The global choice principle in Gödel-Bernays set theory


I’d like to follow up on several posts I made recently on MathOverflow (see here, here and here), which engaged several questions of Gérard Lang that I found interesting. Specifically, I’d like to discuss a number of equivalent formulations of the global choice principle in Gödel-Bernays set theory. Let us adopt the following abbreviations for the usually considered theories:

  • GB is the usual Gödel-Bernays set theory without any choice principle.
  • GB+AC is GB plus the axiom of choice for sets.
  • GBC is GB plus the global choice principle.

The global choice principle has a number of equivalent characterizations, as proved in the theorem below, but for definiteness let us take it as the assertion that there is a global choice function, that is, a class $F$ which is a function such that $F(x)\in x$ for every nonempty set $x$.

Note in particular that I do not use the set version of choice AC in the equivalences, since most of the statements imply AC for sets outright (except in the case of statement 7, where it is stated specifically in order to make the equivalence).

Theorem. The following are equivalent over GB.

  1. The global choice principle. That is, there is a class function $F$ such that $F(x)\in x$ for every nonempty set $x$.
  2. There is a bijection between $V$ and $\Ord$.
  3. There is a global well-ordering of $V$. That is, there is a class relation $\triangleleft$ on $V$ that is a linear order, such that every nonempty set has a $\triangleleft$-least member.
  4. There is a global set-like well-ordering of $V$. There is a class well-ordering $\triangleleft$ as above, such that all $\triangleleft$-initial segments are sets.
  5. Every proper class is bijective with $\Ord$.
  6. Every class injects into $\Ord$.
  7. AC holds for sets and $\Ord$ injects into every proper class.
  8. $\Ord$ surjects onto every class.
  9. Every class is comparable with $\Ord$ by injectivity; that is, one injects into the other.
  10. Any two classes are comparable by injectivity.


($1\to 2$) Assume that $F$ is a global choice class function. Using the axiom of replacement, we may recursively define a class sequence of sets $\langle x_\alpha\mid\alpha\in\Ord\rangle$, where $x_\alpha=F(X_\alpha)$, where $X_\alpha$ is the set of minimal-rank sets $x$ not among $\{x_\beta\mid\beta<\alpha\}$. That is, we use $F$ to choose the next element among the minimal-rank sets not yet chosen. Thus, we have an injection of $V$ into $\Ord$. If a set $x$ does not appear as some $x_\alpha$ on this sequence, then no set of that rank or higher can appear, since we always add sets of the minimal rank not yet having appeared; thus, in this case we will have injected $\Ord$ into some $V_\beta$. But this is impossible by Hartog’s theorem, and so in fact we have bijection between $\Ord$ and $V$.

($2\to 3$) If there is a bijection between $\Ord$ and $V$, then we may define a global well-ordering by $x<y$ if $x$ appears before $y$ in that enumeration.

($3\to 1$) Let $F(x)$ be the least element of $x$ with respect to a fixed global well-ordering.

($3\to 4$) If there is a global well-ordering $<$, then we may refine it to a set-like well-ordering, by defining $x\triangleleft y$ just in case the rank of $x$ is less than the rank of $y$, or they have the same rank and $x<y$. This relation is still a well-order, since the least member of any nonempty set $X$ will be the $\triangleleft$-least member of the set of members of $X$ having minimal rank. The relation $\triangleleft$ is set-like, because the $\triangleleft$-predecessors of any set $x$ are amongst the sets having rank no larger than $x$, and this is a set.

($4\to 5$) If there is a global set-like well-ordering $<$ of $V$ and $X$ is a proper class, then $<$ on $X$ is a well-ordering of $X$, and we may map any ordinal $\alpha$ to the $\alpha^{th}$ member of $X$. This will be a bijection of $\Ord$ with $X$.

($5\to 6$) If every proper class is bijective with $\Ord$, then $V$ is bijective with $\Ord$, and so every set injects into $\Ord$ by restriction.

($6\to 7$) If every class injects into $\Ord$, then in particular, $V$ injects into $\Ord$. The image of this injection is a proper class subclass of $\Ord$, and all such classes are bijective with $\Ord$ by mapping $\alpha$ to the $\alpha^{th}$ member of the class, and so every proper class is bijective with $\Ord$. So $\Ord$ injects the other way, and also AC holds.

($7\to 3$) Suppose that AC holds and $\Ord$ injects into every proper class. Let $W$ be the class of all well-orderings of some rank-initial segment $V_\alpha$ of the set-theoretic universe $V$. Since for each $\alpha$ there are only a set number of such well-orderings of $V_\alpha$, if we inject $\Ord$ into $W$, then there must be well-orderings of unboundedly many $V_\alpha$ in the range of the injection. From this, we may easily construct a global well-ordering of $V$, by defining $x<y$ just in case $x$ has lower rank than $y$, or they have the same rank and $x<y$ in the first well-ordering of a sufficiently large $V_\alpha$ to appear in the range of the injection.

($5\to 8$) Immediate.

($8\to 3$) If $\Ord$ surjects onto $V$, then there is a global well-ordering, defined by $x<y$ if the earliest appearance of $x$ in the surjection is earlier than that of $y$.

($6\to 9$) Immediate.

($9\to 3$) Assume every class is comparable with $\Ord$ via injectivity. It follows that AC holds for sets, since $\Ord$ cannot inject into a set, and if a set injects into $\Ord$ then it is well-orderable. Now, if $\Ord$ injects into the class $W$ used above, consisting of all well-orderings of a $V_\alpha$, then we saw before that we can build a well-ordering of $V$. And if $W$ injects into $\Ord$, then $W$ is well-orderable and we can also in this case build a well-ordering of $V$.

($5\to 10$) If $V$ is bijective with $\Ord$, then every class is bijective with $\Ord$ or with an ordinal, and these are comparable by injections. So any two classes are comparable by injections.

($10\to 9$) Immediate.


Let’s notice a few things.

First, we cannot omit the AC assertion in statement 7. To see this, consider the model $V=L(\R)$, in a case where it does not satisfy AC. I claim that in this model, $\Ord$ injects into every proper class that is definable from parameters. The reason is that every object in $L(\R)$ is definable from ordinal and real parameters, and indeed, definable in some $V_\alpha^{L(\R)}$ by some real and ordinal parameters. Indeed, one needs only one ordinal and real parameter. If $W$ is any proper class, then there
must be a proper subclass $W_0\subset W$ whose elements are all defined by the same definition in this way. And by partitioning further, we may find a single real that works with various ordinal parameters using that definition to define a proper class of
elements of $W$. Thus, we may inject $\Ord$ into $W$, even though AC fails in $L(\R)$.

Second, the surjectivity analogues of a few of the statements are not equivalent to global choice. Indeed, ZF proves that every proper class surjects onto $\Ord$, with no choice at all, since if $W$ is a proper class, then there are unboundedly many ordinals
arising as the rank of an element of $W$, and so we may map each element $x\in W$ to $\alpha$, if the rank of $x$ is the $\alpha^{th}$ ordinal that is the rank of any element of $W$.

An introduction to the theory of infinite games, with examples from infinite chess, University of Connecticut, December 2014

This will be a talk for the interdisciplinary Group in Philosophical and Mathematical Logic at the University of Connecticut in Storrs, on December 5, 2014.

Value omega cubedAbstract. I shall give a general introduction to the theory of infinite games, with a focus on the theory of transfinite ordinal game values. These ordinal game values can be used to show that every open game — a game that, when won for a particular player, is won after finitely many moves — has a winning strategy for one of the players. By means of various example games, I hope to convey the extremely concrete game-theoretic meaning of these game values for various particular small infinite ordinals. Some of the examples will be drawn from infinite chess, which is chess played on a chessboard stretching infinitely without boundary in every direction, and the talk will include animations of infinite chess positions having large numbers of pieces (or infinitely many) with hundreds of pieces making coordinated attacks on the chessboard. Meanwhile, the exact value of the omega one of chess, denoted $\omega_1^{\mathfrak{Ch}_{\!\!\!\!\sim}}$, is not currently known.

