Parallels in universality between the universal algorithm and the universal finite set, Oxford Math Logic Seminar, October 2018

This will be a talk for the Logic Seminar in Oxford at the Mathematics Institute in the Andrew Wiles Building on October 9, 2018, at 4:00 pm, with tea at 3:30.

Abstract. The universal algorithm is a Turing machine program $e$ that can in principle enumerate any finite sequence of numbers, if run in the right model of PA, and furthermore, can always enumerate any desired extension of that sequence in a suitable end-extension of that model. The universal finite set is a set-theoretic analogue, a locally verifiable definition that can in principle define any finite set, in the right model of set theory, and can always define any desired finite extension of that set in a suitable top-extension of that model. Recent work has uncovered a $\Sigma_1$-definable version that works with respect to end-extensions. I shall give an account of all three results, which have a parallel form, and describe applications to the model theory of arithmetic and set theory.

Topological models of arithmetic

  • A. Enayat, J. D. Hamkins, and B. Wcisło, “Topological models of arithmetic,” ArXiv e-prints, 2018. (under review)  
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Abstract. Ali Enayat had asked whether there is a nonstandard model of Peano arithmetic (PA) that can be represented as $\newcommand\Q{\mathbb{Q}}\langle\Q,\oplus,\otimes\rangle$, where $\oplus$ and $\otimes$ are continuous functions on the rationals $\Q$. We prove, affirmatively, that indeed every countable model of PA has such a continuous presentation on the rationals. More generally, we investigate the topological spaces that arise as such topological models of arithmetic. The reals $\newcommand\R{\mathbb{R}}\R$, the reals in any finite dimension $\R^n$, the long line and the Cantor space do not, and neither does any Suslin line; many other spaces do; the status of the Baire space is open.

 

The first author had inquired whether a nonstandard model of arithmetic could be continuously presented on the rational numbers.

Main Question. (Enayat, 2009) Are there continuous functions $\oplus$ and $\otimes$ on the rational numbers $\Q$, such that $\langle\Q,\oplus,\otimes\rangle$ is a nonstandard model of arithmetic?

By a model of arithmetic, what we mean here is a model of the first-order Peano axioms PA, although we also consider various weakenings of this theory. The theory PA asserts of a structure $\langle M,+,\cdot\rangle$ that it is the non-negative part of a discretely ordered ring, plus the induction principle for assertions in the language of arithmetic. The natural numbers $\newcommand\N{\mathbb{N}}\langle \N,+,\cdot\rangle$, for example, form what is known as the standard model of PA, but there are also many nonstandard models, including continuum many non-isomorphic countable models.

We answer the question affirmatively, and indeed, the main theorem shows that every countable model of PA is continuously presented on $\Q$. We define generally that a topological model of arithmetic is a topological space $X$ equipped with continuous functions $\oplus$ and $\otimes$, for which $\langle X,\oplus,\otimes\rangle$ satisfies the desired arithmetic theory. In such a case, we shall say that the underlying space $X$ continuously supports a model of arithmetic and that the model is continuously presented upon the space $X$.

Question. Which topological spaces support a topological model of arithmetic?

In the paper, we prove that the reals $\R$, the reals in any finite dimension $\R^n$, the long line and Cantor space do not support a topological model of arithmetic, and neither does any Suslin line. Meanwhile, there are many other spaces that do support topological models, including many uncountable subspaces of the plane $\R^2$. It remains an open question whether any uncountable Polish space, including the Baire space, can support a topological model of arithmetic.

Let me state the main theorem and briefly sketch the proof.

Main Theorem. Every countable model of PA has a continuous presentation on the rationals $\Q$.

Proof. We shall prove the theorem first for the standard model of arithmetic $\langle\N,+,\cdot\rangle$. Every school child knows that when computing integer sums and products by the usual algorithms, the final digits of the result $x+y$ or $x\cdot y$ are completely determined by the corresponding final digits of the inputs $x$ and $y$. Presented with only final segments of the input, the child can nevertheless proceed to compute the corresponding final segments of the output.

\begin{equation*}\small\begin{array}{rcr}
\cdots1261\quad & \qquad & \cdots1261\quad\\
\underline{+\quad\cdots 153\quad}&\qquad & \underline{\times\quad\cdots 153\quad}\\
\cdots414\quad & \qquad & \cdots3783\quad\\
& & \cdots6305\phantom{3}\quad\\
& & \cdots1261\phantom{53}\quad\\
& & \underline{\quad\cdots\cdots\phantom{253}\quad}\\
& & \cdots933\quad\\
\end{array}\end{equation*}

This phenomenon amounts exactly to the continuity of addition and multiplication with respect to what we call the final-digits topology on $\N$, which is the topology having basic open sets $U_s$, the set of numbers whose binary representations ends with the digits $s$, for any finite binary string $s$. (One can do a similar thing with any base.) In the $U_s$ notation, we include the number that would arise by deleting initial $0$s from $s$; for example, $6\in U_{00110}$. Addition and multiplication are continuous in this topology, because if $x+y$ or $x\cdot y$ has final digits $s$, then by the school-child’s observation, this is ensured by corresponding final digits in $x$ and $y$, and so $(x,y)$ has an open neighborhood in the final-digits product space, whose image under the sum or product, respectively, is contained in $U_s$.

