IDEALS OVER UNCOUNTABLE SETS 13
For each n, there exists a function f F such that X = dom(f ).
n n n n
If a 0 X , then because F F . .. , we have f_(a) f0(a) ...,
A
n
rr.
U
1 0 1
n=0 °°
a contradiction. Hence ( 1 X = 0.
n=0
b) Let us assume that I is not precipitous. Let S be a set of
positive measure and let W_ W_ ... W ... be I-partitions of S
r
0 1 n
CO
such that 0 X is empty whenever L £ X1 3 ... £ X 2.• is a sequence
n=0
of sets such that X ( W for each n. We shall construct functionals on S
n n
such that F„ F, ... F . . . .
0 1 n
Without loss of generality let us assume that if X W , Y £ W and
X £ Y, then X 4 Y. Let T = U W^; note that the partially ordered set
(T,c) is an upside down tree.
n n=n
0
For each z S, let us consider the set T = {X T : z X}. Since
z
every descending sequence Xn D X- D ... 3 X 3 ... in T has empty inter-
section, it follows that for every z 6 S, T has no infinite descending
sequence Xrt o X- D ... 3 X 3 ...; thus the relation c on T is well
n 0 1 n z
founded.
For each z S, let p be the rank function for the well founded
z
relation c on T ; if X,Y T and X c Y, then p (X) p (Y). Hence
for each n, if X W .. . , Y W and z 6 X c Y, then p (X) p (Y).
n+l n z z
Thus we define, for each X T, a function f on X as follows:
f(z) « p (X) (all z X).
X z
For each n, we let F ^{f^rX^W}. It follows that each F is an
n X n n
ordinal valued functional on S, and that F^ F. , .. . F ... . a
0 1 n
Corollary. If I is precipitous then I is weakly normal.
Proof. Assume that I is not weakly normal. Then there exists a set S of
positive measure such that for every unbounded I-function f with dom(f) £ S
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