Slides | Transfinite game values in infinite chess | The mate-in-n problem of infinite chess is decidable

Upward closure in the toy multiverse of all countable models of set theory

The Multiverse by KaeltykThe toy multiverse of all countable models of set theory is upward closed under countably many successive forcing extensions of bounded size…

I’d like to explain a topic from my recent paper

G. Fuchs, J. D. Hamkins, J. ReitzSet-theoretic geology, to appear in the Annals of Pure and Applied Logic.

We just recently made the final revisions, and the paper is available if you follow the title link through to the arxiv. Most of the geology article proceeds from a downward-oriented focus on forcing, looking from a universe $V$ down to its grounds, the inner models $W$ over which $V$ might have arisen by forcing $V=W[G]$. Thus, the set-theoretic geology project arrives at deeper and deeper grounds and the mantle and inner mantle concepts.

One section of the paper, however, has an upward-oriented focus, namely, $\S2$ A brief upward glance, and it is that material about which I’d like to write here, because I find it to be both interesting and comparatively accessible, but also because the topic proceeds from a different perspective than the rest of the geology paper, and so I am a little fearful that it may get lost there.

First is the observation that I first heard from W. Hugh Woodin in the early 1990s.


Observation. If $W$ is a countable model of ZFC set theory, then there are forcing extensions $W[c]$ and $W[d]$, both obtained by adding a Cohen real, which are non-amalgamable in the sense that there can be no model of ZFC with the same ordinals as $W$ containing both $W[c]$ and $W[d]$. Thus, the family of forcing extensions of $W$ is not upward directed.

Proof. Since $W$ is countable, let $z$ be a real coding the entirety of $W$. Enumerate the dense subsets $\langle D_n\mid n<\omega\rangle$ of the Cohen forcing $\text{Add}(\omega,1)$ in $W$. We construct $c$ and $d$ in stages. We begin by letting $c_0$ be any element of $D_0$. Let $d_0$ consist of exactly as many $0$s as $|c_0|$, followed by a $1$, followed by $z(0)$, and then extended to an element of $D_0$. Continuing, $c_{n+1}$ extends $c_n$ by adding $0$s until the length of $d_n$, and then a $1$, and then extending into $D_{n+1}$; and $d_{n+1}$ extends $d_n$ by adding $0$s to the length of $c_{n+1}$, then a $1$, then $z(n)$, then extending into $D_{n+1}$. Let $c=\bigcup c_n$ and $d=\bigcup d_n$. Since we met all the dense sets in $W$, we know that $c$ and $d$ are $W$-generic Cohen reals, and so we may form the forcing extensions $W[c]$ and $W[d]$. But if $W\subset U\models\text{ZFC}$ and both $c$ and $d$ are in $U$, then in $U$ we may reconstruct the map $n\mapsto\langle c_n,d_n\rangle$, by giving attention to the blocks of $0$s in $c$ and $d$. From this map, we may reconstruct $z$ in $U$, which reveals all the ordinals of $W$ to be countable, a contradiction if $U$ and $W$ have the same ordinals. QED

Most of the results here concern forcing extensions of an arbitrary countable model of set theory, which of course includes the case of ill-founded models. Although there is no problem with forcing extensions of ill-founded models, when properly carried out, the reader may prefer to focus on the case of countable transitive models for the results in this section, and such a perspective will lose very little of the point of our observations.

The method of the observation above is easily generalized to produce three $W$-generic Cohen reals $c_0$, $c_1$ and $c_2$, such that any two of them can be amalgamated, but the three of them cannot. More generally:

Observation. If $W$ is a countable model of ZFC set theory, then for any finite $n$ there are $W$-generic Cohen reals $c_0,c_1,\ldots,c_{n-1}$, such that any proper subset of them are mutually $W$-generic, so that one may form the generic extension $W[\vec c]$, provided that $\vec c$ omits at least one $c_i$, but there is no forcing extension $W[G]$ simultaneously extending all $W[c_i]$ for $i<n$. In particular, the sequence $\langle c_0,c_1,\ldots,c_{n-1}\rangle$ cannot be added by forcing over $W$.

Let us turn now to infinite linearly ordered sequences of forcing extensions. We show first in the next theorem and subsequent observation that one mustn’t ask for too much; but nevertheless, after that we shall prove the surprising positive result, that any increasing sequence of forcing extensions over a countable model $W$, with forcing of uniformly bounded size, is bounded above by a single forcing extension $W[G]$.

Theorem. If $W$ is a countable model of ZFC, then there is an increasing sequence of set-forcing extensions of $W$ having no upper bound in the generic multiverse of $W$. $$W[G_0]\of W[G_1]\of\cdots\of W[G_n]\of\cdots$$

Proof. Since $W$ is countable, there is an increasing sequence $\langle\gamma_n\mid n<\omega\rangle$ of ordinals that is cofinal in the ordinals of $W$. Let $G_n$ be $W$-generic for the collapse forcing $\text{Coll}(\omega,\gamma_n)$, as defined in $W$. (By absorbing the smaller forcing, we may arrange that $W[G_n]$ contains $G_m$ for $m<n$.) Since every ordinal of $W$ is eventually collapsed, there can be no set-forcing extension of $W$, and indeed, no model with the same ordinals as $W$, that contains every $W[G_n]$. QED

But that was cheating, of course, since the sequence of forcing notions is not even definable in $W$, as the class $\{\gamma_n\mid n<\omega\}$ is not a class of $W$. A more intriguing question would be whether this phenomenon can occur with forcing notions that constitute a set in $W$, or (equivalently, actually) whether it can occur using always the same poset in $W$. For example, if $W[c_0]\of W[c_0][c_1]\of W[c_0][c_1][c_2]\of\cdots$ is an increasing sequence of generic extensions of $W$ by adding Cohen reals, then does it follow that there is a set-forcing extension $W[G]$ of $W$ with $W[c_0]\cdots[c_n]\of W[G]$ for every $n$? For this, we begin by showing that one mustn’t ask for too much:

Observation. If $W$ is a countable model of ZFC, then there is a sequence of forcing extensions $W\of W[c_0]\of W[c_0][c_1]\of W[c_0][c_1][c_2]\of\cdots$, adding a Cohen real at each step, such that there is no forcing extension of $W$ containing the sequence $\langle c_n\mid n<\omega\rangle$.

Proof. Let $\langle d_n\mid n<\omega\rangle$ be any $W$-generic sequence for the forcing to add $\omega$ many Cohen reals over $W$. Let $z$ be any real coding the ordinals of $W$. Let us view these reals as infinite binary sequences. Define the real $c_n$ to agree with $d_n$ on all digits except the initial digit, and set $c_n(0)=z(n)$. That is, we make a single-bit change to each $d_n$, so as to code one additional bit of $z$. Since we have made only finitely many changes to each $d_n$, it follows that $c_n$ is an $W$-generic Cohen real, and also $W[c_0]\cdots[c_n]=W[d_0]\cdots [d_n]$. Thus, we have $$W\of W[c_0]\of W[c_0][c_1]\of W[c_0][c_1][c_2]\of\cdots,$$ adding a generic Cohen real at each step. But there can be no forcing extension of $W$ containing $\langle c_n\mid n<\omega\rangle$, since any such extension would have the real $z$, revealing all the ordinals of $W$ to be countable. QED

We can modify the construction to allow $z$ to be $W$-generic, but collapsing some cardinals of $W$. For example, for any cardinal $\delta$ of $W$, we could let $z$ be $W$-generic for the collapse of $\delta$. Then, if we construct the sequence $\langle c_n\mid n<\omega\rangle$ as above, but inside $W[z]$, we get a sequence of Cohen real extensions $$W\of W[c_0]\of W[c_0][c_1]\of W[c_0][c_1][c_2]\of\cdots$$ such that $W[\langle c_n\mid n<\omega\rangle]=W[z]$, which collapses $\delta$.

