1.1. ULTRAPOWER S 11
Now w e sho w ji\Ord = j®\Ord. Sinc e MQ
1
(jL Mo, w e canno t us e th e
previous argument . Instead , w e mus t buil d a n isomorphis m
TT: (Or^°( X l )) M o /j
0
(Mi) - Ord Xl/fn.
First w e defin e a ma p
TT: Ordjo{Xl) f l M
0
- Ord Xl
as follows . Suppos e tha t g i s a functio n i n M o fro m jo(Xi) t o th e ordinals .
Define 7r(g): X
x
- Ord b y lettin g ir(g)(a) = g(jo{a)).
- JoOo)
Note tha t jo(^i ) = ^ l (sinc e n i s strongl y inaccessibl e an d greate r tha n K,Q)
and tha t jo(f^i) i s a n ultrafilte r o n jo(Xi) i n Mo .
Claim 1. Suppose that f and g are functions in M o from jo(Xi) to the
ordinals and {a G jo(Xi) | f(a) = g(a)} G jo(^i) Then
{a G Xi | 7r(/)(a) - 7r(s)(a) } G //i.
Claim 2 . Suppose that f and g are functions in M o from jo(Xi) to the
ordinals and {a G jo(Xi) \ f(a) g(a)} G jo(/ii). Then
{a G Xi | TT(/)(O ) 7r(5)(a) } G m.
Claims 1 and 2 sho w tha t TT induce s a n orde r preservin g injectio n
TT : (Or^°( X l ) ) M o /jo(Mi) - O r d X l / / i i .
The proo f wil l the n conclud e b y showin g tha t 7 f is onto .
The proof s o f Claim s 1 and 2 us e th e followin g claim .
Claim 3 . Suppose that i d i , A G //i, C G Mo an d C G jo(fii). Then
{j0(a)\aeA}r\C^(D.
Proof o f Clai m 3 : Her e w e us e th e fac t tha t Xo G V K1. On e difficult y
is tha t {jo(a) \ a G A} i s no t i n genera l i n Mo . Le t d o b e th e elemen t o f
MQ correspondin g t o [io]/x 0? wher e i$ i s th e identit y functio n o n XQ. The n
do £
JO(XQ).
A s w e hav e seen ,
M0 = {j 0(h)(do) \ h: X
0
^ V]
and
M = {HcX
0
\d0e j
0
(H)}.
K\.
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