Mereology, plural quantification and free lunches

It is sometimes claimed that arbitrary mereological fusions and plural quantification are a metaphysical free lunch, just a new way of talking without any deep philosophical (or at least metaphysical) commitments.

I think this is false.

Consider this Axiom of Choice schema for mereology:

1. If for every x and y such that ϕ(x) and ϕ(y), either x = y or x and y don’t overlap, and if every x such that ϕ(x) has a part y such that ψ(y), then there is a z such that for every x such that ϕ(x), there is common part y of x and z such that ψ(y).

Or this Axiom of Choice schema for pluralities:

2. If for all xx and yy such that ϕ(xx) and ϕ(yy) either xx and yy are the same or have nothing in common, then there are zz that have exactly one thing in common with every xx such that ϕ(xx).

If arbitrary mereological fusions and plural quantification are a metaphysical free lunch, just a handy way of talking, then whether (1) or (2) is correct is just a verbal question.

But (1) and (2) respectively imply mereological and plural Banach-Tarski paradoxes:

3. If z is a solid ball made of points, then it has five pairwise non-overlapping parts, of which the first two can be rigidly moved to be pairwise non-overlapping and compose another ball of the same size as z, and the last three can likewise be so moved.

4. If the xx are the points of a solid ball, then there are aa, bb, cc, dd and ee which have nothing pairwise in common and such that together they make up xx, and there are rigid motions that allow one to move aa and bb into pluralities that have nothing in common but make up a solid ball of the same size as xx and to move cc, dd and ee into pluralities that have nothing in common and make up another solid ball of the same size.

Conversely, assuming ZF set theory is consistent, there is no way to prove (3) and (4) if we do not have some extension to the standard axioms of mereology or plurals like the Axiom of Choice. The reason is that we can model pluralities and mereological objects with sets of points in three-dimensional space, and either (3) or (4) in that setting will imply the Banach-Tarski paradox for sets, while the Banach-Tarski paradox for sets is known not to be provable from ZF set theory without Choice.

But whether (3) or (4) is true is not a purely verbal question.

One reason it’s not a purely verbal question is intuitive. Banach-Tarski is too paradoxical for it or its negation to be a purely verbal thing.

Another is a reason that I gave in a previous post with a similar argument. Whether the Banach-Tarski paradox holds for sets is not a purely verbal question. But assuming that the Axiom of Separation can take formulas involving mereological terminology or plural quantification, each of (3) and (4) implies the Banach-Tarski paradox for sets.

"The whole is bigger than the part"

Some people don’t like Cantorian ways of comparing the sizes of sets because they want to have a “whole is bigger than the (proper) part” principle, denying which they consider to be counterintuitive.

Suppose that there is a relation ≤ which provides a way of comparing the sizes of sets of real numbers (or just the sizes of countable sets of real numbers) such that:

1. the comparison satisfies the “the whole is bigger than the part” principle, so that if A is a proper subset of B, then A < B

2. there are no incommensurable sets: given any A and B, at least one of A ≤ B and B ≤ A holds

3. the relation ≤ is transitive and reflexive.

Then the Banach-Tarski paradox follows from (a)–(c) without any use of the Axiom of Choice: there is a way to decompose a ball into a finite number of pieces and move them around to form two balls of the same size as the original. And Banach-Tarski feels like a direct violation of the "whole is bigger" principle!

Thus, intuitive as the “whole is bigger” principle is, the price of being able to compare the sizes of sets of real numbers in conformity with the principle is quite high. I suspect that most people who think that denying the “whole is bigger” principle also think Banach-Tarski is super problematic.

For our next observation, let’s add one more highly plausible condition:

4. the relation ≤ is weakly invariant under reflections of the real line: for any reflection ρ, we have A ≤ B if and only if ρA ≤ ρB.

Proposition: Conditions (a)–(d) are contradictory.

So, I think we should deny that, in the context of comparing the number of elements of a set, the whole is bigger than the proper part.

Proof of Proposition: Write A ∼ B iff A ≤ B and B ≤ A. Then I claim we have A ∼ ρA for any reflection ρ. For otherwise we either have A < ρA or ρA < A by (b). If we have A < ρA, then we also have ρA < ρ²A by (d), and since ρ²A = A, we have ρA < A, a contradiction. If we have ρA < A, then we have ρ²A < ρA by (d), and hence A < ρA, again a contradiction.

Since any translation τ can be made out of two reflections, it follows that A ∼ τA as well. Let τ be translation by one unit to the right. Then {0, 1, 2, ...} ∼ τ{0, 1, 2, ...} = {1, 2, 3, ...}, which contradicts (a).

Two kinds of non-measurable events

Non-measurable events are ones to which the probability function in the situation assigns no probability. Philosophically speaking, non-measurable events come in two varieties:

1. Non-measurable events that should not have any probability assignment.

2. Non-measurable events that should have a probability assignment.

Type (1) non-measurable events are the kinds of weird events that can be constructed from the Hausdorff and Banach-Tarski paradoxes, as well as perhaps (this is less clear) the Vitali non-measurable sets.