But of course, the question of whether the models $W[c_0][c_1]\cdots[c_n]$ have an upper bound is not the same question as whether one can add the sequence $\langle c_n\mid n<\omega\rangle$, since an upper bound may not have this sequence. And in fact, this is exactly what occurs, and we have a surprising positive result:

Theorem. Suppose that $W$ is a countable model of \ZFC, and $$W[G_0]\of W[G_1]\of\cdots\of W[G_n]\of\cdots$$ is an increasing sequence of forcing extensions of $W$, with $G_n\of\Q_n\in W$ being $W$-generic. If the cardinalities of the $\Q_n$’s in $W$ are bounded in $W$, then there is a set-forcing extension $W[G]$ with $W[G_n]\of W[G]$ for all $n<\omega$.

Proof. Let us first make the argument in the special case that we have $$W\of W[g_0]\of W[g_0][g_1]\of\cdots\of W[g_0][g_1]\cdots[g_n]\of\cdots,$$ where each $g_n$ is generic over the prior model for forcing $\Q_n\in W$. That is, each extension $W[g_0][g_1]\cdots[g_n]$ is obtained by product forcing $\Q_0\cross\cdots\cross\Q_n$ over $W$, and the $g_n$ are mutually $W$-generic. Let $\delta$ be a regular cardinal with each $\Q_n$ having size at most $\delta$, built with underlying set a subset of $\delta$. In $W$, let $\theta=2^\delta$, let $\langle \R_\alpha\mid\alpha<\theta\rangle$ enumerate all posets of size at most $\delta$, with unbounded repetition, and let $\P=\prod_{\alpha<\theta}\R_\alpha$ be the finite-support product of these posets. Since each factor is $\delta^+$-c.c., it follows that the product is $\delta^+$-c.c. Since $W$ is countable, we may build a filter $H\of\P$ that is $W$-generic. In fact, we may find such a filter $H\of\P$ that meets every dense set in $\bigcup_{n<\omega}W[g_0][g_1]\cdots[g_n]$, since this union is also countable. In particular, $H$ and $g_0\cross\cdots\cross g_n$ are mutually $W$-generic for every $n<\omega$. The filter $H$ is determined by the filters $H_\alpha\of\R_\alpha$ that it adds at each coordinate.

Next comes the key step. Externally to $W$, we may find an increasing sequence $\langle \theta_n\mid n<\omega\rangle$ of ordinals cofinal in $\theta$, such that $\R_{\theta_n}=\Q_n$. This is possible because the posets are repeated unboundedly, and $\theta$ is countable in $V$. Let us modify the filter $H$ by surgery to produce a new filter $H^*$, by changing $H$ at the coordinates $\theta_n$ to use $g_n$ rather than $H_{\theta_n}$. That is, let $H^*_{\theta_n}=g_n$ and otherwise $H^*_\alpha=H_\alpha$, for $\alpha\notin\{\theta_n\mid n<\omega\}$. It is clear that $H^*$ is still a filter on $\P$. We claim that $H^*$ is $W$-generic. To see this, suppose that $A\of\P$ is any maximal antichain in $W$. By the $\delta^+$-chain condition and the fact that $\text{cof}(\theta)^W>\delta$, it follows that the conditions in $A$ have support bounded by some $\gamma<\theta$. Since the $\theta_n$ are increasing and cofinal in $\theta$, only finitely many of them lay below $\gamma$, and we may suppose that there is some largest $\theta_m$ below $\gamma$. Let $H^{**}$ be the filter derived from $H$ by performing the surgical modifications only on the coordinates $\theta_0,\ldots,\theta_m$. Thus, $H^*$ and $H^{**}$ agree on all coordinates below $\gamma$. By construction, we had ensured that $H$ and $g_0\cross\cdots\cross g_m$ are mutually generic over $W$ for the forcing $\P\cross\Q_0\cross\cdots\cross\Q_m$. This poset has an automorphism swapping the latter copies of $\Q_i$ with their copy at $\theta_i$ in $\P$, and this automorphism takes the $W$-generic filter $H\cross g_0\cross\cdots\cross g_m$ exactly to $H^{**}\cross H_{\theta_0}\cross\cdots \cross H_{\theta_m}$. In particular, $H^{**}$ is $W$-generic for $\P$, and so $H^{**}$ meets the maximal antichain $A$. Since $H^*$ and $H^{**}$ agree at coordinates below $\gamma$, it follows that $H^*$ also meets $A$. In summary, we have proved that $H^*$ is $W$-generic for $\P$, and so $W[H^*]$ is a set-forcing extension of $W$. By design, each $g_n$ appears at coordinate $\theta_n$ in $H^*$, and so $W[g_0]\cdots[g_n]\of W[H^*]$ for every $n<\omega$, as desired.

Finally, we reduce the general case to this special case. Suppose that $W[G_0]\of W[G_1]\of\cdots\of W[G_n]\of\cdots$ is an increasing sequence of forcing extensions of $W$, with $G_n\of\Q_n\in W$ being $W$-generic and each $\Q_n$ of size at most $\kappa$ in $W$. By the standard facts surrounding finite iterated forcing, we may view each model as a forcing extension of the previous model $$W[G_{n+1}]=W[G_n][H_n],$$ where $H_n$ is $W[G_n]$-generic for the corresponding quotient forcing $\Q_n/G_n$ in $W[G_n]$. Let $g\of\text{Coll}(\omega,\kappa)$ be $\bigcup_n W[G_n]$-generic for the collapse of $\kappa$, so that it is mutually generic with every $G_n$. Thus, we have the increasing sequence of extensions $W[g][G_0]\of W[g][G_1]\of\cdots$, where we have added $g$ to each model. Since each $\Q_n$ is countable in $W[g]$, it is forcing equivalent there to the forcing to add a Cohen real. Furthermore, the quotient forcing $\Q_n/G_n$ is also forcing equivalent in $W[g][G_n]$ to adding a Cohen real. Thus, $W[g][G_{n+1}]=W[g][G_n][H_n]=W[g][G_n][h_n]$, for some $W[g][G_n]$-generic Cohen real $h_n$. Unwrapping this recursion, we have $W[g][G_{n+1}]=W[g][G_0][h_1]\cdots[h_n]$, and consequently $$W[g]\of W[g][G_0]\of W[g][G_0][h_1]\of W[g][G_0][h_1][h_2]\of\cdots,$$ which places us into the first case of the proof, since this is now product forcing rather than iterated forcing. QED

Definition. A collection $\{W[G_n]\mid n<\omega\}$ of forcing extensions of $W$ is finitely amalgamable over $W$ if for every $n<\omega$ there is a forcing extension $W[H]$ with $W[G_m]\of W[H]$ for all $m\leq n$. It is amalgamable over $W$ if there is $W[H]$ such that $W[G_n]\of W[H]$ for all $n<\omega$.

The next corollary shows that we cannot improve the non-amalgamability result of the initial observation to the case of infinitely many Cohen reals, with all finite subsets amalgamable.

Corollary. If $W$ is a countable model of ZFC and $\{W[G_n]\mid n<\omega\}$ is a finitely amalgamable collection of forcing extensions of $W$, using forcing of bounded size in $W$, then this collection is fully amalgamable. That is, there is a forcing extension $W[H]$ with $W[G_n]\of W[H]$ for all $n<\omega$.

Proof. Since the collection is finitely amalgamable, for each $n<\omega$ there is some $W$-generic $K$ such that $W[G_m]\of W[K]$ for all $m\leq n$. Thus, we may form the minimal model $W[G_0][G_1]\cdots[G_n]$ between $W$ and $W[K]$, and thus $W[G_0][G_1]\cdots [G_n]$ is a forcing extension of $W$. We are thus in the situation of the theorem, with an increasing chain of forcing extensions. $$W\of W[G_0]\of W[G_0][G_1]\of\cdots\of W[G_0][G_1]\cdots[G_n]\of\cdots$$ Therefore, by the theorem, there is a model $W[H]$ containing all these extensions, and in particular, $W[G_n]\of W[H]$, as desired. QED

Please go to the paper for more details and discussion.

