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If kappa are stationary greater equal aleph subsets sub 2 of .... uncountable is regular comma spaces S subset which equaldo not kappa reflect is stationary§ .0 . commaThere .... then are theremany is useful some and alpha interesting less kappa stationary reflection princi - ples such that S capformulated alpha is stationary and studied in alpha in the period current research concerning new existence \S Then0 . many\quad stationaryThereaxioms ofare reflection set many theory principles useful and combinatorial of and the following interesting aspects form of have infinity stationary . reflection princi − plesbeen formulated formulated and and studiedNotably studied period , one inof the the first current reflection research principles studied concerning i s the following new existence : open parenthesis(A) B closingIf κ parenthesis≥ ℵ2 is .... regular If kappa,S ⊆ greaterκ is stationary equal aleph , sub then 2comma there is S some subsetα < equal κ open square\noindent bracketaxioms kappa closing of set square theory bracket andsuch to the combinatorialthat powerS ∩ α ofis aleph stationary subaspects 0 ....in isα. of stationary infinity comma . .... then there is X in open square bracketThen many kappa stationary closing square reflection bracket principles to the power of the of following aleph sub form 1 .... have such been that formu- \ centerlineS cap open{ squareNotablylated bracket and , studied one X closing of . the square first bracket reflection to the power principles of aleph sub 0 studied .. is stationary i s the in the following space : } ℵ0 ℵ1 open square bracket(B) X closingIf κ square≥ ℵ2,S bracket⊆ [κ] to theis stationarypower of aleph , subthen 0 there period is X ∈ [κ] such that \noindentApparently(A) comma\ ..h f open i l l parenthesisI f $ S\∩kappa[X B]ℵ closing0 is\geq stationaryparenthesis\aleph in i sthe much space{ 2 more}[X$ ] powerfulℵ0\. h f i l la principleis regular than open $ , parenthesisS \subseteq A closing parenthesis\kappaApparently$ is , isperiod ( stationary B ..) i A s natmuch hyphen more , \ h powerful f i l l then a principle there than is ( some A ) is . $ \ Aalpha nat - < \kappaural question$ wouldural question b e : .... canwould there b e be : some useful can generalizations there be some useful of open generalizations parenthesis B ofclosing ( B ) parenthesis t o t o uncountable spaces ? \ centerlineuncountable{ spacessuch ? that $We S show\cap in this\ notealpha that$ one is must stationary b e very careful in in $ formulating\alpha such. $ } We show in thisa notething that . one must b e very careful in formulating such Thena thing many period stationaryOur reflection set - theoretic principles usage i s standard of the . followingAll undefined form t erminology have i s taken beenOur set formulated hyphenfrom theoretic and [ 4 ] . studiedusage i s standard . period .. All undefined t erminology i s taken from open square bracket 4 closing§1 . squareLet bracketκ > period ω b e a regular cardinal . For any cardinal \noindentS 1 period(B) .. Letλ ..\≥h kappa f iκ, l l greaterIlet f $ omega\kappa .. b e\ ageq regular\aleph cardinal period{ 2 } .. For,S any ..\ cardinalsubseteq .. lambda[ \kappa ] ˆ{\aleph { 0 }}$ \ h f i l l is stationary , \ h f i l l then there is $X \ in greater equal kappaPκ comma(λ) = .. let{x ⊆ λ | | x | < κ}. By a nontrivial filter on Pκ(λ) we mean [ P\ subkappa kappa open]a ˆ{\ parenthesisaleph lambda{ 1 }} closing$ \ h parenthesis f i l l such = open that brace x subset equal lambda bar bar x bar less kappa closingcollection brace periodF of .... subsets By a of ....Pκ nontrivial(λ) satisfying filter the on following P sub kappa conditions open: parenthesis lambda closing parenthesis\ centerline ....{ we$ mean S a\cap [ X ] ˆ{\aleph { 0 }}$ \quad is stationary in the space $ [ X ] ˆ{\aleph { 0 }} . $ } collection F of subsets of(1) P∀ subx ∈ kappaPκ(λ)( openPκ(λ) parenthesis− {x}) ∈ F lambda; ∅element closing− parenthesisnegationslashF, satisfying the following conditions : ( 2 ) if A, B ∈ F and A ∩ B ⊆ C, then C ∈ F. \ hspaceopen parenthesis∗{\ f i l l } 1Apparently closing parenthesis , \quad forall( x in B P ) sub i kappa s much open more parenthesis powerful lambda a closing principle parenthesis than open ( A ) is . \quad A nat − parenthesis P sub kappa open parenthesis lambda closing parenthesis minus open brace x closing brace closing parenthesis\noindent inural F semicolon question varnothing would element-negationslash b e : \ h f i l l can F comma there be some useful generalizations of ( B ) t o open parenthesis 2 closing parenthesis if A comma B in F and A cap B subset equal C comma then C in F period\noindent uncountable spaces $ ? $ \ hspace ∗{\ f i l l }We show in this note that one must b e very careful in formulating such

\noindent a thing .

\ hspace ∗{\ f i l l }Our s e t − theoretic usage i s standard . \quad All undefined t erminology i s taken

\noindent from [ 4 ] .

\ hspace ∗{\ f i l l }\S 1 . \quad Let \quad $ \kappa > \omega $ \quad b e a regular cardinal . \quad For any \quad c a r d i n a l \quad $ \lambda \geq \kappa , $ \quad l e t

\noindent $ P {\kappa } ( \lambda ) = \{ x \subseteq \lambda \mid \mid x \mid < \kappa \} . $ \ h f i l l By a \ h f i l l nontrivial filter on $ P {\kappa } ( \lambda ) $ \ h f i l l we mean a

\noindent collection $ F $ of subsets of $ P {\kappa } ( \lambda ) $ satisfying the following conditions :

\ [ ( 1 ) \ f o r a l l x \ in P {\kappa } ( \lambda )(P {\kappa } ( \lambda ) − \{ x \} ) \ in F; \ varnothing element−negationslash F, \ ]

