Goodman and Quine and transitive closure

In the previous post, I showed that Goodman and Quine’s counting method fails for objects that have too much overlap. I think (though the technical parts here are more difficult) that the same is true for their definition of the ancestral or transitive closure of a relation.

GQ showed how to define ancestors in terms of offspring. We can try to extend this definition to the transitive closure of any relation R over any kind of entities:

1. x stands in the transitive closure of R to y iff for every object u that has y as a part and that has as a part anything that stands in R to a part of u, there is a z such that Rxz and both x and z are parts of R.

This works fine if no relatum of R overlaps any other relatum of R. But if there is overlap, it can fail. For instance, suppose we have three atoms a, b and c, and a relation R that holds between a + b and a + b + c and between a and a + b. Then any object u that has a + b + c as a part has c as a part, and so (1) would imply that c stands in the transitive closure of R to a + b + c, which is false.

Can we find some other definition of transitive closure using the same theoretical resources (namely, mereology) that works for overlapping objects? No. Nor even if we add the “bigger than” predicate of GQ’s attempt to define “more”. We can say that x and y are equinumerous provided that neither is bigger than the other.

Let’s work in models made of an infinite number of mereological atoms. Write u ∧ v for the fusion of the common parts of both u and v (assuming u and v overlap), u ∨ v for the fusion of objects that are parts of one or the other, and u − v for the fusion of all the parts of u that do not overlap v (assuming u is not a part of v). Write |x| for the number of atomic parts of x when x is finite. Now make these definitions:

2. x is finite iff an atom is related to x by the transitive closure (with respect to the kind object) of the relation that relates an object to that object plus one atom.

3. Axyw iff x and y are finite and whenever x′ is equinumerous with x and does not overlap y, then x′ ∨ y is equinumerous with w. (This says |x| + |y| = |w|.)

4. Say that D_yuv iff A(u−y,u−y,v−y) (i.e., |v−y| = 2|u−y|) and either v does not overlap y or and u ∧ y is an atom or v and y overlap and u ∧ y consists of v ∧ y plus one atom. (This treats u and v as basically ordered pairs (u−y,u∧y) and (v−y,v∧y), and it makes sure that from the first pair to the second, the first component is doubled in size and the second component is decreased by one.)

5. Say that Q⁰yx iff y is finite and for some atom z not overlapping y we have y ∧ z related to something not overlapping x by the transitive closure of D_y. (This takes the pair (z,y), and applies the double first component and decrease second component relation described in (4) until the second component goes to zero. Thus, it is guaranteed that |x| = 2^|y|.)

6. Say that Qyx iff y is finite and Q⁰yx′ for some non-overlapping x′ that does not overlap y and that is equinumerous with x.

If I got all the details right, then Qyx basically says that |x| = 2^|y|.

Thus, we can define use transitive closure to define binary powers of finite cardinalities. But the results about the expressive power of monadic second-order logic with cardinality comparison say that we can only define semi-linear relations between finite cardinalities, which doesn’t allow defining binary powers.

Remark: We don’t need equinumerosity to be defined in terms of a primitive “bigger”. We can define equinumerosity for non-overlapping finite sets by using transitive closure (and we only need it for finite sets). First let T_yuv iff v − y exists and consists of u − y minus one atom and v ∧ y exists and consists of v ∧ y minus one atom. Then finite x and y are equinumerous₀ iff they are non-overlapping and x ∨ y has exactly two atoms or is related to an object with exactly two atoms by the transitive closure of T_yuv. We now say that x and y are equinumerous provided that they are finite and either x = y (i.e., they have the same atoms) or both x − y and y − x are defined and equinumerous₀.

No fix for Goodman and Quine's counting

In yesterday’s post, I noted that Goodman and Quine’s nominalist mereological definition of what it is to say that there are more cats than dogs fails if there are cats that are conjoint twins. This raises the question whether there is some other way of using the same ontological resources to generate a definition of “more” that works for overlapping objects as well.

I think the answer is negative. First, note that GQ’s project is explicitly meant to be compatible with there being a finite number of individuals. In particular, thus, it needs to be compatible with the existence of mereological atoms, individuals with no proper parts, which every individual is a fusion of. (Otherwise, there would have to be no individuals or infinitely many. For every individual has an atom as a part, since otherwise it has an infinite regress of parts. Furthermore, every individual must be a fusion of the atoms it has as parts, otherwise the supplementation axiom will be violated.) Second, GQ’s avail themselves of one non-mereological tool: size comparisons (which I think must be something like volumes). And then it is surely a condition of adequacy on their theory that it be compatible with the logical possibility that there are finitely many individuals, every individual is a fusion of its atoms and the atoms are all the same size. I will call worlds like that “admissible”.

So, here are GQ’s theoretical resources for admissible worlds. There are individuals, made of atoms, and there is a size comparison. The size comparison between two individuals is equivalent to comparing the cardinalities of the sets of atoms the individuals are made of, since all the atoms are the same size. In terms of expressive power, their theory, in the case of admissible worlds, is essentially that of monadic second order logic with counting, MSO(#), restricted to finite models. (I am grateful to Allan Hazen for putting me on to the correspondence between GQ and MSO.) The atoms in GQ correspond to objects in MSO(#) and the individuals correspond to (extensions of) monadic predicates. The differences are that MSO(#) will have empty predicates and will distinguish objects from monadic predicates that have exactly one object in their extension, while in GQ the atoms are just a special (and definable) kind of individual.

Suppose now that GQ have some way of using their resources to define “more”, i.e., find a way of saying “There are more individuals satisfying F than those satisfying G.” This will be equivalent to MSO(#) defining a second-order counting predicate, one that essentially says “The set of sets of satisfiers of F is bigger than the set of sets of satisfiers of G”, for second-order predicates F and G.

But it is known that the definitional power of MSO(#) over finite models is precisely such as to define semi-linear sets of numbers. However, if we had a second-order counting predicate in MSO(#), it would be easy to define binary exponentiation. For the number of objects satisfying predicate F is equal to two raised to the power of the number of objects satisfying G just in case the number of singleton subsets of F is equal to the number of subsets of G. (Compare in the GQ context: the number of atoms of type F is equal to two the power of the number of atoms of type G provided that the number of atoms of type F is one plus the number of individuals made of the atoms of type G.) And of course equinumerosity can be defined (over finite models) in terms of “more”, while the set of pairs (n,2ⁿ) is clearly not semi-linear.