Transfinite recursion as a fundamental principle in set theory

InfinityAt the Midwest PhilMath Workshop this past weekend, I heard Benjamin Rin (UC Irvine) speak on transfinite recursion, with an interesting new perspective.  His idea was to consider transfinite recursion as a basic principle in set theory, along with its close relatives, and see how they relate to the other axioms of set theory, such as the replacement axiom. In particular, he had the idea of using our intuitions about the legitimacy of transfinite computational processes as providing a philosophical foundation for the replacement axiom.

This post is based on what I learned about Rin’s work from his talk at the workshop and in our subsequent conversations there about it.  Meanwhile, his paper is now available online:

Benjamin Rin, Transfinite recursion and the iterative conception of set, Synthese, October, 2014, p. 1-26. (preprint).

Since I have a little different perspective on the proposal than Rin did, I thought I would like to explain here how I look upon it.  Everything I say here is inspired by Rin’s work.

To begin, I propose that we consider the following axiom, asserting that we may undertake a transfinite recursive procedure along any given well-ordering.

The Principle of Transfinite Recursion. If $A$ is any set with well-ordering $<$ and $F:V\to V$ is any class function, then there is a function $s:A\to V$ such that $s(b)=F(s\upharpoonright b)$ for all $b\in A$, where $s\upharpoonright b$ denotes the function $\langle s(a)\mid a<b\rangle$.

We may understand this principle as an infinite scheme of statements in the first-order language of set theory, where we make separate assertions for each possible first-order formula defining the class function $F$, allowing parameters. It seems natural to consider the principle in the background theory of first-order Zermelo set theory Z, or the Zermelo theory ZC, which includes the axiom of choice, and in each case let me also include the axiom of foundation, which apparently is not usually included in Z.   (Alternatively, it is also natural to consider the principle as a single second-order statement, if one wants to work in second-order set theory.)

Theorem. (ZC) The principle of transfinite recursion is equivalent to the axiom of replacement. In other words,

ZC + transfinite recursion  =  ZFC.

Proof. Work in the Zermelo set theory ZC. The converse implication amounts to the well-known observation in ZF that transfinite recursion is legitimate. Let us quickly sketch the argument. Suppose we are given an instance of transfinite recursion, namely, a well-ordering $\langle A,<\rangle$ and a class function $F:V\to
V$. I claim that for every $b\in A$, there is a unique function $s:\{a\in A\mid a\leq b\}\to V$ obeying the recursive rule $s(d)=F(s\upharpoonright d)$ for all $d\leq b$. The reason is that there can be no least $b$ without such a unique function. If all $a<b$ have such a unique function, then by uniqueness they must cohere with one another, since any difference would show up at a least stage and thereby violate the recursion rule, and so by the replacement axiom of ZFC we may assemble these smaller functions into a single function $t$ defined on all $a<b$, and satisfying the recursion rule for those values. We may then extend this function $t$ to be defined on $b$ itself, simply by defining $u(b)=F(t)$ and $u\upharpoonright b=t$, which thereby satisfies the recursion at $b$. Uniqueness again follows from the fact that there can be no least place of disagreement. Finally, using replacement again, let $s(b)$ be the unique value that arises at $b$ during the recursions that work up to and including $b$, and this function $s:A\to V$ satisfies the recursive definition.

Conversely, assume the Zermelo theory ZC plus the principle of transfinite recursion, and suppose that we are faced with an instance of the replacement axiom. That is, we have a set $A$ and a formula $\varphi$, where every $b\in A$ has a unique $y$ such that $\varphi(b,y)$. By the axiom of choice, there is a well-ordering $<$ of the set $A$. We shall now define the function $F:V\to V$. Given a function $s$, where $\dom(s)=\{a\in A\mid a<b\}$ for some $b\in A$, let $F(s)=y$ be the unique $y$ such that $\varphi(b,y)$; and otherwise let $F(s)$ be anything you like. By the principle of transfinite recursion, there is a function $s:A\to V$ such that $s(b)=F(s\upharpoonright b)$ for every $b\in A$. In this case, it follows that $s(b)$ is the unique $y$ such that $\varphi(a,b)$. Thus, since $s$ is a set, it follows in ZC that $\ran(s)$ is a set, and so we’ve got the image of $A$ under $\varphi$ as a set, which verifies replacement. QED

In particular, it follows that the principle of transfinite recursion implies that every well-ordering is isomorphic to a von Neumann ordinal, a principle Rin refers to as ordinal abstraction. One can see this as a consequence of the previous theorem, since ordinal abstraction holds in ZF by Mostowski’s theorem, which for any well-order $\langle A,<\rangle$ assigns an ordinal to each node $a\mapsto \alpha_a$ according to the recursive rule $\alpha_a=\{\alpha_b\mid b<a\}$. But one can also argue directly as follows, without using the axiom of choice. Assume Z and the principle of transfinite recursion. Suppose that $\langle A,<\rangle$ is a well-ordering. Define the class function $F:V\to V$ so that $F(s)=\ran(s)$, whenever $s$ is a function. By the principle of transfinite recursion, there is a function $s:A\to V$ such that $s(b)=F(s\restrict b)=\ran(s\restrict b)$. One can now simply prove by induction that $s(b)$ is an ordinal and $s$ is an isomorphism of $\langle A,<\rangle$ with $\ran(s)$, which is an ordinal.

Let me remark that the principle of transfinite recursion allows us also to perform proper-class length recursions.

Observation. Assume Zermelo set theory Z plus the principle of transfinite recursion. If $A$ is any particular class with $<$ a set-like well-ordering of $A$ and $F:V\to V$ is any class function, then there is a class function $S:A\to V$ such that $S(b)=F(S\upharpoonright b)$ for every $b\in A$.

Proof. Since $\langle A,<\rangle$ is set-like, the initial segment $A\upharpoonright d=\{a\in A\mid a<d\}$, for any particular $d\in A$, is a set. It follows that the principle of transfinite recursion shows that there is a function $s_d:(A\upharpoonright d)\to V$ such that $s_d(b)=F(s_d\upharpoonright b)$ for every $b<d$. It is now easy to prove by induction that these $s_d$ must all cohere with one another, and so we may define the class $S(b)=s_d(b)$ for any $d$ above $b$ in $A$. (We may assume without loss that $A$ has no largest element, for otherwise it would be a set.) This provides a class function $S:A\to V$ satisfying the recursive definition as desired. QED

Although it appears explicitly as a second-order statement “there is a class function $S$…”, we may actually take this observation as a first-order theorem scheme, if we simply strengthen the conclusion to provide the explicit definition of $S$ that the proof provides. That is, the proof shows exactly how to define $S$, and if we make the observation state that that particular definition works, then what we have is a first-order theorem scheme. So any first-order definition of $A$ and $F$ from parameters leads uniformly to a first-order definition of $S$ using the same parameters.

Thus, using the principle of transfinite recursion, we may also take proper class length transfinite recursions, using any set-like well-ordered class that we happen to have available.

Let us now consider a weakening of the principle of transfinite recursion, where we do not use arbitrary well-orderings, but only the von Neumann ordinals themselves.

Principle of transfinite recursion on ordinals. If $F:V\to V$ is any class function, then for any ordinal $\gamma$ there is a function $s:\gamma\to V$ such that $s(\beta)=F(s\upharpoonright\beta)$ for all $\beta<\gamma$.

This is a weakening of the principle of transfinite recursion, since every ordinal is well-ordered, but in Zermelo set theory, not every well-ordering is necessarily isomorphic to an ordinal. Nevertheless, in the presence of ordinal abstraction, then this ordinal version of transfinite recursion is clearly equivalent to the full principle of transfinite recursion.

Observation. Work in Z. If every well-ordering is isomorphic to an ordinal, then the principle of transfinite recursion is equivalent to its restriction to ordinals.

Meanwhile, let me observe that in general, one may not recover the full principle of transfinite recursion from the weaker principle, where one uses it only on ordinals.

Theorem. (ZFC) The structure $\langle V_{\omega_1},\in\rangle$ satisfies Zermelo set theory ZC with the axiom of choice, but does not satisfy the principle of transfinite recursion. Nevertheless, it does satisfy the principle of transfinite recursion on ordinals.

