Showing posts with label arity. Show all posts
Showing posts with label arity. Show all posts

Friday, September 13, 2019

Multigrade relations

One strategy for avoiding ontological commitment to sets is to deal with pluralities and multigrade relations. Multigrade relations are relations that can be had by a variable number of things. Instead of, say, saying of the books on my shelf that there is a set of them whose total number of pages is exactly 800, one says that there are xs such that each of them is a book on my shelf and the xs stand in the multigrade relation of jointly having 800 pages. Let’s say these books are x1, x2 and x3. Then we express their jointly having 800 pages as:

  • Has800Pages(x1, x2, x3).

We do not need a set of them to express this. And the Has800Pages(x1, ...) predicate flexibly can take as many arguments as one wishes, corresponding to the multigradeness of the property it expresses.

But now consider a different statement: there are two pluralities of books on my shelf, having no books in common, where each plurality has the same total number of pages as the other. Can we make sense of this using multigrade relations instead of sets?

I don’t see how. Let’s say that the plurality x1, x2, x3 and the plurality y1, y2 of my books each have the same total number of pages. So we introduce a predicate with variable arity and say:

  • HasSameTotalNumberOfPages(x1, x2, x3, y1, y2).

But that doesn’t work! For how can we tell if it is says that x1, x2, x3 have the same number of pages as y1, y2 rather than saying that x1, x2 have the same number of pages as x3, y1, y3?

We could multiply predicates with fixed arity and say:

  • TheFirst3HaveTheSameTotalNumberOfPagesAsTheLast2(x1, x2, x3, y1, y2).

But that won’t work with quantification, since we don’t know ahead of time how many xs and how many ys we are dealing with.

Maybe we should do this:

  • Count(3,x1, x2, x3) and Count(2,y1, y2) and SameNumberOfPages(3,x1, x2, x3,2,y1, y2)

where the SameNumberOfPages variable arity predicate takes a number, then a plurality of that number of objects, then another number, and then another plurality of that number of objects.

But these kinds of solutions won’t work for infinite pluralities. For instance, suppose we want to say that the xs cause the ys, where there are ℵ2 xs and ℵ3 ys. Then I guess we say something like:

  • Cause(ℵ2, x1, x2, ..., ℵ3, y1, y2, ...).

There are serious technical problems here, however. I will leave it to the reader to explore these.