# The rearrangement number: how many rearrangements of a series suffice to validate absolute convergence? Warwick Mathematics Colloquium, October 2018

This will be a talk for the Mathematics Colloquium at the University of Warwick, to be held October 19, 2018, 4:00 pm in Lecture Room B3.02 at the Mathematics Institute. I am given to understand that the talk will be followed by a wine and cheese reception. Abstract. The Riemann rearrangement theorem asserts that a series $\sum_n a_n$ is absolutely convergent if and only if every rearrangement $\sum_n a_{p(n)}$ of it is convergent, and furthermore, any conditionally convergent series can be rearranged so as to converge to any desired extended real value. How many rearrangements $p$ suffice to test for absolute convergence in this way? The rearrangement number, a new cardinal characteristic of the continuum, is the smallest size of a family of permutations, such that whenever the convergence and value of a convergent series is invariant by all these permutations, then it is absolutely convergent. The exact value of the rearrangement number turns out to be independent of the axioms of set theory. In this talk, I shall place the rearrangement number into a discussion of cardinal characteristics of the continuum, including an elementary introduction to the continuum hypothesis and an account of Freiling’s axiom of symmetry.

This talk is based in part on joint work with Andreas Blass, Will Brian, myself, Michael Hardy and Paul Larson.

# Models of set theory with the same reals and the same cardinals, but which disagree on the continuum hypothesis I’d like to describe a certain interesting and surprising situation that can happen with models of set theory.

Theorem. If $\newcommand\ZFC{\text{ZFC}}\ZFC$ set theory is consistent, then there are two models of $\ZFC$ set theory $M$ and $N$ for which

• $M$ and $N$ have the same real numbers $$\newcommand\R{\mathbb{R}}\R^M=\R^N.$$
• $M$ and $N$ have the ordinals and the same cardinals $$\forall\alpha\qquad \aleph_\alpha^M=\aleph_\alpha^N$$
• But $M$ thinks that the continuum hypothesis $\newcommand\CH{\text{CH}}\CH$ is true, while $N$ thinks that $\CH$ is false.

This is a little strange, since the two models have the set $\R$ in common and they agree on the cardinal numbers, but $M$ thinks that $\R$ has size $\aleph_1$ and $N$ will think that $\R$ has size $\aleph_2$.  In particular, $M$ can well-order the reals in order type $\omega_1$ and $N$ can do so in order-type $\omega_2$, even though the two models have the same reals and they agree that these order types have different cardinalities.

Another abstract way to describe what is going on is that even if two models of set theory, even transitive models, agree on which ordinals are cardinals, they needn’t agree on which sets are equinumerous, for sets they have in common, even for the reals.

Let me emphasize that it is the requirement that the models have the same cardinals that makes the problem both subtle and surprising. If you drop that requirement, then the problem is an elementary exercise in forcing: start with any model $V$, and first force $\CH$ to fail in $V[H]$ by adding a lot of Cohen reals, then force to $V[G]$ by collapsing the continuum to $\aleph_1$. This second step adds no new reals and forces $\CH$, and so $V[G]$ and $V[H]$ will have the same reals, while $V[H]$ thinks $\CH$ is true and $V[G]$ thinks $\CH$ is false. The problem becomes nontrivial and interesting mainly when you insist that cardinals are not collapsed.

In fact, the situation described in the theorem can be forced over any given model of $\ZFC$.

Theorem. Every model of set theory $V\models\ZFC$ has two set-forcing extensions $V[G]$ and $V[H]$ for which

• $V[G]$ and $V[H]$ have the same real numbers $$\newcommand\R{\mathbb{R}}\R^{V[G]}=\R^{V[H]}.$$
• $V[G]$ and $V[H]$ have the same cardinals $$\forall\alpha\qquad \aleph_\alpha^{V[G]}=\aleph_\alpha^{V[H]}$$
• But $V[G]$ thinks that the continuum hypothesis $\CH$ is true, while $V[H]$ thinks that $\CH$ is false.

Proof. Start in any model $V\models\ZFC$, and by forcing if necessary, let’s assume $\CH$ holds in $V$. Let $H\subset\text{Add}(\omega,\omega_2)$ be $V$-generic for the forcing to add $\omega_2$ many Cohen reals. So $V[H]$ satisfies $\neg\CH$ and has the same ordinals and cardinals as $V$.

Next, force over $V[H]$ using the forcing from $V$ to collapse $\omega_2$ to $\omega_1$, forming the extension $V[H][g]$, where $g$ is the generic bijection between those ordinals. Since we used the forcing in $V$, which is countably closed there, it makes sense to consider $V[g]$.  In this extension, the forcing $\text{Add}(\omega,\omega_1^V)$ and $\text{Add}(\omega,\omega_2^V)$ are isomorphic. Since $H$ is $V[g]$-generic for the latter, let $G=g\mathrel{“}H$ be the image of this filter in $\text{Add}(\omega,\omega_1)$, which is therefore $V[g]$-generic for the former. So $V[g][G]=V[g][H]$. Since the forcing $\text{Add}(\omega,\omega_1)$ is c.c.c., it follows that $V[G]$ also has the same cardinals as $V$ and hence also the same as in $V[H]$.

If we now view these extensions as $V[G][g]=V[H][g]$ and note that the coutable closure of $g$ in $V$ implies that $g$ adds no new reals over either $V[G]$ or $V[H]$, it follows that $\R^{V[G]}=\R^{V[H]}$. So the two models have the same reals and the same cardinals. But $V[G]$ has $\CH$ and $V[H]$ has $\neg\CH$, in light of the forcing, and so the proof is complete. QED

Let me prove the following surprising generalization.