Let us make several elementary observations about the topology. The sets $U_s$ do indeed form the basis of a topology, because $U_s\cap U_t$ is empty, if $s$ and $t$ disagree on some digit (comparing from the right), or else it is either $U_s$ or $U_t$, depending on which sequence is longer. The topology is Hausdorff, because different numbers are distinguished by sufficiently long segments of final digits. There are no isolated points, because every basic open set $U_s$ has infinitely many elements. Every basic open set $U_s$ is clopen, since the complement of $U_s$ is the union of $U_t$, where $t$ conflicts on some digit with $s$. The topology is actually the same as the metric topology generated by the $2$-adic valuation, which assigns the distance between two numbers as $2^{-k}$, when $k$ is largest such that $2^k$ divides their difference; the set $U_s$ is an open ball in this metric, centered at the number represented by $s$. (One can also see that it is metric by the Urysohn metrization theorem, since it is a Hausdorff space with a countable clopen basis, and therefore regular.) By a theorem of Sierpinski, every countable metric space without isolated points is homeomorphic to the rational line $\Q$, and so we conclude that the final-digits topology on $\N$ is homeomorphic to $\Q$. We’ve therefore proved that the standard model of arithmetic $\N$ has a continuous presentation on $\Q$, as desired.

But let us belabor the argument somewhat, since we find it interesting to notice that the final-digits topology (or equivalently, the $2$-adic metric topology on $\N$) is precisely the order topology of a certain definable order on $\N$, what we call the final-digits order, an endless dense linear order, which is therefore order-isomorphic and thus also homeomorphic to the rational line $\Q$, as desired.

Specifically, the final-digits order on the natural numbers, pictured in figure 1, is the order induced from the lexical order on the finite binary representations, but considering the digits from right-to-left, giving higher priority in the lexical comparison to the low-value final digits of the number. To be precise, the final-digits order $n\triangleleft m$ holds, if at the first point of disagreement (from the right) in their binary representation, $n$ has $0$ and $m$ has $1$; or if there is no disagreement, because one of them is longer, then the longer number is lower, if the next digit is $0$, and higher, if it is $1$ (this is not the same as treating missing initial digits as zero). Thus, the even numbers appear as the left half of the order, since their final digit is $0$, and the odd numbers as the right half, since their final digit is $1$, and $0$ is directly in the middle; indeed, the highly even numbers, whose representations end with a lot of zeros, appear further and further to the left, while the highly odd numbers, which end with many ones, appear further and further to the right. If one does not allow initial $0$s in the binary representation of numbers, then note that zero is represented in binary by the empty sequence. It is evident that the final-digits order is an endless dense linear order on $\N$, just as the corresponding lexical order on finite binary strings is an endless dense linear order.

The basic open set $U_s$ of numbers having final digits $s$ is an open set in this order, since any number ending with $s$ is above a number with binary form $100\cdots0s$ and below a number with binary form $11\cdots 1s$ in the final-digits order; so $U_s$ is a union of intervals in the final-digits order. Conversely, every interval in the final-digits order is open in the final-digits topology, because if $n\triangleleft x\triangleleft m$, then this is determined by some final segment of the digits of $x$ (appending initial $0$s if necessary), and so there is some $U_s$ containing $x$ and contained in the interval between $n$ and $m$. Thus, the final-digits topology is the precisely same as the order topology of the final-digits order, which is a definable endless dense linear order on $\N$. Since this order is isomorphic and hence homeomorphic to the rational line $\Q$, we conclude again that $\langle \N,+,\cdot\rangle$ admits a continuous presentation on $\Q$.