But I think there are also type (2) non-measurable events relative to standard choices of probability functions. For instance, suppose that in each universe of an infinite multiverse a fair coin is tossed countably infinitely often.

How likely is it that in at least one universe all the coin tosses are heads? If the universes form a countable infinity, classical probability theory gives an answer: zero. But if the universes form an uncountable infinity, classical probability theory gives no answer at all—the standard completed product measure makes the event be non-measurable. However, intuitively, there should be an answer in at least some cases. If the number of universes is much larger than the number of possible countable sequences of coin tosses (i.e., is much larger than 2^ω), we would expect the probability to be 1 or close to it. We can coherently extend the standard probability function to give that answer. But we can also coherently extend it to give a different answer, including the answer that the probability of an all-heads universe is zero, even if the number of universes is a gigantic infinite cardinality.

We don’t want to just make up an answer here. We want the answer to be derivable in some way resembling the proof of the theorem that if you toss a coin infinitely many times, you’ve got probability 1 of getting heads at least once.

I suppose we could take it to be a metaphysical axiom that if you have K disjoint collections each with M coin tosses, then if K and M are infinite and K > M, then with probability one at least one collection yields all heads. But it would be nice to have more than just intuition here, and in similar problems.

Set size and paradox

Some people want to be able to compare the sizes of sets in a way that respects the principle:

1. If A is a proper subset of B, then A ≤ B but not B ≤ A.

They do this in order to escape what they think are paradoxical consequences of the Cantorian way of comparing sizes. But from one paradox they fall into another. For the following can be proved without the Axiom of Choice:

2. If there is a transitive and reflexive relation ≤ between sets of reals (or just countable sets of reals) that satisfies (1), then the Banach-Tarski Paradox holds.

And the Banach-Tarski Paradox is arguably more paradoxical than the paradoxes of infinity that (1) is supposed to avoid.

Dutch Books and infinite sequences of coin tosses

Suppose we have an infinite sequence of independent and fair coins. A betting portfolio is a finite list of subsets of the space of outcomes (heads-tails sequences) together with a payoff for each subset. Assume:

1. Permutation: If a rational agent would be happy to pay x for a betting portfolio, and A is one of the subsets in the betting portfolio, then she would also be happy to pay x for a betting portfolio that is exactly the same but with A replaced by A*, where A* is isomorphic to A under a permutation of the coins.
2. Equivalence: A rational agent who is happy to pay x for one betting scenario, will be willing to accept an equivalent betting scenario---one that is certain to give the same payoff for each outcome---for the same price.
3. Great Deal: A rational agent will be happy to pay $1.00 for a betting scenario where she wins $1.25 as long as the outcome is not all-heads or all-tails.

Leveraging the Axiom of Choice and using the methods of the Banach-Tarski Paradox, one can then find two betting portfolios that the agent would be happy to accept for $1.00 each such that it is certain that if she accepts them both, she will win at most $1.25; hence she will accept a sure loss of at least $0.75. For details of a closely related result, see Chapter 6 in Infinity, Causation and Paradox (draft temporarily here).

So what to do? I think one should accept Causal Finitism, the doctrine that causal antecedents are always finite. Given Causal Finitism, one can't have a real betting scenario based an infinite number of coin tosses. Moreover, the only known way to operationalize the use of the Axiom of Choice in the proof in a betting scenario also involves infinite causal dependencies.

Another impossibility result for finitely additive probabilities and invariance

Consider a countably infinite sequence of fair and independent coin tosses. Given the Axiom of Choice, there is no finitely additive probability measure that satisfies these conditions:

1. It is defined for all sets of outcomes.
2. It agrees with the classical probabilities where these are defined.
3. It is invariant under permutations of coins.

(Sketch of proof: Index the coins with members of the free group of rank two. The members of the group then induce permutations of coins, and hence act on the space of outcomes. The set of non-trivial fixed points under that action has classical probability zero. Throwing that out, we can use a standard paradoxical decomposition of the free group of rank two to generate a paradoxical decomposition of the rest of our space of results--here Choice will be used--and that rules out the possibility of a finitely additive probability measure.)

Betting on paradoxical sets

Suppose that a point z will be uniformly randomly chosen on the surface of a sphere S and you are asked to place bets as to which set z is in. Then, plausibly:

1. If two sets A and B are equivalent under rotations about the center of the sphere, you should accept this offer: get three dollars if z is in A and pay two dollars if z is in B.

But now consider a paradoxical decomposition of the whole sphere S, by a version of the Banach-Tarski Paradox[note 1]. In this, the sphere is partitioned into two subsets C and D, each of which can be decomposed into a finite number of subsets that can be rotated to form the whole sphere. Applying (1) to each set in the decomposition of C and its rotation, you will accept a sequence of deals that adds up to:

2. If z is in C, you get three dollars and if z is in S you pay two dollars.

Repeating this with D's decomposition, you get a sequence of deals that adds up to:

3. If z is in D, you get three dollars and if z is in S you pay two dollars.

But of course if z is in C or D, it is also in S, and if it's in S, then it's in exactly one of C or D. It follows that the deal adds up to:

4. No matter what, you get three dollars and you pay four dollars.

So, repeated application of (1) yields an unacceptable conclusion.