One now wants to ask a more general question. Could GQ define counting of individuals using some other predicates on individuals besides size comparison? I don’t know. My guess would be no, but my confidence level is not that high, because this deals in logic stuff I know little about.

Goodman and Quine and shared bits

Goodman and Quine have a clever way of saying that there are more cats than dogs without invoking sets, numbers or other abstracta. The trick is to say that x is a bit of y if x is a part of y and x is the same size as the smallest of the dogs and cats. Then you’re supposed to say:

1. Every object that has a bit of every cat is bigger than some object that has a bit of every dog.

This doesn’t work if there is overlap between cats. Imagine there are three cats, one of them a tiny embryonic cat independent of the other two cats, and the other two are full-grown twins sharing a chunk larger than the embryonic cat, while there are two full-grown dogs that are not conjoined. Then a bit is a part the size of the embryonic cat. But (assuming mereological universalism along with Goodman and Quine) there is an object that has a bit of every cat that is no bigger than any object has a bit of every dog. For imagine an object that is made out of the embryonic cat together with a bit that the other two cats have in common. This object is no bigger than any object that has a bit of each of the dogs.

It’s easy to fix this:

2. Every object that has an unshared bit of every cat is bigger than some object that has an unshared bit of every dog,

where an unshared bit is a bit x not shared between distinct cats or distinct dogs.

But this fix doesn’t work in general. Suppose the following atomistic thesis is true: all material objects are made of equally-sized individisible particles. And suppose I have two cubes on my desk, A and B, with B having double the number of particles as A. Consider this fact:

4. There are more pairs of particles in A than particles in B.

(Again, Goodman and Quine have to allow for objects that are pairs of particles by their mereological universalism.) But how do we make sense of this? The trick behind (1) and (2) was to divide up our objects into equally-sized pieces, and compare the sizes. But any object made of the parts of all the particles in B will be the same size as B, since it will be made of the same particles as B, and hence will be bigger than any object made of parts of A.

Comparing sizes of infinite sets

Some people want to be able to compare the sizes of infinite sets while preserving the proper subset principle that holds for finite sets:

1. If A is a proper subset of B, then A < B.

We also want to make sure that our comparison agrees with how we compare finite sets:

2. If A and B are finite, then A ≤ B if and only if A has no more elements than B.

For simplicity, let’s just work with sets of natural numbers. Then there is a total preorder (total, reflexive and transitive relation) ≤ on the sets of natural numbers (or on subsets of any other set) that satisfies (1) and (2). Moreover, we can require the following plausible weak translation invariance principle in addition to (1) and (2):

3. A ≤ B if and only if 1 + A ≤ 1 + B,

where 1 + C is the set C translated one unit to the right: 1 + C = {1 + n : n ∈ C}. (See the Appendix for the existence proofs.) So far things are sounding pretty good.

But here is another plausible principle, which we can call discreteness:

4. If A and C differ by a single element, then there is no B such that A < B < C.

(I write A < B provided that A ≤ B but not B ≤ A.) When two sets differ by a single element, intuitively their sizes should differ by one, and sizes should be multiples of one.

Fun fact: There is no total preorder on the subsets of the natural numbers that satisfies the proper subset principle (1), the weak translation invariance principle (3) and the discreteness principle (4).

The proof will be given in a bit.

One way to try to compare sets that respects the subset principle (1) would be to use hypernatural numbers (which are the extension of the natural numbers to the context of hyperreals).

Corollary 1: There is no way to assign a hypernatural number s(A) to every set A of natural numbers such that (a) s(A) < s(B) whenever A ⊂ B, (b) s(A) − s(B) = s(1+A) − s(1+B), and (c) if A and B differ by one element, then |s(A)−s(B)| = 1.

For if we had such an assignment, we could define A ≤ B if and only if s(A) ≤ s(B), and we would have (1), (3) and (4).

Corollary 2: There is no way to assign a hyperreal probability P for a lottery with tickets labeled with the natural numbers such that (a) each individual ticket has equal non-zero probability of winning α, (b) P(A) − P(B) and P(1+A) − P(1+B) are always either both negative, both zero, or both positive, and (c) no two distinct probabilities of events differ by less than α.

Again, if we had such an assignment, we could define A ≤ B if and only if P(A) ≤ P(B), and we would have (1), (3) and (4).

I will now prove the fun fact. The proof won’t be the simplest possible one, but is designed to highlight how wacky a total preorder that satisfies (1) and (4) must be. Suppose we have such a total preorder ≤. Let A_n be the set {n, 100 + n, 200 + n, 300 + n, ...}. Observe that A₁₀₀ = {100, 200, 300, 400, ...} $ is a proper subset of A₀ = {0, 100, 200, 300, ...}, and differs from it by a single element. Now let’s consider how the elegant sequence of shifted sets A₀, A₁, ..., A₁₀₀ behaves with respect to the preorder ≤. Because A_n + 1 = 1 + A_n, if we had (3), the order relationship between successive sets in the series would always be the same. Thus we would have exactly one of these three options:

1. A₀ ≈ A₁ ≈ ... ≈ A₁₀₀

2. A₀ < A₁ < ... < A₁₀₀

3. A₀ > A₁ > ... > A₁₀₀,

where A ≈ B means that A ≤ B and B ≤ A. But (i) and (ii) each contradict (1), since A₁₀₀ is a proper subset of A₀, while (iii) contradicts (4) since A₀ and A₁₀₀ differ by one element.

This completes the proof. But we can now think a little about what the ordering would look like if we didn’t require (3). The argument in the previous paragraph would still show that (i), (ii) and (iii) are impossible. Similarly, A₀ ≥ A₁ ≥ ... ≥ A₁₀₀ is impossible, since A₁₀₀ < A₀ by (1). That means we have two possibilities.

