1.1. ULTRAPOWER S 1 1 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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