We now complete the proof by considering an arbitrary countable model $M$ of PA. Let $\triangleleft^M$ be the final-digits order as defined inside $M$. Since the reasoning of the above paragraphs can be undertaken in PA, it follows that $M$ can see that its addition and multiplication are continuous with respect to the order topology of its final-digits order. Since $M$ is countable, the final-digits order of $M$ makes it a countable endless dense linear order, which by Cantor’s theorem is therefore order-isomorphic and hence homeomorphic to $\Q$. Thus, $M$ has a continuous presentation on the rational line $\Q$, as desired. $\Box$

The executive summary of the proof is: the arithmetic of the standard model $\N$ is continuous with respect to the final-digits topology, which is the same as the $2$-adic metric topology on $\N$, and this is homeomorphic to the rational line, because it is the order topology of the final-digits order, a definable endless dense linear order; applied in a nonstandard model $M$, this observation means the arithmetic of $M$ is continuous with respect to its rational line $\Q^M$, which for countable models is isomorphic to the actual rational line $\Q$, and so such an $M$ is continuously presentable upon $\Q$.

Let me mention the following order, which it seems many people expect to use instead of the final-digits order as we defined it above. With this order, one in effect takes missing initial digits of a number as $0$, which is of course quite reasonable.

The problem with this order, however, is that the order topology is not actually the final-digits topology. For example, the set of all numbers having final digits $110$ in this order has a least element, the number $6$, and so it is not open in the order topology. Worse, I claim that arithmetic is not continuous with respect to this order. For example, $1+1=2$, and $2$ has an open neighborhood consisting entirely of even numbers, but every open neighborhood of $1$ has both odd and even numbers, whose sums therefore will not all be in the selected neighborhood of $2$. Even the successor function $x\mapsto x+1$ is not continuous with respect to this order.

Finally, let me mention that a version of the main theorem also applies to the integers $\newcommand\Z{\mathbb{Z}}\Z$, using the following order.

Go to the article to read more.

  • A. Enayat, J. D. Hamkins, and B. Wcisło, “Topological models of arithmetic,” ArXiv e-prints, 2018. (under review)  
    @ARTICLE{EnayatHamkinsWcislo2018:Topological-models-of-arithmetic,
    author = {Ali Enayat and Joel David Hamkins and Bartosz Wcisło},
    title = {Topological models of arithmetic},
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    month = {},
    note = {under review},
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A new proof of the Barwise extension theorem, without infinitary logic

I have found a new proof of the Barwise extension theorem, that wonderful yet quirky result of classical admissible set theory, which says that every countable model of set theory can be extended to a model of $\text{ZFC}+V=L$.

Barwise Extension Theorem. (Barwise 1971) $\newcommand\ZF{\text{ZF}}\newcommand\ZFC{\text{ZFC}}$ Every countable model of set theory $M\models\ZF$ has an end-extension to a model of $\ZFC+V=L$.

The Barwise extension theorem is both (i) a technical culmination of the pioneering methods of Barwise in admissible set theory and infinitary logic, including the Barwise compactness and completeness theorems and the admissible cover, but also (ii) one of those rare mathematical theorems that is saturated with significance for the philosophy of mathematics and particularly the philosophy of set theory. I discussed the theorem and its philosophical significance at length in my paper, The multiverse perspective on the axiom of constructibility, where I argued that it can change how we look upon the axiom of constructibility and whether this axiom should be considered ‘restrictive,’ as it often is in set theory. Ultimately, the Barwise extension theorem shows how wrong a model of set theory can be, if we should entertain the idea that the set-theoretic universe continues growing beyond it.

Regarding my new proof, below, however, what I find especially interesting about it, if not surprising in light of (i) above, is that it makes no use of Barwise compactness or completeness and indeed, no use of infinitary logic at all! Instead, the new proof uses only classical methods of descriptive set theory concerning the representation of $\Pi^1_1$ sets with well-founded trees, the Levy and Shoenfield absoluteness theorems, the reflection theorem and the Keisler-Morley theorem on elementary extensions via definable ultrapowers. Like the Barwise proof, my proof splits into cases depending on whether the model $M$ is standard or nonstandard, but another interesting thing about it is that with my proof, it is the $\omega$-nonstandard case that is easier, whereas with the Barwise proof, the transitive case was easiest, since one only needed to resort to the admissible cover when $M$ was ill-founded. Barwise splits into cases on well-founded/ill-founded, whereas in my argument, the cases are $\omega$-standard/$\omega$-nonstandard.

To clarify the terms, an end-extension of a model of set theory $\langle M,\in^M\rangle$ is another model $\langle N,\in^N\rangle$, such that the first is a substructure of the second, so that $M\subseteq N$ and $\in^M=\in^N\upharpoonright M$, but further, the new model does not add new elements to sets in $M$. In other words, $M$ is an $\in$-initial segment of $N$, or more precisely: if $a\in^N b\in M$, then $a\in M$ and hence $a\in^M b$.

