satisfying it.

• The universal algorithm,
• J. D. Hamkins and H. W. Woodin, “The universal finite set,” Mathematics ArXiv, p. 1–16, 2017.
[Bibtex]
@ARTICLE{HamkinsWoodin:The-universal-finite-set,
author = {Joel David Hamkins and W. Hugh Woodin},
title = {The universal finite set},
journal = {Mathematics ArXiv},
year = {2017},
volume = {},
number = {},
pages = {1--16},
month = {},
note = {Manuscript under review},
abstract = {},
keywords = {under-review},
source = {},
doi = {},
eprint = {1711.07952},
archivePrefix = {arXiv},
primaryClass = {math.LO},
url = {http://jdh.hamkins.org/the-universal-finite-set},
}
• The modal logic of arithmetic potentialism,
• J. D. Hamkins, “The modal logic of arithmetic potentialism and the universal algorithm,” Mathematics ArXiv, p. 1–35, 2018.
[Bibtex]
@ARTICLE{Hamkins:The-modal-logic-of-arithmetic-potentialism,
author = {Joel David Hamkins},
title = {The modal logic of arithmetic potentialism and the universal algorithm},
journal = {Mathematics ArXiv},
year = {2018},
volume = {},
number = {},
pages = {1--35},
month = {},
eprint = {1801.04599},
archivePrefix = {arXiv},
primaryClass = {math.LO},
note = {Under review},
url = {http://wp.me/p5M0LV-1Dh},
abstract = {},
keywords = {under-review},
source = {},
doi = {},
}
• A new proof of the Barwise extension theorem
• Kameryn’s blog post about the paper

# When does every definable nonempty set have a definable element?

• F. G. Dorais and J. D. Hamkins, “When does every definable nonempty set have a definable element?,” Mathematical Logic Quarterly, vol. 65, iss. 4, pp. 407-411, 2019.
[Bibtex]
@ARTICLE{DoraisHamkins:When-does-every-definable-nonempty-set-have-a-definable-element,
author = {Dorais, François G. and Hamkins, Joel David},
title = {When does every definable nonempty set have a definable element?},
journal = {Mathematical Logic Quarterly},
volume = {65},
number = {4},
pages = {407-411},
doi = {10.1002/malq.201700035},
abstract = {Abstract The assertion that every definable set has a definable element is equivalent over ZF to the principle V=HOD, and indeed, we prove, so is the assertion merely that every Π2-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying V≠HOD in which every Σ2-definable set has an ordinal-definable element. Similar results hold for HOD(R) and HOD(Ordω) and other natural instances of HOD(X).},
year = {2019},
keywords = {},
source = {},
doi = {},
eprint = {1706.07285},
archivePrefix = {arXiv},
primaryClass = {math.LO},
url = {http://jdh.hamkins.org/definable-sets-with-definable-elements},
}

Abstract. The assertion that every definable set has a definable element is equivalent over ZF to the principle $V=\newcommand\HOD{\text{HOD}}\HOD$, and indeed, we prove, so is the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying $V\neq\HOD$ in which every $\Sigma_2$-definable set has an ordinal-definable element. Similar results hold for $\HOD(\mathbb{R})$ and $\HOD(\text{Ord}^\omega)$ and other natural instances of $\HOD(X)$.

It is not difficult to see that the models of ZF set theory in which every definable nonempty set has a definable element are precisely the models of $V=\HOD$. Namely, if $V=\HOD$, then there is a definable well-ordering of the universe, and so the $\HOD$-least element of any definable nonempty set is definable; and conversely, if $V\neq\HOD$, then the set of minimal-rank non-OD sets is definable, but can have no definable element.

In this brief article, we shall identify the limit of this elementary observation in terms of the complexity of the definitions. Specifically, we shall prove that $V=\HOD$ is equivalent to the assertion that every $\Pi_2$-definable nonempty set contains an ordinal-definable element, but that one may not replace $\Pi_2$-definability here by $\Sigma_2$-definability.

