Over the phone the other day, Peter L. suggested the following objection to the bundle-of-universals theory of ordinary particulars, 'BT' hereafter. (I leave out of consideration for the nonce bundle-of-tropes bundle theories.) I am not sure I understood what Peter was driving at. But here is the gist of what I thought he was saying.

1. Suppose x is a proper (spatial) part of y, y being a physical thing. On BT, both y and x are bundles of universals. Now it often happens that a whole has a property that is not had by all its parts. Think of a rubber ball. The ball is spherical (or spheroid, if you insist). But it has proper parts that are not spherical. For example, its hemispheres are not spherical. Nor are the cubes of rubber internal to it spherical. (They too are proper parts of it on classical mereology. These cubes could be 'liberated' by appropriate cutting of the ball.) The ball is red, let us say, but beneath the surface it is black. And so on. in sum, wholes often have properties that their parts do not have.

2. On BT, property-possession is understood, not in terms of the asymmetrical relation of exemplification, but in terms of the symmetrical relation of bundling. Accordingly, for a property to be possessed by something is not for it to be exemplified by this thing, but for it to be bundled with other logically and nomologically compossible properties. Exemplification, the asymmetrical relation that connects a substratum to a first-level property is replaced by bundling which is a symmetrical relation that connects sufficiently many properties (which we are assuming to be universals) so as to form a particular. When the universals are bundled, the result is a whole of which the universals are ontological constituents, with the bundling relation taking over the unifying job of the substratum. While bundling is symmetrical -- if U1 is bundled with U2, then U2 is bundled with U1-- ontological constituency is asymmetrical: if U is an ontological constituent of B, then B is not an ontological constituent of U.

3. Given that the ball is a bundle of universals, and that the ball is spherical, it follows that the ball has as one of its ontological 'parts' the universal, sphericality. Now sphericality and cubicality are not broadly-logically compossible. Hence they cannot be bundled together to form an individual. But our ball has a proper part internal to it which is a cube. That proper part has cubicality as a constituent universal. So it seems a broadly-logical contradiction ensues: the ball has as constituents both sphericality and cubicality, universals that are not compossible.

4. An interesting objection! But note that it assumes Transitivity of Bundling: it assumes that if sphericality is bundled with sufficiently many other Us to form a complete individual, and cubicality is bundled with one of these Us -- say being made of rubber -- then sphericality is bundled with cubicality. But it is well-known that bundling is not transitive. Suppose roundness and redness are bundled in our ball, and redness and stickiness are bundled in a numerically distinct disk, but there is nothing that is both round and sticky. That's a possible scenario which shows that Transitivity of Bundling fails. From the fact that U1 is bundled with U2, and U2 with U3, one cannot infer that U1 is bundled with U3. So from the fact that sphericality is bundled with rubberness, and rubberness with cubicality, it does not follow that sphericality is bundled with cubicality.

The bundle theory can accommodate the fact that a property of a whole needn't be a property of all its proper parts. Or am I missing something?

Hi Bill

You mean: "So from the fact that sphericality is bundled with rubberness, and rubberness with cubicality, it does not follow that sphericality is bundled with CUBICALITY".

But what explains transivity in a BT? There is maybe a similar way to see the problem: assume we have 2 red things, A and B, and transitivity holds, now since the universal property red is the same one which is bound with all the other properties of A and B, it follows that all the properties of A would be bound with all the properties of B. But how can we count A and B as 2 distinct objects? If one replies: thanks to time and space. Then my objections would be: how? In fact if time and space were also properties then A's time-space properties would be connected to B's time and space properties by transitivity again.

Posted by: aresh v. | Wednesday, October 13, 2010 at 04:13 AM

Aresh,

Thanks for the correction. Yes, that's what I meant.

But I don't understand what the rest of of your comment means.

Posted by: Bill Vallicella | Wednesday, October 13, 2010 at 11:08 AM

I've got a question for you. Suppose there were no red objects in existence. Then, you imagine a red car. Does "redness" exist in the universe in this case?

Posted by: Paul | Wednesday, October 13, 2010 at 12:06 PM

According to your argument, if one accepted that "bundling" were not only symmetrical but also transitive the consequence would be e.g. that "the ball has as constituents both sphericality and cubicality, universals that are not compossible". This may happen when we consider a whole individual and its part. My point is that this problem can arise also between individual and individual: in fact if according to a BT theory of universals 2 complete individuals must be at least disntinguishable in virtue of space and time and again space and time were properties, it follows that

all the properties of one individual would be bundled with all the properties of the other individual, included space and time properties by transitivity. And this result makes me wonder why then we talk about 2 individuals (e.g. 2 red apples existing in different time/space) instead of one individual (which is the bundle of all the properties the 2 apples share along with the properties they don't share since the latter are bundled together by means of the former). Sorry for my bad english.

Posted by: aresh v. | Wednesday, October 13, 2010 at 12:22 PM

Paul asks: " Suppose there were no red objects in existence. Then, you imagine a red car. Does "redness" exist in the universe in this case?"

Are you assuming that universals exist only if exemplified? I think you are, and that's a possible view. Another view is that universals can exist unexemplified. If universals exist only if exemplified, and there are no red particulars, then redness does not exist. Now if there never were any red particulars, then I don't think one could imagine a red car or a red anything. So let's suppose that there were some red particulars and I remember one or more of them. Then I presumably could now imagine a red car even if now there is nothing outside the mind that is a red particular.

Your question, then, is this: given that universals exist only if exemplified, and the only red particular is an object of imagination, would the universal redness then exist?

I would say no. For those who maintain that universals exist only if exemplified mean that they exist only if exemplified by particulars that exist 'outside' the mind.

Posted by: Bill Vallicella | Wednesday, October 13, 2010 at 03:30 PM

A,

That the bundling relation cannot be transitive is obvious if you think about it. All you need to do is find a possible case in which three universals satisfy these conditions: U1 is bundled with U2; U2 is bundled with U3; U1 is not bundled with U3. I gave an example above.

This is not controversial. All agree that the bundling relation is nontransitive.

Posted by: Bill Vallicella | Wednesday, October 13, 2010 at 03:40 PM