Every ordinal has only finitely many order-types for its final segments

Posted on August 1, 2015 by Joel David Hamkins

droste_effect_clock_by_kiluncle-d4ya2gn I was recently asked an interesting elementary question about the number of possible order types of the final segments of an ordinal, and in particular, whether there could be an ordinal realizing infinitely many different such order types as final segments. Since I found it interesting, let me write here how I replied.

The person asking me had noted that every nonempty final segment of the first infinite ordinal $\omega$ is isomorphic to $\omega$ again, since if you start counting from $5$ or from a million, you have just as far to go in the natural numbers. Thus, if one includes the empty final segment, there are precisely two order-types that arise as final segments of $\omega$, namely, $0$ and $\omega$ itself. A finite ordinal $n$, in contrast, has precisely $n+1$ many final segments, corresponding to each of the possible cuts between any of the elements or before all of them or after all of them, and these final segments, considered as orders themselves, all have different sizes and hence are not isomorphic.

He wanted to know whether an ordinal could have infinitely many different order-types for its tails.

Question. Is there an ordinal having infinitely many different isomorphism types for its final segments?

The answer is no, and I’d like to explain why. I’ll discuss two different arguments, the first being an easy direct argument aimed only at this answer, and the second being a more careful analysis aimed at understanding exactly how many and which order-types arise as the order type of a final segment of an ordinal $\alpha$.

Theorem. Every ordinal has only finitely many order types of its final segments.

Proof: Suppose that $\alpha$ is an ordinal, and consider the order types of the final segments $[\eta,\alpha)$, for $\eta\leq\alpha$. Note that as $\eta$ increases, the final segment $[\eta,\alpha)$ becomes smaller as a suborder, and so it’s order type does not go up. And since these are well-orders, it can go down only finitely many times. So only finitely many order types arise, and the theorem is proved. QED

But let’s figure out exactly how many and which order types arise.

Theorem. The number of order types of final segments of an ordinal $\alpha$ is precisely $n+1$, where $n$ is the number of terms in the Cantor normal form of $\alpha$, and one can describe those order types in terms of the normal form of $\alpha$.

Cantor proved that every ordinal $\alpha$ can be uniquely expressed as a finite sum $$\alpha=\omega^{\beta_n}+\cdots+\omega^{\beta_0},$$ where $\beta_n\geq\cdots\geq\beta_0$, and this is called the Cantor normal form of the ordinal. There are alternative forms, where one allows terms like $\omega^\beta\cdot n$ for finite $n$, but in my favored formulation, one simply expands this into $n$ terms with $\omega^\beta+\cdots+\omega^\beta$. In particular, the ordinal $\omega=\omega^1$ has exactly one term in its Cantor normal form, and a finite number $n=\omega^0+\cdots+\omega^0$ has exactly $n$ terms in its Cantor normal form. So the statement of the theorem agrees with the calculations that we had made at the very beginning.

Proof: First, let’s observe that every nonempty final segment of an ordinal of the form $\omega^\beta$ is isomorphic to $\omega^\beta$ again. This amounts to the fact that ordinals of the form $\omega^\beta$ are additively indecomposable, or in other words, closed under ordinal addition, since the final segments of an ordinal $\alpha$ are precisely the ordinals $\zeta$ such that $\alpha=\xi+\zeta$ for some $\xi$. If $\alpha$ is additively indecomposable, then it cannot be that $\zeta<\alpha$, and so all final segments would be isomorphic to $\alpha$. So let’s prove that $\omega^\beta$ is additively indecomposable. This is clear if $\beta=0$, since the only ordinal less than $\omega^0=1$ is $0$ and $0+0<1$. If $\beta$ is a limit ordinal, then the ordinals $\omega^\eta$ for $\eta<\beta$ are unbounded in $\omega^\beta$, and adding them stays below because $\omega^\eta+\omega^\eta=\omega^\eta\cdot 2\leq\omega^\eta\cdot\omega=\omega^{\eta+1}<\omega^\beta$. If $\beta=\delta+1$ is a successor ordinal, then $\omega^\beta=\omega^{\delta+1}=\omega^\delta\cdot\omega=\sup_{n<\omega}\omega^\delta\cdot n$, but again adding them stays below because $\omega^\delta\cdot n+\omega^\delta\cdot m=\omega^\delta\cdot(n+m) < \omega^\delta\cdot\omega=\omega^\beta$.

To prove the theorem, consider any ordinal $\alpha$ with Cantor normal form $\alpha=\omega^{\beta_n}+\cdots+\omega^{\beta_0}$, where $\beta_n\geq\cdots\geq\beta_0$. So as an order type, $\alpha$ consists of finitely many pieces, the first of type $\omega^{\beta_n}$, the next of type $\omega^{\beta_{n-1}}$ and so on up to $\omega^{\beta_0}$. Any final segment of $\alpha$ therefore consists of a final segment of one of these segments, together with all the segments after that segment (and omitting any segments prior to it, if any). But since these segments all have the form $\omega^{\beta_i}$, they are additively indecomposable and therefore are isomorphic to all their nonempty final segments. So any final segment of $\alpha$ is order-isomorphic to an ordinal whose Cantor normal form simply omits some (or none) of the terms from the front of the Cantor normal form of $\alpha$. Since we may start with any of the $n$ terms (or none), this gives precisely $n+1$ many order types of the final segments of $\alpha$, as claimed.

The argument shows, furthermore, that the possible order types of the final segments of $\alpha$, where $\alpha=\omega^{\beta_n}+\cdots+\omega^{\beta_0}$, are precisely the ordinals of the form $\omega^{\beta_k}+\cdots+\omega^{\beta_0}$, omitting terms only from the front, where $k\leq n$. QED

The axiom of determinacy for small sets

Posted on July 3, 2015 by Joel David Hamkins

I should like to argue that the axiom of determinacy is true for all games having a small payoff set. In particular, the size of the smallest non-determined set, in the sense of the axiom of determinacy, is the continuum; every set of size less than the continuum is determined, even when the continuum is enormous.

We consider two-player games of perfect information. Two players, taking turns, play moves from a fixed space $X$ of possible moves, and thereby together build a particular play or instance of the game $\vec a=\langle a_0,a_1,\ldots\rangle\in X^\omega$. The winner of this instance of the game is determined according to whether the play $\vec a$ is a member of some fixed payoff set $U\subset X^\omega$ specifying the winning condition for this game. Namely, the first player wins in the case $\vec a\in U$.

A strategy in such a game is a function $\sigma:X^{<\omega}\to X$ that instructs a particular player how to move next, given the sequence of partial play, and such a strategy is a winning strategy for that player, if all plays made against it are winning for that player. (The first player applies the strategy $\sigma$ only on even-length input, and the second player only to the odd-length inputs.) The game is determined, if one of the players has a winning strategy.

It is not difficult to see that if $U$ is countable, then the game is determined. To see this, note first that if the space of moves $X$ has at most one element, then the game is trivial and hence determined; and so we may assume that $X$ has at least two elements. If the payoff set $U$ is countable, then we may enumerate it as $U=\{s_0,s_1,\ldots\}$. Let the opposing player now adopt the strategy of ensuring on the $n^{th}$ move that the resulting play is different from $s_n$. In this way, the opposing player will ensure that the play is not in $U$, and therefore win. So every game with a countable payoff set is determined.

Meanwhile, using the axiom of choice, we may construct a non-determined set even for the case $X=\{0,1\}$, as follows. Since a strategy is function from finite binary sequences to $\{0,1\}$, there are only continuum many strategies. By the axiom of choice, we may well-order the strategies in order type continuum. Let us define a payoff set $U$ by a transfinite recursive procedure: at each stage, we will have made fewer than continuum many promises about membership and non-membership in $U$; we consider the next strategy on the list; since there are continuum many plays that accord with that strategy for each particular player, we may make two additional promises about $U$ by placing one of these plays into $U$ and one out of $U$ in such a way that this strategy is defeated as a winning strategy for either player. The result of the recursion is a non-determined set of size continuum.

So what is the size of the smallest non-determined set? For a lower bound, we argued above that every countable payoff set is determined, and so the smallest non-determined set must be uncountable, of size at least $\aleph_1$. For an upper bound, we constructed a non-determined set of size continuum. Thus, if the continuum hypothesis holds, then the smallest non-determined set has size exactly continuum, which is $\aleph_1$ in this case. But what if the continuum hypothesis fails? I claim, nevertheless, that the smallest non-determined set still has size continuum.

Theorem. Every game whose winning condition is a set of size less than the continuum is determined.

Proof. Suppose that $U\subset X^\omega$ is the payoff set of the game under consideration, so that $U$ has size less than continuum. If $X$ has at most one element, then the game is trivial and hence determined. So we may assume that $X$ has at least two elements. Let us partition the elements of $X^\omega$ according to whether they have exactly the same plays for the second player. So there are at least continuum many classes in this partition. If $U$ has size less than continuum, therefore, it must be disjoint from at least one (and in fact from most) of the classes of this partition (since otherwise we would have an injection from the continuum into $U$). So there is a fixed sequence of moves for the second player, such that any instance of the game in which the second player makes those moves, the result is not in $U$ and hence is a win for the second player. This is a winning strategy for the second player, and so the game is determined. QED

This proof generalizes the conclusion of the diagonalization argument against a countable payoff set, by showing that for any winning condition set of size less than continuum, there is a fixed play for the opponent (not depending on the play of the first player) that defeats it.