Theorem. The following are equivalent in any model $M$ of ZF:

1. $M$ is a model of $\text{ZFC}+\text{V}=\text{HOD}$.
2. $M$ thinks there is a definable well-ordering of the universe.
3. Every definable nonempty set in $M$ has a definable element.
4. Every definable nonempty set in $M$ has an ordinal-definable element.
5. Every ordinal-definable nonempty set in $M$ has an ordinal-definable element.
6. Every $\Pi_2$-definable nonempty set in $M$ has an ordinal-definable element.

Theorem. Every model of ZFC has a forcing extension satisfying $V\neq\HOD$, in which every $\Sigma_2$-definable set has a definable element.

The proof of this latter theorem is reminiscent of several proofs of the maximality principle (see A simple maximality principle), where one undertakes a forcing iteration attempting at each stage to force and then preserve a given $\Sigma_2$ assertion.

This inquiry grew out of a series of questions and answers posted on MathOverflow and the exchange of the authors there.

# The definable cut of a model of set theory can be changed by small forcing

If $M$ is a model of ZFC set theory, let $I$ be the definable cut of its ordinals, the collection of ordinals that are below an ordinal $\delta$ of $M$ that is definable in $M$ without parameters. This would include all the ordinals of $M$, if the definable ordinals happen to be unbounded in $M$, but one can also construct examples where the definable cut is bounded in $M$.  Let $M_I$ be the corresponding definable cut of $M$ itself, the rank-initial segment of $M$ determined by $I$, or in other words, the collection of all sets $x$ in $M$ of rank below a definable ordinal of $M$. Equivalently, $$M_I=\bigcup_{\delta\in I} V_\delta^M.$$ It is not difficult to see that this is an elementary substructure $M_I\prec M$, because we can verify the Tarski-Vaught criterion as follows. If $M\models\exists y\ \varphi(x,y)$, where $x\in M_I$, then let $\delta$ be a definable ordinal above the rank of $x$. In this case, the ordinal $\theta$, which is the supremum over all $a\in V_\delta$ of the minimal rank of a set $y$ for which $\varphi(a,y)$, if there is such a $y$. This supremum $\theta$ is definable, and so since $x\in V_\delta$, the minimal rank of a $y$ such that $\varphi(x,y)$ is at most $\theta$. Consequently, since $\theta\in I$, such a $y$ can be found in $M_I$. So we have found the desired witness inside the substructure, and so it is elementary $M_I\prec M$. Note that in the general case, one does not necessarily know that $I$ has a least upper bound in $M$. Under suitable assumptions, it can happen that $I$ is unbounded in $M$, that $I$ is an ordinal of $M$, or that $I$ is bounded in $M$, but has no least upper bound.

What I am interested in for this post is how the definable cut might be affected by forcing. Of course, it is easy to see that if $M$ is definable in $M[G]$, then the definable cut of $M[G]$ is at least as high as the definable cut of $M$, simply because the definable ordinals of $M$ remain definable in $M[G]$.

A second easy observation is that if the definable cut of $M$ is bounded in $M$, then we could perform large collapse forcing, collapsing a cardinal above $I$ to $\omega$, which would of course make every cardinal of $I$ countable in the extension $M[G]$. In this case, since $\omega_1^{M[G]}$ is definable, it would change the definable cut. So this kind of very large forcing can change the definable cut, making it larger.

But what about small forcing? Suppose that the forcing notion $\newcommand\P{\mathbb{P}}\P$ we intend to forcing with is small in the sense that it is in the definable cut $M_I$. This would be true if $\P$ itself were definable, for example, but really we only require that $\P$ has rank less than some definable ordinal of $M$. Can this forcing change the definable cut?

Let me show at least that the definable cut can never go up after small forcing.