\ centerline {( 2 ) i f $ A , B \ in F $ and $ A \cap B \subseteq C , $ then $ C \ in F . $ } 1 76 .. Q period F e n g and M period .. Magi d o r \noindentUnless otherwise1 76 \ spquad ecifiedQ.FengandM. comma all filters are assumed\quad t oMagi be nontrivial d o r period A filter F on P sub kappa open parenthesis lambda closing parenthesis i s kappa hyphen complete if F i s closed\ centerline under intersections{ Unless otherwiseof sp ecified , all filters are assumed t o be nontrivial . } less than kappa many of it s members period F is normal if F i s closed under diagonal \ hspaceintersections∗{\ f i comma l l }Afilter i period e period $F$ comma on whenever $P {\ angbracketleftkappa } ( A sub\lambda alpha bar alpha) $ less i lambda s $ right\kappa −angbracket$ complete i s a sequence1 if 76 $ Qfrom F . F $ F e n comma i g ands closed M then . Magi under d o r intersections of vartriangle A sub alphaUnless = open otherwise brace x in sp P ecified sub kappa , all filters open parenthesis are assumed lambda t o be closing nontrivial parenthesis . bar forall \noindent l e s s than $ \kappa $ many of it s members $ . F$ is normal if $F$ alpha in x x in A sub alpha closingA filter braceF inon FPκ period(λ) i s κ− complete if F i s closed under intersections of i sIf closedF is a filter underless on P than diagonal subκ kappamany open of it s parenthesis members .F lambdais normal closingif parenthesisF i s closed comma under we diagonal denote by F to the power of plus all Fintersections hyphen positive , i . sete . ,s whenever comma i periodhAα | α e < period λi i s comma a sequence from F, then \noindentF to the powerintersections of plus = open , brace i . A e subset . , wheneverequal P sub kappa $ \ langle open parenthesisA {\ lambdaalpha closing}\mid parenthesis\alpha < \lambda \rangle $ i s a sequence from $F , $ then bar forall C in F A cap C negationslash-equalM Aα varnothing= {x ∈ Pκ( closingλ) | ∀α brace∈ xx ∈ periodAα} ∈ F. Then a filter F is normal if and only if for each A in F to+ the power of plus and for every choice If F is a filter on Pκ(λ), we denote by F all F − positive set s , i . e . , \ [ function\ vartriangle f on A open A parenthesis{\alpha i period} = e period\{ commax \ fin open parenthesisP {\kappa x closing} ( parenthesis\lambda in x for) \mid \ f o r a l l \alpha \ in x x \ in A {\alpha }\}\ in F. \ ] each x in A closing parenthesis there is a B+ in F to the power of plus such that F = {A ⊆ Pκ(λ) | ∀C ∈ FA ∩ C 6= ∅}. f is constant on B period A filter F is fineThen if for a filter eachF alphais normal in lambda if and we only have if for open each braceA ∈ xF in+ Pand sub for kappa every open choice parenthesis function f lambda closing\noindent parenthesisIfon $F$barA( alpha i . e is in., f x( ax closing) filter∈ x for brace each on inx F $P∈ periodA) there{\kappa is a B ∈} F +( such\ thatlambdaf is constant), on $B. wedenote by $ F ˆ{ + }$ a l l $ F − $ positive set s , i . e . , Also a filter F i s called anA ultrafilter filter F is iffine F =if F for to the each powerα ∈ λ ofwe plus have period{x ∈ Pκ(λ) | α ∈ x} ∈ F. For a subset AAlso subset a filterequalF Pi sub s called kappa an openultrafilter parenthesisif F = lambdaF +. closing parenthesis comma A i s directed if \ [ F ˆ{ + } = \{ A \subseteq P {\kappa } ( \lambda ) \mid \ f o r a l l for any x comma y inFor A comma a subset thereA ⊆ i sPκ some(λ),A i s directed if for any x, y ∈ A, there i s some z ∈ A so C \ in FA \cap C \not= \ varnothing \} . \ ] z in A so that xthat cupx y∪ subsety ⊆ z; equalA is ω z− semicolondirected Aif is for omega any sequence hyphen directedhXn : n if< for ωi anyfrom sequenceA, there angbracketleft i s an S X sub n : n less omegaX ∈ A rightsuch angbracket that n<ω Xn ⊆ X; A is unbounded if for from A commaeach therex i∈ sP anκ( Xλ) in there A such is y that∈ A unionso that ofx sub⊆ ny; lessA i omega s c losed X subif for n subset any directed equal X subset semicolon A S is\noindent unboundedThen if forD a⊆ filterA with $ F| D $| is< normal κ we have if andD ∈ onlyA; ifA fori s a eachc lu b $Aif A is\ bothin F ˆ{ + }$ and for every choice each x in P subclosed kappa and open unbounded parenthesis.A lambda⊆ Pκ( closingλ) is tight parenthesisif for each thereω− isdirected y in A soD that⊆ A xif subsetω < cf equal y function $f$ on $A ($ i.eS $. , f ( x ) \ in x $ f o r each semicolon A i s c losed(| D | if) for≤ any | D directed| < κ then D ∈ A. It i s well known that for a subset A ⊆ Pκ(λ) $ xsubset\ in D subsetAto equal b )$ e a A club with thereisa it .. is bar necessary D bar $B less and kappa sufficient\ in we have thatF ˆA{ union+is unbounded} of$ D such in A semicolonand thatA i s closed A i s a under .. c lu b if A $ f $ is constanton $B . $ is both unions of sequences of length less than κ. We say that S ⊆ Pκ(λ) i s stationary if it closed and unboundedhas nonempty period Aintersection subset equal with P subeach kappa club . open parenthesis lambda closing parenthesis is tight if\ hspace for each∗{\ omegaf i l hyphen l }ABy filter a directed theorem$F$ D of subset Jech equal[is 3 ] fine , A all if the if clubs for generate each a$ normal\alphaκ− complete\ in fine\lambda filter $ we have $ \{ x \ in P {\kappa } ( \lambda ) \mid \alpha \ in x \}\ in omega less cf openon P parenthesisκ(λ). We bar use DF to bar denote closing the parenthesis club filter less on P orκ( equalλ). barAlso D let barD lessdenote kappa the then filter union of FD in A . period $ .. Itgenerated i s well known by the that t ight for unbounded a subset subsets . Then one can A subset equalcheck P sub that kappaD i opens a κ− parenthesiscomplete filter lambda extending closing the parenthesis club filter to . b e a club it is necessary and \noindent Also a filter $F$ i s called an ultrafilter if $F = Fˆ{ + } . $ sufficient that A is unboundedR e m a r k . If κ = ω2, then being t ight unbounded i s the same as being and A i s closedω1 under− closed unions and of unbounded sequences of ( length i . e . less , than closed kappa under period unions We say of strictly that increasing For a subset $ A \subseteq P {\kappa } ( \lambda ) , A$ i s directed if for any S subset equalsequences P sub kappa of length openω parenthesis1). Also if lambda cf (λ) closing≥ ω2, parenthesisand A ⊆ Pω2 i( sλ) stationary i s defined if so it that has for nonempty $ x , y \ in A ,$ there i ssome intersection with eacheach clubx ∈ periodPω2 (λ) we have x ∈ A if and only if the order type of x i s a limit ordinal and $ z \ in A $ so that $ x \cup y \subseteq z ; A $ i s $ \omega By a theorem ofx is Jechω− openclosed square , then bracketA i s a ω 31 closing− clubsquare . bracket comma all the clubs generate a normal −kappa$ hyphen directed complete ifThe for fine following any sequence theorem answers $ \ langle the questionX of Balogh{ n } [ 1 ]: of whether n < a supercompact\omega \rangle $ fromfilter on $A P sub ,$kappacardinal thereopen always parenthesis i assumes san lambda a $X strong closing form\ in reflection parenthesisA$ such. period that .. We $ use\bigcup F to denote{ then club< filter\omega } Xon P{ subn kappa}\subseteq openTheorem parenthesisX lambda ;1 . A$ 1closing . isAssume parenthesis unbounded periodthat ifκ .. for Alsois letλ D− supercompact with denote the filterλ generated≥ κ byregular the t ight( i unbounded . e . , there subsets is a periodκ− complete .. Then one normal can fine ultrafilter on \noindent each $ x \ in P {\kappa } ( \lambda )$ there is $y \ in check that D iP sκ a(λ kappa)). hyphenThen for complete every stationary filter extendingS ⊆ P theω1 (λ club) and filter for period every tight and unbounded A $ so that $ x \subseteq y ; A$ i s c losed if for any directed R e m a r k periodA ⊆ ..Pκ If(λ kappa), there =omega is an X sub∈ 2A commasuch that thenS being∩ Pω1 t(X ight) unboundedis stationary i s in theP sameω1 (X as). being subsetomega sub $ 1 D hyphen\subseteq closedWe and are unbounded goingA $ t witho prove .. open\quad two parenthesis lemmas$ \mid which i period willD e give period\mid the comma theorem< .. .closed\kappa under$ unions we have ..$ of\bigcup strictly increasingD \ in A ; A $ i s a \quad c lub if $A$ is both closedsequences and of unbounded length omega $sub . 1 closing A \ parenthesissubseteq periodP ..{\ Alsokappa if cf} open( parenthesis\lambda lambda) $ closing is tight if for each parenthesis$ \omega greater− $ equal directed omega sub $D 2 comma\subseteq and A subsetA equal $ Pi f sub omega sub 2 open parenthesis lambda closing$ \omega parenthesis< i$ s defined c f $so ( \mid D \mid ) \ leq \mid D \mid < \kappa $ thenthat $for\ eachbigcup x in P subD omega\ in subA 2 open . $parenthesis\quad lambdaIt i s closing well parenthesis known that we have for x a in subset A if and only if the$ A order\ typesubseteq of x i s P {\kappa } ( \lambda ) $ to b e a club it is necessary and sufficient that $ Aa limit $ is ordinal unbounded and x is omega hyphen closed comma then A i s a omega sub 1 hyphen club period andThe following $ A $ theorem i s closed answers under the question unions of Balogh of sequences open square of bracket length 1 closing less square than bracket $ \kappa of whether. $ Wea say that $supercompact S \subseteq cardinal alwaysP {\ assumeskappa a} strong( form\lambda reflection period) $ i s stationary if it has nonempty intersection with each club . Theorem .. 1 period 1 period .. Assume .. that kappa .. is .. lambda hyphen supercompact .. with .. lambda Bygreater a theorem equal kappa of .. Jech regular [ 3 ] , all the clubs generate a normal $ \kappa − $ complete fine f iopen l t e r parenthesis on $ P i period{\kappa e period} comma( \ therelambda is a kappa) hyphen . $ \ completequad We normal use fine $ Fultrafilter $ to on denote P sub the club filter on kappa$ P open{\kappa parenthesis} ( lambda\lambda closing parenthesis) . $ closing\quad parenthesisAlso l period e t $ .. D Then $ for every denotestationary the S subset filter equal generated P sub omega by thesub 1 t open ight parenthesis unbounded lambda subsets closing . parenthesis\quad Then .. and one for can every tight and unbounded A subset equal P sub kappa open parenthesis lambda closing parenthesis comma there \noindentis an X incheck A such that that S cap $D$ P sub omega i sa sub $ 1\ openkappa parenthesis− $ X complete closing parenthesis filter .. extending is stationary the in P club filter . sub omega sub 1 open parenthesis X closing parenthesis period R eWe m are a goingr k . t o\quad prove twoI f lemmas $ \kappa which will= give\omega the theorem{ 2 period} , $ then being t ight unbounded i s the same as being $ \omega { 1 } − $ closed and unbounded \quad ( i . e . , \quad closed under unions \quad of strictly increasing sequences of length $ \omega { 1 } ) . $ \quad Also i f c f $ ( \lambda ) \geq \omega { 2 } , $ and $ A \subseteq P {\omega { 2 }} ( \lambda ) $ i s defined so that for each $ x \ in P {\omega { 2 }} ( \lambda ) $ we have $ x \ in A$ if and only if the order type of $ x $ i s a limit ordinal and $ x $ is $ \omega − $ closed ,then $A$ i sa $ \omega { 1 } − $ club .

