Monday, May 28, 2012

Why do sets have their members essentially?

It is often said that sets have their members essentially. Why? Normally the finger gets pointed at the axiom of extensionality which says that two sets that have exactly the same members are equal. The axiom of extensionality does not seem to imply by itself that sets have their members essentially. Consider a set theory without ur-elements (i.e., every member of a set is itself a set). Suppose that in the actual world there are two sets, Bob and Jane, where Bob is has no members and Jane's only member is Bob. Then imagine that in every other worlds all the sets are exactly like in the actual world, except that Jane has no members and Bob has Jane as its only member. I do not see—though I could be missing something—how this violates any of the axioms of set theory. If the axiom of extensionality were extended to a transworld context (if A in w1 has the same members as B does in w2, then A=B), we would get a contradiction, but the axiom of extensionality is an intraworld axiom.

But nonetheless I have reason to think that sets have their members essentially if the axiom of extensionality is correct, namely that what seem to me to be the two most plausible intuitive kinds of reasons to think sets need not have their members essentially conflict with the axiom of extensionality. We can argue for this by considering paradigmatic cases of the two reasons.

First kind of reasons: Losing members. Let S be the set { Quine, Wittgenstein, Socrates }. Then, the intuition goes, in a world w* where Quine never comes into existence, S only has two members, namely Wittgenstein and Socrates.

But in the actual world there is also the set T={ Wittgenstein, Socrates }. And we have every reason to think that T will exist in w* and have the same two members there that it actually has. Thus, in w*, the sets S and T have the same members, namely Wittgenstein and Socrates, and so in w* we have S=T. Since identity is necessary, it follows that actually S=T. But that's absurd since s has three members and T has two. So we need to reject

Second kind of reasons: Sets defined by descriptions. Let M be the set of all mathematicians. Intuitively, in a world where Hilbert becomes a biologist instead of a mathematician, M still exists, but Hilbert isn't a member of M. The intuition here is that in every world, or at least every world with mathematicians, M has as its members all and only the mathematicians.

But by the same token, if B is the set of all biologists, then in every world where there are biologist B will have as its members all and only the biologists. But there is a world where all and only biologists are mathematicians. In that world, then M=B. But since identity is necessary, it follows that actually M=B, and hence that Darwin was a mathematician and Hilbert a biologist, which is absurd.

Maybe you have some other reasons to doubt the essentiality of set membership?

2 comments:

James said...

When you say "identity is necessary," do you mean something like "If A in w1 has the same members as B does in w2, then in every possible world, A=B"? Or perhaps, "then in the actual world, A=B"? I'm not sure how you're reasoning from S=T in w* to S=T in the actual world.

It's somewhat confusing going from sets described according to the possible worlds in which they reside into a statement like "A=B" that lacks any explicit possible world context. It seems like there's a kind of equivocation going on; if A-in-w1 has the same elements as B-in-w2, then isn't it true that A-in-w1 = B-in-w2, even if "A=B" is false? What set is being referred to by "A"?


Isn't it sufficient to say that S=S, and hence the result of S 'losing a member' must be a set different from S, since they contain different members?

Alexander R Pruss said...

I mean: If x=y is true in one world, then x=y is true in every world in which x exists and y exists.

"Isn't it sufficient to say that S=S, and hence the result of S 'losing a member' must be a set different from S, since they contain different members?"

I think not, become the result of S losing a member might not exist in the same world as S does.