Theorem. If $G\subset\P$ is $M$-generic for forcing $\P$ in the definable cut of $M$, then the definable cut of $M[G]$ is below or the same in the ordinals as it was in $M$.

Proof. Suppose that $G\subset\P$ is $M$-generic, and we consider the forcing extension $M[G]$. We have already proved that $M_I\prec M$ is an elementary submodel. I claim that this relation lifts to the forcing extension $M_I[G]\prec M[G]$. Note first that since $\P\in M_I$ and $M_I$ is a rank initial segment of $M$, it follows that $M_I$ has all the subsets of $\P$ in $M$, and so $G$ is $M_I$-generic. So the extension $M_I[G]$ makes sense. Next, suppose that $M[G]\models\varphi(a)$ for some $a\in M_I[G]$. If $\dot a$ is a name for $a$ in $M_I$, then there is some condition $p\in G$ forcing $\varphi(\dot a)$ over $M$. Since $M_I\prec M$, this is also forced by $p$ over $M_I$, and thus $M_I[G]\models\varphi(a)$ as well, as desired. So $M_I[G]\prec M[G]$, and from this it follows that every definable ordinal of $M[G]$ is in the cut $I$. So the definable cut did not get higher. QED

But can it go down? Not if the model $M$ is definable in $M[G]$, by our earlier easy observation. Consequently,

Theorem. If $M$ is definable in $M[G]$, where $G\subset\P$ is $M$-generic for forcing $\P$ below the definable cut of $M$, then the definable cut of $M[G]$ is the same as the definable cut of $M$.

Proof. It didn’t go down, since $M$ is definable in $M[G]$; and it didn’t go up, since $\P$ was small. QED

What if $M$ is not definable in $M[G]$? Can we make the definable cut go down after small forcing? The answer is yes.

Theorem. If ZFC is consistent, then there is a model $M\models\text{ZFC}$ with a definable notion of forcing $\P$ (hence in the definable cut of $M$), such that if $G\subset\P$ is $M$-generic, then the definable cut of the forcing extension $M[G]$ is strictly shorter than the definable cut of $M[G]$.

Proof. Start with a model of $\text{ZFC}+V=L$, whose definable ordinals are bounded by a cardinal $\delta$. Let’s call it $L$, and let $I$ be the definable cut of $L$, which we assume is bounded by $\delta$. Let $M=L[G]$ be the forcing extension of $L$ obtained by performing an Easton product, adding a Cohen subset to every regular cardinal above $\delta$ in $L$. Since this forcing adds no sets below $\delta$, but adds a Cohen set at $\delta^+$, it follows that $\delta$ becomes definable in $L[G]$. In fact, since the forcing is homogeneous and definable from $\delta$, it follows that the definable ordinals of $L[G]$ are precisely the ordinals that are definable in $L$ with parameter $\delta$. These may be bounded or unbounded in $L[G]$. Now, let $\newcommand\Q{\mathbb{Q}}\Q$ be the Easton product forcing at the stages below $\delta$, and suppose that $G\subset\Q$ is $L[G]$-generic. Consider the model $L[G][H]$. Note that the forcing $\Q$ is definable in $L[G]$, since $\delta$ is definable there. This two-step forcing can be combined into one giant Easton product in $L$, the product that simply forces to add a Cohen subset to every regular cardinal. Since this version of the forcing is homogeneous and definable in $L$, it follows that the definable ordinals of $L[G][H]$ are precisely the definable ordinals of $L$, which are bounded by $I$. In summary, the definable cut of $L[G]$ is strictly above $\delta$, since $\delta$ is definable in $L[G]$, and the forcing $\Q$ has size and rank $\delta$; but the forcing extension $L[G][H]$ has definable cut $I$, which is strictly bounded by $\delta$. So the definable cut was made smaller by small forcing, as claimed. QED

This post is an account of some ideas that Alexander Block and I had noted today during the course of our mathematical investigation of another matter.