Theorem. If $V$ is any model of $\ZFC$ and $V[G]$ is the forcing extension obtained by adding $\kappa$ many Cohen reals, for some uncountable $\kappa$, then for any other uncountable cardinal $\lambda$, there is another forcing extension $V[H]$ where $H$ is $V$-generic for the forcing to add $\lambda$ many Cohen reals, yet $\R^{V[G]}=\R^{V[H]}$.

Proof. Start in $V[G]$, and let $g$ be $V[G]$-generic to collapse $\lambda$ to $\kappa$, using the collapse forcing of the ground model $V$. This forcing is countably closed in $V$ and therefore does not add reals over $V[G]$. In $V[g]$, the two forcing notions $\text{Add}(\omega,\kappa)$ and $\text{Add}(\omega,\lambda)$ are isomorphic. Thus, since $G$ is $V[g]$-generic for the former poset, it follows that the image $H=g\mathrel{“}G$ is $V[g]$-generic for the latter poset. So $V[H]$ is generic over $V$ for adding $\lambda$ many Cohen reals. By construction, we have $V[G][g]=V[H][g]$, and since $g$ doesn’t add reals, it follows that $\R^{V[G]}=\R^{V[H]}$, as desired. QED

I have a vague recollection of having first heard of this problem many years ago, perhaps as a graduate student, although I don’t quite recall where it was or indeed what the construction was — the argument above is my reconstruction (which I have updated and extended from my initial post). If someone could provide a reference in the comments for due credit, I’d be appreciative.  The problem appeared a few years ago on MathOverflow.

# The rearrangement number: how many rearrangements of a series suffice to verify absolute convergence? Mathematics Colloquium at Penn, September 2016

This will be a talk for the Mathematics Colloquium at the University of Pennsylvania, Wednesday, September 14, 2016, 3:30 pm, tea at 3 pm, in the mathematics department. Abstract. The well-known Riemann rearrangement theorem asserts that a series $\sum_n a_n$ is absolutely convergent if and only if every rearrangement $\sum_n a_{p(n)}$ of it is convergent, and furthermore, any conditionally convergent series can be rearranged so as to converge to any desired extended real value. But how many rearrangements $p$ suffice to test for absolute convergence in this way? The rearrangement number, a new cardinal characteristic of the continuum, is the smallest size of a family of permutations, such that whenever the convergence and value of a convergent series is invariant by all these permutations, then it is absolutely convergent. The exact value of the rearrangement number turns out to be independent of the axioms of set theory. In this talk, I shall place the rearrangement number into a discussion of cardinal characteristics of the continuum, including an elementary introduction to the continuum hypothesis and, time permitting, an account of Freiling’s axiom of symmetry.

This talk is based in part on current joint work with Jörg Brendle, Andreas Blass, Will Brian, myself, Michael Hardy and Paul Larson.

Related MathOverflow post: How many rearrangements must fail to alter the value of a sum before you conclude that none do?

# Freiling’s axiom of symmetry, or throwing darts at the real line, Graduate Student Colloquium, April 2016

This will be a talk I’ll give at the CUNY Graduate Center Graduate Student Colloquium on Monday, April 11 (new date!), 2016, 4-4:45 pm.  The talk will be aimed at a general audience of mathematics graduate students. Abstract. I shall give an elementary presentation of Freiling’s axiom of symmetry, which is the principle asserting that if $x\mapsto A_x$ is a function mapping every real $x\in[0,1]$ in the unit interval to a countable set of such reals $A_x\subset[0,1]$, then there are two reals $x$ and $y$ for which $x\notin A_y$ and $y\notin A_x$.  To argue for the truth of this principle, Freiling imagined throwing two darts at the real number line, landing at $x$ and $y$ respectively: almost surely, the location $y$ of the second dart is not in the set $A_x$ arising from that of the first dart, since that set is countable; and by symmetry, it shouldn’t matter which dart we imagine as being first. So it may seem that almost every pair must fulfill the principle. Nevertheless, the principle is independent of the axioms of ZFC and in fact it is provably equivalent to the failure of the continuum hypothesis.  I’ll introduce the continuum hypothesis in a general way and discuss these foundational matters, before providing a proof of the equivalence of $\neg$CH with the axiom of symmetry. The axiom of symmetry admits natural higher dimensional analogues, such as the case of maps $(x,y)\mapsto A_{x,y}$, where one seeks a triple $(x,y,z)$ for which no member is in the set arising from the other two, and these principles also have an equivalent formulation in terms of the size of the continuum.

# The rearrangement number: how many rearrangements of a series suffice to verify absolute convergence? Vassar Math Colloquium, November 2015

This will be a talk for the Mathematics Colloquium at Vassar College, November 10, 2015, tea at 4:00 pm, talk at 4:15 pm, Rockefeller Hall 310 Abstract. The Riemann rearrangement theorem asserts that a series $\sum_n a_n$ is absolutely convergent if and only if every rearrangement $\sum_n a_{p(n)}$ of it is convergent, and furthermore, any conditionally convergent series can be rearranged so as to converge to any desired extended real value. How many rearrangements $p$ suffice to test for absolute convergence in this way? The rearrangement number, a new cardinal characteristic of the continuum introduced just recently, is the smallest size of a family of permutations, such that whenever the convergence and value of a convergent series is invariant by all these permutations, then it is absolutely convergent. The exact value of the rearrangement number turns out to be independent of the axioms of set theory. In this talk, I shall place the rearrangement number into a discussion of cardinal characteristics of the continuum, including an elementary introduction to the continuum hypothesis and an account of Freiling’s axiom of symmetry.

This talk is based in part on current joint work with Andreas Blass, Will Brian, myself, Michael Hardy and Paul Larson.

My Lecture Notes are available.

# The axiom of determinacy for small sets 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.

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