The proof of the theorem uses the axiom of choice in the step where we deduce that $U$ must be disjoint from a piece of the partition, since there are continuum many such pieces and $U$ had size less than the continuum. Without the axiom of choice, this conclusion does not follow. Nevertheless, what the proof does show without AC is that every set that does not surject onto $\mathbb{R}$ is determined, since if $U$ contained an element from every piece of the partition it would surject onto $\mathbb{R}$. Without AC, the assumption that $U$ does not surject onto $\mathbb{R}$ is stronger than the assumption merely that it has size less the continuum, although these properties are equivalent in ZFC. Meanwhile, these issues are relevant in light of the model suggested by Asaf Karagila in the comments below, which shows that it is consistent with ZF without the axiom of choice that there are small non-determined sets. Namely, the result of Monro shows that it is consistent with ZF that $\mathbb{R}=A\sqcup B$, where both $A$ and $B$ have cardinality less than the continuum. In particular, in this model the continuum injects into neither $A$ nor $B$, and consequently neither player can have a strategy to force the play into their side of this partition. Thus, both $A$ and $B$ are non-determined, even though they have size less than the continuum.

Determinacy for proper-class clopen games is equivalent to transfinite recursion along proper-class well-founded relations

Posted on July 2, 2015 by Joel David Hamkins

I’d like to continue a bit further my exploration of some principles of determinacy for proper-class games; it turns out that these principles have a surprising set-theoretic strength. A few weeks ago, I explained that the determinacy of proper-class open games and even clopen games implies Con(ZFC) and much more. Today, I’d like to prove that clopen determinacy is exactly equivalent over GBC to the principle of transfinite recursion along proper-class well-founded relations. Thus, GBC plus either of these principles is a strictly intermediate set theory between GBC and KM.

The principle of clopen determinacy for class games is the assertion that in any two-player infinite game of perfect information whose winning condition is a clopen class, there is a winning strategy for one of the players. Players alternately play moves in a playing space $X$, thereby creating a particular play $\vec a\in X^\omega$, and the winner is determined according to whether $\vec a$ is in a certain fixed payoff class $U\subset X^\omega$ or not. One has an open game when this winning condition class $U$ is open in the product topology (using the discrete topology on $X$). A game is open for a player if and only if every winning play for that player has an initial segment, all of whose extensions are also winning for that player. So the game is won for an open player at a finite stage of play. A clopen game, in contrast, has a payoff set that is open for both players. Clopen games can be equivalently cast in terms of the game tree, consisting of positions in the game where the winner is not yet determined, and where play terminates when the winner is known. Namely, a game is clopen exactly when this game tree is well-founded, so that in every play, the outcome is known already at a finite stage.

A strategy is a class function $\sigma:X^{<\omega}\to X$ that instructs the player what to play next, given a position of partial play, and the strategy is winning for a player if all plays that accord with it satisfy the winning condition for that player.

The principle of transfinite recursion along well-founded class relations is the assertion that we may undertake recursive definitions along any class well-founded partial order relation. That is, suppose that $\lhd$ is a class well-founded partial order relation on a class $A$, and suppose that $\varphi(F,a,y)$ is a formula, using only first-order quantifiers but having a class variable $F$, which is functional in the sense that for any class $F$ and any set $a\in A$ there is a unique $y$ such that $\varphi(F,a,y)$. The idea is that $\varphi(F,a,y)$ expresses the recursive rule to be iterated, and a solution of the recursion is a class function $F$ such that $\varphi(F\upharpoonright a,a,F(a))$ holds for every $a\in A$, where $F\upharpoonright a$ means the restriction of $F$ to the class $\{ b\in A\mid b\lhd a\}$. Thus, the value $F(a)$ is determined by the class of previous values $F(b)$ for $b\lhd a$. The principle of transfinite recursion along class well-founded relations is the assertion scheme that for every such well-founded partial order class $\langle A,\lhd\rangle$ and any recursive rule $\varphi$ as above, there is a solution.

In the case that the relation $\lhd$ is set-like, which means that the predecessors $\{b\mid b\lhd a\}$ of any point $a$ form a set (rather than a proper class), then GBC easily proves that there is a unique solution class, which furthermore is definable from $\lhd$. Namely, one can show that every $a\in A$ has a partial solution that obeys the recursive rule at least up to $a$, and furthermore all such partial solutions agree below $a$, because there can be no $\lhd$-minimal violation of this. It follows that the class function $F$ unifying these partial solutions is a total solution to the recursion. Similarly, GBC can prove that there are solutions to other transfinite recursion instances where the well-founded relation is not necessarily set-like, such as a recursion of length $\text{Ord}+\text{Ord}$ or even much longer.

Meanwhile, if GBC is consistent, then it cannot in general prove that transfinite recursions along non-set-like well-founded relations always succeed, since this principle would imply that there is a truth-predicate for first-order truth, as the Tarskian conditions are precisely such a recursion on a well-founded relation based on the complexity of formulas. (That relation is not set-like, since when considering the truth of $\exists x\,\psi(x,\vec a)$, we want to consider the truth of $\psi(b,\vec a)$ for any parameter $b$, and there are a proper class of such $b$.) Thus, GBC plus transfinite recursion (or plus clopen determinacy) is strictly stronger than GBC, although it is provable in Kelley-Morse set theory KM essentially the same as GBC proves the set-like special case.

Theorem. Assume GBC. Then the following are equivalent.

Clopen determinacy for class games. That is, for any two-player game of perfect information whose payoff class is both open and closed, there is a winning strategy for one of the players.
Transfinite recursion for proper class well-founded relations (not necessarily set-like).

Proof. ($2\to 1$) Assume the principle of transfinite recursion for proper class well-founded relations, and suppose we are faced with a clopen game. Consider the game tree $T$, consisting of positions arising during play, up to the moment that a winner is known. This tree is well-founded because the game is clopen. Let us label the terminal nodes of the tree with I or II according to who has won the game in that position, and more generally, let us label all the nodes of the tree with I or II according to the following transfinite recursion: if a node has I to play, then it will have label I if there is a move to a node labeled I, and otherwise II; and similarly when it is II to play. By the principle of transfinite recursion, there is a labeling of the entire tree that accords with this recursive rule. It is now easy to see that if the initial node is labeled with I, then player I has a winning strategy, which is simply to stay on the nodes labeled I. Note that player II cannot play in one move from a node labeled I to one labeled II. Similarly, if the initial node is labeled II, then player II has a winning strategy; and so the game is determined, as desired.

($1\to 2$) Conversely, let us assume the principle of clopen determinacy for class games. Suppose we are faced with a recursion along a class relation $\lhd$ on a class $A$, using a recursion rule $\varphi(F,a,y)$. We shall define a certain clopen game, and prove that any winning strategy for this game will produce a solution for the recursion.

It will be convenient to assume that $\varphi(F,a,y)$ is strongly functional, meaning that not only does it define a function as we have mentioned in $V$, but also that $\varphi(F,a,y)$ defines a function $(F,a)\mapsto y$ when used over any model $\langle V_\theta,\in,F\rangle$ for any class $F\subset V_\theta$. The strongly functional property can be achieved simply by replacing the formula with the assertion that $\varphi(F,a,y)$, if $y$ is unique such that this holds, and otherwise $y=\emptyset$.

At first, let us consider a slightly easier game, which will be open rather than clopen; a bit later, we shall revise this game to a clopen game. The game is the recursion game, which will be very much like the truth-telling game of my previous post, Open determinacy for proper class games implies Con(ZFC) and much more. Namely, we have two players, the challenger and the truth-teller. The challenger will issues challenges about truth in a structure $\langle V,\in,\lhd,F\rangle$, where $\lhd$ is the well-founded class relation and $F$ is a class function, not yet specified. Specifically, the challenger is allowed to ask about the truth of any formula $\varphi(\vec a)$ in this structure, and to inquire as to the value of $F(a)$ for any particular $a$. The truth-teller, as before, will answer the challenges by saying either that $\varphi(\vec a)$ is true or false, and in the case $\varphi(\vec a)=\exists x\,\psi(x,\vec a)$ and the formula was declared true, by also giving a witness $b$ and declaring $\psi(b,\vec a)$ is true; and the truth-teller must specify a specific value for $F(a)$ for any particular $a$. The truth-teller loses immediately, if she should ever violate Tarski’s recursive definition of truth; and she also loses unless she declares the recursive rules $\varphi(F\upharpoonright a,a,F(a))$ to be true. Since these violations occur at a finite stage of play if they do at all, the game is open for the challenger.

Lemma. The challenger has no winning strategy in the recursion game.

Proof. Suppose that $\sigma$ is a strategy for the challenger. So $\sigma$ is a class function that instructs the challenger how to play next, given a position of partial play. By the reflection theorem, there is an ordinal $\theta$ such that $V_\theta$ is closed under $\sigma$, and using the satisfaction class that comes from clopen determinacy, we may actually also arrange that $\langle V_\theta,\in,\lhd\cap V_\theta,\sigma\cap V_\theta\rangle\prec\langle V,\in,\lhd,\sigma\rangle$. Consider the relation $\lhd\cap V_\theta$, which is a well-founded relation on $A\cap V_\theta$. The important point is that this relation is now a set, and in GBC we may certainly undertake transfinite recursions along well-founded set relations. Thus, there is a function $f:A\cap V_\theta\to V_\theta$ such that $\langle V_\theta,\in,f\rangle$ satisfies $\varphi(f\upharpoonright a,a,f(a))$ for all $a\in V_\theta$, where $f\upharpoonright a$ means restricting $f$ to the predecessors of $a$ in $V_\theta$, and this may not be all the predecessors of $a$ with respect to $\lhd$, which may not be set-like. Note that this is the place where we use our assumption that $\varphi$ was strongly functional, since we want to ensure that it can still be used to define a valid recursion over $\lhd\cap V_\theta$. (We are not claiming that $\langle V_\theta,\in,\lhd\cap V_\theta,f\rangle$ models $\text{ZFC}(\lhd,f)$.)

