Universality and embeddability amongst the models of set theory, CTFM 2015, Tokyo, Japan

Posted on by
2

Tokyo Institute of TechnologyThis will be a talk for the Computability Theory and Foundations of Mathematics conference at the Tokyo Institute of Technology, September 7-11, 2015.  The conference is held in celebration of Professor Kazuyuki Tanaka’s 60th birthday.

Abstract. Recent results on the embeddability phenomenon and universality amongst the models of set theory are an appealing blend of ideas from set theory, model theory and computability theory. Central questions remain open.

A surprisingly vigorous embeddability phenomenon has recently been uncovered amongst the countable models of set theory. It turns out, for instance, that among these models embeddability is linear: for any two countable models of set theory, one of them embeds into the other. Indeed, one countable model of set theory $M$ embeds into another $N$ just in case the ordinals of $M$ order-embed into the ordinals of $N$. This leads to many surprising instances of embeddability: every forcing extension of a countable model of set theory, for example, embeds into its ground model, and every countable model of set theory, including every well-founded model, embeds into its own constructible universe.

V to LAlthough the embedding concept here is the usual model-theoretic embedding concept for relational structures, namely, a map $j:M\to N$ for which $x\in^M y$ if and only if $j(x)\in^N j(y)$, it is a weaker embedding concept than is usually considered in set theory, where embeddings are often elementary and typically at least $\Delta_0$-elementary. Indeed, the embeddability result is surprising precisely because we can easily prove that in many of these instances, there can be no $\Delta_0$-elementary embedding.

The proof of the embedding theorem makes use of universality ideas in digraph combinatorics, including an acyclic version of the countable random digraph, the countable random $\mathbb{Q}$-graded digraph, and higher analogues arising as uncountable Fraïssé limits, leading to the hypnagogic digraph, a universal homogeneous graded acyclic class digraph, closely connected with the surreal numbers. Thus, the methods are a blend of ideas from set theory, model theory and computability theory.

Results from Incomparable $\omega_1$-like models of set theory show that the embedding phenomenon does not generally extend to uncountable models. Current joint work of myself, Aspero, Hayut, Magidor and Woodin is concerned with questions on the extent to which the embeddings arising in the embedding theorem can exist as classes inside the models in question. Since the embeddings of the theorem are constructed externally to the model, by means of a back-and-forth-style construction, there is little reason to expect, for example, that the resulting embedding $j:M\to L^M$ should be a class in $M$. Yet, it has not yet known how to refute in ZFC the existence of a class embedding $j:V\to L$ when $V\neq L$. However, many partial results are known. For example, if the GCH fails at an uncountable cardinal, if $0^\sharp$ exists, or if the universe is a nontrivial forcing extension of some ground model, then there is no embedding $j:V\to L$. Meanwhile, it is consistent that there are non-constructible reals, yet $\langle P(\omega),\in\rangle$ embeds into $\langle P(\omega)^L,\in\rangle$.

CFTM 2015 extended abstract | Article | CFTM | Slides

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

Posted on by
8

droste_effect_clock_by_kiluncle-d4ya2gnI 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 by
4

Lewis ChessmenI 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 by
5

The Infinite Combat - Philipp Klinger

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.

  1. 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.
  2. 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 by
5

1000px-Apollonian_gasket.svg

$\newcommand\Tr{\text{Tr}}$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!

The hypnagogic digraph, with applications to embeddings of the set-theoretic universe, JMM Special Session on Surreal Numbers, Seattle, January 2016

Posted on by
1

JMM 2016 SeattleThis will be an invited talk for the AMS-ASL special session on Surreal Numbers at the 2016 Joint Mathematics Meetings in Seattle, Washington, January 6-9, 2016.