First, we might have A₀ ≤ A₁ ≤ ... ≤ A₁₀₀. But because A₀ and A₁₀₀ differ by one element, by (4) it follows that exactly one of these ≤ is actually strict. Thus, in the sequence A₀, A₁, ..., A₁₀₀ suddenly there is exactly one point at which the size of the set goes up by one. This is really counterintuitive. We are generating our sequence of sets by starting with A₀ and then shifting the set over to the right by one (since $A_{n+1}=1+A_n), and suddenly the size jumps.

The second option is we don’t have monotonicity at all. This means that at some point in the sequence we go up and at some other point we go down: there are m and n between 0 and 99 such that A_m < A_m + 1 and A_n > A_n + 1. This again is really counterintuitive. All these sets look alike: they consist in an infinite sequence of points 100 units apart, just with a different starting point. But yet the sizes wobble up and and down. This is weird!

This suggests to me that the problem lies with the subset principle (1) or possibly with discreteness (4), not with the details of how to formulate the translation invariance principle (3). If we have (1) and (4) things are just too weird. I think discreteness is hard to give up on: counting should be discrete—two sets can’t differ in size by, say, 1/100 or 1/2. And so we are pressed to give up the subset principle (1).

Appendix: Existence proofs

Let U be any set. Let ∼ be the equivalence relation on subsets of U defined by A ∼ B if and only if either A = B or A and B are finite and of the same cardinality. The subset relation yields a partial order on the ∼-equivalence classes, and by the Szpilrajn extension theorem extends to a total order. We can use this total order on the equivalence classes of subsets to define a total preorder on the subsets, and this will satisfy (1) and (2).

If we want (3), let U be the integers, and instead of the Szpilrajn extension theorem, use Theorem 2 of this paper.

The proof of the “Fun Fact” is really easy. Suppose we have such a total preorder ≤. Let A = {2, 4, 6, ...}, B = {1, 2, 3, 4, ...} and C = {0, 2, 4, 6, 8, ...}. By (1), we have A < C. Suppose first that B ≤ A. Then C = 1 + B ≤ 1 + A = B by (3). Hence C ≤ A by transitivity, contradicting A < C. So A < B by totality. Thus B < C by (3). Since A and C differ by one element, this contradicts (4).

Counting with plural quantification

I’ve been playing with the question of what if anything we can say with plural quantification that we can’t say with, say, sets and classes.

Here’s an example. Plural quantification may let us make sense of cardinality comparisons that go further than standard methods. For instance, if our mathematical ontology consists only of sets, we can still define cardinality comparisons for pluralities of sets:

1. Suppose the xx and the yy are pluralities of sets. Then |xx| ≤ |yy| iff there are zz that are an injective function from the xx to the yy.

What is an injective function from the xx to the yy? It is a plurality, the zz, such that each of the zz is an ordered pair of classes, and such that for any a among the xx there is unique b among the yy such that (a,b) is among the zz and for any b among the yy there is at most one a among the xx such that [a,b] is among the zz.

This lets us say stuff like:

2. There are more sets than members of any set.

Or if our mathematical ontology includes sets and classes, we can compare the cardinalities of pluralities of classes using (1), as long as we can define an ordered pair of classes—which we can, e.g., by identifying the ordered pair of a and b with the class of all ordered pairs (i,x) where i = 0 and x ∈ a or where i = 1 and x ∈ b.

This would then let us say (and prove using a variant of Cantor’s diagonal argument, assuming Comprehension for pluralities):

3. There are more classes than sets.

An argument that there are more positive odd numbers than positive even ones

Suppose that we embrace these intuitive theses about sets of natural numbers:

1. If A is a proper subset of B, then B has more members than A.

2. If A has at least as many members as B, then 1 + A has at least as many members as 1 + B, where 1 + C = {1 + c : c ∈ C} is C shifted over by one to the right.

3. Either there are more odd positive numbers than even positive numbers or there are at least as many even positive numbers as odd positive numbers.

Let O be the positive odd numbers and E be the positive even numbers. Write A ≲ B to mean that B has at least as many members as A, and write A ∼ B to mean that they have the same number of members. For a reductio, suppose that O ≲ E. Then 1 + O ≲ 1 + E by (2). But 1 + O = E. Thus, O ≲ E ≲ 1 + E. But 1 + E is all the odd numbers starting with three, which is a proper subset of O, which contradicts (1). So we do not have O ≲ E, and hence by (3):

4. There are more odd positive numbers than even positive numbers!

Of course, the usual Cantorian way of comparing sizes of sets rejects (1).

I think the non-Cantorian’s best bet is either to embrace the conclusion (4) or to deny the special case of totality of comparison in (3). In either case, the non-Cantorian needs to deny the intuitive claim that:

5. There are equally many odd positive numbers and even positive numbers.

Note that (2) does not say or imply that 1 + A has the same number of members as A. Since that would imply that {1, 2, 3, ...} and {2, 3, 4, ...} have the same number of members, that would beg the question against the typical non-Cantorian for whom (1) is a central intuition. One might also wonder whether there is a way of comparing sets of natural numbers that satisfies (1)–(3): the answer is yes (even with (3) generalized to all pairs of sets of naturals).

More on presentism and decisions

You have seven friends, isolated from each other for a week. And you have a choice between these three options:

1. In four days, all of your friends will experience an innocent pleasure P at the same time.

2. Over the next week, each day a different one of your friends will experience P.

3. You presently experience an innocent pleasure whose magnitude is twice that of P.

It seems like a good idea to go for options 1 or 2 over option 3. But there is very little reason to prefer 1 over 2 or 2 over 1.

On eternalism, the parity between 1 and 2 makes perfect sense: in both cases, reality will contain seven copies of P, and the only difference is between how the copies are arranged in spacetime. And it also makes perfect sense that 1 or 2 is a better choice than 3: reality on 1 or 2 contains 3.5 times as much innocent pleasure.

But on presentism, I think it is difficult to explain these judgments. First, it’s difficult to explain why the sacrifice of 3 is worth it: a real, because present, pleasure is being sacrificed for a bunch of unreal, because future, pleasures. (Growing block has this problem, too.)

Now, if the choice is between 1 and 3, then at least the presentist can say this:

• On option 1, there will be an occurrence of 3.5 times the pleasure that would have occurred on option 3.