Consider now the play of the recursion game in $V$, where the challenger uses the strategy $\sigma$ and the truth-teller plays in accordance with $\langle V_\theta,\in,\lhd\cap V_\theta,f\rangle$. Since $V_\theta$ was closed under $\sigma$, the challenger will never issue challenges outside of $V_\theta$. And since the function $f$ fulfills the recursion $\varphi(f\upharpoonright a,a,f(a))$ in this structure, the truth-teller will not be trapped in any violation of the Tarski conditions or the recursion condition. Thus, the truth-teller will win this instance of the game, and so $\sigma$ was not a winning strategy for the challenger, as desired. QED

Lemma. The truth-teller has a winning strategy in the recursion game if and only if there is a solution of the recursion.

Proof. If there is a solution $F$ of the recursion, then by clopen determinacy, we also get a satisfaction class for the structure $\langle V,\in,\lhd,F\rangle$, and the truth-teller can answer all queries of the challenger by referring to what is actually true in this structure. This will be winning for the truth-teller, since the actual truth obeys the Tarskian conditions and the recursive rule.

Conversely, suppose that $\tau$ is a winning strategy for the truth-teller in the recursion game. We claim that the truth assertions made by $\tau$ do not depend on the order in which challenges are made by the challenger; they all cohere with one another. This is easy to see for formulas not involving $F$ by induction on formulas, for if the truth of a formula $\psi(\vec a)$ is independent of play, then also the truth of $\neg\psi(\vec a)$ is as well, and similarly if $\exists x\psi(x,\vec a)$ is declared true with witness $\psi(b,\vec a)$, then by induction $\psi(b,\vec a)$ is independent of the play, in which case $\exists x\psi(x,\vec a)$ must always be declared true by $\tau$ independently of the order of play by the challenger (although the particular witness $b$ provided by $\tau$ may depend on the play). Now, let us also argue that the values of $F(a)$ declared by $\tau$ are also independent of the order of play. If not, there is some $\lhd$-least $a$ where this fails. (Note that such an $a$ exists, since $\tau$ is a class, and we can define from $\tau$ the class of $a$ for which the value of $F(a)$ declared by $\tau$ depends on the order of play; without $\tau$, one might have expected to need $\Pi^1_1$-comprehension to find a minimal $a$ where the recursion fails.) As in the truth-telling game, the truth assertions made by $\tau$ about $\langle V,\in,\lhd,F\upharpoonright a\rangle$, where $F\upharpoonright a$ is the class function of values that are determined by $\tau$ on $b\lhd a$, must not depend on the order of play. Since the recursion rule $\varphi(F\upharpoonright a,a,y)$ is functional, there is only one value $y=F(a)$ for which this formula can be truthfully held, and so if some play causes $\tau$ to play a different value for $F(a)$, the challenger can in finitely many additional moves (bounded by the syntactic complexity of $\varphi$) trap the truth-teller in a violation of the Tarskian conditions or the recursion condition. Thus, the values of $F(a)$ declared by $\tau$ must in fact all cohere independently of the order of play, and so $\tau$ is describing a class function $F:A\to V$ such that $\varphi(F\upharpoonright a,a,F(a))$ is true for every $a\in A$. So the recursion has a solution, as desired. QED

So far, we have established that the principle of open determinacy implies the principle of transfinite recursion along well-founded class relations. In order to improve this implication to use only clopen determinacy rather than open determinacy, we modify the game to become a clopen game rather than an open game.

Consider the clopen form of the recursion game, where we insist also that the challenger announce on the first move a natural number $n$, such that the challenger loses if the truth-teller survives for at least $n$ moves. This is now a clopen game, since the winner will be known by that time, either because the truth-teller will violate the Tarski conditions or the recursion condition, or else the challenger’s limit on play will expire.

Since the modified version of the game is even harder for the challenger, there can still be no winning strategy for the challenger. So by the principle of clopen determinacy, there is a winning strategy $\tau$ for the truth-teller. This strategy is allowed to make decisions based on the number $n$ announced by the challenger on the first move, and it will no longer necessarily be the case that the theory declared true by $\tau$ will be independent of the order of play. Nevertheless, it will be the case, we claim, that the theory declared true by $\tau$ for all plays with sufficiently large $n$ (and with sufficiently many remaining moves) will be independent of the order of play. One can see this by observing that if an assertion $\psi(\vec a)$ is independent in this sense, then also $\neg\psi(\vec a)$ will be independent in this sense, for otherwise there would be plays with large $n$ giving different answers for $\neg\psi(\vec a)$ and we could then challenge with $\psi(\vec a)$, which would have to give different answers or else $\tau$ would not win. Similarly, since $\tau$ is winning, one can see that allowing the challenger to specify a bound on the total length of play does not prevent the arguments above showing that $\tau$ describes a coherent solution function $F:A\to V$ satisfying the recursion $\varphi(F\upharpoonright a,a,F(a))$, provided that one looks only at plays in which there are sufficiently many moves remaining. There cannot be a $\lhd$-least $a$ where the value of $F(a)$ is not determined in this sense, and so on as before.

Thus, we have proved that the principle of clopen determinacy for class games is equivalent to the principle of transfinite recursion along well-founded class relations. QED

The material in this post will become part of a joint project with Victoria Gitman and Thomas Johnstone. We are currently investigating several further related issues.

Open determinacy for proper class games implies Con(ZFC) and much more

Posted on June 17, 2015 by Joel David Hamkins

$\newcommand\Tr{\text{Tr}}$One of the intriguing lessons we have learned in the past half-century of set-theoretic developments is that there is a surprisingly robust connection between infinitary game theory and fundamental set-theoretic principles, including large cardinals. Assertions about the existence of strategies in infinite games often turn out to have an unexpected set-theoretic power. In this post, I should like to give another specific example of this, which Thomas Johnstone and I hit upon yesterday in an enjoyable day of mathematics.

Specifically, I’d like to prove that if we generalize the open-game concept from sets to classes, then assuming consistency, ZFC cannot prove that every definable open class game is determined, and indeed, over Gödel-Bernays set theory GBC the principle of open determinacy (and even just clopen determinacy) implies Con(ZFC) and much more.

To review a little, we are talking about games of perfect information, where two players alternately play elements from an allowed space $X$ of possible moves, and together they build an infinite sequence $\vec x=\langle x_0,x_1,x_2,\ldots\rangle$ in $X^\omega$, which is the resulting play of this particular instance of the game. We have a fixed collection of plays $A\subset X^\omega$ that is used to determine the winner, namely, the first player wins this particular instance of the game if the resulting play $\vec x$ is in $A$, and otherwise the second player wins. A strategy for a player is a function $\sigma:X^{<\omega}\to X$, which tells a player how to move next, given a finite position in the game. Such a strategy is winning for that player, if he or she always wins by following the instructions, no matter how the opponent plays. The game is determined, if one of the players has a winning strategy.

It is a remarkable but elementary fact that if the winning condition $A$ is an open set, then the game is determined. One can prove this by using the theory of ordinal game values, and my article on transfinite game values in infinite chess contains an accessible introduction to the theory of game values. Basically, one defines that a position has game value zero (for player I, say), if the game has already been won at that stage, in the sense that every extension of that position is in the winning payoff set $A$. A position with player I to play has value $\alpha+1$, if player I can move to a position with value $\alpha$, and $\alpha$ is minimal. The value of a position with player II to play is the supremum of the values of all the positions that he or she might reach in one move, provided that those positions have a value. The point now is that if a position has a value, then player I can play so as strictly to decrease the value, and player II cannot play so as to increase it. So if a position has a value, then player I has a winning strategy, which is the value-reducing strategy. Conversely, if a position does not have a value, then player II can maintain that fact, and player I cannot play so as to give it a value; thus, in this case player II has a winning strategy, the value-maintaining strategy. Thus, we have proved the Gale-Stewart theorem: every open game is determined.

That proof relied on the space of moves $X$ being a set, since we took a supremum over the values of the possible moves, and if $X$ were a proper class, we couldn’t be sure to stay within the class of ordinals and the recursive procedure might break down. What I’d like to do is to consider more seriously the case where $X$ is a proper class. Similarly, we allow the payoff collection $A$ to be a proper class, and the strategies $\sigma:X^{<\omega}\to X$ are also proper classes. Can we still prove the Gale-Steward theorem for proper classes? The answer is no, unless we add set-theoretic strength. Indeed, even clopen determinacy has set-theoretic strength.

Theorem. (GBC) Clopen determinacy for proper classes implies Con(ZFC) and much more. Specifically, there is a clopen game, such the existence of a winning strategy is equivalent to the existence of a satisfaction class for first-order truth.