Abstract. The hypnagogic digraph, a proper-class analogue of the countable random $\mathbb{Q}$-graded digraph, is a surreal-numbers-graded acyclic digraph exhibiting the set-pattern property (a form of existential-closure), making it set-homogeneous and universal for all class acyclic digraphs. A natural copy of this canonical structure arises during the course of the usual construction of the surreal number line, using as vertices the surreal-number numerals $\{\ A \mid B\ \}$.  I shall explain the construction and elementary theory of the hypnagogic digraph and describe recent uses of it in connection with embeddings of the set-theoretic universe, such as in the proof that the countable models of set theory are linearly pre-ordered by embeddability.

Slides | schedule | related article | surreal numbers (Wikipedia)

Erin Carmody

Posted on by
9

Erin Carmody successfully defended her dissertation under my supervision at the CUNY Graduate Center on April 24, 2015, and she earned her Ph.D. degree in May, 2015. Her dissertation follows the theme of killing them softly, proving many theorems of the form: given $\kappa$ with large cardinal property $A$, there is a forcing extension in which $\kappa$ no longer has property $A$, but still has large cardinal property $B$, which is very slightly weaker than $A$. Thus, she aims to enact very precise reductions in large cardinal strength of a given cardinal or class of large cardinals. In addition, as a part of the project, she developed transfinite meta-ordinal extensions of the degrees of hyper-inaccessibility and hyper-Mahloness, giving notions such as $(\Omega^{\omega^2+5}+\Omega^3\cdot\omega_1^2+\Omega+2)$-inaccessible among others.

Erin Carmody

G+ profile | math genealogy | MathOverflow profile | NY Logic profile | ar$\chi$iv

Erin Carmody, “Forcing to change large cardinal strength,”  Ph.D. dissertation for The Graduate Center of the City University of New York, May, 2015.  ar$\chi$iv | PDF

Erin has accepted a professorship at Nebreska Wesleyan University for.the 2015-16 academic year.

 

Erin is also an accomplished artist, who has had art shows of her work in New York, and she has pieces for sale. Much of her work has an abstract or mathematical aspect, while some pieces exhibit a more emotional or personal nature. My wife and I have two of Erin’s paintings in our collection:
OceanIMG_0597

Transfinite Nim

Posted on by
7

Wooden blocksShall 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 by
27

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!

DSC00078

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.

DSC00076

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.

DSC00074

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.

 

The absolute truth about non-absolute truth, JAF – Weak Arithmetics Days, New York, July 2015

Posted on by
Reply

This will be a talk for the Journées sur les Arithmétiques Faibles – Weak Arithmetics Days conference, held in New York at the CUNY Graduate Center, July 7 – 9, 2015.

Abstract. I will discuss several fun theorems and folklore results illustrating that the satisfaction relation of first-order logic is less absolute than one might have expected. Two models of set theory, for example, can have the same natural numbers $\langle\mathbb{N},+,\cdot,0,1,<\rangle$, yet disagree on their theories of arithmetic truth; two models of set theory can have the same natural numbers and a computable linear order in common, yet disagree on whether it is a well-order and hence disagree about $\omega_1^{CK}$; two models of set theory can have the same natural numbers and the same reals, yet disagree on projective truth; two models of set theory can have a rank initial segment of the universe $\langle V_\delta,{\in}\rangle$ in common, yet disagree about whether it is a model of ZFC. These theorems and others can be proved with elementary classical model-theoretic methods. Indefinite arithmetic truthOn the basis of these observations, Ruizhi Yang (Fudan University, Shanghai) and I have argued that the definiteness of the theory of truth for a structure, even in the case of arithmetic, cannot be seen as arising solely from the definiteness of the structure itself in which that truth resides, but rather is a higher-order ontological commitment.

Slides |  Main article: Satisfaction is not absolute

The weakly compact embedding property, Apter-Gitik celebration, CMU 2015

Posted on by
2

This will be a talk at the Conference in honor of Arthur W. Apter and Moti Gitik at Carnegie Mellon University, May 30-31, 2015.  I am pleased to be a part of this conference in honor of the 60th birthdays of two mathematicians whom I admire very much.

