FOM: Refuting the Power Shift Axiom Scheme
JoeShipman@aol.com
JoeShipman at aol.com
Sat Apr 3 13:18:18 EST 1999
Consider the following set-theoretic formula:
Phi(X): The largest integer n such that |X|=|P^n(Y)| for some set Y is even.
In ZC one can prove that such an integer exists (of all the cardinals <|X|
who have an iterated powerset with cardinality = |X|, take the least).
Whenever X satisfies Phi, P(X) fails to satisfy it, refuting the Power Shift
Axiom Scheme.
This argument uses full Choice (since you're assuming sets of cardinals have
least elements). More interestingly, if you're willing to use Replacement
rather than Comprehension (that is, go from Z to ZF) you only need (a Class
version of) AC_2 (choice for collections of two-element sets). Define two
cardinals j and k to be equivalent if there exist integers m and n with
|P^m(j)|=|P^n(k)|. This partitions the Directed Graph (I'm using capitals to
show this is a proper Class) whose vertices are cardinals and whose edges are
given by the exponentiation operation into connected components. Within each
component, declare two cardinals to be equivalent if the number of times you
have to exponentiate each one to reach their lub has the same parity. Now
select from each component one of the two equivalence classes.
If a choice function for collections of 2-element sets could be given by a
formula, then the above could (within ZF) turned into a formula which refutes
the Power Shift Axiom Scheme.
How much Choice is needed to refute NF over Z and ZF? If you have
Replacement can you get a weaker form of Choice to refute NF?
-- Joe Shipman
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