Proof. It is easy to verify all the Zermelo axioms in $V_{\omega_1}$, as well as the axiom of choice, provided choice holds in $V$. Notice that there are comparatively few ordinals in $V_{\omega_1}$—only the countable ordinals exist there—but $V_{\omega_1}$ has much larger well-orderings. For example, one may find a well-ordering of the reals already in $V_{\omega+k}$ for small finite $k$, and well-orderings of much larger sets in $V_{\omega^2+17}$ and so on as one ascends toward $V_{\omega_1}$. So $V_{\omega_1}$ does not satisfy the ordinal abstraction principle and so cannot satisfy replacement or the principle of transfinite recursion. But I claim nevertheless that it does satisfy the weaker principle of transfinite recursion on ordinals, because if $F:V_{\omega_1}\to V_{\omega_1}$ is any class in this structure, and $\gamma$ is any ordinal, then we may define by recursion in $V$ the function $s(\beta)=F(s\restrict\beta)$, which gives a class $s:\omega_1\to V_{\omega_1}$ that is amenable in $V_{\omega_1}$. In particular, $s\restrict\gamma\in V_{\omega_1}$ for any $\gamma<\omega_1$, simply because $\gamma$ is countable and $\omega_1$ is regular. QED

My view is that this example shows that one doesn’t really want to consider the weakened principle of transfinite recursion on ordinals, if one is working in the Zermelo background ZC, simply because there could be comparatively few ordinals, and this imposes an essentially arbitrary limitation on the principle.

Let me point out, however, that there was a reason we had to go to $V_{\omega_1}$, rather than considering $V_{\omega+\omega}$, which is a more-often mentioned model of the Zermelo axioms. It is not difficult to see that $V_{\omega+\omega}$ does not satisfy the principle of transfinite recursion on the ordinals, because one can define the function $s(n)=\omega+n$ by recursion, setting $s(0)=\omega$ and $s(n+1)=s(n)+1$, but this function does not exist in $V_{\omega+\omega}$. This feature can be generalized as follows:

Theorem. Work in the Zermelo set theory Z. The principle of transfinite recursion on ordinals implies that if $\langle A,<\rangle$ is a well-ordered set, and $A$ is bijective with some ordinal, then $\langle A,<\rangle$ is order-isomorphic with an ordinal.

In other words, we get ordinal abstraction for well-orderings whose underlying set is bijective with an ordinal.

First, the proof of the first theorem above actually shows the following local version:

Lemma. (Z) If one has the principle of transfinite recursion with respect to a well-ordering $\langle A,<\rangle$, then $A$-replacement holds, meaning that if $F:V\to V$ is any class function, then the image $F”A$ is a set.

Proof of theorem. Suppose that $\langle A,<\rangle$ is a well-ordering, and that $A$ is bijective with some ordinal $\kappa$, and that $F:V\to V$ is a class function. Assume the principle of transfinite recursion for $\kappa$. We prove by induction on $d\in A$ that there is a unique function $s_d$ with $\dom(s)=\{a\in A\mid a\leq d\}$ and satisfying the recursive rule $s(b)=F(s\upharpoonright b)$. If this statement is true for all $d<d’$, then because the size of the predecessors of $d’$ in $\langle A,<\rangle$ is at most $\kappa$, we may by the lemma form the set $\{s_d\mid d<d’\}$, which is a set by $\kappa$-replacement. These functions cohere, and the union of these functions gives a function $t:(A\upharpoonright d’)\to V$ satisfying the recursion rule for $F$. Now extend this function one more step by defining $s(d’)=F(t)$ and $s\upharpoonright d’=t$, thereby handling the existence claim at $d’$. As in the main theorem, all these functions cohere with one another, and by $\kappa$-replacement we may form the set $\{s_d\mid d\in A\}$, whose union is the desired function $s:A\to V$ satisfying the recursion rule given by $F$, as desired. QED

For example, if you have the principle of transfinite recursion for ordinals, and $\omega$ exists, then every countable well-ordering is isomorphic to an ordinal. This explains why we had to go to $\omega_1$ to find a model satisfying transfinite recursion on ordinals. One can understand the previous theorem as showing that although the principle of transfinite recursion on ordinals does not prove ordinal abstraction, it does prove many instances of it: for every ordinal $\kappa$, every well-ordering of cardinality at most $\kappa$ is isomorphic to an ordinal.

It is natural also to consider the principle of transfinite recursion along a well-founded relation, rather than merely a well-ordered relation.

The principle of well-founded recursion. If $\langle A,\lhd\rangle$ is a well-founded relation and $F:V\to V$ is any class function, then there is a function $s:A\to V$ such that $s(b)=F(s\restrict b)$ for all $b\in A$, where $s\restrict b$ means the function $s$ restricted to the domain of elements $a\in A$ that are hereditarily below $b$ with respect to $\lhd$.

Although this principle may seem more powerful, in fact it is equivalent to transfinite recursion.

Theorem. (ZC) The principle of transfinite recursion is equivalent to the principle of well-founded recursion.

Proof. The backward direction is immediate, since well-orders are well-founded. For the forward implication, assume that transfinite recursion is legitimate. It follows by the main theorem above that ZFC holds. In this case, well-founded recursion is legitimate by the familiar arguments. For example, one may prove in ZFC that for every node in the field of the relation, there is a unique solution of the recursion defined up to and including that node, simply because there can be no minimal node without this property.  Then, by replacement, one may assemble all these functions together into a global solution. Alternatively, arguing directly from transfinite recursion, one may put an ordinal ranking function for any given well-founded relation $\langle A,\lhd\rangle$, and then prove by induction on this rank that one may construct functions defined up to and including any given rank, that accord with the recursive rule. In this way, one gets the full function $s:A\to V$ satisfying the recursive rule. QED

Finally, let me conclude this post by pointing out how my perspective on this topic differs from the treatment given by Benjamin Rin. I am grateful to him for his idea, which I find extremely interesting, and as I said, everything here is inspired by his work.

One difference is that Rin mainly considered transfinite recursion only on ordinals, rather than with respect to an arbitrary well-ordered relation (but see footnote 17 in his paper). For this reason, he had a greater need to consider whether or not he had sufficient ordinal abstraction in his applications. My perspective is that transfinite recursion, taken as a basic principle, has nothing fundamentally to do with the von Neumann ordinals, but rather has to do with a general process undertaken along any well-order. And the theory seems to work better when one undertakes it that way.

Another difference is that Rin stated his recursion principle as a principle about iterating through all the ordinals, rather than only up to any given ordinal. This made the resulting functions $S:\text{Ord}\to V$ into class-sized objects for him, and moved the whole analysis into the realm of second-order set theory. This is why he was led to prove his main equivalence with replacement in second-order Zermelo set theory. My treatment shows that one may undertake the whole theory in first-order set theory, without losing the class-length iterations, since as I explained above the class-length iterations form classes, definable from the original class functions and well-orders. And given that a completely first-order account is possible, it seems preferable to undertake it that way.

MathOverflow, the eternal fountain of mathematics: reflections on a hundred kiloreps

profile for Joel David Hamkins at MathOverflow, Q&A for professional mathematiciansIt seems to appear that I have somehow managed to pass  the 100,000 score milestone for reputation on MathOverflow.  A hundred kiloreps!  Does this qualify me for micro-celebrity status?  I have clearly been spending an inordinate amount of time on MO…  Truly, it has been a great time.

MathOverflow, an eternal fountain of mathematics, overflows with fascinating questions and answers on every imaginable mathematical topic, drawing unforeseen connections, seeking generalizations, clarification, or illustrative examples, questioning assumptions, or simply asking for an explanation of a subtle mathematical point.  The mathematics is sophisticated and compelling.  How could a mathematician not immediately plunge in?

I first joined MathOverflow in November 2009, when my colleague-down-the-hall Kevin O’Bryant dropped into my office and showed me the site.  He said that it was for “people like us,” research mathematicians who wanted to discuss mathematical issues with other professionals, and he was completely right.  Looking at the site, I found Greg Kuperberg’s answer to a question on the automorphism tower problem in group theory, which was one of the first extremely popular questions at that time, the top-rated question.  I was hooked immediately, and I told Kevin on that very first day that it was clear that MathOverflow was going to take a lot of time.