# Being HOD-of-a-set is invariant throughout the generic multiverse

$\newcommand\HOD{\text{HOD}}$The axiom $V=\HOD$, introduced by Gödel, asserts that every set is ordinal definable. This axiom has a subtler foundational aspect than might at first be expected. The reason is that the general concept of “object $x$ is definable using parameter $p$” is not in general first-order expressible in set theory; it is of course a second-order property, which makes sense only relative to a truth predicate, and by Tarski’s theorem, we can have no first-order definable truth predicate. Thus, the phrase “definable using ordinal parameters” is not directly meaningful in the first-order language of set theory without further qualification or explanation. Fortunately, however, it is a remarkable fact that when we allow definitions to use arbitrary ordinal parameters, as we do with $\HOD$, then we can in fact make such qualifications in such a way that the axiom becomes first-order expressible in set theory. Specifically, we say officially that $V=\HOD$ holds, if for every set $x$, there is an ordinal $\theta$ with $x\in V_\theta$, for which which $x$ is definable by some formula $\psi(x)$ in the structure $\langle V_\theta,{\in}\rangle$ using ordinal parameters. Since $V_\theta$ is a set, we may freely make reference to first-order truth in $V_\theta$ without requiring any truth predicate in $V$. Certainly any such $x$ as this is also ordinal-definable in $V$, since we may use $\theta$ and the Gödel-code of $\psi$ also as parameters, and note that $x$ is the unique object such that it is in $V_\theta$ and satisfies $\psi$ in $V_\theta$. (Note that inside an $\omega$-nonstandard model of set theory, we may really need to use $\psi$ as a parameter, since it may be nonstandard, and $x$ may not be definable in $V_\theta$ using a meta-theoretically standard natural number; but fortunately, the Gödel code of a formula is an integer, which is still an ordinal, and this issue is the key to the issue.) Conversely, if $x$ is definable in $V$ using formula $\varphi(x,\vec\alpha)$ with ordinal parameters $\vec\alpha$, then it follows by the reflection theorem that $x$ is defined by $\varphi(x,\vec\alpha)$ inside some $V_\theta$. So this formulation of $V=HOD$ is expressible and exactly captures the desired second-order property that every set is ordinal-definable.

Consider next the axiom $V=\HOD(b)$, asserting that every set is definable from ordinal parameters and parameter $b$. Officially, as before, $V=\HOD(b)$ asserts that for every $x$, there is an ordinal $\theta$, formula $\psi$ and ordinals $\vec \alpha<\theta$, such that $x$ is the unique object in $V_\theta$ for which $\langle V_\theta,{\in}\rangle\models\psi(x,\vec\alpha,b)$, and the reflection argument shows again that this way of defining the axiom exactly captures the intended idea.

The axiom I actually want to focus on is $\exists b\,\left( V=\HOD(b)\right)$, asserting that the universe is $\HOD$ of a set. (I assume ZFC in the background theory.) It turns out that this axiom is constant throughout the generic multiverse.

Theorem. The assertion $\exists b\, (V=\HOD(b))$ is forcing invariant.

• If it holds in $V$, then it continues to hold in every set forcing extension of $V$.
• If it holds in $V$, then it holds in every ground of $V$.

Thus, the truth of this axiom is invariant throughout the generic multiverse.

Proof. Suppose that $\text{ZFC}+V=\HOD(b)$, and $V[G]$ is a forcing extension of $V$ by generic filter $G\subset\mathbb{P}\in V$. By the ground-model definability theorem, it follows that $V$ is definable in $V[G]$ from parameter $P(\mathbb{P})^V$. Thus, using this parameter, as well as $b$ and additional ordinal parameters, we can define in $V[G]$ any particular object in $V$. Since this includes all the $\mathbb{P}$-names used to form $V[G]$, it follows that $V[G]=\HOD(b,P(\mathbb{P})^V,G)$, and so $V[G]$ is $\HOD$ of a set, as desired.