I am dubious that it makes sense to compare the future pleasure to the present one on presentism, but let’s grant that for the sake of the argument.

But now suppose the choice is between 2 and 3. Then, one cannot say there will be 3.5 times the pleasure. Rather:

• On option 2, on seven occasions, there will be half of the pleasure of option 3.

But the locution “on seven occasions” is misleading. For it makes it sound like there will be seven of something valuable. But there won’t be seven of something. Rather:

• There will be one of P to friend 1, and there will be one of P to friend 2, and so on.

But one cannot conjoin these “will be” claims into a single:

• There will be one of P to friend 1 and one of P to friend 2, and so on.

For that will never happen.

The deep point here is this. Cross-time counting on presentism is logically quite different from synchronic counting. In fact, in a sense it’s not “counting” at all, for there won’t be and has not been that number of items. One way to see the point is to compare the logical analysis of synchronic and cross-time counting claims on presentism:

• “There are (presently) two unicorns”: There exist x and y such that x is a unicorn and y is a unicorn and x ≠ y and for all z if z is a unicorn, then z = x or z = y.

• “There are (cross-time) two unicorns”: It was, is or will be the case that: There exists x such that x is a unicorn and it was, is or will be the case that there exists y such that y is a unicorn and x ≠ y, and it was, is and will be the case that for every z if z is unicorn, then z = x or z = y.

These are logically very different claims.

(I am also a little worried about the technical details of the cross-time identity claims on presentism, by the way.)

Fake counting

When someone’s walking speed is two miles per hour, there are not two things, “one mile per hour walkings”, that are present.

When we say that a sculpture has three dimensions, we are not saying there are exactly three things—dimensions?—that are present in it. But are there not width, height and depth? In a way. But rotate the sculpture by 45 degrees, and “width”, “height” and “depth” refer to measurement along three other axes. There are, it seems, infinitely many axes along which the sculpture can be non-trivially measured.

These are examples of what one might call “fake counting”. We speak as if there were n of something, but the following argument is invalid:

1. There are n Fs.

2. n ≥ 1.

3. So, there are some Fs.

And, similarly, this is invalid:

4. There are exactly two Fs.

5. So, ∃x∃y(F(x)&F(y)&∀z(F(z)→(z = x ∨ z = y))).

In fake counting of Fs, there is counting involved, but it is not counting of Fs. For instance, when we say that the sculpture has three dimensions, we mean something like this:

• there are three mutually perpendicular axes such that the sculpture has non-zero extent along each of them, but there are no four such axes.

So, there is a counting of axes, but it is not a counting of dimensions. If we were counting dimensions, we would have to have say what the first one is, what the second one is and what the third one is, and as the rotation thought experiment shows, that doesn’t work. And the counting of axes doesn’t involve counting axes overall, but rather axes in a particular set of them.

We need to beware of fake counting when making metaphysical arguments for the existence of entities of some sort. For instance, topologists have ways of “counting holes”. But topological properties are invariant under deformations. Now, imagine a pancake with, as we would say, “one hole in the middle”. Well, however we distort the pancake, it has one topological hole. But if we ask where that hole is, there is no topological answer to it (in the animation below, is the hole outlined in red or in blue?). So, topological hole counting is fake counting.

1+1=3 or 2+2=4

On numerical-sameness-without-identity views, two entities that share their matter count as one when we are counting objects.

Here is a curious consequence. Suppose I have a statue of Plato made of bronze with the nose broken off and lost. I make up a batch of playdough, sculpt a nose out of it and stick it on. The statue of Plato survives the restoration, and a new thing has been added, a nose. But now notice that I have three things, counting by sameness:

• The statue of Plato

• The lump of bronze

• The lump of playdough.

Yet I only added one thing, the lump of playdough or the nose that is numerically the same (without being identical) as it. So, it seems, 1+1=3.

Now, it is perfectly normal to have cases where by adding one thing to another I create an extra thing. Thus, I could have a lump of bronze and a lump of playdough and they could come together to form a statue, with neither lump being a statue on its own. A new entity can be created by the conjoining of old entities. But that’s not what happens in the case of the statue of Plato. I haven’t created a new entity. The statue was already there at the outset. And I added one thing.

Maybe, though, what should be said is this: I did create a new thing, a lump of bronze-and-playdough. This thing didn’t exist before. It is now numerically the same as the statue of Plato, which isn’t new, but it is still itself a new thing. I am sceptical, however, whether the lump of bronze-and-playdough deserves a place in our ontology. We have unification qua statue, but qua lump it’s a mere heap.

Suppose we do allow, however, that I created a lump of bronze-and-playdough. Then we get another strange consequence. After the restoration, counting by sameness:

• There are two things that I created: the nose and the lump of bronze-and-playdough

• There are two things that I didn’t create: the statue of Plato and the lump of bronze.

But there are only three things. Which makes it sound like 2+2=3. That’s perhaps not quite fair, but it does seem strange.

Sameness without identity

Mike Rea’s numerical-sameness-without-identity solution to the problem of material constitution holds that the statue and the lump have numerical sameness but do not have identity. Rea explicitly says that numerical sameness implies sharing of all parts but not identity.

Does Rea here mean: sharing of all parts, proper or improper? It had better not be so. For improper parthood is transitive.

Proposition. If improper parthood is transitive and x and y share all their parts (proper and improper), then x = y.

Proof: But suppose that x and y share all parts. Then since x is a part of x, x is a part of y, and since y is a part of y, y is a part of x. Moreover, if x ≠ y, then x is a proper part of y and y is a proper part of x. Hence by transitivity, x would be a proper part of x, which is absurd, so we cannot have x ≠ y. □

So let’s assume charitably that Rea means the sharing of all proper parts. This is perhaps coherent, but it doesn’t allow Rea to preserve common sense in Tibbles/Tib cases. Suppose Tibbles the cat loses everything below the neck and becomes reduced to a head in a life support unit. Call the head “Head”. Then Head is a proper part of Tibbles. The two are not identical: the modal properties of heads and cats are different. (Cats can have normal tails; heads can’t.) This is precisely the kind of case where Rea’s sameness without identity mechanism should apply, so that Head and Tibbles are numerically the same without identity. But Tibbles has Head as a proper part and Head does not have Head as a proper part. But that means Tibbles and Head do not share all their proper parts.