I was pleased to find right from the beginning that, although there were not yet many logicians participating on MO, there were nevertheless many logic questions, revealing an unexpectedly broad interest in math logic issues amongst the general mathematical community.  I found questions about definability, computability, undecidability, logical independence, about the continuum hypothesis and the axiom of choice and about large cardinals, asked by mathematicians in diverse research areas, who seemed earnestly to want to know the answer.  How pleased I was to find such a level of interest in the same issues that fascinated me; and how pleased I was also to find that I was often able to answer.

In the early days, I may have felt a little that I should be a kind of ambassador for logic, introducing the subject or aspects of it to those who might not know all about it yet; for example, in a few answers I explained and introduced the topic of cardinal characteristics of the continuum and the subject of Borel equivalence relation theory, since I had felt that mathematicians outside logic might not necessarily know much about it, even when it offered connections to things they did know about.  I probably wouldn’t necessarily answer the same way today, now that MO has many experts in those subjects and a robust logic community.  What a pleasure it has become.

A while back I wrote a post The use and value of MathOverflow in response to an inquiry of François Dorais, and I find the remarks I made then are as true for me today as ever.

I feel that mathoverflow has enlarged me as a mathematician.  I have learned a huge amount here in the past few years, particularly concerning how my subject relates to other parts of mathematics.  I’ve read some really great answers that opened up new perspectives for me.  But just as importantly, I’ve learned a lot when coming up with my own answers.  It often happens that someone asks a question in another part of mathematics that I can see at bottom has to do with how something I know about relates to their area, and so in order to answer, I must learn enough about this other subject in order to see the connection through.  How fulfilling it is when a question that is originally opaque to me, because I hadn’t known enough about this other topic, becomes clear enough for me to have an answer.  Meanwhile, mathoverflow has also helped me to solidify my knowledge of my own research area, often through the exercise of writing up a clear summary account of a familiar mathematical issue or by thinking about issues arising in a question concerning confusing or difficult aspects of a familiar tool or method.

Mathoverflow has also taught me a lot about good mathematical exposition, both by the example of other’s high quality writing and by the immediate feedback we all get on our posts.  This feedback reveals what kind of mathematical explanation is valued by the general mathematical community, in a direct way that one does not usually get so well when writing a paper or giving a conference talk.  This kind of knowledge has helped me to improve my mathematical writing in general.

Thanks very much again, MathOverflow!  I am grateful.

A few posts come to mind:

There have been so many more great questions and posts.  If you are inclined, feel free to post comments below linking to your favorite MO posts!

Concerning the MO reputation system, I suppose some might suspect me of harboring unnatural thoughts on reputation — after all, I once proposed (I can’t find the link now) that the sole basis of tenure and promotion decisions for mathematics faculty, as well as choice of premium office space, should be:  MO reputation, ha! — but in truth, I look upon it all as a good silly game.  One may take reputation as seriously as one takes any game seriously, and many mathematicians can indeed take a game seriously.  My honest opinion is that the reputation and badge system is an ingenious piece of social engineering.  The designers must have had a good grasp on human psychology, an understanding of the kinds of reasons that might motivate a person to participate in such a site; one thinks, for example, of the intermittent reward theory.  I find it really amazing what the stackexchange designers have created, and who doesn’t love a good game?

Announcement on History of MathOverflow

Does definiteness-of-truth follow from definiteness-of-objects? NY Philosophical Logic Group, NYU, November 2014

This will be a talk for the New York Philosophical Logic Group, November 10, 2014, 5-7pm, at the NYU Philosophy Department, 5 Washington Place, Room 302.

Indefinite arithmetic truth

Abstract. This talk — a mix of mathematics and philosophy — concerns the extent to which we may infer definiteness of truth in a mathematical context from definiteness of the underlying objects and structure of that context. The philosophical analysis is based in part on the mathematical observation that the satisfaction relation for model-theoretic truth is less absolute than often supposed.  Specifically, two models of set theory can have the same natural numbers and the same structure of arithmetic in common, yet disagree about whether a particular arithmetic sentence is true in that structure. In other words, two models can have the same arithmetic objects and the same formulas and sentences in the language of arithmetic, yet disagree on their corresponding theories of truth for those objects. Similarly, two models of set theory can have the same natural numbers, the same arithmetic structure, and the same arithmetic truth, yet disagree on their truths-about-truth, and so on at any desired level of the iterated truth-predicate hierarchy.  These mathematical observations, for which I shall strive to give a very gentle proof in the talk (using only elementary classical methods), suggest that a philosophical commitment to the determinate nature of the theory of truth for a structure cannot be seen as a consequence solely of the determinateness of the structure in which that truth resides. The determinate nature of arithmetic truth, for example, is not a consequence of the determinate nature of the arithmetic structure N = {0,1,2,…} itself, but rather seems to be an additional higher-order commitment requiring its own analysis and justification.

This work is based on my recent paper, Satisfaction is not absolute, joint with Ruizhi Yang (Fudan University, Shanghai).

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.

My research collaborators

CollaborationI have been very fortunate in my research to have had the opportunity to work closely with a number of insightful researchers. I’ve learned a great deal from them, and I’m truly grateful.

So I’ve gathered here a list of my collaborators. In almost all cases, the collaboration resulted in a published joint research article, which you can find on my list of publications (in a few instances, for collaborations currently underway, a paper is not necessarily yet available).  Several of my collaborations have been sustained long-term affairs, leading to a series of joint publications on various topics over several years.  Naturally, I am hopeful that all my collaborations will continue to be fruitful for many years into the future.

The rule-making game

They said a king once ruled the forest, by Lizzie ThomasLet me tell you about a new game that we’ve been playing in our family, the rule-making game.  It is a talking game, requiring no pieces or objects of any kind, and it can easily be played whilst walking or traveling.  My children and I recently played several rounds of it walking around London on a recent visit there.

The game has no rules, initially, nor even any definite procedure — it is different every time — but things usually become clear soon enough.  It usually makes a better game to cooperate on the first several turns to lay the groundwork.

Let me explain how to play simply by example:

Papa:  The first rule is that the players shall take turns making rules, and that every rule shall have a rule number, which is incremented on each turn.

Horatio:  The second rule is that the players must state their rules in the form, “The first rule is…” or “the second rule is…” and so on, and that players are not allowed to ask what is the current rule number, or they lose.

Hypatia:  The third rule is that the other players must say, “thank you” after another player makes a rule.

     (… “thank you”…. “thank you”….)

Papa: The fourth rule is that the rules must not contradict each other, and no rule is allowed that abrogates an earlier rule.

     (… “thank you”…. “thank you”….)

Horatio:  The fifth rule is that after making an odd-numbered rule, the player must stomp on the ground.

     (STOMP… “thank you”…. “thank you”….)

Hypatia: The sixth rule is that no player may win immediately after their own rule.

     (… “thank you”…. “thank you”….)

Papa:  The seventh rule is that right after a player stomps according to rule five, the other two players must hop.

     (STOMP … “thank you”…. “thank you”….HOP….HOP…)

Horatio:  The eighth rule is that if a player loses, then the game continues without that person.

     (… “thank you”…. “thank you”….)

Hypatia: The ninth rule is that after stating a rule, the other two players must state a different color.

     (STOMP … “thank you”…. “thank you”….HOP…HOP… “blue”… “green”…)

Papa:  The tenth rule is that furthermore, those colors must never repeat, and they must be stated simultaneously, on the count of 1-2-3.

     (… “thank you”…. “thank you”…. “1-2-3: neon green / violet”)

Horatio: The eleventh rule is that if there is only one player left, then that player wins.

     (STOMP … “thank you”…. “thank you”….HOP…HOP… “1-2-3: red/orange”)

Hypatia:  The twelfth rule is that every player must jump up and down (…jump…) while stating their rule. (….jump jump jump…)

     (… “thank you”…. “thank you”…. “1-2-3: pink/turquoise”)

Papa: (jump jump…) The thirteenth rule is that (…jump…) in the case of dispute (…jump…), the question of whether or not someone has violated or followed a rule shall be decided by majority vote (…jump…).