Proof. Let me first describe a certain open game, the truth-telling game, with those features, and I shall later modify it to a clopen game. The truth-telling game will have two players, which I call the challenger and the truth-teller. At any point in the game, the challenger plays by making an inquiry about a particular set-theoretic formula $\varphi(\vec a)$ with parameters. The truth-teller must reply to the inquiry by stating either true or false, and in the case that the formula $\varphi$ is an existential assertion $\exists x\,\psi(x,\vec a)$ declared to be true, then the truth teller must additionally identify a particular witness $b$ and assert that $\psi(b,\vec a)$ is true. So a play of the game consists of a sequence of such inquires and replies.

The truth-teller wins a play of the game, provided that she never violates the recursive Tarskian truth conditions. Thus, faced with an atomic formula, she must state true or false in accordance with the actual truth or falsity of that atomic formula, and similarly,
she must say true to $\varphi\wedge\psi$ just in case she said true to both $\varphi$ and $\psi$ separately (if those formulas had been issued by the challeger), and she must state opposite truth values for $\varphi$ and $\neg\varphi$, if both are issued as challenges.

This is an open game, since the challenger will win, if at all, at a finite stage of play, when the violation of the Tarskian truth conditions is first exhibited.

Lemma 1. The truth-teller has a winning strategy in the truth-telling game if and only if there is a satisfaction class for first-order truth.

Proof. Clearly, if there is a satisfaction class for first-order truth, then the truth-teller has a winning strategy, which is simply to answer all questions about truth by consulting the
satisfaction class. Since that class obeys the Tarskian conditions, she will win the game, no matter which challenges are issued.

Conversely, suppose that the truth-teller has a winning strategy $\tau$ in the game. I claim that we may use $\tau$ to build a satisfaction class for first-order truth. Specifically, let $T$ be the collection of formulas $\varphi(\vec a)$ that are asserted to be true by $\tau$ in any play according to $\tau$. I claim that $T$ is a satisfaction class. We may begin by noting that since $T$ must correctly state the truth of all atomic formulas, it follows that the particular answers that $\tau$ gives on the atomic formulas does not depend on the order of the challenges issued by the challenger. Now, we argue by induction on formulas that the truth values issued by $\tau$ does not depend on the order of the challenges. For example, if all plays in which $\varphi(\vec a)$ is issued as a challenge come out true, then all plays in which $\neg\varphi(\vec a)$ is challenged will result in false, or else we would have a play in which $\tau$ would violate the Tarskian truth conditions. Similarly, if $\varphi$ and $\psi$ always come out the same way, then so does $\varphi\wedge\psi$. We don’t claim that $\tau$ must always issue the same witness $b$ for an existential $\exists x\,\psi(x,\vec a)$, but if it ever says true to this statement, then it will provide some witness $b$, and for that statement $\psi(b,\vec a)$, the truth value stated by $\tau$ is independent of the order of play by the challenger, by induction. Thus, by induction on formulas, the answers provided by the truth-teller strategy $\tau$ gives us a satisfaction predicate for first-order truth. QED

Lemma 2. The challenger has no winning strategy in the truth-telling game.

Proof. Suppose that $F$ is a strategy for the challenger. So $F$ is a proper class function that directs the challenger to issue certain challenges, given the finite sequence of previous challenges and truth-telling answers. By the reflection theorem, there is a closed unbounded proper class of cardinals $\theta$, such that $F”V_\theta\subset V_\theta$. That is, $V_\theta$ is closed under $F$, in the sense that if all previous challenges and responses come from $V_\theta$, then the next challenge will also come from $V_\theta$. Since $\langle V_\theta,\in\rangle$ is a set, we have a satisfaction predicate on it. Consider the play, where the truth-teller replies to all inquires by consulting truth in $V_\theta$, rather than truth in $V$. The point is that if the challenger follows $\tau$, then all the inquiries will involve only parameters $\vec a$ in $V_\theta$, provided that the truth-teller also always gives witnesses in $V_\theta$, which in this particular play will be the case. Since the satisfaction predicate on $V_\theta$ does satisfy the Tarskian truth conditions, it follows that the truth-teller will win this instance of the game, and so $F$ is not a winning strategy for the challenger. QED

Thus, if open determinacy holds for classes, then there is a satisfaction predicate for first-order truth.

This implies Con(ZFC) for reasons I explained on my post KM implies Con(ZFC) and much more, by appealing to the fact that we have the collection axiom relative to the class for the satisfaction predicate itself, and this is enough to verify that the nonstandard instances of collection also must be declared true in the satisfaction predicate.

But so far, we only have an open game, rather than a clopen game, since the truth-teller wins only by playing the game out for infinitely many steps. So let me describe how to modify the game to be clopen. Specifically, consider the version of the truth-telling game, where the challenger must also state on each move a specific ordinal $\alpha_n$, which descend during play $\alpha_0>\alpha_1>\cdots>\alpha_n$. If the challenger gets to $0$, then the truth-teller is declared the winner. For this modified game, the winner is known in finitely many moves, because either the truth-teller violates the Tarskian conditions or the challenger hits zero. So this is a clopen game. Since we made the game harder for the challenger, it follows that the challenger still can have no winning strategy. One can modify the proof of lemma 1 to say that if $\tau$ is a winning strategy for the truth teller, then the truth assertions made by $\tau$ in response to all plays with sufficiently large ordinals for the challenger all agree with one another independently of the order of the formulas issued by the challenger. Thus, there is a truth-telling strategy just in case there is a satisfaction class for first-order truth.

So clopen determinacy for class games implies the existence of a satisfaction class for first-order truth, and this implies Con(ZFC) and much more. QED

One may easily modify the game by allowing a fixed class parameter $B$, so that clopen determinacy implies that there is a satisfaction class relative to truth in $\langle V,\in,B\rangle$.

Furthermore, we may also get iterated truth predicates. Specifically, consider the iterated truth-telling game, which in addition to the usual language of set theory, we have a hierarchy of predicates $\Tr_\alpha$ for any ordinal $\alpha$. We now allow the challenger to ask about formulas in this expanded language, and the truth teller is required to obey not only the usual Tarskian recursive truth conditions, but also the requirements that $\Tr_\alpha(\varphi(\vec a))$ is declared true just in case $\varphi(\vec a)$ uses only truth predicates $\Tr_\beta$ for $\beta<\alpha$ and also $\varphi(\vec a)$ is declared true (if this challenge was issued).

The main arguments as above generalize easily to show that the challenger cannot have a winning strategy in this iterated truth-telling game, and the truth-teller has a strategy just in case there is a satisfaction predicate for truth-about-truth iterated through the ordinals. Thus, the principle of open determinacy for proper class games implies Con(Con(ZFC)) and $\text{Con}^\alpha(\text{ZFC})$ and so on.

Let me finish by mentioning that Kelley-Morse set theory is able to prove open determinacy for proper class games in much the same manner as we proved the Gale-Stewart theorem above, using well-ordered class meta-ordinals, rather than merely set ordinals, as well as in other ways. If there is interest, I can make a further post about that, so just ask in the comments!

Transfinite Nim

Posted on May 27, 2015 by Joel David Hamkins

Shall we have a game of transfinite Nim? One of us sets up finitely many piles of wooden blocks, each pile having some ordinal height, possibly transfinite, and the other of us decides who shall make the first move. Taking turns, we each successively remove a top part of any one pile of our choosing, making it strictly shorter. Whoever takes the very last block wins. (It is fine to remove an entire pile on a turn or to remove blocks from a different pile on a later turn.)

In my challenge problem last week, for example, I set up six piles with heights:
$$1\qquad \omega+3\qquad \omega^\omega+5 \qquad \omega^{\omega+3}+\omega^\omega\cdot3+\omega\cdot 5+7\qquad \epsilon_0\qquad \omega_1$$Would you want to go first or second? What is the best move? In general, we can start with any finite number of piles of arbitrary ordinal heights — what is the general winning strategy?

Before proceeding with the transfinite case, however, let’s review the winning strategy in ordinary finite Nim, which I explained in my post last week concerning my visit to the 7th/8th grade Math Team at my son’s school. To say it quickly again, a finite Nim position is balanced, if when you consider the binary representations of the pile heights, there are an even number of ones in each binary place position. Another way to say this, and this is how I explained it to the school kids, is that if you think of each pile height as a sum of distinct powers of two, then any power of two that arises in any pile does so an even number of times overall for all the piles. The mathematical facts to establish are that (1) any move on a balanced position will unbalance it; and (2) any unbalanced position admits a balancing move. Since the winning move of taking the very last block is a balancing move, it follows that the winning strategy is to balance whatever position with which you are faced. At the start, if the position is unbalanced, then you should go first and balance it; if it is already balanced, then you should go second and adopt the balancing strategy. It may be interesting to note that this winning strategy is unique in the sense that any move that does not balance the position is a losing move, since the opposing player can adopt the balancing strategy from that point on. But of course there is often a choice of balancing moves.

