[FOM] Predicative Powerset Construction
steve newberry
stevnewb at ix.netcom.com
Thu Feb 6 12:12:55 EST 2003
CONSTRUCTION OF PREDICATIVE POWERSET
by R. Stephen Newberry [February 5, 2003]
Outline of procedure:
(1) Take a countable set, any countable set; call it S.
(2) Define a functor ' * ' which converts S to a sequence, any
sequence whose elements are all and only the elements of S.
(If this is considered as invoking the Countable Axiom of Choice,
so be it.) Call the sequence *S .
(3) Define a functor ' $', which takes as argument a countable sequence,
let's call it X, and returns a "sequence of ordinal structure" isomorphic
with omega. Thus if X were <0, 1, 2, 3, . . . > then $(X) would be
< {0}, {0, 1}, {0, 1, 2}, {0, 1, 2, 3 } . . . >; if X is a set rather
than a sequence,
then $(X) = X.
(4) Define a "map function" '^Pwr' which takes as argument an infinite
sequence X of sets and returns an infinite sequence of powersets, one for
each component set of the given sequence; if X is a set rather than a
sequence of sets, then ^Pwr(X) = Pwr(X). ['Pwr' denotes standard powerset.]
(5) Defined a special union 'U, which takes an infinite sequence of
sets as argument and returns the union of that sequence of sets ;
if X is a set rather than a sequence of sets, then 'U(X) = X.
(6) Define the Predicative Powerset functor 'PPwr' as
PPwr(S) =df= 'U(^Pwr($(*S)))
E.g., if $(*S) were < {0}, {0, 1}, {0, 1, 2}, {0, 1, 2, 3 } . . . >; then
^Pwr($(*S)) would be < Pwr{0}, Pwr{0, 1}, Pwr{0,1, 2} , Pwr{0, 1, 2 },
Pwr{0, 1, 2, 3 } . . . >
and PPwr(S) =df= 'U(^Pwr($(*S))) is U(< Pwr{0}, Pwr{0, 1}, Pwr{0,1, 2} ,
Pwr{0, 1, 2 }, Pwr{0, 1, 2, 3 } . . . >),
which differs from the standard powerset Pwr(S) in only two respects:
(1): Every element of PPwr(S) is predicatively defined; and
(2): PPwr(S) is a union of countably many countable sets, and is therefore
itself countable. Think of
PPwr(S) as being just exactly the standard powerset Pwr(S) **after** all
the IMpredicative subsets
(such as the Diagonal Function) have been removed.
Please comment.
R. Stephen Newberry
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