     (STOMP … “thank you”…. “thank you”….HOP…HOP… “1-2-3: yellow/brown”)

Horatio: (jump….) The fourteenth rule is that (…jump…) before stating their rule, the players must state a country, and that whoever repeats a country loses (…jump…)

     (… “thank you”…. “thank you”…. “1-2-3: black/gray”)

Hypatia:  (jump…)  Germany.  The fifteenth rule is that (…jump…) there can be at most twenty-five rules.

(STOMP … “thank you”…. “thank you”….HOP…HOP… “1-2-3: sky blue / peach”)

Papa:  (jump…)  United States.  The sixteenth rule is that (…jump…) if all current players lose at the same time after a rule, then the player previous to that rule-maker is declared the “honorary winner”.  (…jump…)

(… “thank you”…. “thank you”…. “1-2-3: white / white”)

Oh no! Since both Horatio and Hypatia said “white”, they both lose.  And then Papa also loses in light of rule six. So we’ve all lost!  But then, in light of rule sixteen, Hypatia is declared the honorary winner! Hooray for Hypatia!

I hope you all get the idea.  Please enjoy!  And report your crazy or interesting rules in the comments below.

The theory of infinite games: how to play infinite chess and win, VCU Math Colloquium, November 2014

Releasing the hordesI shall speak at the Virginia Commonwealth University Math Colloquium on November 21, 2014.

Abstract. I shall give a general introduction to the theory of infinite games, using infinite chess—chess played on an infinite chessboard stretching without bound in every direction—as a central example. Since chess, when won, is always won at a finite stage of play, infinite chess is an example of what is known technically as an open game, and such games admit the theory of transfinite ordinal game values, which provide a measure in a position of the distance remaining to victory. I shall exhibit several interesting positions in infinite chess with very high transfinite ordinal game values. Some of these positions involve large numbers of pieces, and the talk will include animations of infinite chess in play, with hundreds of pieces (or infinitely many) making coordinated attacks on the board. Meanwhile, the precise ordinal value of the omega one of chess is an open mathematical question.

Slides | Transfinite game values in infinite chess | The mate-in-n problem of infinite chess is decidable

The span of infinity, roundtable discussion at The Helix Center, October 2014

I shall be a panelist at The Span of Infinity, a roundtable discussion held at The Helix Center, at the New York Psychoanalytic Society & Institute, 247 E 82nd Street, on October 25, 2014, 2:30 – 4:30 pm.

The Helix Center describes the discussion topic as:

Perhaps no thing conceived in the mind has enjoyed a greater confluence of cosmological, mathematical, philosophical, psychological, and theological inquiry than the notion of the infinite. The epistemological tension between the concrete and the ideal, between the phenomenological and the ontological, is nowhere clearer in outline yet more obscure in content. These inherent paradoxes limn the vital, eternal questions we will explore about humankind’s place in the universe and the comprehensibility of existence.

The Helix Center Roundtable Series is described by:

Our roundtable format is designated the Theaetetus Table, an extempore discussion among five participants, all leaders in their respective fields, and named for the classical Greek mathematician and eponym for the Platonic dialogue investigating the nature of knowledge, who proved that there are five regular convex polyhedra, or Platonic solids. Each Theaetetus Table aspires to emulate the dialogue’s unhurried search for wisdom; and, like the five Platonic solids held to be the fundamental building blocks of the classical elements, the contributions of our five participants become the fundamental constituents of interdisciplinary insights emerging in the alchemy of the roundtable, insights that, in turn, transform the elemental thinking of those participants. The gathering of five discussants also symbolizes the five interrelated qualities of mind our interdisciplinary forums are intended to facilitate in our participants, and inculcate in our audience: curiosity, playfulness, inspiration, reflection, and wonder.

The pluralist perspective on the axiom of constructibility, MidWest PhilMath Workshop, Notre Dame, October 2014

University of Notre DameThis will be a featured talk at the Midwest PhilMath Workshop 15, held at Notre Dame University October 18-19, 2014.  W. Hugh Woodin and I will each give one-hour talks in a session on Perspectives on the foundations of set theory, followed by a one-hour discussion of our talks.

Abstract. I shall argue that the commonly held $V\neq L$ via maximize position, which rejects the axiom of constructibility V = L on the basis that it is restrictive, implicitly takes a stand in the pluralist debate in the philosophy of set theory by presuming an absolute background concept of ordinal. The argument appears to lose its force, in contrast, on an upwardly extensible concept of set, in light of the various facts showing that models of set theory generally have extensions to models of V = L inside larger set-theoretic universes.

Set-theorists often argue against the axiom of constructibility V=L on the grounds that it is restrictive, that we have no reason to suppose that every set should be constructible and that it places an artificial limitation on set-theoretic possibility to suppose that every set is constructible. Penelope Maddy, in her work on naturalism in mathematics, sought to explain this perspective by means of the MAXIMIZE principle, and further to give substance to the concept of what it means for a theory to be restrictive, as a purely formal property of the theory. In this talk, I shall criticize Maddy’s proposal, pointing out that neither the fairly-interpreted-in relation nor the (strongly) maximizes-over relation is transitive, and furthermore, the theory ZFC + `there is a proper class of inaccessible cardinals’ is formally restrictive on Maddy’s account, contrary to what had been desired. Ultimately, I shall argue that the V≠L via maximize position loses its force on a multiverse conception of set theory with an upwardly extensible concept of set, in light of the classical facts that models of set theory can generally be extended to models of V=L. I shall conclude the talk by explaining various senses in which V=L remains compatible with strength in set theory.

This talk will be based on my paper, A multiverse perspective on the axiom of constructibility.


Large cardinals need not be large in HOD

  • Y. Cheng, S. Friedman, and J. D. Hamkins, “Large cardinals need not be large in HOD.” (under review)  
    author = {Yong Cheng and Sy-David Friedman and Joel David Hamkins},
    title = {Large cardinals need not be large in {HOD}},
    journal = {},
    year = {},
    volume = {},
    number = {},
    pages = {},
    month = {},
    note = {under review},
    eprint = {1407.6335},
    url = {\url{}},
    abstract = {},
    keywords = {},
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Abstract. We prove that large cardinals need not generally exhibit their large cardinal nature in HOD. For example, a supercompact cardinal $\kappa$ need not be weakly compact in HOD, and there can be a proper class of supercompact cardinals in $V$, none of them weakly compact in HOD, with no supercompact cardinals in HOD. Similar results hold for many other types of large cardinals, such as measurable and strong cardinals.

In this article, we prove that large cardinals need not generally exhibit their large cardinal nature in HOD, the inner model of hereditarily ordinal-definable sets, and there can be a divergence in strength between the large cardinals of the ambient set-theoretic universe $V$ and those of HOD. Our general theme concerns the questions:


1. To what extent must a large cardinal in $V$ exhibit its large cardinal properties in HOD?

2. To what extent does the existence of large cardinals in $V$ imply the existence of large cardinals in HOD?

For large cardinal concepts beyond the weakest notions, we prove, the answers are generally negative. In Theorem 4, for example, we construct a model with a supercompact cardinal that is not weakly compact in HOD, and Theorem 9 extends this to a proper class of supercompact cardinals, none of which is weakly compact in HOD, thereby providing some strongly negative instances of (1). The same model has a proper class of supercompact cardinals, but no supercompact cardinals in HOD, providing a negative instance of (2). The natural common strengthening of these situations would be a model with a proper class of supercompact cardinals, but no weakly compact cardinals in HOD. We were not able to arrange that situation, however, and furthermore it would be ruled out by Conjecture 13, an intriguing positive instance of (2) recently proposed by W. Hugh Woodin, namely, that if there is a supercompact cardinal, then there is a measurable cardinal in HOD. Many other natural possibilities, such as a proper class of measurable cardinals with no weakly compact cardinals in HOD, remain as open questions.