The following theorem answers the question of Balogh [ 1 ] of whether a supercompact cardinal always assumes a strong form reflection .

Theorem \quad 1 . 1 . \quad Assume \quad that $ \kappa $ \quad i s \quad $ \lambda − $ supercompact \quad with \quad $ \lambda \geq \kappa $ \quad r e g u l a r (i .e. ,there isa $ \kappa − $ complete normal fine ultrafilter on $ P {\kappa } ( \lambda ) ) . $ \quad Then for every stationary $ S \subseteq P {\omega { 1 }} ( \lambda ) $ \quad and for every tight and unbounded $ A \subseteq P {\kappa } ( \lambda ) , $ the re i s an $ X \ in A$ such that $S \cap P {\omega { 1 }} ( X ) $ \quad is stationary in $ P {\omega { 1 }} ( X ) . $

\ centerline {We are going t o prove two lemmas which will give the theorem . } Reflection of stationary sets 1 77 Lemma 1 . 1 . Assume that λ ≥ κ and κ is λ− supercompact . Let U be a κ− complete normal fine ultrafilter on Pκ(λ). If S ⊆ Pω1 (λ) is stationary

, then there is a set A ∈ U such that S ∩ Pω1 (X) is stationary in Pω1 (X) for

eachX ∈ A.

P r o o f . Assume not . Let S ⊆ Pω1 (λ) be a counterexample . Then define ω A ⊆ Pκ(λ) to be the following set . For x ∈ Pκ(λ), let x ∈ A if and only if S ∩ [x] i s nonstationary in [x]ω. It follows that A ∈ U. ω For x ∈ A, fix a club Cx ⊆ [x] disj oint from S. Let

ω C = {y ∈ [λ] | {x ∈ A | y ∈ Cx} ∈ U}. # ω Claim .C is a c lub in [λ First notice the following fact , which follows . from the normality and

ℵ1 − completenessofU. Fact . Let D ∈ U and let f : D → [λ]ω be such that for each x ∈ D we have f(x) ⊆ x. Then for some D0 ∈ U, f is constant on D0. To see that C i s closed , let hyn : n < ωi be from C such that yn ⊆ yn + 1. Let T Dn = {x ∈ Pκ(λ) | yn ∈ Cx} ∈ U. Then D = n<ω Dn i s in U. Now for each x ∈ D, we have ∀n < ωyn ∈ Cx. Since each Cx i s closed ,

[ [ ∈ Cx.Hence ∈ C. n<ωyn n<ωyn To see that C i s unbounded , let y0 ∈ [λ]ω; we need t o find some y ∈ C such that y0 ⊆ y. ω Applying the above fact , we inductively define Dn ∈ U, yn, fn : Dn → [λ] so that T ∀x ∈ Dnfn(x) ∈ Cx, yn ⊆ fn(x) and ∀x ∈ Dn+1fn(x) = yn + 1. Let D = Dn. S n<ω Then for each x ∈ D we have y0 ⊆ y = n<ωyn

Cx. ∈ Since C is a club and S i s stationary , let Y ∈ C ∩ S. Then {x ∈ A | y ∈ Cx} ∈ U and ∀x ∈ ACx ∩ S = ∅, a contradiction . Lemma 1 . 2 . Assume that λ ≥ κ is regular and κ is λ− supercompact . If U is a κ− complete normal fine ultrafilter on Pκ(λ), then every tight and unbounded subset A ⊆ Pκ(λ) is in U. P r o o f . Let U be a κ− complete normal fine ultrafilter on Pκ(λ). Let j : V → M be the canonical embedding , where M is the transitive collapse P of V κ(λ)/U. By the standard theory of supercompact cardinals [ 4 ] , we have