Here may be what Rea should say: if x and y are numerically the same, then any part of the one is numerically the same as a part of the other. This does, however, have the cost that the sharing-of-parts condition now cannot be understood by someone who doesn’t already understand sameness without identity.

Counting goods

Suppose I am choosing between receiving two goods, A and B, or one, namely C, where all the goods are equal. Obviously, I should go for the two. But why?

Maybe what we should say is this. Since A is at least as good as C, and B is non-negative, I have at least as good reason to go for the two goods as to go for the one. This uses the plausible assumption that if one adds a good to a good, one gets something at least as good. (It would be plausible to say that one gets something better, but infinitary cases provide a counterexample.) But there is no parallel argument that it is at least as good to go for the one good as to go for the two. Hence, it is false that I have at least as good reason to go for the one as to go for the two. Thus, I have better reason to go for the two.

This line of thought might actually solve the puzzles in these two posts: headaches and future sufferings. And it's very simple and obvious. But I missed it. Or am I missing something now?

Counting infinitely many headaches

If the worries in this post work, then the argument in this one needs improvement.

Suppose there are two groups of people, the As and the Bs, all of whom have headaches. You can relieve the headaches of the As or of the Bs, but not both. You don’t know how many As or Bs there are, or even whether the numbers are finite or finite. But you do know there are more As than Bs.

Obviously:

1. You should relieve the As’ headaches rather than the Bs’, because there are more As than Bs.

But what does it mean to say that there are more As than Bs? Our best analysis (simplifying and assuming the Axiom of Choice) is something like this:

2. There is no one-to-one function from the As to the Bs.

So, it seems:

3. You should relieve the As’ headache rather than the Bs’, because there is no one-to-one function from the As to the Bs.

For you should be able to replace an explanation by its analysis.

But that’s strange. Why should the non-existence of a one-to-one function from one set or plurality to another set or plurality explain the existence of a moral duty to make a particular preferential judgment between them?

If the number of As and Bs is finite, I think we can do better. We can then express the claim that there are more As than Bs by an infinite disjunction of claims of the form:

4. There exist n As and there do not exist n Bs,

which claims can be written as simple existentially quantified claims, without any mention of functions, sets or pluralities.

Any such claim as (4) does seem to have some intuitive moral force, and so maybe their disjunction does.

But in the infinite case, we can’t find a disjunction of existentially quantified claims that analysis the claim that there are more As than Bs.

Maybe what we should say is that “there are more As than Bs” is primitive, and the claim about there not being a one-to-one function is just a useful mathematical equivalence to it, rather than an analysis?

The thoughts here are also related to this post.

Ersatz objects and presentism

Let Q be a set of all relevant unary predicates (relative to some set of concerns). Let PQ be the powerset of Q. Let T be the set of abstract ersatz times (e.g., real numbers or maximal tensed propositions). Then an ersatz pre-object is a partial function f from a non-empty subset of T to PQ. Let b be a function from the set of ersatz pre-objects to T such that b(f) is a time in the domain of f (this uses the Axiom of Choice; I think the use of it is probably eliminable, but it simplifies the presentation). For any ersatz pre-object f, let n(f) be the number of objects o that did, do or will exist at b(f) and that are such that:

1. o did, does or will exist at every time in the domain of f

2. o did not, does not and will not exist at any time not in the domain of f

3. for every time t in the domain of f and every predicate F in Q, o did, does or will satisfy F at t if and only if F ∈ f(t).

Then let the set of all ersatz objects relative to Q be:

• O_Q = { (i,f) : i < n(f) },

where i ranges over ordinals and f over ersatz pre-objects. We then say that an ersatz object (i, f) ersatz-satisfies a predicate F at a time t if and only if F ∈ f(t).

The presentist can then do with ersatz objects anything that the eternalist can do with non-ersatz objects, as long as we stick to unary predicates. In particular, she can do cross-time counting, being able to say things like: “There were more dogs than cats born in the 18th century.”

Extending this construction to non-unary predicates is a challenging project, however.

Presentism and counting future sufferings

I find it hard to see why on presentism or growing block theory it’s a bad thing that I will suffer, given that the suffering is unreal. Perhaps, though, the presentist or growing blocker can say that is a primitive fact that it is bad for me that a bad thing will happen to me.

But there is now a second problem for the presentist. Suppose I am comparing two states of affairs:

1. Alice will suffer for an hour in 10 hours.
2. Bob will suffer for an hour in 5 hours and again for an hour in 15 hours.

Other things being equal, Alice is better off than Bob. But why?

The eternalist can say:

1. There are more one-hour bouts of suffering for Bob than for Alice.

Maybe the growing blocker can say:

2. It will be the case in 16 hours that there are more bouts of suffering for Bob than for Alice.

(I feel that this doesn’t quite explain why it’s B is twice as bad, given that the difference between B and A shouldn’t be grounded in what happens in 16 hours, but nevermind that for this post.)

But what about the presentist? Let’s suppose preentism is true. We might now try to explain our comparative judgment by future-tensing (1):

3. There will be more bouts of suffering for Bob than for Alice.

But what does that mean? Our best account of “There are more Xs than Ys” is that the set of Xs is bigger than the set of Ys. But given presentism, the set of Bob’s future bouts of suffering is no bigger than the set of Alice’s future bouts of suffering, because if presentism is true, then both sets are empty as there are no future bouts of suffering. So (3) cannot just mean that there are more future bouts of suffering for Bob than for Alice. Perhaps it means that:

4. It will be the case that the set of Bob’s bouts of suffering is larger than the set of Alice’s.

This is true. In 5.5 hours, there will presently be one bout of suffering for Bob and none for Alice, so it will then be the case that the set of Bob’s bouts of suffering is larger than the set of Alice’s. But while it is true, it is similarly true that:

5. It will be the case that the set of Alice’s bouts of suffering is larger than the set of Bob’s.

For in 10.5 hours, there will presently be one bout for Alice and none for Bob. If we read (3) as (4), then, we have to likewise say that there will be more bouts of suffering for Alice than for Bob, and so we don’t have an explanation of why Alice is better off.

