[FOM] paradox and circularity

[Stephen Yablo]

Thomas Forster writes (in a paper he mentioned recently here) "There 
is a more or less received modern view of the paradoxes which one 
associates with the names of Russell and Tarski to the effect that 
paradox can be evaded if one recognizes that every sentence belongs 
to a particular level in an infinite hierarchy of linguistic levels. 
The idea being that truth-predicates for languages lower down in the 
hierarchy are to be found only higher up.  It seems to me that the 
real significance of Yablo's observation is that this is not enough 
by itself.  Evidently the sentences in Yablo's example can be _typed_ 
in the appropriate sense...Nevertheless we still have a paradox."


It seems to be also part of this more or less received view that set 
paradoxes can be evaded if one recognizes that every set belongs to a 
particular level in an infinite hierarchy of types, with sets of type 
n can allowed to belong only to sets of type n+1 (or alternative type 
m higher than n).    If that is right  then, just as one would expect 
there to be no semantical paradoxes in a Tarskian hierarchy typed by 
the integers,  one would expect there to be no set paradoxes in a 
Russellian hierarchy typed by the integers.  Here is an apparent 
counterexample, or at least I am wondering if it is a 
ccounterexample. It's a variant of Mirimanoff's paradox making no 
play (apparently) with the idea of self-membership.

A set S of  type k is well-founded if there are no sets S_(k-1), 
S_(k-2), etc. such that S contains S_(k-1) contains S_(k-2) and so on 
forever.

For each integer n, let G_n be the set of well-founded sets of type (n-1).

On the one hand, each G_n must be well-founded, because an infinite 
descending membership chain starting from it would include an 
infinite descending membership chain starting from one of its 
members, and its members are one and all well-founded.

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.

Does this work?