Set theory, of course, overflows with instances of end-extensions. For example, the rank-initial segments $V_\alpha$ end-extend to their higher instances $V_\beta$, when $\alpha<\beta$; similarly, the hierarchy of the constructible universe $L_\alpha\subseteq L_\beta$ are end-extensions; indeed any transitive set end-extends to all its supersets. The set-theoretic universe $V$ is an end-extension of the constructible universe $L$ and every forcing extension $M[G]$ is an end-extension of its ground model $M$, even when nonstandard. (In particular, one should not confuse end-extensions with rank-extensions, also known as top-extensions, where one insists that all the new sets have higher rank than any ordinal in the smaller model.)

Let’s get into the proof.

Proof. Suppose that $M$ is a model of $\ZF$ set theory. Consider first the case that $M$ is $\omega$-nonstandard. For any particular standard natural number $k$, the reflection theorem ensures that there are arbitrarily high $L_\alpha^M$ satisfying $\ZFC_k+V=L$, where $\ZFC_k$ refers to the first $k$ axioms of $\ZFC$ in a fixed computable enumeration by length. In particular, every countable transitive set $m\in L^M$ has an end-extension to a model of $\ZFC_k+V=L$. By overspill (that is, since the standard cut is not definable), there must be some nonstandard $k$ for which $L^M$ thinks that every countable transitive set $m$ has an end-extension to a model of $\ZFC_k+V=L$, which we may assume is countable. This is a $\Pi^1_2$ statement about $k$, which will therefore also be true in $M$, by the Shoenfield absolutenss theorem. It will also be true in all the elementary extensions of $M$, as well as in their forcing extensions. And indeed, by the Keisler-Morley theorem, the model $M$ has an elementary top extension $M^+$. Let $\theta$ be a new ordinal on top of $M$, and let $m=V_\theta^{M^+}$ be the $\theta$-rank-initial segment of $M^+$, which is a top-extension of $M$. Let $M^+[G]$ be a forcing extension in which $m$ has become countable. Since the $\Pi^1_2$ statement is true in $M^+[G]$, there is an end-extension of $\langle m,\in^{M^+}\rangle$ to a model $\langle N,\in^N\rangle$ that $M^+[G]$ thinks satisfies $\ZFC_k+V=L$. Since $k$ is nonstandard, this theory includes all the $\ZFC$ axioms, and since $m$ end-extends $M$, we have found an end-extension of $M$ to a model of $\ZFC+V=L$, as desired.

It remains to consider the case where $M$ is $\omega$-standard. By the Keisler-Morley theorem, let $M^+$ be an elementary top-extension of $M$. Let $\theta$ be an ordinal of $M^+$ above $M$, and consider the corresponding rank-initial segment $m=V_\theta^{M^+}$, which is a transitive set in $M^+$ that covers $M$. If $\langle m,\in^{M^+}\rangle$ has an end-extension to a model of $\ZFC+V=L$, then we’re done, since such a model would also end-extend $M$. So assume toward contradiction that there is no such end-extension of $m$. Let $M^+[G]$ be a forcing extension in which $m$ has become countable. The assertion that $m$ has no end-extension to a model of $\ZFC+V=L$ is actually true and hence true in $M^+[G]$. This is a $\Pi^1_1$ assertion there about the real coding $m$. Every such assertion has a canonically associated tree, which is well-founded exactly when the statement is true. Since the statement is true in $M^+[G]$, this tree has some countable rank $\lambda$ there. Since these models have the standard $\omega$, the tree associated with the statement is the same for us as inside the model, and since the statement is actually true, the tree is actually well founded. So the rank $\lambda$ must come from the well-founded part of the model.

If $\lambda$ happens to be countable in $L^{M^+}$, then consider the assertion, “there is a countable transitive set, such that the assertion that it has no end-extension to a model of $\ZFC+V=L$ has rank $\lambda$.” This is a $\Sigma_1$ assertion, since it is witnessed by the countable transitive set and the ranking function of the tree associated with the non-extension assertion. Since the parameters are countable, it follows by Levy reflection that the statement is true in $L^{M^+}$. So $L^{M^+}$ has a countable transitive set, such that the assertion that it has no end-extension to a model of $\ZFC+V=L$ has rank $\lambda$. But since $\lambda$ is actually well-founded, the statement would have to be actually true; but it isn’t, since $L^{M^+}$ itself is such an extension, a contradiction.