Conversely, suppose that $W$ is a ground of $V$, so that $V=W[G]$ for some $W$-generic filter $G\subset\mathbb{P}\in W$, and $V=\HOD(b)$ for some set $b$. Let $\dot b$ be a name for which $\dot b_G=b$. Every object $x\in W$ is definable in $W[G]$ from $b$ and ordinal parameters $\vec\alpha$, so there is some formula $\psi$ for which $x$ is unique such that $\psi(x,b,\vec\alpha)$. Thus, there is some condition $p\in\mathbb{P}$ such that $x$ is unique such that $p\Vdash\psi(\check x,\dot b,\check{\vec\alpha})$. If $\langle p_\beta\mid\beta<|\mathbb{P}|\rangle$ is a fixed enumeration of $\mathbb{P}$ in $W$, then $p=p_\beta$ for some ordinal $\beta$, and we may therefore define $x$ in $W$ using ordinal parameters, along with $\dot b$ and the fixed enumeration of $\mathbb{P}$. So $W$ thinks the universe is $\HOD$ of a set, as desired.

Since the generic multiverse is obtained by iteratively moving to forcing extensions to grounds, and each such movement preserves the axiom, it follows that $\exists b\, (V=\HOD(b))$ is constant throughout the generic multiverse. QED

Theorem. If $V=\HOD(b)$, then there is a forcing extension $V[G]$ in which $V=\HOD$ holds.

Proof. We are working in ZFC. Suppose that $V=\HOD(b)$. We may assume $b$ is a set of ordinals, since such sets can code any given set. Consider the following forcing iteration: first add a Cohen real $c$, and then perform forcing $G$ that codes $c$, $P(\omega)^V$ and $b$ into the GCH pattern at uncountable cardinals, and then perform self-encoding forcing $H$ above that coding, coding also $G$ (see my paper on Set-theoretic geology for further details on self-encoding forcing). In the final model $V[c][G][H]$, therefore, the objects $c$, $b$, $P(\omega)^V$, $G$ and $H$ are all definable without parameters. Since $V\subset V[c][G][H]$ has a closure point at $\omega$, it satisfies the $\omega_1$-approximation and cover properties, and therefore the class $V$ is definable in $V[c][G][H]$ using $P(\omega)^V$ as a parameter. Since this parameter is itself definable without parameters, it follows that $V$ is parameter-free definable in $V[c][G][H]$. Since $b$ is also definable there, it follows that every element of $\HOD(b)^V=V$ is ordinal-definable in $V[c][G][H]$. And since $c$, $G$ and $H$ are also definable without parameters, we have $V[c][G][H]\models V=\HOD$, as desired. QED

Corollary. The following are equivalent.

1. The universe is $\HOD$ of a set: $\exists b\, (V=\HOD(b))$.
2. Somewhere in the generic multiverse, the universe is $\HOD$ of a set.
3. Somewhere in the generic multiverse, the axiom $V=\HOD$ holds.
4. The axiom $V=\HOD$ is forceable.

Proof. This is an immediate consequence of the previous theorems. $1\to 4\to 3\to 2\to 1$. QED

Corollary. The axiom $V=\HOD$, if true, even if true anywhere in the generic multiverse, is a switch.

Proof. A switch is a statement such that both it and its negation are necessarily possible by forcing; that is, in every set forcing extension, one can force the statement to be true and also force it to be false. We can always force $V=\HOD$ to fail, simply by adding a Cohen real. If $V=\HOD$ is true, then by the first theorem, every forcing extension has $V=\HOD(b)$ for some $b$, in which case $V=\HOD$ remains forceable, by the second theorem. QED

# Different models of set theory with the same subset relation

Recently Makoto Kikuchi (Kobe University) asked me the following interesting question, which arises very naturally if one should adopt a mereological perspective in the foundations of mathematics, placing a focus on the parthood relation rather than the element-of relation. In set theory, this perspective would lead one to view the subset or inclusion relation $\subseteq$ as the primary fundamental relation, rather than the membership $\in$ relation.