Moti GitikArthur W. Apter

 

 

 

 

 

 

 

 

Abstract. The weakly compact embedding property for a cardinal $\kappa$ is the assertion that for every transitive set $M$ of size $\kappa$ with $\kappa\in M$, there is a transitive set $N$ and an elementary embedding $j:M\to N$ with critical point $\kappa$. When $\kappa$ is inaccessible, this property is one of many equivalent characterizations of $\kappa$ being weakly compact, along with the weakly compact extension property, the tree property, the weakly compact filter property and many others. When $\kappa$ is not inaccessible, however, these various properties are no longer equivalent to each other, and it is interesting to sort out the relations between them. In particular, I shall consider the embedding property and these other properties in the case when $\kappa$ is not necessarily inaccessible, including interesting instances of the embedding property at cardinals below the continuum, with relations to cardinal characteristics of the continuum.

This is joint work with Brent Cody, Sean Cox, myself and Thomas Johnstone.

Slides | Article | Conference web site

Carnegie Mellon University, College of Fine Arts

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

Posted on by
5

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!

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

 

The continuum hypothesis and other set-theoretic ideas for non-set-theorists, CUNY Einstein Chair Seminar, April, 2015

Posted on by
9

At Dennis Sullivan’s request, I shall speak on set-theoretic topics, particularly the continuum hypothesis, for the Einstein Chair Mathematics Seminar at the CUNY Graduate Center, April 27, 2015, in two parts:

  • An introductory background talk at 11 am, Room GC 6417
  • The main talk at 2 – 4 pm, Room GC 6417

I look forward to what I hope will be an interesting and fruitful interaction. There will be coffee/tea and lunch between the two parts.

Abstract. I shall present several set-theoretic ideas for a non-set-theoretic mathematical audience, focusing particularly on the continuum hypothesis and related issues.

At the introductory background talk, in the morning (11 am), I shall discuss and prove the Cantor-Bendixson theorem, which asserts that every closed set of reals is the union of a countable set and a perfect set (a closed set with no isolated points), and explain how it led to Cantor’s development of the ordinal numbers and how it establishes that the continuum hypothesis holds for closed sets of reals. We’ll see that there are closed sets of arbitrarily large countable Cantor-Bendixson rank. We’ll talk about the ordinals, about $\omega_1$, the long line, and, time permitting, we’ll discuss Suslin’s hypothesis.

At the main talk, in the afternoon (2 pm), I’ll begin with a discussion of the continuum hypothesis, including an explanation of the history and logical status of this axiom with respect to the other axioms of set theory, and establish the connection between the continuum hypothesis and Freiling’s axiom of symmetry. I’ll explain the axiom of determinacy and some of its applications and its rich logical situation, connected with large cardinals. I’ll briefly mention the themes and goals of the subjects of cardinal characteristics of the continuum and of Borel equivalence relation theory.  If time permits, I’d like to explain some fun geometric decompositions of space that proceed in a transfinite recursion using the axiom of choice, mentioning the open questions concerning whether there can be such decompositions that are Borel.

Dennis has requested that at some point the discussion turn to the role of set theory in the foundation for mathematics, compared for example to that of category theory, and I would look forward to that. I would be prepared also to discuss the Feferman theory in comparison to Grothendieck’s axiom of universes, and other issues relating set theory to category theory.

I know that you know that I know that you know…., CSI Undergraduate Conference on Research, Scholarship, and Performance, April 2015

Posted on by
3

UGCI shall give the plenary talk at the CSI Undergraduate Conference on Research, Scholarship, and Performance, April 30, 2015. My presentation will be followed by a musical performance.

This is a conference where undergraduate students show off their various scholarly and creative research projects, spanning all disciplines.

In my talk, I’ll present various logic puzzles that involve reasoning about knowledge, including knowledge of knowledge or knowledge of the lack of knowledge.  I’ll discuss the solution of Cheryl’s birthday problem, recently in the news, as well as other classic puzzles and some new ones.

It will be fun!

Slides

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

Posted on by
86

 

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.