Does this balancing strategy idea continue to apply to transfinite Nim? Yes! All we need to do is to develop a little of the theory of transfinite binary representation. Let me assume that you are all familiar with the usual ordinal arithmetic, for which $\alpha+\beta$ is the ordinal whose order type is isomorphic to a copy of $\alpha$ followed by a copy of $\beta$, and $\alpha\cdot\beta$ is the ordinal whose order type is isomorphic to $\beta$ many copies of $\alpha$. Consider now ordinal exponentiation, which can be defined recursively as follows:
$$\alpha^0=1$$ $$\alpha^{\beta+1}=\alpha^\beta\cdot\alpha$$ $$\alpha^\lambda=\sup_{\beta<\lambda} \alpha^\beta\qquad\lambda\text{ limit}$$ It turns out that $\alpha^\beta$ is the order-type of the finite-support functions from $\beta$ to $\alpha$, under the suitable lexical order. Ordinal exponentiation should not be confused with cardinal exponentiation, since they are very different. For example, with ordinal exponentiation, one has $$2^\omega=\sup_{n<\omega}2^n=\omega,$$which of course is not the case with cardinal exponentiation. In this post, I use only ordinal exponentiation.

Theorem. Every ordinal $\beta$ has a unique representation as a decreasing finite sum of ordinal powers of two. $$\beta=2^{\beta_n}+\cdots+2^{\beta_0}, \qquad \beta_n>\cdots>\beta_0$$

The proof is easy! We simply prove it by transfinite induction on $\beta$. If the theorem holds below an ordinal $\beta$, first let $2^\alpha$ be the largest power of two that is at most $\beta$, so that $\beta=2^\alpha+\gamma$ for some ordinal $\gamma$. It follows that $\gamma<2^\alpha$, for otherwise we could have made $2^{\alpha+1}\leq\beta$. Thus, by induction, $\gamma$ has a representation with powers of two, and so we may simply add $2^\alpha$ at the front to represent $\beta$. To see that the representations are unique, first establish that any power of two is equal to or more than the supremum of the finite decreasing sums of any strictly smaller powers of two. From this, it follows that any representation of $\beta$ as above must have used $2^\alpha$ just as we did for the first term, because otherwise it couldn’t be large enough, and then the representation of the remaining part $\gamma$ is unique by induction, and so we get uniqueness for the representation of $\beta$. QED

Thus, the theorem shows that every ordinal has a unique binary representation in the ordinals, with finitely many nonzero bits. Suppose that we are given a position in transfinite Nim with piles of ordinal heights $\eta_0,\ldots,\eta_n$. We define that such a position is balanced, if every power of two appearing in the representation of any of the piles appears an even number of times overall for all the piles.

The mathematical facts to establish are (1) any move on a balanced position will unbalance it; and (2) every unbalanced position has a balancing move. These facts can be proved in the transfinite case in essentially the same manner as the finite case. Namely, if a position is balanced, then any move affects only one pile, changing the ordinal powers of two that appear in it, and thereby destroy the balanced parity of whichever powers of two are affected. And if a position is unbalanced, then look at the largest unbalanced ordinal power of two appearing, and make a move on any pile having such a power of two in its representation, reducing it so as exactly to balance all the smaller powers of two appearing in the position.

Finally, those two facts again imply that the balancing strategy is a winning strategy, since the winning move of taking the last block or blocks is a balancing move, down to the all-zero position, which is balanced.

In the case of my challenge problem above, we may represent the ordinals in binary. We know how to do that in the case of 1, 3, 5 and 7, and actually those numbers are balanced. Here are some other useful binary representations:

$\omega+3=2^\omega+2+1$

$\omega^\omega+5 = (2^\omega)^\omega+5=2^{\omega^2}+4+1$

$\omega^{\omega+3}=(2^\omega)^{\omega+3}=2^{\omega^2+\omega\cdot 3}$

$\omega^\omega\cdot3=(2^\omega)^\omega\cdot 3=2^{\omega^2}\cdot 2+2^{\omega^2}=2^{\omega^2+1}+2^{\omega^2}$

$\omega\cdot 5+7 =2^{\omega}\cdot 2^2+2^\omega+7=2^{\omega+2}+2^\omega+4+2+1$

$\epsilon_0 = 2^{\epsilon_0}$

$\omega_1=2^{\omega_1}$

I emphasize again that this is ordinal exponentiation. The Nim position of the challenge problem above is easily seen to be unbalanced in several ways. For example, the $\omega_1$ term among others appears only once. Thus, we definitely want to go first in this position. And since $\omega_1$ is the largest unbalanced power of two and it appears only once, we know that we must play on the $\omega_1$ pile. Once one represents all the ordinals in terms of their powers of two representation, one sees that the unique winning move is to reduce the $\omega_1$ pile to have ordinal height
$$\epsilon_0+\omega^{\omega+3}+\omega^\omega\cdot 2+\omega\cdot 4.$$This will exactly balance all the smaller powers of two in the other piles and therefore leaves a balanced position overall. In general, the winning strategy in transfinite Nim, just as for finite Nim, is always to leave a balanced position.

Special honors to Pedro Sánchez Terraf for being the only one to post the winning move in the comments on the other post!

Win at Nim! The secret mathematical strategy for kids (with challange problems in transfinite Nim for the rest of us)

Posted on May 22, 2015 by Joel David Hamkins

Welcome to my latest instance of Math for Kids!

Today I had the pleasure to make an interactive mathematical presentation at my son’s school to the 7th / 8th grade Math Team, about 30 math-enthusiastic kids (twelve and thirteen years old) along with their math teachers and the chair of the school math department.

The topic was the game of Nim! This game has a secret mathematical strategy enabling anyone with that secret knowledge to win against those without it. It is a great game for kids, because with the strategy they can realistically expect to beat their parents, friends, siblings and parent’s friends almost every single time!

To play Nim, one player sets up a number of piles of blocks, and the opponent chooses whether to go first or second. The players take turns removing blocks — each player may remove any number of blocks (at least one) from any one pile, and it is fine to take a whole pile — whichever player takes the last block wins.

For the math team, we played a few demonstration games, in which I was able to beat all the brave challengers, and then the kids paired off to play each other and gain familiarity with the game. Then, it was time for the first strategy discussion.

What could the secret winning strategy be? I explained to the kids a trick that mathematicians often use when approaching a difficult problem, namely, to consider in detail some very simple special cases or boundary instances of the problem. It often happens that these special cases reveal a way of thinking about the problem that applies much more generally.

Perhaps one of the easiest special cases of Nim occurs when there is only one pile. If there is only one pile, then clearly one wants to go first, in order to make the winning move: take the entire pile!

Two balanced piles

A slightly less trivial and probably more informative case arises when there are exactly two piles. If the stacks have the same height, then the kids realized that the second player could make copying moves so as to preserve this balanced situation. The key insight now is that this copying strategy is a winning strategy, because if one can always copy, then in particular one will have a move whenever the opponent did, and so the opponent will never take the last block. With two piles, therefore, one wants always to make them balanced. If they are initially unbalanced, then choose to go first and follow the balancing strategy. If they are initially balanced, then choose to go second, and copy whatever moves your opponent makes to rebalance them.

A balanced position

With that insight, it is not difficult to see that it is winning to leave a position with any number of pairs of balanced piles. One can in effect play on each pair separately, because whenever the opponent makes a move on one of the piles, one can copy the move with the corresponding partner pile. In this way, we may count such a position overall as balanced. The more fundamental game-theoretic observation to make is that balanced piles in effect cancel each other out in any position, and one can ignore them when analyzing a position. When two balanced piles are present in a possibly more complicated position, one can pretend that they aren’t there, precisely because whenever your opponent plays on one of them, you can copy the move on the other, and so any winning strategy for the position in which those piles are absent can be converted into a winning strategy in which the balanced piles are present.

This idea now provides a complete winning strategy in the case that all piles have height one or two at most. One wants to leave a position with an even number of piles of each height. If only one height has an odd number of piles, then take a whole pile of that height. And if there are odd numbers of piles both of height one and two, then turn a height-two pile into a pile of height one, and this will make them both even. So any unbalanced position can be balanced, and any move on a balanced position will unbalance it.

1+2+3 counts as balanced

Let’s now consider that there may be piles of height three. For example, consider the basic position with piles of height one, two and three. The observation to make here is that any move on this position can be replied to with a move that leaves it balanced (check it yourself to be sure!). It follows that this position is winning to leave for the other player (and so one should go second on $1+2+3$). It would be nice if we could consider this position itself as already balanced in some sense. Indeed, we may incorporate this situation into the balancing idea if we think of the pile of height three as really consisting of two subpiles, one of height two and one of height one. In this way, the Nim position 1+2+3 counts as balanced, since the 3 counts as 2+1, which balances the other stacks. The 1+2+3 position has two stacks of height two and two of height one, when one regards the stack of height three as having a substack of height two and a substack of height one.

This way of thinking produces a complete winning strategy for Nim positions involving piles of height at most three. (And this is a strategy that can be mastered even by very young children — a few years ago I had talked about Nim with much younger children, Math for six-year-olds: Win at Nim!, first-graders at my daughter’s school, and at that time we concentrated on posititions with piles of height at most three. Older kids, however, can handle the full strategy.) Namely, the winning strategy in this case is to strive to balance the position, to make an even number overall of piles of height one and two, where we count piles of height three as one each of one and two. If you always give your opponent a balanced position, then you will win! Faced with an unbalanced position, it is a fact that you can always find a balancing move, and any move on an balanced position will unbalance it. If the game is just starting, and you are deciding whether to go first or second, you should determine whether it is balanced yet or not. If it unbalanced, then you should go first and make the balancing move; if it is already balanced, then you should go second and adopt the copying strategy, in which you re-balance the position with each move.