Question. Can there be two different models of set theory, with the same inclusion relation?

We spent an evening discussing it, over delicious (Rokko-michi-style) okonomiyaki and bi-ru, just like old times, except that we are in Tokyo at the CTFM 2015, and I’d like to explain the answer, which is yes, this always happens in every model of set theory.

Theorem. In any universe of set theory $\langle V,\in\rangle$, there is a definable relation $\in^*$, different from $\in$, such that $\langle V,\in^*\rangle$ is a model of set theory, in fact isomorphic to the original universe $\langle V,\in\rangle$, for which the corresponding inclusion relation $$u\subseteq^* v\iff \forall a\, (a\in^* u\to a\in^* v)$$ is identical to the usual inclusion relation $u\subseteq v$.

Proof. Let $\theta:V\to V$ be any definable non-identity permutation of the universe, and let $\tau:u\mapsto \theta[u]=\{\ \theta(a)\mid a\in u\ \}$ be the function determined by pointwise image under $\theta$. Since $\theta$ is bijective, it follows that $\tau$ is also a bijection of $V$ to $V$, since every set is the $\theta$-image of a unique set. Furthermore, $\tau$ is an automorphism of $\langle V,\subseteq\rangle$, since $$u\subseteq v\iff\theta[u]\subseteq\theta[v]\iff\tau(u) \subseteq\tau(v).$$ I had used this idea a few years ago in my answer to the MathOverflow question, Is the inclusion version of Kunen inconsistency theorem true?, which shows that there are nontrivial $\subseteq$ automorphisms of the universe. Note that since $\tau(\{a\})=\{\theta(a)\}$, it follows that any instance of nontriviality $\theta(a)\neq a$ in $\theta$ leads immediately to an instance of nontriviality in $\tau$.

Using the map $\tau$, define $a\in^* b\iff\tau(a)\in\tau(b)$. By definition, therefore, $\tau$ is an isomorphism of $\langle V,\in^*\rangle\cong\langle V,\in\rangle$. Let us show that $\in^*\neq \in$. Since $\theta$ is nontrivial, there is an $\in$-minimal set $a$ with $\theta(a)\neq a$. By minimality, $\theta[a]=a$ and so $\tau(a)=a$. But as mentioned, $\tau(\{a\})=\{\theta(a)\}\neq\{a\}$. So we have $a\in\{a\}$, but $\tau(a)=a\notin\{\theta(a)\}=\tau(\{a\})$ and hence $a\notin^*\{a\}$. So the two relations are different.

Meanwhile, consider the corresponding subset relation. Specifically, $u\subseteq^* v$ is defined to mean $\forall a\,(a\in^* u\to a\in^* v)$, which holds if and only if $\forall a\, (\tau(a)\in\tau(u)\to \tau(a)\in\tau(v))$; but since $\tau$ is surjective, this holds if and only if $\tau(u)\subseteq \tau(v)$, which as we observed at the beginning of the proof, holds if and only if $u\subseteq v$. So the corresponding subset relations $\subseteq^*$ and $\subseteq$ are identical, as desired.

Another way to express what is going on is that $\tau$ is an isomorphism of the structure $\langle V,{\in^*},{\subseteq}\rangle$ with $\langle V,{\in},{\subseteq}\rangle$, and so $\subseteq$ is in fact that same as the corresponding inclusion relation $\subseteq^*$ that one would define from $\in^*$. QED

Corollary. One cannot define $\in$ from $\subseteq$ in a model of set theory.

Proof. The map $\tau$ is a $\subseteq$-automorphism, and so it preserves every relation definable from $\subseteq$, but it does not preserve $\in$. QED

Nevertheless, I claim that the isomorphism type of $\langle V,\in\rangle$ is implicit in the inclusion relation $\subseteq$, in the sense that any other class relation $\in^*$ having that same inclusion relation is isomorphic to the $\in$ relation.