Perhaps, though, instead of counting bouts of suffering, the presentist can count intervals of time during which there is suffering. For instance:

6. The set of hour-long periods of time during which Bob is suffering is bigger than the set of hour-long periods of time during which Alice is suffering.

Notice that the times here need to be something like abstract ersatz times. For the presentist does not think there are any future real concrete times, and so if the periods were real and concrete, the two sets in (6) would be both empty.

And now we have a puzzle. How can fact (6), which is just a fact about sets of abstract ersatz times, explain the fact about how Bob is (or is going to be) worse off than Alice? I can see how a comparative fact about sets of sufferings might make Bob worse off than Alice. But a comparative fact about sets of abstract times should not. It is true that (6) entails that Bob is worse off than Alice. But (6) isn’t the explanation of why.

Our best explanation of why Bob is worse off than Alice is, thus, (1). But the presentist can’t accept (1). So, presentism is probably false.

More remarks on Aristotelian set theory

If we have an Aristotelian picture of abstracta, we should expect that what mathematical objects exist differs between possible worlds.

For the Aristotelian, abstract objects are abstractions from concrete things. So we shouldn’t expect the same full panoply of sets regardless of what concrete things there are. For instance, suppose that the universe contains exactly three point particles, A, B and C. Then we can immediately abstract from these particle positions distance ratios like AB : BC, AC : AB and AC : BC. These ratios are then represented by real numbers. So we are going to have these real numbers. More sophisticated abstractive processes may well generate other real numbers: for instance, we will have a real number representing the ratio of the height of the triangle drawn from A to the base BC. And given a real number, we might be able to use purely abstract processes to generate further real numbers: given a and b, we may generate a + b and ab, say. But there is no reason to think that these abstract processes will generate the same collection of real numbers regardless of what the three particle positions we start with are.

So, what real numbers exist should vary between possible worlds. But every real number defines a subset of the natural numbers (just write the real number in binary, and let the nth bit decide if n is in the subset or not). If the real numbers vary between possible worlds, so do the subsets of the natural numbers. In particular, we should expect that in different possible worlds, a different set counts as ``the power set’’ of the natural numbers.

Furthermore, what bijections there are between sets will vary between possible worlds. Thus, if we see the question of whether two sets have the same count of members as having the same answer in every world where the two sets exist, we cannot take the standard Cantorian account of the size of a set. Instead, we may want to generate the concept of sameness of size from bijections in different worlds. Thus, we may try to say that two sets A and B are the same size at level 0 provided that there is a bijection between A and B. Then we say that A and B are the same size at level n provided that possibly there is a set C that is the same size as A at level p and the same size as B at level q and n ≥ 1 + p + q. Finally, we say that A and B are the same size simpliciter provided that they are the same size at some finite level. This is complicated, and I haven’t checked under what assumptions it generates a transitive relation (it’s plausibly reflexive and symmetric).

Anyway, the point is this: It is an interesting and not easy philosophical project to work out the set-theoretic consequences of Aristotelianism. This could make a good dissertation.

Universal countable numerosity: A hypothesis worth taking seriously?

Here’s a curious tale about sets and possible worlds: What sets there are varies between metaphysically possible worlds and for any possible world w₁, the sets at w₁ satisfy the full ZFC axioms and there is also a possible world w₂ at which there exists a set S such that:

1. At w₂, there is a bijection of S onto the natural numbers (i.e., a function that is one-to-one and whose range is all of the natural numbers).

2. The members of S are precisely the sets that exist at w₁.

Suppose that this tale is true. Then assume S5 and this further principle:

3. If two sets A and B are such that possibly there is a bijection between them, then they have the same numerosity.

(Here I distinguish between “numerosity” and “cardinality”: to have the same cardinality, they need to actually have a bijection.) Then:

4. Necessarily, all infinite sets have the same numerosity, and in particular necessarily all infinite sets have the same numerosity as the set of natural numbers.

For if A and B are infinite sets in w₁, then at w₂ they are subsets of the countable-at-w₂ set S, and hence at w₂ they have a bijection with the naturals, and so by (3) they have the same numerosity.

Given the tale, there is then an intuitive sense in which all infinite sets are the same size. But it gets more fun than that. Add this principle:

5. If two pluralities are such that possibly there is a bijection between them, then the two pluralities have the same numerosity.

(Here, a bijection between the xs and the ys is a binary relation R such that each of the xs stands in R to a unique one of the ys, and vice versa.) Then:

6. Necessarily, the plurality of sets has the same numerosity as the plurality of natural numbers.

For if the xs are the plurality of sets of w₁, then there will be a world w₂ and a countable-at-w₂ set S such that the xs are all and only the members of S. Hence, there will be a bijection between the xs and the natural numbers at w₂, and hence at w₁ they will have the same numerosity by (5).

So if my curious tale is true, not only does each infinite set have the same numerosity, but the plurality of sets has the same numerosity as each of these infinite sets.

We can now say that a set or plurality has countable numerosity provided that it is either finite or has the same numerosity as the naturals. Then the conclusion of the tale is that each set (finite and infinite), as well as the plurality of sets, has countable numerosity. I.e., universal countable numerosity.

But hasn’t Cantor proved this is all false? Not at all. Cantor proved that this is false if we put “cardinality” in place of “numerosity”, where cardinality is defined in terms of actual bijections while numerosity is defined in terms of possible bijections. And I think that possible bijections are a better way to get at the intuitive concept of the count of members.

Still, is my curious tale mathematically consistent? I think nobody knows. Will Brian, a colleague in the Mathematics Department, sent me a nice proof which, assuming my interpretation of its claims is correct, shows that if ZFC + “there is an inaccessible cardinal” is consistent, then so is my tale. And we have no reason to doubt that ZFC + “there is an inaccessible cardinal” is consistent. So we have no reason to doubt the consistency of the tale.

As for its truth, that's a different matter. One philosophically deep question is whether there could in fact be so much variation as to what the sets are in different metaphysically possible worlds.