The general winning strategy, of course, goes beyond three. The key idea is to realize that what is really going on when we represent $3$ as $2+1$ is that we are using the binary representation of the number $3$. To explain, I wrote the following numbers on the chalkboard $$1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ \cdots$$ and was very pleased when the kids immediately shouted out, “The powers of two!” I explained that any natural number can be expressed uniquely as a sum of distinct powers of two. Asked for a favorite number less than one hundred, one student suggested $88$, and together we calculated $$88=64+16+8,$$ which means that the binary representation of $88$ is $1011000$, which I read off as, “one $64$, no $32$s, one $16$, one $8$, no $4$s, no $2$s and no $1$s. This is just the same as thinking of $9572$ as 9 thousands, 5 hundreds, 7 tens and 2 ones, using the powers of ten. It is interesting to learn that one may easily count very high on one hand using binary, up to 1023 on two hands!

The general strategy is to view every Nim pile as consisting of subpiles whose height is a power of two, and to make sure that one leaves a position that is balanced in the sense that every power of two has an even number of such instances in the position. So we think of $3$ as really $2+1$ for the purposes of balancing; $4$ counts as itself because it is a power of two, but $5$ counts as $4+1$ and $6$ counts as $4+2$ and $7$ as $4+2+1$. Another way to describe the strategy is that we express all the pile heights in binary, and we want an even number of $1$s in each binary place position.

The mathematical facts to verify are (1) any move on a balanced position in this powers-of-two sense will cause it to become unbalanced, and (2) any unbalanced position can be balanced in one move. It follows that leaving balanced positions is a winning strategy, because the winning move of taking the last block is a balancing move rather than an unbalancing move.

One can prove statement (1) by realizing that when you move a single stack, the binary representation changes, and so whichever binary digits changed will now become unbalanced. For statement (2), consider the largest unbalanced power of two $2^k$ and move on any stack that contains a $2^k$ size substack. Since $2^k-1=111\cdots11$ in binary, one can attain any binary pattern for the smaller height stacks by removing between $1$ and $2^k$ many blocks. So one can balance the position.

As a practical matter, the proof of (2) also shows how one can find a (winning) balancing move, which can otherwise be difficult in some cases: look for the largest unbalanced power of two, and move on any pile containing such a subpile, making sure to leave a balanced position.

In most actual instances of Nim, the pile heights are rarely very tall, and so one is usually considering just $1$, $2$ and $4$ as the powers of two that arise. A traditional starting configuration has piles of height 1, 3, 5, and 7, and this position is balanced, because one may view it as: $1, 2+1, 4+1, 4+2+1$, and there are an even number of 1s, 2s and 4s.

It is interesting to consider also the Misère form of Nim, where one wants NOT to take the last block. This version of the game also has a secret mathematical strategy, which I shall reveal later on.

Challenge 1. What is the winning strategy in Misère Nim?

If you figure it out, please post a comment! I’ll post the solution later. One might naively expect that the winning strategy of Misère Nim is somehow totally opposite to the winning strategy of regular Nim, but in fact, the positions $1,2,3$ and $1,3,5,7$ are winning for the second player both in Nim and also in Misère Nim. Indeed, I claim that all nontrivial Nim positions that are winning for regular Nim (with a suitable meaning of “nontrivial”) are also winning for Misère Nim. Can you prove it?

Another interesting generalization, for the set-theorists, is to consider transfinite Nim, where the piles can have transfinite ordinal height. So we have finitely many piles of ordinal height, perhaps infinite, and a move consists of making any one pile strictly shorter. Since there are no infinite descending sequence of ordinals, the game will terminate in finitely many moves, and the winner is whowever removes the last block.

Challenge 2. Who wins the transfinite Nim game with piles of heights: $$1\qquad \omega+3\qquad \omega^\omega+5\qquad \omega^{\omega+3}+\omega^\omega\cdot3+\omega\cdot 5+7\qquad \epsilon_0\qquad \omega_1$$ and what are the winning moves? What is the general winning strategy for transfinite Nim?

Post your solutions! You can also see my solution and further discussion.

Solution to my transfinite epistemic logic puzzle, Cheryl’s Rational Gifts

Posted on April 25, 2015 by Joel David Hamkins

Thanks so much to everyone for trying out my transfinite epistemic logic puzzle, which I have given the name Cheryl’s Rational Gifts, on account of her gifts to Albert and Bernard. I hope that everyone enjoyed the puzzle. See the list of solvers and honorable mentions at the bottom of this post. Congratulations!

As I determine it, the solution is that

Albert has the number $100\frac38$, and

Bernard has the number $100\frac7{16}$.

Let me explain my reasoning and address a few issues that came up in the comments.

First, let’s understand the nature of the space of possible numbers that Cheryl describes, those of the form the form $$n-\frac{1}{2^k}-\frac{1}{2^{k+r}},$$ where $n$ and $k$ are positive integers and $r$ is a non-negative integer. Although this may seem complicated at first, in fact this set consists simply of a large number of increasing convergent sequences, one after the other. Specifically, the smallest of the numbers is $0=1-\frac12-\frac12$, and then $\frac14$, $\frac38$, $\frac7{16}$, and so on, converging to $\frac12$, which itself arises as $\frac12=1-\frac14-\frac14$. So the numbers begin with the increasing convergent sequence $$0 \quad\frac14\quad \frac38\quad \frac7{16}\quad\cdots\quad\to\quad \frac12.$$Immediately after this comes another sequence of points of the form $1-\frac14-\frac1{2^{2+r}}$, which converge to $\frac34$, which itself arises as $1-\frac18-\frac18$. So we have $$\frac12\quad \frac58\quad \frac{11}{16}\quad\frac{23}{32}\quad\cdots\quad\to\quad \frac34.$$Following upon this, there is a sequence converging to $\frac78$, and then another converging to $\frac{15}{16}$, and so on. Between $0$ and $1$, therefore, what we have altogether is an increasing sequence of increasing sequences of rational numbers, where the start of the next sequence is precisely the limit of the previous sequence.

Cheryl's numbers

The same pattern recurs between $1$ and $2$, between $2$ and $3$, and indeed between any positive integer $n$ and its successor $n+1$, for the numbers the occur between $n$ and $n+1$ are simply a translation of the numbers between $0$ and $1$. Thus, for every positive integer $k$ we have $n-\frac1{2^k}$ as the limit of the numbers $n-\frac{1}{2^k}-\frac{1}{2^{k+r}}$, as $r$ increases. Between any two non-negative integers, therefore, we have an increasing sequence of converging increasing sequences. Altogether, we have infinitely many copies of the picture between $0$ and $1$, which was infinitely many increasing convergent sequences, one after the other.

For those readers who are familiar with the ordinals, what this means is that the set of numbers forms a closed set of order type exactly $\omega^3$. We may associate the number $n-\frac{1}{2^k}-\frac{1}{2^{k+r}}$ with the ordinal number $\omega^2\cdot (n-1)+\omega\cdot (k-1)+r$, and observe that this correspondence is a (continuous) order-isomorphism of our numbers with the ordinals below $\omega^3$. In this way, we could replace all talk of the specific rational numbers in this puzzle with their corresponding ordinals below $\omega^3$ and imagine that Cheryl has actually given her friends ordinals rather than rationals. But to explain the solution, allow me to stick with the rational numbers.

The fact that Albert initially does not know who has the larger number implies that Albert does not have $0$, the smallest number overall, since if he were to have had $0$, then he would have known that Bernard’s must have been larger. Since then Bernard does not know, it follows that his number is neither $0$ nor $\frac14$, which is the next number, since otherwise he would have known that Albert’s number must have been larger. Since Albert continues not to know, we rule out the numbers up to $\frac38$ for him. And then similarly ruling out the numbers up to $\frac7{16}$ for Bernard. In this way, each step of the back-and-forth continuing denials of knowing eliminates the lowest remaining numbers from possibility.

Consequently, when Cheryl interrupts the first time, we learn that Albert and Bernard cannot have numbers on the first increasing sequence (below $\frac12$), since otherwise they would at some point come to know in that back-and-forth procedure who has the larger number, and so it wouldn’t be true that they wouldn’t know no matter how long they continued the back-and-forth, as Cheryl stated. Thus, after her remark, both Albert and Bernard know that both numbers are at least $\frac12$, which is the first limit point of the set of possible numbers.

Since at this point Albert states that he still doesn’t know who has the larger number, it cannot be that he has $\frac12$ himself, since otherwise he would have known that he had the smaller number. And then next since Bernard still doesn’t know, it follows that Bernard cannot have either $\frac12$ or $\frac58$, the next number. Thus, if they were to continue the iterative not-knowing-yet pronouncements, they would systematically eliminate the numbers on the second increasing sequence, which converges to $\frac34$. Because of Cheryl’s second interruption remark, therefore, it follows that their numbers do not appear on that second sequence, for otherwise they would have known by continuing that pattern long enough. Thus, after her remark, they both know that both numbers are at least $\frac34$.

And since Albert and Bernard in succession state that they still do not know, they have begun to eliminate numbers from the third sequence.

Consider now Cheryl’s contentful exasperated remark. What she says in the first part is that no matter how many times the three of them repeat that pattern, they will still not know. The content of this remark is precisely that neither of the two numbers can be on next sequence (the third), nor the fourth, nor the fifth and so on; they cannot be on any of the first $\omega$ many sequences (that is, below $1$), because if one of the numbers occurred on the $k^{th}$ sequence below $1$, as $1-\frac1{2^k}-\frac1{2^{k+r}}$, for example, then after $k-1$ repetitions of the three-way-pattern, it would no longer be true for Cheryl to say that no matter how long Albert and Bernard continued their back-and-forth they wouldn’t know, since they would indeed know after $r$ steps of that at that point. Thus, the first part of Cheryl’s remark implies that the numbers must both be at least $1$.