One might say that this is an artifact of the fact that there is no finitely additive rotation-invariant probability measure on the sphere. But I think the above formulation is a little bit more telling. I make no reference to probabilities here. All I assume is (1), which is a very intuitive rationality judgment, namely that when one has two equivalent scenarios, one should accept an unequal bet between them that is in one's favor.

What to conclude? One conclusion might be that a single application of (1) is fine, but the sequence of applications needed to yield (4) is not.

My own conclusion, however, is that it is metaphysically impossible to have a betting scenario like the above. But why not? What's wrong with it? Well, one possibility is that space is necessarily discrete, but that doesn't seem very plausible to me.

My own preference, however, is to conclude that it is impossible to have anything causally depend on whether a random point (or a particle or the like) is in one of these weird sets that are found in the paradoxical decomposition of the sphere. Why is that? I think it's because it would in effect be a violation of causal finitism, the thesis that no event can causally depend on infinitely many things. But the full story here requires significant amounts of work to complete.

Regular probability comparisons imply the Banach-Tarski Paradox

Paper posted here (forthcoming in Synthese). Among goodies in the paper is a proof that the order extension principle (even in a weak form) implies the Banach-Tarski paradox, and a new argument against commensurability in decision theory. This is a very technical paper, so reader beware.

There is no regular approximately invariant finitely additive probability measure on all subsets of a cube or ball

For a totally ordered field K, say a hyperreal one, write x≈y (and say that they are approximately equal) provided that x−y is 0 or infinitesimal. A K-valued probability P defined for all subsets of Ω is said to be regular provided that P(A)>0 whenever A is non-empty. It is approximately rigid motion invariant provided that P(A)≈P(gA) for every rigid motion g and set A such that A∪gA⊆Ω. The following can be proved in Zermelo-Fraenkel (ZF) set theory without any Axiom of Choice:

Theorem 1. There is no totally ordered field K and a regular K-valued approximately rigid motion invariant finitely additive probability on all subsets of a ball or cube Ω.

If we delete "approximately", this follows from this.

The result follows from this post. Given such a regular measure we can define a preorder ≤ by letting A≤B if and only if P(A)≤P(B). By the Theorem from that post, it follows in ZF that Banach-Tarski is true. But Banach-Tarski implies that there is no approximately rigid motion invariant finitely additive probability on all subsets of a ball or cube.

(Why ball or cube? This saves me from having to worry about some edge effects given our definition of invariance.)

Another result, proved by similar methods:

Theorem 2. Let Ω be a subset of three-dimensional Euclidean space invariant under rotations about the origin 0. If K is a totally ordered field and P is a regular K-valued finitely additive probability on all subsets of Ω approximately invariant under rotations about the origin, then P({0})≈1.

Suppose now that we have a particle undergoing Brownian motion released at time t₀ at the origin, and then observed at time t₁. The probability of its being in some set at t₁ should be at least approximately invariant under rotations, and of course it is unacceptable to say that the probability that it is at the origin is approximately one—on the contrary, with approximately unit probability it is going to be away from the origin.

Update: Similar things hold for full-conditional probabilities, where approximate invariance is replaced with
invariance conditionally on the whole space (but there is no requirement of invariance conditionally on subsets of the space).

More details on Banach-Tarski and Axiom of Choice

Earlier I claimed that even without the Axiom of Choice, the proof of the Banach-Tarski Paradox yields paradoxical results, so that the Axiom of Choice isn't the source of paradoxicality.

Here is something more precise. Without the Axiom of Choice, one can prove the following.

Almost-BT. Suppose B is a punctured ball, i.e., a solid ball without a center point, and v is a translation sufficiently large that vB does not intersect B. Then there are rigid motions (combinations of rotations and translations) α and β such that B can be partitioned into an infinite collection U of countable subsets with the property that each member A of the partition U can be further partitioned into four subsets A₁, A₂, A₃ and A₄, such that (a) A₁ and αA₂ are a partition of A and (b) vA₃ and βA₄ are a partition of vA. (This follows from the method of proof of Theorem 4.5 in the Wagon book on Banach-Tarski.)

In other words, one can partition B into a bunch of sets, each of which can then be divided into four pieces and reassembled into two copies of itself oriented the same way. Moreover, all the reassembly can be done using the same rigid motions. When we add the Axiom of Choice to the mix, we can basically take B_i to be the union of "the" sets A_i ("the" is in quotation marks because the sets A_i aren't unique given A; if they were, we wouldn't need Choice), and then B=B₁∪B₂∪B₃∪B₄=B₁∪αB₂ and vB=vB₃∪βB₄, and we have Banach-Tarski for punctured balls. (It's a bit of extra work to get Banach-Tarski for non-punctured balls, but that extra work doesn't need Choice.) But even without the Axiom of Choice, Almost-BT seems pretty counterintuitive.