Might all infinities be the same size?

A lot of people find Cantor's discovery that there are different infinities paradoxical. To be honest, there are many counterintuitive things involving infinities, but this one doesn't strike me as particularly counterintuitive. Nonetheless, I want to explore the possibility that while Cantor's Theorem is of course true, it doesn't actually show that infinity comes in different sizes. Cantor's Theorem says that if A is a set, then there is no pairing (i.e., bijection) between the members of A and those of the power set PA. It follows that there are different (cardinal, but in an intuitive rather than mathematical sense) sizes of infinity given this Pairing Principle:

1. PP: Two sets A and B have the same size if and only if there is a pairing between them.

Given PP and Cantor's Theorem, if A is an infinite set, then A is a different size from PA. But of course PA is infinite if A is, so there are infinite sets of different size.

A number of people have disputed the sufficiency part of PP, since it gives rise to the counterintuitive consequence that the set of primes and the set of integers have the same size as you can pair them up. But you really shouldn't both complain that there are different infinities and that PP makes the primes and the integers have the same infinite size. I am going to leave the sufficiency of PP untouched, but suggest that the pairing condition might not be necessary for sameness of size, and I will offer an alternative. That alternative seems to leave open the possibility that all infinities are the same.

To think about this, start with this thought experiment. Imagine that there is a possible world w that has some but not all of the actual world's sets, but that it still has enough sets to satisfy the ZFC axioms just as (I shall suppose) the sets of the actual world do. The set membership relation in w will be the same as in the actual world in the sense that if A is a set that exists both w and the actual world, then A has exactly the same members in both worlds (and in particular, all the actual world members of A exist in w). Then here is something that might well happen. We have two sets A and B that exist both in the actual world and in w. In the actual world, there is a pairing f between A and B. A pairing is just a set of ordered pairs satisfying some additional constraints (the first element is always from A and the second is always from B, and each element of A occurs as the first element of exactly one pair, and each element of B occurs as the second element of exactly one pair). It might, then, be the case that although A and B exist in w, f does not--it exists in the actual world but not in the impoverished world w. It might even be the case that no pairing between A and B exists in the impoverished world. In that case, we have something very interesting: A and B satisfy the pairing condition in PP in the actual world but fail to satisfy it in w. If we are to satisfy the ZFC in w, this can only happen if both A and B are infinite.

Things might go even further. We might suppose that w only contains sets that are countable in the actual world. The mathematical (much less metaphysical!) possibility of such a scenario cannot be proved from ZFC if ZFC is consistent, but it follows from the Standard Model Hypothesis which a lot of set theorists find plausible. If w only contains sets that are actually countable, then any infinite sets in w will have a pairing in the actual world. There is, thus, an important sense in which from the broader point of view of the actual world, all infinite sets in w have the same size. But w is impoverished. There are pairings that exist in the actual world but don't exist in w, and so applying PP inside w will yield the conclusion that the infinite sets in w come in different sizes. However, intuitively, it still seems true to say that these sets in w are all the same size, but w just doesn't have enough pairings to see this.

Here's one way to argue for this interpretation of the hypothesis. Plausibly:

2. Pairing-Sufficiency: If there is a pairing between sets A and B, they are the same size.
3. Absolute-Size: If two sets are the same size in one possible world, they are the same size in any world in which they both exist.

Pairing-Sufficiency is one half of PP. Given Pairing-Sufficiency and Absolute-Size, if two sets have a pairing in any possible world, including the actual one, they have the same size in every world, including w. Thus, in w all the infinite sets in fact have the size, but you wouldn't know that if your tools were restricted to the pairings in w.

Thinking about the above scenario suggests a modification of PP to a Possible Pairing Principle:

4. PPP: Two sets A and B have the same size if and only if possibly there is a pairing between them.

Given that the members of a set cannot vary between possible worlds, I think PPP is at least as plausible as PP. Moreover, I think that if there are any cases (like my hypothesis that a world like w is possible) where PPP and PP come apart, we should side with PPP. Here's why. I think we go for PP as an abstraction from our general method of comparing the sizes of pluralities by pairing. (One imagines a pre-numerate people trading goats and spears in 1:1 ratio by lining up each goat with a spear.) But the natural abstraction from our general method is that if one could pair up the two sets, then and only then they are the same size. So PPP is the natural hypothesis. The only reason to go for PP is, I think, acceptance of PPP plus an additional hypothesis such as that what pairings there are doesn't vary between possible worlds.

If PPP (or just (2) and (3)) is true and my w hypothesis is a genuine metaphysical possibility, then it is metaphysically possible that all infinite sets are the same size--i.e., it could be that the actual world is relevantly like w. Furthermore, we clearly don't have relevant empirical evidence to the contrary. So, if all this works, it is epistemically possible that all infinite sets are of the same size. (Of course, the most controversial part of all this is the idea that what sets there are might differ--even in the case of pure sets--between worlds.)

But perhaps this won't satisfy the people who find size differences between infinities paradoxical. For they might find it paradoxical enough that there could be infinities of different sizes, something that was definitely a part of my story (remember that I started with two worlds, one in which there were differently sized infinities and an impoverished w with all infinities of the same size according to PPP).

I think I might be able to do something to satisfy them, while at the same avoiding the biggest problem with the above story, namely the assumption that what pure sets there are differs between worlds. Here's my trick. In the above, I assumed that pairings were all sets. But in line with the Platonism suffusing all of the above arguments, let's try something. Let's allow that there are pairings that aren't sets. Those pairings would be binary relations satisfying the right formal axioms. But here I mean "relations" in the Platonic philosopher's sense, not in the mathematician's sense where a relation is a set of ordered pairs. Let's suppose, further, that corresponding to any set of ordered pairs, there is a relation which relates all and only those pairs which are found in the set. In my earlier story, I made sense of the idea that two sets in w might not have a pairing in w and yet might be the same size by adverting to pairings that exist in another world (the actual one--and then at the end I flipped things around so that w was actual). But now we do the same thing by distinguishing between mathematical pairings--namely, sets of ordered pairs satisfying the right axioms--and metaphysical pairings--namely, Platonic binary relations satisfying analogous axioms. If my earlier story is coherent (I mean that to be a weaker condition than "metaphysically possible"), then so is this one: In w, there are infinite sets that do not have a mathematical pairing, but every pair of infinite sets possibly has a metaphysical pairing. But now this story doesn't rely on varying what pure sets exist between worlds. The story appears compatible with the idea that pure sets are the same in every world. But there are, nonetheless, metaphysical pairings that do not correspond to mathematical pairings, and PPP should be interpreted with respect to the metaphysical pairings, not just the mathematical ones. Note, too, that what metaphysical pairings hold between sets might differ between possible worlds, without any variation in sets. For some of the pairings may correspond to extrinsic relations. Here is an extrinsic relation that could turn out to be a metaphysical pairing, depending on what I actually was thinking yesterday: x is related to y if and only if x and y came up in one of my thoughts yesterday in this order.