Theorem. Assume ZFC in the universe $\langle V,\in\rangle$. Suppose that $\in^*$ is a class relation for which $\langle V,\in^*\rangle$ is a model of set theory (a weak set theory suffices), such that the corresponding inclusion relation $$u\subseteq^* v\iff\forall a\,(a\in^* u\to a\in^* v)$$is the same as the usual inclusion relation $u\subseteq v$. Then the two membership relations are isomorphic $$\langle V,\in\rangle\cong\langle V,\in^*\rangle.$$

Proof. Since the singleton set $\{a\}$ has exactly two subsets with respect to the usual $\subseteq$ relation — the empty set and itself — this must also be true with respect to the inclusion relation $\subseteq^*$ defined via $\in^*$, since we have assumed $\subseteq^*=\subseteq$. Thus, the object $\{a\}$ is also a singleton with respect to $\in^*$, and so there is a unique object $\eta(a)$ such that $x\in^* a\iff x=\eta(a)$. By extensionality and since every object has its singleton, it follows that $\eta:V\to V$ is both one-to-one and onto. Let $\theta=\eta^{-1}$ be the inverse permutation.

Observe that $a\in u\iff \{a\}\subseteq u\iff \{a\}\subseteq^* u\iff\eta(a)\in^* u$. Thus, $$b\in^* u\iff \theta(b)\in u.$$

Using $\in$-recursion, define $b^*=\{\ \theta(a^*)\mid a\in b\ \}$. The map $b\mapsto b^*$ is one-to-one by $\in$-recursion, since if there is no violation of this for the elements of $b$, then we may recover $b$ from $b^*$ by applying $\theta^{-1}$ to the elements of $b^*$ and then using the induction assumption to find the unique $a$ from $a^*$ for each $\theta(a^*)\in b^*$, thereby recovering $b$. So $b\mapsto b^*$ is injective.

I claim that this map is also surjective. If $y_0\neq b^*$ for any $b$, then there must be an element of $y_0$ that is not of the form $\theta(b^*)$ for any $b$. Since $\theta$ is surjective, this means there is $\theta(y_1)\in y_0$ with $y_1\neq b^*$ for any $b$. Continuing, there is $y_{n+1}$ with $\theta(y_{n+1})\in y_n$ and $y_{n+1}\neq b^*$ for any $b$. Let $z=\{\ \theta(y_n)\mid n\in\omega\ \}$. Since $x\in^* u\iff \theta(x)\in u$, it follows that the $\in^*$-elements of $z$ are precisely the $y_n$’s. But $\theta(y_{n+1})\in y_n$, and so $y_{n+1}\in^* y_n$. So $z$ has no $\in^*$-minimal element, violating the axiom of foundation for $\in^*$, a contradiction. So the map $b\mapsto b^*$ is a bijection of $V$ with $V$.

Finally, we observe that because $$a\in b\iff\theta(a^*)\in b^*\iff a^*\in^* b^*,$$ it follows that the map $b\mapsto b^*$ is an isomorphism of $\langle V,\in\rangle$ with $\langle V,\in^*\rangle$, as desired. QED

The conclusion is that although $\in$ is not definable from $\subseteq$, nevertheless, the isomorphism type of $\in$ is implicit in $\subseteq$, in the sense that any other class relation $\in^*$ giving rise to the same inclusion relation $\subseteq^*=\subseteq$ is isomorphic to $\in$.

Meanwhile, I do not yet know what the situation is when one drops the assumption that $\in^*$ is a class with respect to the $\langle V,\in\rangle$ universe.

Question. Can there be two models of set theory $\langle M,\in\rangle$ and $\langle M,\in^*\rangle$, not necessarily classes with respect to each other, which have the same inclusion relation $\subseteq=\subseteq^*$, but which are not isomorphic?

