[FOM] clarification

Stephen Yablo yablo at mit.edu
Sun Sep 15 22:59:35 EDT 2002

Some (I think) private messages show that some aspects of the proposed
type-respecting set paradox were not very well explained.  Sorry about
that. Here are some questions/objections and proposed answers.

Q:  "Are you intending to let k range over all integers, including the
negative ones? (otherwise the definition of well-foundedness would be
trivial, for all sequences like k, k-1, k-2, ..., k-k would be finite.)
This could hardly be a hierarchy. Since the "hierarchy" has no lowest
level, to which type would the empty set belong?"

A: Yes, k ranges over all integers.  As in traditional type theory, there's
a comprehension axiom for each type n. The naive form  says "for each
property P of sets of type n-1,  that there's a set of type n containing
all sets with that property."  This gives you an empty set at each level.

Q:  ">>On the other hand, if each G_n is well-founded, then it belongs to
>> the set of well-founded sets one level up, that is G_n belongs to
>> G_(n+1).  Since n here ranges over the integers this gives us an
>> infinite descending chain: each G_k contains G_(k-1) contains G_(k-2)
>> etc.  So no G_n is well-founded.  Contradiction.
>I don't see. Suppose each G_n is well-founded. Then each G_n belongs to
>G_(n+1). This gives us at most an infinite ascending membership chain:
>G_k, G_(k+1), G_(k+2),..., each G_n of which has only a finite descending
>membership chain, stopping at G_0 or some G_m, which is empty, for

A: If it holds for all k that G_k belongs  to G_(k+1),  then it holds for k
= 0, k = -1, k = -2, etc.  Therefore .,,G_(-n-1) is a member of G_(-n) is a
member of...G_(-2) is a member of G_(-1) is a member of G_0. (Analogy:
since it holds for all integers k that k < k+1, then there's an infinite
descending chain of integers 0 > -1 > -2.........

Q: "what are the sets of type -1, -2, ...?"

A: The theory doesn't say, not any more than naive set theory says in so
many words what its sets are; both theories let comprehension do the

Q:  "Presumably, types start at 0, in which case there is no infinite
descending chain."

A: The types can't start at 0, because there's comprehension for each type,
which gives you at  a minimum for each type n (positive or negative) (i) a
null set of  type n, (ii) a singleton of the type (n-1) null set, (iii) a
doubleton of the type (n-2) null set , and so on. Hao Wang has a 1952
article  "Negative Types" which points out that an  axiom of infinity seems
no longer needed when you posit negative types;   you automatically get an
infinity of pure sets at each level. (Unfortunately as he also points out
all the argument shows is that  infinity is true in all intended models;
you can't prove infinity unless you give the theory a way of talking about
its own types.)

Perhaps these clarifications make the whole thing seem more crazy than
ever.  Let me stress, the suggestion is not that negative types are
attractive (not any more than an infinite descending cascade of
metalanguages is attractive), but that their adherence (debatable, I
realize) to Russell's  No Vicious Circles rule does not secure them from

Steve Yablo

PS.  It seems arguable that the intended structure is ruled out already by
Cantor's theorem, since  each type ought it seems to be the power set of
the one below it.  But the same can be said of the intended structure for
naive set theory; the universal set has to contain all its subsets.  What
makes Russell's paradox interesting (I am tempted to think) is not that the
set of non-self-membered sets destroys what would otherwise have been a
naively satisfying arrangement,  but that it brings out a problem that was
there anyway in a  memorable and arresting way.

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