We can now suppose that this story works in every possible world. Thus, assuming the coherence of the Standard Model Hypothesis, we have a mathematically coherent story--whether the metaphysics works is another question (the story is too Platonic for my taste, and I don't share the motivation anyway)--on which (a) all infinite sets are really of the same size (and hence of the same size as the natural numbers), (b) what pure sets there are does not differ between worlds, and (c) Cantor's Theorem and all the axioms of ZF or ZFC are true. If we were to go for such a view, we would want to distinguish between sets being of the size metaphysically speaking and their having the same mathematical cardinality. The latter relationship would be defined by a version of the PP with "pairings" restricted to the mathematical ones. And then mathematics could go on as usual.

Presentist counting

In a posthumous paper, David Lewis shows that one can find a presentist paraphrase of sentences like "There have ever been, are or ever will be n Fs" for any finite n. But his method doesn't work for infinite counting.

It turns out that there is a solution that works for finite and infinite counts, using a bit of set theory. For any set S of times, say that an object x exactly occupies S provided that at every time in S it was, is or will be the case that x exists and at no time outside of S it was, is or will be the case that x exists. For any non-empty set S of times, let n_F(S) be a cardinality such that at every time t in S it was, is or will be the case that there are exactly n_F(S) objects exactly occupying S. This is a presentist-friendly definition. Let N be any set of abstracta with cardinality n_F(S) (e.g., if we have the Axiom of Choice, we should have an initial ordinal of that cardinality) and let e_F(S) be the set of ordered pairs { <S,x> : x∈N }. We can think of the members of e_F(S) as the ersatz Fs exactly occupying S. Let e_F be the union of all the e_F(S) as S ranges over all subsets of times. (It's quite possible that I'm using the Axiom of Choice in the above constructions.) Then "There have ever been, are or ever will be n Fs" can be given the truth condition |e_F|=n.

This ersatzist construction suggests a general way in which presentists can talk of ersatz past, present or future objects. For instance, "There were, are or ever will be more Fs than Gs" gets the truth condition: |e_G|≤|e_F|. "Most Fs that have ever been, are or will be were, are or will be Gs" gets the truth condition |e_FG|>(1/2)|e_F|, where FG is the conjunction of F with G. I don't know just how much can be paraphrased in such ways, but I think quite a lot. Consequently, just as I think the B-theory can't be rejected on linguistic grounds, it's going to be hard to reject presentism on linguistic grounds.

How many zebras lived in the 19th century?

Here is a puzzle for a presentist. There seems to be a determinate answer to the question "How many zebras lived (at least in part) in the 19th century?" or at least it is quite possible there is a determinate answer.[note 1] But can the presentist make any sense of the question?

This puzzle is somewhat different from the general puzzle about truths about the past. I am willing to grant for the sake of argument that the presentist can make sense of questions like: "Did Napoleon win at Waterloo?" For the presentist can take the proposition p that Napoleon wins at Waterloo, and say that p was false at the relevant time, and hence the answer is negative.

But the question how many zebras lived in the 19th century is much tougher. Given any time t in the 19th century, the presentist can make sense of the question how many zebras there were alive at t.[note 2] That question is the question of what number z(t) is such that it was true at t that there are z(t) zebras. But the answer to the question of how many zebras lived in the 19th century does not supervene on the values of z(t) as t ranges over the 19th century.

If the presentist has haecceities in her ontology, she can probably make sense of the question. For then the question is: "How many haecceities h are there such that h is a haecceity of a zebra, and h was instantiated in the 19th century?" So the haecceitist presentist seems to be out of trouble.

Can a non-haecceitist presentist do the job? Yes, if she is a closed-future presentist. (A closed-future presentist accepts bivalence for claims about the future.) But it is surprisingly tricky (at least if we want to take into account the possibility that a zebra might have a temporally gappy existence). Here is the simplest way I have. Let T be the set of times in the 19th century. Let S be a non-empty subset of T. Let z(S) be defined as follows. Choose any t in S. Let z(S) be the unique number n such that it was true at t that there exist exactly n zebras z such that P_S(z). Here, P_S(z) is the claim that for every time t' in S, z exists, existed or will exist at t', and for no time t' in T−S is it the case that z exists, existed or will exist at t'. (A−B is the set of all members of A that are not members of B.) (This a definition apparently compatible with presentism, but since P_S(z) partly concerns the then-future, only a closed-future presentist will have no qualms about it.) Then the number of zebras that lived in the 19th century is equal to the sum of z(S) as S ranges over all non-empty subsets of T.

Maybe there is a simpler way of counting 19th century zebras on presentism. But I can't think of one. More obvious solutions fail (thus one might keep track of when zebras come into existence, and count the comings into existence, but this doesn't work very well on presentist grounds for zebras that come into existence on an interval of times open at the bottom end).

There may be a clever way to do this within the confines of open-future presentism. But it's going to be tricky and messy. If it can't be done, then we have an argument why an open-future presentist should be a haecceitist.

I wonder if how complicated the answer to the question is does not give an argument against presentism. For, intuitively, the claim that there were exactly n₁ zebras at noon on January 18, 1855 should be made true similarly to the way the claim that there were n₂ zebras in the 19th cenutry is made true. But the non-haecceitist presentist will have to use very different counting methods for the two cases.