foreachX ⊆ Pκ(λ), X ∈ U ⇔ j00λ ∈ j(X). Reflection of stationary sets .. 1 77 \ hspaceLemma∗{\ ..f 1 i lperiod l } Reflection 1 period .. of Assume stationary that lambda sets greater\quad equal1 77 kappa .. and kappa is lambda hyphen supercompact period .. Let U .. be a Lemmakappa\quad hyphen1 complete . 1 . \ normalquad Assume fine ultrafilter that on $ P\ sublambda kappa open\geq parenthesis\kappa lambda$ \ closingquad and parenthesis $ \kappa $ periodi s $ ..\ Iflambda S subset equal− $ P sub supercompact omega sub 1 open . \quad parenthesisLet lambda $ U $ closing\quad parenthesisbe a .. is stationary comma $then\kappa there is− a set$ A complete in U such normal that S cap fine P sub ultrafilter omega sub 1 on open $ parenthesis P {\kappa X closing} ( parenthesis\lambda is ) . $ \quad I f $ S \subseteq P {\omega { 1 }} ( \lambda ) $ \quad is stationary , stationary in P subLet omegaA ⊆ subPκ 1( openλ) bparenthesis e t ight and X closing unbounded parenthesis . We for want t o show that A ∈ U. theneach thereX in A period isHence a set it suffices $A to show\ in thatU$j00λ ∈ suchj(A). that $S \cap P {\omega { 1 }} (P X r o o f)$ period is ..Let Assume stationaryD = not{j( periodp) in| Letp $P∈ SA subset}{\. omegaBy equal a theorem P{ sub1 omega}} of Solovay( sub 1 X open , we ) parenthesis have $λ f<κ o r = lambdaλ. So closing parenthesis be a counterexampleλ = | D | < period j(κ). ..By Then the define closure property of M, we find that D ∈ M \ beginA subset{ a l i equal g n ∗} P sub kappa open parenthesis lambda closing parenthesis to be the following set period .. For xeach in P sub X kappa\ openin parenthesisA. lambda closing parenthesis comma let x in A if and only if \endS cap{ a l i open g n ∗} square bracket x closing square bracket to the power of omega i s nonstationary in open square bracket x closing square bracket to the power of omega period .. It follows that A in U period P rFor o x o in f A . comma\quad fixAssume a club C not sub x . subset Let equal $ S open\subseteq square bracketP x closing{\omega square{ bracket1 }} to the( power\lambda of) omega $ be disj a counterexampleoint from S period .. . Let\quad Then d e f i n e $C A = open\subseteq brace y in openP square{\kappa bracket} lambda( closing\lambda square bracket) $ to to the be power the following of omega bar setopen .brace\quad For x$ in x A bar\ in y in CP sub{\ x closingkappa brace} ( in U\ closinglambda brace) period , $ l e t $ x \ in A$ if and only if $Claim S period\cap C is[ a c lub x in ] open ˆ{\ squareomega bracket}$ lambda i s nonstationary Case 1 omega Case in 2 period $ [ x ] ˆ{\omega } . $First\quad .. noticeIt the follows following that fact comma $ A .. which\ in followsU from . $ the normality .. and aleph sub 1 hyphen completeness of U period \ centerlineFact period{ LetFor D in $ U x and\ letin f : DA right ,$ arrow fixaclub open square bracket $C lambda{ x }\ closingsubseteq square bracket[ to x the ] ˆ{\omega }$ powerdisj of oint omega from .. be such $S that . for $ each\quad x in DLet we } have f open parenthesis x closing parenthesis subset equal x period .. Then for some D sub 0 in U comma f is constant\ [ Con = D sub\{ 0 periody \ in [ \lambda ] ˆ{\omega }\mid \{ x \ in A \mid y To\ seein thatC C{ i sx closed}\}\ comma letin angbracketleftU \} y. n\ :] n less omega right angbracket be from C such that y n subset equal y n plus 1 period Let .. D sub n = open brace x in P sub kappa open parenthesis lambda closing parenthesis bar y n in C sub x\noindent closing braceClaim in U period $. .. Then C$ .. D isaclubin = intersection of sub $[ n less\lambda omega D sub\ l e n f .. t ] i\ sbegin in .. U{ perioda l i g n e d } & \omegaNow for\\ each x in D comma we have forall n less omega y n in C sub x period .... Since each C sub x i s closed comma&. \end{ a l i g n e d }\ right . $ Funion i r s t \ ofquad sub nnotice less omega the to the following power of y fact n in C ,sub\quad x periodwhich Hence follows union of sub from n less the omega normality to the power\quad and of y n in C period \ beginTo see{ a that l i g n C∗} i s unbounded comma let y 0 in open square bracket lambda closing square bracket to the power of\ omegaaleph semicolon{ 1 } we − needcompleteness t o find some y in C of U . \endsuch{ a thatl i g n y∗} 0 subset equal y period Applying the above fact comma we inductively define D sub n in U comma y n comma f sub n : D sub n right arrowFact . Let $ D \ in U$ andlet $f : D \rightarrow [ \lambda ] ˆ{\omega }$ \quadopenbe square such bracket that lambda for each closing $ square x bracket\ in toD the$ powerwe of omega so that forall x in D sub n f sub nhave open parenthesis $ f ( x closing x ) parenthesis\subseteq in C subx comma . $ y n\quad subsetThen equal ffor sub some n open parenthesis$ D { 0 x}\ closingin Uparenthesis , f$ and forall is constantonx in D sub n plus $D 1 f sub{ n0 open} parenthesis. $ x closing parenthesis = y n plus 1 period Let D = intersection of sub n less omega D sub n period .. Then for each x in D we have y 0 subset equal y To= union see of that sub n less$C$ omega i to s the closed power of , y let n $ \ langle y n : n < \omega \rangle $ befromEquation: $C$ in .. C sub such x period that $y n \subseteq y n + 1 . $ LetSince\quad C is a club$ D and{ Sn i s} stationary= \{ commax let\ in Y in CP cap{\ S periodkappa ..} Then( open\lambda brace x in A) bar \mid yy inn C sub\ in x closingC { bracex }\}\ in U and forallin x inU A C . sub $ x\ capquad S =Then varnothing\quad comma$ D a contradiction = \bigcap period{ n < Lemma\omega .. 1} periodD { 2 periodn }$ ..\ Assumequad i .. s that in lambda\quad greater$ U equal . $ kappa .. is .. regular .. and kappa .. is lambda hyphen supercompact period \noindentIf U is a kappaNow for hyphen each complete $ x normal\ in fineD ultrafilter , $ we on P have sub kappa $ \ f openo r a l lparenthesisn < lambda\omega closing parenthesisy n \ commain C .. then{ x every} . tight $ and\ h f i l l Since each $ C { x }$ i s c l o s e d , unbounded subset A subset equal P sub kappa open parenthesis lambda closing parenthesis .. is in U period \ beginP r o{ oa f l period i g n ∗} .. Let .. U be a kappa hyphen complete normal fine ultrafilter on P sub kappa open parenthesis lambda\bigcup closing{ parenthesisn < period\omega .. Letˆ{ y n }}\ in C { x } . Hence \bigcup { n < j :\ Vomega right arrowˆ{ y M be n the}}\ canonicalin embeddingC. comma where M is the transitive collapse \endof{ Va l to i g then ∗} power of P kappa open parenthesis lambda closing parenthesis sub slash U period .... By the standard theory of supercompact cardinals open square bracket 4 closing square bracket comma we have Tofor see each that X subset $C$ equal P i sub s kappaunbounded open parenthesis , let $ylambda closing0 \ in parenthesis[ comma\lambda X in U] Leftrightarrow ˆ{\omega } j; to $ the we power need of prime t o find prime some lambda $in jy open\ in parenthesisC $ X closing parenthesis period suchLet A that subset $y equal P 0 sub\ kappasubseteq open parenthesisy . $ lambda closing parenthesis .. b e t ight and unbounded period .. We want t o show that A in U period ApplyingHence it suffices the above to show fact that j , to we the inductively power of prime prime define lambda $ D in j{ openn }\ parenthesisin U A closing , parenthesis y n period, f { n } :D { n }\rightarrow $ $Let [ D =\lambda open brace j] open ˆ{\ parenthesisomega }$ p closing so that parenthesis $ \ f obar r a p l l in Ax closing\ in braceD period{ n ..} By af theorem{ n } of( Solovay x comma ) \ in we haveC lambda{ x } to the, power y of n less\ kappasubseteq = lambdaf period{ n } ( x ) $ and $ \ f o r a l l x So\ lambdain D = bar{ n D bar + less 1 j open} f parenthesis{ n } kappa( closingx ) parenthesis = y period n .. + By the 1 closure . $ property of MLet comma $ ..D we =find that\bigcap D in M { n < \omega } D { n } . $ \quad Then for each $ x \ in D$ wehave $y 0 \subseteq y = \bigcup { n < \omega ˆ{ y n }}$

\ begin { a l i g n ∗} \ tag ∗{$ \ in $} C { x } . \end{ a l i g n ∗}

\ hspace ∗{\ f i l l } Since $C$ is a club and $S$ i s stationary , let $Y \ in C \cap S . $ \quad Then $ \{ x \ in A \mid $

\noindent $ y \ in C { x }\}\ in U $ and $ \ f o r a l l x \ in AC { x } \cap S = \ varnothing , $ a contradiction .

Lemma \quad 1 . 2 . \quad Assume \quad that $ \lambda \geq \kappa $ \quad i s \quad r e g u l a r \quad and $ \kappa $ \quad i s $ \lambda − $ supercompact . I f $ U $ i s a $ \kappa − $ complete normal fine ultrafilter on $ P {\kappa } ( \lambda ) , $ \quad then every tight and unbounded subset $ A \subseteq P {\kappa } ( \lambda ) $ \quad i s in $ U . $

P r o o f . \quad Let \quad $ U $ be a $ \kappa − $ complete normal fine ultrafilter on $ P {\kappa } ( \lambda ) . $ \quad Let $ j : V \rightarrow M $ be the canonical embedding , where $ M $ is the transitive collapse

\noindent o f $ V ˆ{ P }\kappa ( \lambda ) { /U. }$ \ h f i l l By the standard theory of supercompact cardinals [ 4 ] , we have

\ begin { a l i g n ∗} f o r each X \subseteq P {\kappa } ( \lambda ), \\ X \ in U \Leftrightarrow j ˆ{\prime \prime }\lambda \ in j ( X ) . \end{ a l i g n ∗}

\noindent Let $ A \subseteq P {\kappa } ( \lambda ) $ \quad b e t ight and unbounded . \quad We want t o show that $ A \ in U . $ Hence it suffices to show that $ j ˆ{\prime \prime }\lambda \ in j ( A ) . $