CUNY talkRutgers talk | Luminy talk

A common forcing extension obtained via different forcing notions

I’d like to write about the situation that occurs in set theory when a forcing extension $V[G]=V[H]$ arises over a ground model $V$ in two different ways simultaneously, using generic filters over two different forcing notions $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$. The general fact, stated in theorem 1, is that in this case, the two forcing notions are actually isomorphic on a cone $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, with the isomorphism carrying the one generic filter to the other. In other words, below these respective conditions $b$ and $c$, the forcing notions and the respective generic filters are not actually different.

I have always assumed that this fact was part of the classical forcing folklore results, but it doesn’t seem to be mentioned explicitly in the usual forcing literature (it appears as lemma 25.5 in Jech’s book), and so I am writing an account of it here. Victoria Gitman and I have need of it in a current joint project. (Bob Solovay mentions in the comments below that the result is due to him, and provides a possible 1975 reference.)

Theorem 1. If $V[G]=V[H]$, where $G\subset \mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{B}\upharpoonright b$ is isomorphic to $\mathbb{C}\upharpoonright c$ by an isomorphism carrying $G$ to $H$.

The proof will also establish the following related result, concerning the situation where one extension is merely contained in the other.

Theorem 2. If $V[H]\subset V[G]$, where $G\subset\mathbb{B}$ and $H\subset\mathbb{C}$ are $V$-generic filters on the complete Boolean algebras $\mathbb{B}$ and $\mathbb{C}$, respectively, then there are conditions $b\in\mathbb{B}$ and $c\in\mathbb{C}$ such that $\mathbb{C}\upharpoonright c$ is isomorphic to a complete subalgebra of $\mathbb{B}\upharpoonright b$.

By $\mathbb{B}\upharpoonright b$, where $b$ is a condition in $\mathbb{B}$ (that is, a nonzero element of $\mathbb{B}$), what I mean is the Boolean algebra consisting of the interval $[0,b]$ in $\mathbb{B}$, using relative complement $b-a$ as the negation of $a$. This is the complete Boolean algebra that arises when forcing with the conditions in $\mathbb{B}$ below $b$.

Proof: In order to prove theorem 2, let me assume at first only that $V[H]\subset V[G]$. It follows that $H=\dot H_G$ for some $\mathbb{B}$-name $\dot H$, and we may choose a condition $b\in G$ forcing that $\dot H$ is a $\check V$-generic filter on $\check{\mathbb{C}}$.

I claim that there is some $c\in H$ such that every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$. Note that every $d\in H$ has $[\![\check d\in\dot H]\!]\in G$ by the truth lemma, since $H=\dot H_G$, and so $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ for $d\in H$. If $c\in H$ forces that every $d$ in the generic filter has that property, then indeed every $d\leq c$ has $b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}\neq 0$ as claimed.
In other words, from the perspective of the $\mathbb{B}$ forcing, every $d\leq c$ has a nonzero possibility to be in $\dot H$.

Define $\pi:\mathbb{C}\upharpoonright c\to\mathbb{B}$ by $$\pi(d)=b\wedge [\![\check d\in\dot H]\!]^{\mathbb{B}}.$$ Using the fact that $b$ forces that $\dot H$ is a filter, it is straightforward to verify that

  • $d\leq e\implies \pi(d)\leq\pi(e)$, since if $d\leq e$ and $d\in H$, then $e\in H$.
  • $\pi(d\wedge e)=\pi(d)\wedge \pi(e)$, since $[\![\check d\in\dot H]\!]\wedge[\![\check e\in \dot H]\!]=[\![\check{(b\wedge e)}\in\dot H]\!]$.
  • $\pi(d-e)=\pi(d)-\pi(e)$, since $[\![\check{(d-e)}\in\dot H]\!]=[\![\check d\in\dot H]\!]-[\![\check e\in\dot H]\!]$.

Thus, $\pi$ is a Boolean algebra embedding of $\mathbb{C}\upharpoonright c$ into $\mathbb{B}\upharpoonright\pi(c)$.

Let me argue that this embedding is a complete embedding. Suppose that $a=\bigvee A$ for some subset $A\subset\mathbb{C}\upharpoonright c$ with $A\in V$. Since $H$ is $V$-generic, it follows that $a\in H$ just in case $H$ meets $A$. Thus, $[\![\check a\in\dot H]\!]=[\![\exists x\in\check A\, x\in \dot H]\!]=\bigvee_{x\in A}[\![\check x\in\dot H]\!]$, and so $\pi(\bigvee A)=\bigvee_{x\in A}\pi(x)$, and so $\pi$ is complete, as desired. This proves theorem 2.

To prove theorem 1, let me now assume fully that $V[G]=V[H]$. In this case, there is a $\mathbb{C}$ name $\dot G$ for which $G=\dot G_H$. By strengthening $b$, we may assume without loss that $b$ also forces that, that is, that $b$ forces $\Gamma=\check{\dot G}_{\dot H}$, where $\Gamma$ is the canonical $\mathbb{B}$-name for the generic object, and $\check{\dot G}$ is the $\mathbb{B}$-name of the $\mathbb{C}$-name $\dot G$. Let us also strengthen $c$ to ensure that $c$ forces $\dot G$ is $\check V$-generic for $\check{\mathbb{C}}$. For $d\leq c$ define $\pi(d)=[\![\check d\in\dot H]\!]^{\mathbb{B}}$ as above, which provides a complete embedding of $\mathbb{C}\upharpoonright c$ to $\mathbb{B}\upharpoonright\pi(c)$. I shall now argue that this embedding is dense below $\pi(c)$. Suppose that $a\leq \pi(c)$ in $\mathbb{B}$. Since $a$ forces $\check a\in\Gamma$ and also $\check c\in\dot H$, it must also force that there is some $d\leq c$ in $\dot H$ that forces via $\mathbb{C}$ over $\check V$ that $\check a\in\dot G$. So there must really be some $d\leq c$ forcing $\check a\in\dot G$. So $\pi(d)$, which forces $\check d\in\dot H$, will also force $\check a\in\check{\dot G}_{\dot H}=\Gamma$, and so $\pi(d)\Vdash_{\mathbb{B}}\check a\in\Gamma$, which means $\pi(d)\leq a$ in ${\mathbb{B}}$. Thus, the range of $\pi$ on $\mathbb{C}\upharpoonright c$ is dense below $\pi(c)$, and so $\pi$ is a complete dense embedding of ${\mathbb{C}}\upharpoonright c$ to ${\mathbb{B}}\upharpoonright \pi(c)$. Since these are complete Boolean algebras, this means that $\pi$ is actually an isomorphism of $\mathbb{C}\upharpoonright c$ with $\mathbb{B}\upharpoonright \pi(c)$, as desired.

Finally, note that if $d\in H$ below $c$, then since $H=\dot H_G$, it follows that $[\![\check d\in\dot H]\!]\in G$, which is to say $\pi(d)\in G$, and so $\pi$ carries $H$ to $G$ on these cones. So $\pi^{-1}$ is the isomorphism stated in theorem 1.QED

Finally, I note that one cannot get rid of the need to restrict to cones, since it could be that $\mathbb{B}$ and $\mathbb{C}$ are the lottery sums of a common forcing notion, giving rise to $V[G]=V[H]$, together with totally different non-isomorphic forcing notions below some other incompatible conditions. So we cannot expect to prove that $\mathbb{B}\cong\mathbb{C}$, and are content to get merely that $\mathbb{B}\upharpoonright b\cong\mathbb{C}\upharpoonright c$, an isomorphism below respective conditions.

Large cardinals need not be large in HOD, International Workshop on Set Theory, CIRM, Luminy, September 2014

I shall speak at the 13th International Workshop on Set Theory, held at the CIRM Centre International de Rencontres Mathématiques in Luminy near Marseille, France, September 29 to October 3, 2014. 

Abstract.  I shall prove that large cardinals need not generally exhibit their large cardinal nature in HOD. For example, a supercompact cardinal need not be weakly compact in HOD, and there can be a proper class of supercompact cardinals in $V$, none of them weakly compact in HOD, with no supercompact cardinals in HOD. Similar results hold for many other types of large cardinals, such as measurable and strong cardinals. There are many open questions.

This talk will include joint work with Cheng Yong and Sy-David Friedman.

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