But Cheryl also says that the same statement would be true if she said it again. Thus, the numbers must not lie on any of the first $\omega$ many sequences above $1$. Those sequences converge to the limit points $1\frac12$, $1\frac34$, $1\frac78$ and so on. Consequently, after that second statement, everyone would know that the numbers must both be at least $2$. Similarly, after making the statement a third time, everyone knows the numbers must be at least $3$, and after the fourth time, everyone knows the numbers must be at least $4$.

Cheryl says that she could make the statement a hundred times altogether in succession (counting the time she has already said it as amongst the one hundred), and it would be true every time. Since each time she makes the statement, it eliminates precisely the possibility that one of the numbers is on any of the next $\omega$ many sequences, what everyone would know after the one hundredth pronouncement is precisely that both numbers are at least $100$. Even though she didn’t actually make the statement one hundred times, Albert and Bernard are entitled to know exactly that information even so, because she had said that the statement would be true every time, if she were to have said it one hundred times.

Note that it would be perfectly compatible with Cheryl making that statement one hundred times, if one of the numbers had been $100$, since each additional assertion simply eliminates the possibility that one of the numbers occurs on the sequences strictly before the next integer limit point, without eliminating the integer limit point itself.

Next Cheryl makes an additional statement, which it seems to me that some of the commentators did not give sufficient attention. Namely, she says, “And furthermore, even after my having said it altogether one hundred times in succession, you would still not know who has the larger number!” This statement gives additional epistemic information beyond the content of saying that the $100^{th}$ statement would be true. After the $100^{th}$ statement, according to what we have said, both Albert and Bernard would know precisely that both numbers are at least $100$. But Cheryl is telling them that they still would not know, even after the $100^{th}$ statement. Thus, it must be that neither Albert nor Bernard has $100$, since having that number is the only way they could know at that point who has the larger number. (Note also that Cheryl did not say that they would know that the other does not know, but only that they each would not know after the $100^{th}$ assertion, an issue that appeared to trip up some commentators. So she is making a common-knowledge assertion about what their individual epistemic states would be in that case.) The first few numbers after $100$ are: $$100\qquad 100\frac14\qquad 100\frac38\qquad 100\frac7{16}\qquad 100\frac{15}{32}\qquad\cdots\to\quad
100\frac12$$ So putting everything together, what everyone knows after Cheryl’s exasperated remark is that both numbers are at least $100\frac14$.

Since Albert still doesn’t know, it means his number is at least $100\frac 38$. Since Bernard doesn’t know after this, it means that Bernard cannot have either $100\frac14$ or $100\frac38$, since otherwise he would know that Albert’s is larger. So Bernard has at least $100\frac7{16}$.

But now suddenly, finally, Albert knows who has the larger number! How can this be? So far, all we knew is that Albert’s number was at least $100\frac 38$ and Bernard’s is at least $100\frac7{16}$. If Albert had $100\frac 38$, then indeed he would know that Bernard’s number is larger; but note also that if Albert had $100\frac7{16}$, then he would also know that Bernard must have the larger number (since he knows the numbers are different). But if Albert’s number were larger than $100\frac7{16}$, then he couldn’t know whether Bernard’s number was larger or not. So after Albert’s assertion, what we all know is precisely that Albert has either $100\frac38$ or $100\frac7{16}$.

But now, Bernard claims to know both numbers! How could he know which number Albert has? The only way that he can distinguish those two possibilities that we mentioned is if Bernard himself has $100\frac7{16}$, since this is the smallest possibility remaining for Bernard and if Bernard’s number were larger than that then Albert could have consistently had either $100\frac38$ or $100\frac7{16}$. Thus, because Bernard knows the numbers, it must be that Bernard has $100\frac7{16}$, which would eliminate this possibility for Albert.

So Albert has $100\frac38$ and Bernard has $100\frac7{16}$, and that is the solution of the puzzle.

A number of commentators solved the puzzle, coming to the same conclusion that I did, and so let me give some recognition here. Congratulations!

Neel Nanda
evanjhub
Aris Katsaris
Joe Dobrow and Kellen Kirchberg, also on math.SE
MatNat2 Jonas
Warwick Logic Reading Group
Gavi Dov
Paul Crowley (after fixing the fencepost)

Let me also give honorable mentions to the following people, who came very close.

Cheryl’s Rational Gifts: transfinite epistemic logic puzzle challenge!

Posted on April 18, 2015 by Joel David Hamkins

Can you solve my challenge puzzle?

Cheryl's numbers

Cheryl Welcome, Albert and Bernard, to my birthday party, and I thank you for your gifts. To return the favor, as you entered my party, I privately made known to each of you a rational number of the form $$n-\frac{1}{2^k}-\frac{1}{2^{k+r}},$$ where $n$ and $k$ are positive integers and $r$ is a non-negative integer; please consider it my gift to each of you. Your numbers are different from each other, and you have received no other information about these numbers or anyone’s knowledge about them beyond what I am now telling you. Let me ask, who of you has the larger number?

Albert I don’t know.

Bernard Neither do I.

Albert Indeed, I still do not know.

Bernard And still neither do I.

Cheryl Well, it is no use to continue that way! I can tell you that no matter how long you continue that back-and-forth, you shall not come to know who has the larger number.

Albert What interesting new information! But alas, I still do not know whose number is larger.

Bernard And still also I do not know.

Albert I continue not to know.

Bernard I regret that I also do not know.

Cheryl Let me say once again that no matter how long you continue truthfully to tell each other in succession that you do not yet know, you will not know who has the larger number.

Albert Well, thank you very much for saving us from that tiresome trouble! But unfortunately, I still do not know who has the larger number.

Bernard And also I remain in ignorance. However shall we come to know?

Cheryl Well, in fact, no matter how long we three continue from now in the pattern we have followed so far—namely, the pattern in which you two state back-and-forth that still you do not yet know whose number is larger and then I tell you yet again that no further amount of that back-and-forth will enable you to know—then still after as much repetition of that pattern as we can stand, you will not know whose number is larger! Furthermore, I could make that same statement a second time, even after now that I have said it to you once, and it would still be true. And a third and fourth as well! Indeed, I could make that same pronouncement a hundred times altogether in succession (counting my first time as amongst the one hundred), and it would be true every time. And furthermore, even after my having said it altogether one hundred times in succession, you would still not know who has the larger number!

Albert Such powerful new information! But I am very sorry to say that still I do not know whose number is larger.

Bernard And also I do not know.

Albert But wait! It suddenly comes upon me after Bernard’s last remark, that finally I know who has the larger number!

Bernard Really? In that case, then I also know, and what is more, I know both of our numbers!

Albert Well, now I also know them!

Question. What numbers did Cheryl give to Albert and Bernard?

If you can determine the answer, make comments below or post a link to your solution. I have posted my own solution on another post.

See my earlier transfinite epistemic logic puzzles, with solutions. These were inspired by Timothy Gowers’s example.

Now I know!

Posted on April 16, 2015 by Joel David Hamkins

Inspired by Timothy Gowers’s example, here is my transfinite epistemic logic problem.

First, let’s begin with a simple finite example.

Cheryl Hello Albert and Bernard! I have given you each a different natural number ($0, 1, 2, \ldots$). Who of you has the larger number?

Albert   I don’t know.

Bernard   I don’t know either.

Albert    Even though you say that, I still don’t know.

Bernard    And still neither do I.

Albert Alas, I continue not to know.

Bernard And also I do not know.

Albert Yet, I still do not know.

Bernard Aha! Now I know which of us has the larger number.

Albert In that case, I know both our numbers.

Bernard. And now I also know both numbers.

Question: What numbers do Albert and Bernard have?

Click for the solution.

Now, let us consider a transfinite instance. Consider the following conversation.

Cheryl I have given you each a different ordinal number, possibly transfinite, but possibly finite. Who of you has the larger ordinal?

Albert I don’t know.

Bernard I don’t know, either.

Albert Even though you say that, I still don’t know.

Bernard And still neither do I.

Albert Alas, I still don’t know.

Bernard And yet, neither do I.

Cheryl Well, this is becoming boring. Let me tell you that no matter how much longer you continue that back-and-forth, you still will not know the answer.

Albert Well, thank you, Cheryl, for that new information. However, I still do not know who has the larger ordinal.

Bernard And yet still neither do I.

Albert Alas, even now I do not know!

Bernard And neither do I!

Cheryl Excuse me; you two can go back and forth like this again, but let me tell you that no matter how much longer you continue in that pattern, you will not know.

Albert Well, ’tis a pity, since even with this further information, I still do not know.

Bernard Aha! Now at last I know who of us has the larger ordinal.

Albert In that case, I know both our ordinals.

Bernard. And now I also know both ordinals.

Question: What ordinals do they have?

Click for a solution.

See my next transfinite epistemic logic puzzle challenge!

Solutions.