Let $ D = \{ j ( p ) \mid p \ in A \} . $ \quad By a theorem of Solovay , we have $ \lambda ˆ{ < \kappa } = \lambda . $ So $ \lambda = \mid D \mid < j ( \kappa ) . $ \quad By the closure property of $ M , $ \quad we find that $D \ in M $ 1 78 .. Q period F e n g and M period .. Magi d o r \noindentand M thinks1 78 that\quad D is omegaQ.FengandM. hyphen directed and\ jquad openMagi parenthesis d o A r closing parenthesis i s t ight and unboundedand $M$ period thinks .. So that $D$ is $ \omega − $ directedand $j ( A )$ i sEquation: t ight union and unboundedof D in j open . parenthesis\quad So A closing parenthesis period .. Therefore comma j to the power of prime prime lambda in j open parenthesis A closing parenthesis since j to the power of prime prime lambda = union\ begin of{ Da lperiod i g n ∗} \ tagS 2∗{ period$ \bigcup .. It1 i 78 s knownD Q . thatF\ ein n the g and followingj M . ( Magi reflection A d o) r principleand . $M} thinks iTherefore s consistent that D andis ω ,− directed j ˆ{\ andprimej(A) \prime } \lambdahas large cardinal\ in i s t aspectsj ight (and period unbounded A ) . since So jˆ{\prime \prime }\lambda = \bigcup D.Suppose kappa greater equal omega sub 2 .. and S subset equal open square bracket H sub kappa closing square\end{ bracketa l i g n ∗} to the power of omega .. is stationary period .. Then there is an X in 00 00 [ S open square bracket H sub kappaTherefore closing square, j λ ∈ bracketj(A)since to thej λ power= D. of aleph sub 1 .... suchD that∈ j omega(A). sub 1 subset\S 2 equal . \quad X ....It and i S scap known open square that bracket the following X closing square reflection bracket to theprinciple power of omega i s consistent .... is stationary and inhas open large square bracketcardinal§2 X . closing aspectsIt square i s known . bracket that to the the following power of reflection omega comma principle .... iwhere s consistent H sub kappa and has = open brace X barlarge bar cardinal TC open aspects parenthesis . open brace X closing brace closing parenthesis bar less kappa ω closing\ hspace brace∗{\ periodf i l l }SupposeSuppose $ κ\kappa≥ ω2 and\geqS ⊆\omega[Hκ] { is2 stationary}$ \quad .and Then there$ S is an\subseteq [HA natural{\kappa questionX } would∈ ] ˆ be{\ whetheromega we}$ could\quad have ais similar stationary principle . \quad Then there is an $ X ℵ1 ω ω \ infor$ the uncountable[Hκ] spacesuch comma that iω period1 ⊆ X e periodand S comma∩ [X] couldis the stationary following in be[X consistent] , where : Hκ = If kappa greater equal omega sub 3 .. and S subset equal open square bracket H sub kappa closing square bracket\noindent to the power$ [ of H aleph{\ 1kappa .. is stationary} ] ˆ comma{\aleph .. then{ there1 }} is$ an\ Xh in f i open l l such square that bracket $ H\ subomega kappa{ 1 } {X || TC({X}) |< κ}. closing\subseteq square bracketX $ to\ h the f il power l and of aleph $ S 2 \cap [ X ] ˆ{\omega }$ \ h f i l l is stationary in $ [ X ] ˆ{\omega } , $ \ h f i l l where $ H {\kappa } = $ such that omega subA natural 2 subset question equal X wouldand S cap be whether open square wecould bracket have H sub a similar kappa closing principle square for the bracket to the power of aleph sub 1 .. is stationary in open square bracket X closing square bracket to the power of aleph \ begin { a l i g n ∗}uncountable space , i . e . , could the following be consistent : sub 1 period ℵ ℵ \{ X \midIf κ ≥\midω3 andTCS ⊆ ( [H\{κ] 1 Xis stationary\} ) , \midthen there< is an\kappaX ∈ \}[Hκ] 2. Since .. omega sub 1 hyphen closed .. forcing ..ℵ might1 .. kill .. stationaryℵ ..1 set s .. of the .. uncountable such that ω2 ⊆ X and S ∩ [Hκ] is stationary in [X] . \endspace{ a l commai g n ∗} .. one obvious way t o show the consistency does not go through period .. It Since ω1− closed forcing might kill stationary set s of the uncount- turns out this isable not space simply , a tone echnique obvious problem way t o period show .. the We consistency show that doesthis simply not go through . It A naturalcannot b e question done comma would i period be e period whether comma we the could statement have stated a similar above is principle really false for sufficiently for the uncountableturns out this space is not , simply i . e a .t echnique , could problem the following . We show be that consistent this simply cannot : large kappa periodb e done , i . e . , the statement stated above is really false for sufficiently large κ. Theorem 2 period 1 period If kappa greater equalℵ open++ parenthesis 2 to the power of aleph 2 closing parenthesis \ centerline { I fTheorem $ \kappa 2 .\ 1geq . If κ ≥\omega(2 2) { is3 regular}$ \quad, thenand there $ exists S \ asubseteq stationary [ to the power of plus plus ..ℵ is1 regular comma .. then there existsℵ2 a stationary ℵ1 H {\kappa } S ⊆][ ˆH{\κ] alephsuch} that1 $ for\ anyquadX ∈is[H stationaryκ] , ω2 ⊆ X implies , \quad that thenS ∩ [X there] isis an $X S subset equal open square bracket Hℵ sub kappa closing square bracket to the power of aleph sub 1 .... such \ in [H {\notkappa stationary} in][ ˆX{\] 1 .aleph } 2 $ } that for any X in openP rsquare o o f . bracket Toward H sub a contradiction kappa closing , square assume bracket otherwise to the . power Let ofκ alephb e a coun sub 2 - comma omega sub 2 subset equal X implies that S cap open square bracket X closingℵ1 square bracket to the powerℵ2 of aleph \noindent suchterexample that $ .\omega Then for{ any2 stationary}\subseteqS ⊆ [HκX] $there and is $ an SX \∈cap[Hκ][H {\kappa } sub 1 .... is ℵ1 ℵ1 such that ω2 ⊆ X and S ∩ [X] is stationary in [X] . ] ˆnot{\ stationaryaleph { in1 open}}$ square\quad bracketis X stationary closing square in bracket $ [ to the X power ] ˆ of{\ alephaleph sub 1{ period1 }} . $ Claim 1 . The nonstationary ideal NSω has the following property : P r o o f period .. Toward a contradiction comma .. assume otherwise2 period .. Let .. kappa b e a coun hyphen Since \quad $(∗)\omegafor any{ stationary1 } − T$⊆ c l o sω e2 d \andquad for anyf o rs c iequence n g \quadhAαmight| α\quad∈ ωk1 ii l l \quad s t a t i o n a r y \quad s e t s \quad o f the \quad uncountable terexample period .. Then for any stationary S subset equal open square bracketα H sub kappa closing square of maximal antichains below T mod NSω , letting Aα = {X | β < λα} bracketspace to , the\quad powerone of aleph obvious sub 1 .. way there t is o an show X in open the square consistency2 bracket H does sub kappaβ not closing go through square bracket . \quad to I t turns out thisfor is notα < simplyω1, there a is t a stationary echniqueS problem⊆ T such . that\quad for eachWe showα < ω that1 the this set simply the power of aleph sub 2 α Aα  S = {Xβ ∩ S | β < λα} has cardinality at most ℵ2. cannotsuch that b omega e done sub , 2 i subset . e equal . ,the X and statement S cap open squarestated bracket above X closingis really square false bracket for to the sufficiently power To see (∗), let T, hAα | α < ω1i b e given . Consider the following set : ofl alepha r g e sub $ 1\ iskappa stationary. in $ open square bracket X closing square bracket to the power of aleph sub 1 period