(This question is now answered! See my joint paper with Kikuchi at Set-theoretic mereology.)

# Ehrenfeucht’s lemma in set theory

• G. Fuchs, V. Gitman, and J. D. Hamkins, “Ehrenfeucht’s Lemma in Set Theory,” Notre Dame Journal of Formal Logic, vol. 59, iss. 3, p. 355–370, 2018.
[Bibtex]
@ARTICLE{FuchsGitmanHamkins2018:EhrenfeuchtsLemmaInSetTheory,
author = "Fuchs, Gunter and Gitman, Victoria and Hamkins, Joel David",
doi = "10.1215/00294527-2018-0007",
fjournal = "Notre Dame Journal of Formal Logic",
journal = "Notre Dame Journal of Formal Logic",
number = "3",
pages = "355--370",
publisher = "Duke University Press",
title = "Ehrenfeucht’s Lemma in Set Theory",
volume = "59",
year = "2018",
eprint = {1501.01918},
archivePrefix = {arXiv},
primaryClass = {math.LO},
url = {http://jdh.hamkins.org/ehrenfeuchts-lemma-in-set-theory},
}


Abstract. Ehrenfeucht’s lemma asserts that whenever one element of a model of Peano arithmetic is definable from another, then they satisfy different types. We consider here the analogue of Ehrenfeucht’s lemma for models of set theory. The original argument applies directly to the ordinal-definable elements of any model of set theory, and in particular, Ehrenfeucht’s lemma holds fully for models of set theory satisfying $V=\HOD$. We show that the lemma can fail, however, in models of set theory with $V\neq\HOD$, and it necessarily fails in the forcing extension to add a generic Cohen real. We go on to formulate a scheme of natural parametric generalizations of Ehrenfeucht’s lemma, namely, the principles of the form $\Ehrenfeucht(A,P,Q)$, which asserts that whenever an object $b$ is definable in $M$ from some $a\in A$ using parameters in $P$, with $b\neq a$, then the types of $a$ and $b$ over $Q$ in $M$ are different. We also consider various analogues of Ehrenfeucht’s lemma obtained by using algebraicity in place of definability, where a set $b$ is \emph{algebraic} in $a$ if it is a member of a finite set definable from $a$ (as in J. D. Hamkins and C. Leahy, Algebraicity and implicit definability in set theory). Ehrenfeucht’s lemma holds for the ordinal-algebraic sets, we prove, if and only if the ordinal-algebraic and ordinal-definable sets coincide. Using similar analysis, we answer two open questions posed in my paper with Leahy, by showing that (i) algebraicity and definability need not coincide in models of set theory and (ii) the internal and external notions of being ordinal algebraic need not coincide.

# When does every definable set have a definable member? CUNY Set Theory Seminar, October 2014

This will be a talk for the CUNY set theory seminar, October 10, 2014, 12pm  GC 6417.

Abstract. Although the concept of `being definable’ is not generally expressible in the language of set theory, it turns out that the models of ZF in which every definable nonempty set has a definable element are precisely the models of V=HOD.  Indeed, V=HOD is equivalent to the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, this is not true in the case of $\Sigma_2$-definability, because every model of ZFC has a forcing extension satisfying $V\neq\text{HOD}$ in which every $\Sigma_2$-definable set has an ordinal-definable element.

This is joint work with François G. Dorais and Emil Jeřábek, growing out of some questions and answers on MathOverflow, namely,

Definable collections without definable members
A question asked by Ashutosh five years ago, in which François and I gradually came upon the answer together.
Is it consistent that every definable set has a definable member?
A similar question asked last week by (anonymous) user38200
Can $V\neq\text{HOD}$ if every $\Sigma_2$-definable set has an ordinal-definable member?
A question I had regarding the limits of an issue in my answer to the previous question.

In this talk, I shall present the answers to all these questions and place the results in the context of classical results on definability, including a review of basic concepts for graduate students.