For the first problem, with natural numbers, let us call the numbers $a$ and $b$, respectively, for Albert and Bernard. Since Albert doesn’t know at the first step, it means that $a\neq 0$, and so $a$ is at least $1$. And since Bernard can make this conclusion, when he announces that he doesn’t know, it must mean that $b$ is not $0$ or $1$, for otherwise he would know, and so $b\geq 2$. On the next round, since Albert still doesn’t know, it follows that $a$ must be at least $3$, for otherwise he would know; and then, because Bernard still doesn’t know, it follows that $b$ is at least $4$. The next round similarly yields that $a$ is at least $5$ and then that $b$ is at least $6$. Because Albert can undertake all this reasoning, it follows that $a$ is at least $7$ on account of Albert’s penultimate remark. Since Bernard announces at this point that he knows who has the larger number, it must be that Bernard has $6$ or $7$ and that Albert has the larger number. And since Albert now announces that he knows the numbers, it must be because Albert has $7$ and Bernard has $6$.
For the transfinite problem, let us call the ordinal numbers $\alpha$ and $\beta$, respectively, for Albert and Bernard. Since Albert doesn’t know at the first step, it means that $\alpha\neq 0$ and so $\alpha\geq 1$. Similarly, $\beta\geq 2$ after Bernard’s remark, and then $\alpha\geq 3$ and $\beta\geq 4$ and this would continue for some time. Because Cheryl says that no matter how long they continue, they will not know, it follows that both $\alpha$ and $\beta$ are infinite ordinals, at least $\omega$. But since Albert does not know at this stage, it means $\alpha\geq\omega+1$, and then $\beta\geq \omega+2$. Since Cheryl says again that no matter how long they continue that, they will not know, it means that $\alpha$ and $\beta$ must both exceed $\omega+k$ for every finite $k$, and so $\alpha$ and $\beta$ are both at least $\omega\cdot 2$. Since Albert still doesn’t know after that remark, it means $\alpha\geq\omega\cdot 2+1$. But now, since Bernard knows at this point, it must be that $\beta=\omega\cdot 2$ or $\omega\cdot 2+1$, since otherwise he couldn’t know. So Albert’s ordinal is larger. Since at this point Albert knows both the ordinals, it must be because Albert has $\omega\cdot 2+1$ and Bernard has $\omega\cdot 2$.

It is clear that one may continue in this way through larger transfinite ordinals. When the ordinals become appreciable in size, then it will get harder to turn it into a totally finite conversation, by means of Cheryl’s remarks, but one may succeed at this for quite some way with suitably obscure pronouncements by Cheryl describing various limiting processes of the ordinals. To handle any given (possibly uncountable) ordinal, it seems best that we should consider conversations of transfinite length.

A picture of logic, between mathematics and philosophy

Posted on January 29, 2015 by Joel David Hamkins

$A picture of logic between math and philosophy$ Years ago, when I was a student and young professor in Berkeley, one often heard it said that the subject of logic or at least metamathematics, according to the Tarski school, could be divided into four subjects: model theory, set theory, computability theory (then called recursion theory) and proof theory.

I have long felt that this taxonomy has become increasingly inadequate as a description, and so at the start of my course Logic for Philosophers yesterday, I tried to draw a somewhat fuller picture on the whiteboard. You can see my diagram above, showing the main parts of logic, occupying regions both within mathematics, within philosophy and within computer science, as well as filling out regions that might be considered between these subjects.

OK, this picture is a first attempt, and I’ll try to improve it. Please post comments and criticisms below.

It would certainly be possible to flesh out subareas of nearly all the subjects mentioned. We may imagine set theory, for example, broken into descriptive set theory, Borel equivalence relation theory, large cardinals, forcing, cardinal characteristics, and so on; and similarly we may break up the huge subjects of model theory, computability theory and proof theory. In particular, most subjects don’t fit neatly into a small region, since almost all the subjects have parts that touch areas very far away. Computability theory, for example, touches not only complexity theory and computer science, but also model theory in computable model theory, as well as reverse mathematics, foundations of mathematics, proof theory and so on. Category theory is a kind of diffuse superposition onto the entire diagram, with comparatively less direct interaction with these other areas. Proof theory should be closer to reverse mathematics and to model theory, and probably closer to mathematics.

The global choice principle in Gödel-Bernays set theory

Posted on December 3, 2014 by Joel David Hamkins

$\newcommand\Ord{\text{Ord}}
\newcommand\R{\mathbb{R}}
\newcommand\HOD{\text{HOD}}$

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.

The global choice principle. That is, there is a class function $F$ such that $F(x)\in x$ for every nonempty set $x$.
There is a bijection between $V$ and $\Ord$.
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.
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.
Every proper class is bijective with $\Ord$.
Every class injects into $\Ord$.
AC holds for sets and $\Ord$ injects into every proper class.
$\Ord$ surjects onto every class.
Every class is comparable with $\Ord$ by injectivity; that is, one injects into the other.
Any two classes are comparable by injectivity.

Proof.

($1\to 2$) 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.

QED

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 $W$.

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

Posted on November 20, 2014 by Joel David Hamkins

The Multiverse by Kaeltyk The 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. Reitz, Set-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.

$\newcommand\P{\mathbb{P}}\newcommand\Q{\mathbb{Q}}\newcommand\R{\mathbb{R}}\newcommand\of{\subset}\newcommand\cross{\times}$

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

Posted on October 20, 2014 by Joel David Hamkins

$\newcommand\dom{\text{dom}}
\newcommand\ran{\text{ran}}
\newcommand\restrict{\upharpoonright}$
At 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.

Update. (August 17, 2018) I’ve now realized how to eliminate the need for the axiom of choice in the arguments above. Namely, the main argument above shows that the principle of transfinite recursion implies the principle of well-ordered replacement, meaning the axiom of replacement for instances where the domain set is well-orderable. The point now is that in recent work with Alfredo Roque Freire, I have realized that The principle of well-ordered replacement is equivalent to full replacement over Zermelo set theory with foundation. We may therefore deduce:

Corollary. The principle of transfinite recursion is equivalent to the replacement axiom over Zermelo set theory with foundation.

We do not need the axiom of choice for this.

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

Posted on September 18, 2014 by Joel David Hamkins

It 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:

My highest-voted answer is to the question What are some reasonable-sounding statements that are independent of ZFC?, a very early question on MO asked by Anton Geraschenko, the founder of MathOverflow. In my answer, I explained that the the implication $\kappa<\lambda\implies 2^\kappa<2^\lambda$ is independent of ZFC. I have become fascinated by the reception of this answer, which surprised me, and I now often mention it in talks on the philosophy of set theory, for it provides an example of a principle that is (1) extremely appealing to non-set-theorists; (2) affirms many pre-theoretic expectations that one might have for a concept of “size” in mathematics; (3) known to be consistent with the axioms of set theory; and yet (4) almost universally rejected by set-theorists when it is proposed as a fundamental principle. So it provides a counterpoint to what might seem otherwise to be a reasonable philosophical position, that the first three properties would be grounds for general acceptance.
My very first question, Does every set admit a rigid binary relation? led to a paper by Justin Palumbo and myself, where we introduce the rigid relation principle, which asserts that indeed every set has a rigid binary relation, and which, we prove, is neither provable in ZF without the axiom of choice nor is it equivalent to the axiom of choice. Thus, it is a new weak choice principle.
There was a fascinating flock of questions on the topic of infinite chess, including especially Johan Wästlund’s question Checkmate in $\omega$ moves? and Richard Stanley’s question on the decidability of infinite chess, which I pondered for some time. Eventually, Philipp Schlicht and I proved that The mate-in-$n$ problem of infinite chess is decidable, and Cory Evans and I discovered very high Transfinite game values in infinite chess, finding new finite positions with value $\omega$, infinite positions with values up to $\omega^3$, and proving that the omega one of 3D infinite chess is true $\omega_1$.
I became truly obsessed with Ewan Delanoy’s question, who asked for an easy way to see that the countable models of set theory were not linearly ordered by embeddability (on our companion site math.stackexchange). After having worked furiously on the problem, I finally discovered to my astonishment, contrary to all our initial expectations, that this collection of models actually was linearly ordered after all! This work led directly to my theorem showing that every countable model of set theory is isomorphic to a submodel of its own constructible universe, and to a question, Can there can be an embedding $j:V\to L$ from the set-theoretic universe $V$ into the constructible universe $L$, when $V\neq L$?, which remains open.
Cole Leahy’s extremely interesting question on Pointwise algebraic models of set theory led to our joint paper on Algebraicity and implicit definability in set theory, and to further open questions on Imp, the inner model of implicitly constructible sets.

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

The rule-making game

Posted on September 2, 2014 by Joel David Hamkins

Let 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.

Joel David Hamkins

mathematics and philosophy of the infinite

Category Archives: Exposition

Every ordinal has only finitely many order-types for its final segments

The axiom of determinacy for small sets

Determinacy for proper-class clopen games is equivalent to transfinite recursion along proper-class well-founded relations

Open determinacy for proper class games implies Con(ZFC) and much more

Transfinite Nim

Win at Nim! The secret mathematical strategy for kids (with challange problems in transfinite Nim for the rest of us)

Solution to my transfinite epistemic logic puzzle, Cheryl’s Rational Gifts

Cheryl’s Rational Gifts: transfinite epistemic logic puzzle challenge!

Now I know!

A picture of logic, between mathematics and philosophy

The global choice principle in Gödel-Bernays set theory

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

Transfinite recursion as a fundamental principle in set theory

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

The rule-making game

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