Claim 1 period .. The nonstationary ideal NS subℵ1 omega sub 2 .. has the followingα property : S = {N ∈ [Hκ] | ∀α < ω1∃β ∈ NN ∩ ω2 ∈ X }. \ hspaceopen parenthesis∗{\ f i l l }Theorem * closing parenthesis 2 . 1 . If .. for $ any\kappa stationary\geq T subset( equal 2 ˆ omega{\β aleph sub 2} ..2 and for ) ˆany{ + s + }$ \quad i s r e g u l a r , \quad then there exists a stationaryℵ1 equence .. angbracketleft A sub alpha bar alphaFact in0 omega.S is stationary sub 1 right in angbracket[Hκ] . .. of maximal antichainsIt below will T mod suffice NS sub t omega o prove sub 2 comma that ..S lettingi s A weakly sub alpha stationary = open brace , X i sub . beta \noindent $ S \subseteq [H {\<ωkappa } ] ˆ{\aleph { 1 }}$ \ h f i l l such that for any to the power of alphae . bar , beta for less any lambdaf sub:[H alphaκ] closing→ Hκ brace, there i s an N ∈ S such that for each $ Xfor alpha\ in less[H omegae ∈ [N sub]<ω{\, 1 fkappa comma(e) ∈ N.} there] is ˆ{\ a stationaryaleph S{ subset2 }} equal, T\ suchomega that for{ 2 each}\ alphasubseteq less omega X$ implies that $ S \cap [<ω X ] ˆ{\aleph { 1 n}}$ \ h f i l l i s sub 1 .. the Let f :[Hκ] → Hκ and fn = f  [Hκ] . Let set A sub alpha upharpoonright S = open brace X sub beta to the power of alpha cap S bar beta less lambda \noindent not stationary in $ [ X ] ˆ{\aleph { 1 }} . $ sub alpha closing brace has cardinality at most alephH = subhHκ 2, ∈ period, fn, ...in<ω. To see open parenthesis * closing parenthesis comma let T comma angbracketleft A sub alpha bar alpha less Subclaim . For each α < ω there is a club C ⊆ [H ]ℵ1 such that omega\ hspace sub∗{\ 1 rightf i l l angbracket}P r o o b f e given . \quad periodToward .. Consider a1 contradiction the following set :α , \quadκ assume otherwise . \quad Let \quad $ \Skappa = open$ brace b eN ina opencoun square− bracket H sub kappa closing square bracket to the power of aleph sub 1 bar (1)∀N ∈ C N ≺ ; and forall alpha less omega sub 1 exists beta in N N cap omegaα subH 2 in X sub beta to the power of alpha closing brace\noindent period terexample . \quad Then for any stationary $ S \subseteq [H {\kappa } ] ˆFact{\aleph 0 period S{ is1 stationary}}$ \quad in openthere square is bracket an H $X sub kappa\ in closing[H square{\ bracketkappa to the} power] ˆ{\ ofaleph aleph { 2 }}$ sub 1 period \noindentIt .. will ..such suffice that .. t o .. $ prove\omega .. that{ ..2 S ..}\ i s ..subseteq weakly .. stationaryX $ and comma $ .. S i period\cap e period[ comma X ].. ˆ{\aleph { 1 }}$ foris .. stationary any .. f : in $ [ X ] ˆ{\aleph { 1 }} . $ open square bracket H sub kappa closing square bracket to the power of less omega right arrow H sub kappa comma\ centerline there i{ sClaim an N in 1 S such . \quad that forThe each nonstationary e in open square ideal bracket N $ closing NS {\ squareomega bracket{ to2 the}}$ power\quad of has the following property : } less omega comma f open parenthesis e closing parenthesis in N period \noindentLet f : open$ square ( bracket∗ ) H $ sub\quad kappafor closing any square stationary bracket to the $ T power\ ofsubseteq less omega right\omega arrow H{ sub2 }$ kappa\quad andand f sub for n = any f upharpoonright s equence \ openquad square$ \ bracketlangle H subA kappa{\alpha closing square}\mid bracket\ toalpha the power\ in of n period\omega .. Let{ 1 }\rangle $ \quad o f maximalH = angbracketleft antichains H sub below kappa comma $ T $ in mod comma $ f NS sub n{\ commaomega period{ period2 }} period, $ right\quad angbracketl e t t i n sub g n$ less A omega{\alpha period} = \{ X ˆ{\alpha } {\beta }\mid \beta < \lambda {\alpha } \} Subclaim$ period .. For each alpha less omega sub 1 .. there is a club C sub alpha subset equal open square bracket H sub kappa closing square bracket to the power of aleph sub 1 .. such that f o ropen $ parenthesis\alpha 1< closing\omega parenthesis{ 1 forall} N, in $ C there sub alpha is N a prec stationary H semicolon and$ S \subseteq T $ such that for each $ \alpha < \omega { 1 }$ \quad the s e t $ A {\alpha }\ upharpoonright S = \{ X ˆ{\alpha } {\beta }\cap S \mid \beta < \lambda {\alpha }\} $ has cardinality at most $ \aleph { 2 } . $

\ centerline {To s ee $ ( ∗ ) , $ l e t $ T , \ langle A {\alpha }\mid \alpha < \omega { 1 }\rangle $ b e given . \quad Consider the following set : }

\ [ S = \{ N \ in [H {\kappa } ] ˆ{\aleph { 1 }}\mid \ f o r a l l \alpha < \omega { 1 }\ exists \beta \ in NN \cap \omega { 2 }\ in X ˆ{\alpha } {\beta }\} . \ ]

\ centerline { Fact $0 . S$ is stationary in $[ H {\kappa } ] ˆ{\aleph { 1 }} . $ }

I t \quad w i l l \quad s u f f i c e \quad t o \quad prove \quad that \quad $ S $ \quad i s \quad weakly \quad stationary , \quad i . e . , \quad f o r \quad any \quad $ f : $ $ [ H {\kappa } ] ˆ{ < \omega }\rightarrow H {\kappa } , $ there i s an $ N \ in S $ such that for each $ e \ in [ N ] ˆ{ < \omega } , f ( e ) \ in N . $

\ centerline { Let $ f : [ H {\kappa } ] ˆ{ < \omega }\rightarrow H {\kappa }$ and $ f { n } = f \ upharpoonright [ H {\kappa } ] ˆ{ n } . $ \quad Let }

\ [ H = \ langle H {\kappa } , \ in , f { n } ,... \rangle { n < \omega } . \ ]

\ centerline { Subclaim . \quad For each $ \alpha < \omega { 1 }$ \quad there is a club $ C {\alpha }\subseteq [H {\kappa } ] ˆ{\aleph { 1 }}$ \quad such that }