So we may assume $\lambda$ is uncountable in $M^+$. In this case, since $\lambda$ was actually well-ordered, it follows that $L^M$ is well-founded beyond its $\omega_1$. Consider the statement “there is a countable transitive set having no end-extension to a model of $\ZFC+V=L$.” This is a $\Sigma^1_2$ sentence, which is true in $M^+[G]$ by our assumption about $m$, and so by Shoenfield absoluteness, it is true in $L^{M^+}$ and hence also $L^M$. So $L^M$ thinks there is a countable transitive set $b$ having no end-extension to a model of $\ZFC+V=L$. This is a $\Pi^1_1$ assertion about $b$, whose truth is witnessed in $L^M$ by a ranking of the associated tree. Since this rank would be countable in $L^M$ and this model is well-founded up to its $\omega_1$, the tree must be actually well-founded. But this is impossible, since it is not actually true that $b$ has no such end-extension, since $L^M$ itself is such an end-extension of $b$. Contradiction. $\Box$

One can prove a somewhat stronger version of the theorem, as follows.

Theorem. For any countable model $M$ of $\ZF$, with an inner model $W\models\ZFC$, and any statement $\phi$ true in $W$, there is an end-extension of $M$ to a model of $\ZFC+\phi$. Furthermore, one can arrange that every set of $M$ is countable in the extension model.

In particular, one can find end-extensions of $\ZFC+V=L+\phi$, for any statement $\phi$ true in $L^M$.

Proof. Carry out the same proof as above, except in all the statements, ask for end-extensions of $\ZFC+\phi$, instead of end-extensions of $\ZFC+V=L$, and also ask that the set in question become countable in that extension. The final contradictions are obtained by the fact that the countable transitive sets in $L^M$ do have end-extensions like that, in which they are countable, since $W$ is such an end-extension. $\Box$

For example, we can make the following further examples.

Corollaries.

  1. Every countable model $M$ of $\ZFC$ with a measurable cardinal has an end-extension to a model $N$ of $\ZFC+V=L[\mu]$.
  2. Every countable model $M$ of $\ZFC$ with extender-based large cardinals has an end-extension to a model $N$ satisfying $\ZFC+V=L[\vec E]$.
  3. Every countable model $M$ of $\ZFC$ with infinitely many Woodin cardinals has an end-extension to a model $N$ of $\ZF+\text{AD}+V=L(\mathbb{R})$.

And in each case, we can furthermore arrange that every set of $M$ is countable in the extension model $N$.

This proof grew out of a project on the $\Sigma_1$-definable universal finite set, which I am currently undertaking with Kameryn Williams and Philip Welch.


Jon Barwise. Infinitary methods in the model theory of set theory. In Logic
Colloquium ’69 (Proc. Summer School and Colloq., Manchester, 1969), pages
53–66. North-Holland, Amsterdam, 1971.

Paul K. Gorbow, PhD 2018, University of Gothenburg

Paul K. Gorbow successfully defended his dissertation, “Self-similarity in the foundations” on June 14, 2018 at the University of Gothenburg in the Department of Philosophy, Linguistics and Theory of Science, under the supervision of Ali Enayat, with Peter LeFanu Lumsdaine and Zachiri McKenzie serving as secondary supervisors.  The defense opponent was Roman Kossak, with a dissertation committee consisting of Jon Henrik Forssell, Joel David Hamkins (myself) and Vera Koponen, chaired by Fredrik Engström. Congratulations!

University of Gothenburg profilear$\chi$ivResearch Gate

Paul K. Gorbow, “Self-similarity in the foundations,” PhD dissertation for the University of Gothenburg, Acta Philosophica Gothoburgensia 32, June 2018. (arxiv:1806.11310)

Abstract. This thesis concerns embeddings and self-embeddings of foundational structures in both set theory and category theory. 

The first part of the work on models of set theory consists in establishing a refined version of Friedman’s theorem on the existence of embeddings between countable non-standard models of a fragment of ZF, and an analogue of a theorem of Gaifman to the effect that certain countable models of set theory can be elementarily end-extended to a model with many automorphisms whose sets of fixed points equal the original model. The second part of the work on set theory consists in combining these two results into a technical machinery, yielding several results about non-standard models of set theory relating such notions as self-embeddings, their sets of fixed points, strong rank-cuts, and set theories of different strengths.

The work in foundational category theory consists in the formulation of a novel algebraic set theory which is proved to be equiconsistent to New Foundations (NF), and which can be modulated to correspond to intuitionistic or classical NF, with or without atoms. A key axiom of this theory expresses that its structures have an endofunctor with natural properties.

In the Swedish style of dissertation defense, the opponent (in this case Roman Kossak) summarizes the dissertation, placing it in a broader context, and then challenges various parts of it, probing the candidate’s expertise in an extended discussion. What a pleasure it was to see this.  After this, there is a broader discussion, in which the committee is also involved.