\ [ ( 1 ) \ f o r a l l N \ in C {\alpha } N \prec H ; and \ ] Reflection of stationary sets .. 1 79 \ hspaceLine 1∗{\ openf i parenthesis l l } Reflection 2 closing of parenthesis stationary forall sets N in C\quad sub alpha1 79 N cap omega sub 2 in T double stroke right arrow exists beta N cap omega sub 2 in X sub beta to the power of alpha ampersand Sk to the power of H open\ [ \ begin parenthesis{ a l i g Nn e cup d } open( brace 2 beta ) closing\ f o r a brace l l closingN \ parenthesisin C {\ capalpha omega sub} 2N = Line\cap 2 N cap\omega omega { 2 } sub\ in 2 periodT \Rightarrow \ exists \beta N \cap \omega { 2 }\ in X ˆ{\alpha } {\beta } \&Assuming Sk ˆ{ theH subclaim} (N comma\cup we show\{\ that therebeta is an\} N in S) which\cap is closed\omega { 2 } = \\ Nunder\cap f period\ Magidoromega Let{ 2 and} D. =\end intersection{ a l i g n e of d }\ Shelah] alphaReflection open square of stationary bracket sets less omega1 79 sub 1 2 closing square bracket comma D sub alpha for sub some to the power of period to the power of Then D i s a club α H expansion sub H to the power(2) of∀N prime∈ Cα ofN to∩ ω the2 ∈ powerT ⇒ ∃ ofβN period∩ ω2 to∈ X theβ &Sk power(N of∪ By {β} H) ∩ commaω2 = a sub we lemma \ hspace ∗{\ f i l l }Assuming the subclaim , we show that there is an $ N \ in S $ which is closed sub have that to the power of of Foreman sub for any to the power of comma N ∩ ω2. N prec H to the power of primeAssuming if bar theN bar subclaim = aleph , we sub show 1 then that for there each is alpha an N less∈ S omegawhich subis closed 1 comma N cap\ [ omegaunder sub f2 in T . implies{ Magidor there is} betaLet { and } D = \bigcap { Shelah }\alpha{ [ } < \omegainduction{ to1 the}{ power2 } of],D such that N on{\ capalpha alpha}{ omegaf o r sub}ˆ 2{ less. in} omega{ some sub} 1ˆ{ betaThen from} t oD to X to i the s a \ clubsuit { expansion }ˆ{ . } \{ H ˆ{\prime } o f. }ˆThen{ By } H ,. a By { we } ofForeman, power of alpha andunder subf. buildMagidorLetand Sk an to theD power= Shelah of H toα[ the< ω power12],Dα offor primesome openDisa parenthesis♣expansion N0 Nof cupH, inawe openlemmahavethat forany lemma { have } that ˆ{ o f Foreman }ˆ{ , } { f o r any }\ ] H brace S beta closing brace0 so closing parenthesis cap omega sub 2 that = f N prime prime sub open square bracket N ≺ H if | N | = ℵ1 then for each α < ω1,N ∩ ω2 ∈ T implies there is β N closing square bracket sub less omega to the power of cap omega0 sub 2 period Now subset equal N period use suchthat to H ∩ω2. induction N ∩ ω < ∈ β α and Sk ( N∪ { β} )∩ω that = fN Now a simple on α 2 ω1 X build an N ∈ S so 2 00[N] <ω ⊆N. \noindentNow we proceed$ Nuse t o a\ show simpleprec the Now subclaimH we ˆ{\ proceedprime period t o} show$ i thef subclaim $ \mid . N \mid = \aleph { 1 }$ thenAssume for .. each otherwise $Assume\ periodalpha .. otherwise Let< .. alpha\omega . less Let omega{ α1 } sub<,N 1 ω ..1 b eb .. e\cap a ..a counterexample counterexample\omega { period2 .}\ Let ..in Let ..T S $ subimplies 1 .. be there .. a S is1 be $ \beta a witness$ . That is ,S1 is stationary such that for each X ∈ S1 we have X ≺ H and X ∩ ω2 ∈ T and for each β < witness period .. That .. is comma S subα 1 .. is .. stationary .. such .. that .. for .. each X in S sub 1 .. we have\noindent $ inductionλα if X ∩ ω ˆ2{ such∈ Xβ thatthen } N { on }\cap {\alpha }\omega { 2 }{ < } \ inX prec{\ Homega .. and ..{ X1 cap}}\ omegabeta subˆ 2{ int T .. oand} for{ X .. each ˆ{\ ..alpha beta less}} lambdaand sub{ alphab ui ld ..} if XSk cap omega{ an }ˆ{ H ˆ{\prime }} ( { N } N \cup {\ in }\{ { SH}\beta \} { so } ) \cap \omega { 2 }{ that } ={ f } sub 2 in X sub beta to the power of alpha .. thenSk (X ∪ {β}) ∩ ω2 6= X ∩ ω2. N Sk{\ toprime the power\ ofprime H open{ parenthesis[N] X cup}} openˆ{\ bracecap beta\omega closing brace{ 2 closing} . parenthesis} { < \ capomega omega} subNow {\subseteq ℵ2 N.2 negationslash-equal}$ use a Xsimple capBy omega our sub assumption 2 period , let A ∈ [Hκ] , ω2 ⊆ A, A ≺ H NowBy we.. our proceed .. assumptionbe t such o show comma that the .. let subclaim .. A in open . square bracket H sub kappa closing square bracket to the power of aleph sub 2 comma omega sub 2 subset equal A comma A prec H .. be .. such .. that Assume \quad otherwiseℵ . \quad Let \quad $ \alphaS < \omega { 1 }$ \quad b e \quad a \quad counterexample . \quad Let \quad chain to the powerchain ofS1 S∩[A sub] 1 1 cap openisstationary square. bracketWriteA A sub= of toγ <ω the2 powerN 1asthe of closing∪ of squarean elementary bracket to| the of submodels Nγ ≺Awith |N | = ℵ γ . ThenT1 ={Nγ ∩ ω2 N ∈S } power$ S of{ aleph1 }$ sub\quad submodelsbe \ toquad the powera of 1 to the power of is stationaryγ sub N sub gamma to the power ofγ 1 witness . \quad That \quad i s $ , S { 1 }$ \quad i s \quad s t a t i o n a r y \quad such \quad that \quad f o r \quad each period Write precis A stationary with to the in powerω2, and ofC A1 = { subNγ bar| γ < N ω sub2} is gamma a club .to the power of union of sub bar to the $ X \ in S { 1 }$ \quad we have α power of gamma sub = to theFirst power , if ofβ less∈ A omegathen T sub1 ∩ X alephβ is2 nonstationary N sub gamma . 1sub Let periodβ ∈ A. ThenThen to therethe power of $ X \prec H $ \quad and \quad $ X \cap \omega { 2 }\ in T $ \quad and f o r \quad each \quad as the sub T sub 1is to a γ the < powerω2 so that of cup for sub each =η open≥ γ, βbrace∈ Nη N. sub gamma to theNotice power that of for of each sub cap such subη, omega $ \beta < \lambda {\alpha }$ H \quad i f $ X \cap \omega { 2 }\ in X ˆ{\alpha } {\beta }$ sub 2 to the powerif ofN anη ∩ elementaryω2 ∈ T1 barthen sub Sk N sub(Nη gamma∪ {β}) in∩ Sω2 sub= 1 closingNη ∩ ω2 brace, which implies that Nη ∩ \quad then α α is stationary inω omega2element sub− 2negationslashX comma and C subβ . 1Hence = openT1 brace∩ Xβ Nis subbounded gamma . bar gamma less omega sub 2 closing brace is a club period Since T1 ⊆ T is stationary , there i s some β 6∈ A such that \ begin { a l i g n ∗} α First comma ifT beta1 ∩ X in A then T sub 1 cap X sub beta to the power of alpha is nonstationary period .. Let Sk ˆ{ H } (X β \cup \{\beta \} ) \cap \omega { 2 }\not= X beta in A period ..is Then stationary there . Denote this set by T . Let Y = SkH(A ∪ {β}) and N 0 = \cap \omega { 2 } . 2 γ is a gamma less omegaH sub 2 so that for each eta0 greater equal gamma comma beta in N sub eta period .... \end{ a l i g n ∗} Sk (Nγ ∪ {β}). Then C2 = {Nγ ∩ ω2 | γ < ω2} is a club . So T2 ∩ C1 ∩ C2 6= ∅. Notice that for each such eta comma 0 Let Nγ ∩ ω2 be in the intersection . Then Nγ ∩ ω2 = Nγ ∩ ω2. But this if N sub eta cap omega sub 2 in T sub 1 .. then Skα to the power of H open parenthesis N sub eta cup open is a contradiction since Nγ ∩ ω2 ∈ X and Nγ ∈ S1. brace\ hspace beta∗{\ closingf i l l brace}By closing\quad parenthesisour \quad capassumption omegaβ sub 2 = , N\quad sub etal ecap t \ omegaquad sub$ 2 A comma\ in which[H implies {\kappa } ] ˆ{\aleph { 2 }} , \omegaThis{ finishes2 }\ thesubseteq proof of the subclaimA,A . \prec H $ \quad be \quad such \quad that that ℵ2 To finish the proof of the claim , let X ∈ [Hκ] be such that ω2 ⊆ X,X ≺ Hκ N sub eta cap omega sub 2ℵ element-negationslash1 Xℵ1 sub beta to the power of alpha period .. Hence T sub 1 and S ∩ [X] i s stationary in [X] . Write X as a union of ℵ2 many elementary cap\ [X chain sub beta ˆ{ toS the{ power1 }\ ofcap alpha is[A bounded}ˆ period{ ] ˆ{\aleph }} { o f }ˆ{ 1 } { submodels }ˆ{ i s submodels Nγ of siz e ℵ1. Then S0 = {Nγ ∩ ω2 | γ < ω2,Nγ ∈ S} is stationary in ω2. s t aSince t i o n Ta rsub y } 1ˆ{ subset. } equal{ N T{\ is ..gamma stationary}} commaWrite ..{\ thereprec i s .. someA beta with negationslash-element}ˆ{ A } = {\ Amid such N {\gamma }}Sinceˆ{\Nbigcup∩ ω2 ∈ S}0ˆ{\impliesgamma that} for{\ eachmidα <}ˆ ω{1 there< \omega } { = } {\aleph } 2 that .. T sub 1 cap X sub beta to the power of alpha α is some β ∈ X such that N ∩ ω2 ∈ X , it follows immediately that for each N is{\ stationarygamma period}{ 1 } ....{ Denote. Then this set}ˆ by{ Tas sub 2 the periodβ } { ....T Let{ Y1 =}} Skˆ to{\ thecup power} { of= H open\{ parenthesisN {\gamma }}ˆ{ o f } {\cap }ˆ{ an } {\omega { 2 }} elementary {\mid } { N {\gamma } \ in S { 1 }\}}\α < ω1 the] set Aα  S0 has cardinality at most ℵ2. A cup open brace beta closing brace closingThis parenthesis finishes the .... andproof N of sub the gamma claim . to the power of prime = Sk to the power of H open parenthesisTo get a contradiction N sub gamma , we cup are opennow going brace to beta show closing that the brace nonstationary closing parenthesis period .... Then C sub 2 = open brace N sub gamma to the power of prime cap omega sub 2 bar gamma less ideal on ω2 does not have the property (∗). omega\noindent sub 2 closingis stationary brace is a club in period $ \ ....omega So T sub{ 2 2 cap} C, sub $ 1 and cap C $ sub C 2 negationslash-equal{ 1 } = \{ varnothingN {\gamma } Claim 2 . The nonstationary ideal NSω on ω2 does not have the proper - period\mid \gamma < \omega { 2 }\} $ i s a club2 . Let N sub gamma cap omega sub 2 be in the intersection period .. Then N sub gamma to the power of prime cap\ hspace omega∗{\ subf 2i l = l } NF sub i r s t gamma , i f cap $ omega\beta sub 2\ periodin Aty ..( But $∗). thisthen $ T { 1 }\cap X ˆ{\alpha } {\beta }$ isis nonstationary a contradiction since . \ Nquad sub gammaLet $cap\beta omega sub\ in 2 in XA sub beta . $ to the\quad powerThen of alpha the and re N sub gamma Assume that NSω2 does have the property (∗). Let B = P (ω2)/NSω2 . in S sub 1 period \noindentThis finishesi s the a proof $ \gamma of the subclaim< \ periodomega { 2 }$ so that for each $ \eta \geq \gamma , To\ finishbeta the\ proofin ofN the{\ claimeta comma} . let $ X\ inh openf i l l squareNotice bracket that H for sub kappa each closing such square $ \eta bracket, to $ the power of aleph sub 2 .. be such that omega sub 2 subset equal X comma \noindentX prec H subi f kappa $ N and{\ Seta cap open}\ squarecap bracket\omega X closing{ 2 }\ squarein bracketT to{ the1 } power$ \quad of alephthen sub 1 $ i s Sk ˆ{ H } stationary(N {\ in openeta square}\cup bracket\{\ X closingbeta square\} bracket) to the\cap power of\omega aleph sub{ 12 period} = Write N X as{\ aeta union} of\cap aleph sub\omega 2 many{ 2 } , $ which implies that $elementary N {\eta submodels}\cap N sub gamma\omega of siz{ e2 aleph} element sub 1 period−negationslash Then S sub 0 = open X brace ˆ{\alpha N sub gamma} {\ capbeta } omega. $ \ subquad 2 barHence gamma $ less T omega{ 1 sub}\ 2cap commaX N sub ˆ{\ gammaalpha in} S{\ closingbeta brace}$ is bounded . is stationary in omega sub 2 period .. Since N cap omega sub 2 in S sub 0 implies that for each alpha less omega\ hspace sub∗{\ 1 theref i l l } Since $ T { 1 }\subseteq T $ i s \quad stationary , \quad the re i s \quad some $ \isbeta some beta\not in\ Xin suchA that $ N such cap thatomega\ subquad 2 in$ X T sub{ beta1 }\ to thecap powerX of ˆ{\ alphaalpha comma} {\ it followsbeta }$ immediately that for each \noindentalpha less omegais stationary sub 1 the set . A\ subh f ialpha l l Denote upharpoonright this set S sub by 0 has $ cardinality T { 2 } at most. $ aleph\ h f sub i l l 2Let period $ Y =This Sk finishes ˆ{ H } the proof(A of the\cup claim period\{\beta \} ) $ \ h f i l l and $ N ˆ{\prime } {\gamma } = $To get a contradiction comma we are now going to show that the nonstationary ideal on omega sub 2 does not have the property open parenthesis * closing parenthesis period \noindentClaim 2 period$ Sk .. Theˆ{ H nonstationary} (N ideal{\gamma NS sub omega}\cup sub 2 ..\{\ on omegabeta sub 2\} does not) have . $ the\ properh f i l l Then hyphen$ C { 2 } = \{ N ˆ{\prime } {\gamma }\cap \omega { 2 }\mid \gamma < ty\ openomega parenthesis{ 2 }\} * closing$ parenthesis i s a club period . \ h f i l l So $ T { 2 }\cap C { 1 }\cap C Assume{ 2 }\ thatnot NS= sub\ omegavarnothing sub 2 .. does . have $ the property open parenthesis * closing parenthesis period .. Let B = P open parenthesis omega sub 2 closing parenthesis slash NS sub omega sub 2 period \ hspace ∗{\ f i l l } Let $ N {\gamma }\cap \omega { 2 }$ be in the intersection . \quad Then $ N ˆ{\prime } {\gamma }\cap \omega { 2 } = N {\gamma }\cap \omega { 2 } . $ \quad But t h i s

\noindent is a contradiction since $ N {\gamma }\cap \omega { 2 }\ in X ˆ{\alpha } {\beta }$ and $ N {\gamma }\ in S { 1 } . $

\ centerline { This finishes the proof of the subclaim . }

To finish the proof of the claim , let $ X \ in [H {\kappa } ] ˆ{\aleph { 2 }}$ \quad be such that $ \omega { 2 }\subseteq X , $ $ X \prec H {\kappa }$ and $ S \cap [ X ] ˆ{\aleph { 1 }}$ i s stationary in $ [ X ] ˆ{\aleph { 1 }} . $ Write $X$ as a union of $ \aleph { 2 }$ many elementary submodels $ N {\gamma }$ o f s i z e $ \aleph { 1 } . $ Then $ S { 0 } = \{ N {\gamma }\cap \omega { 2 }\mid \gamma < \omega { 2 } , N {\gamma }\ in S \} $ is stationary in $ \omega { 2 } . $ \quad Since $ N \cap \omega { 2 }\ in S { 0 }$ implies that for each $ \alpha < \omega { 1 }$ there

\noindent i s some $ \beta \ in X$ such that $N \cap \omega { 2 }\ in X ˆ{\alpha } {\beta } , $ it follows immediately that for each

\noindent $ \alpha < \omega { 1 }$ the s e t $ A {\alpha }\ upharpoonright S { 0 }$ has cardinality at most $ \aleph { 2 } . $

\ centerline { This finishes the proof of the claim . }

\ hspace ∗{\ f i l l }To get a contradiction , we are now going to show that the nonstationary

\noindent i d e a l on $ \omega { 2 }$ does not have the property $ ( ∗ ) . $

\ hspace ∗{\ f i l l }Claim 2 . \quad The nonstationary ideal $ NS {\omega { 2 }}$ \quad on $ \omega { 2 }$ does not have the proper −

\ begin { a l i g n ∗} ty ( ∗ ). \end{ a l i g n ∗}

\ hspace ∗{\ f i l l }Assume that $ NS {\omega { 2 }}$ \quad does have the property $ ( ∗ ) . $ \quad Let $ B = P ( \omega { 2 } ) / NS {\omega { 2 }} . $ 1 80 .. Q period F e n g and M period .. Magi d o r \noindentLet G subset1 80 equal\quad B bQ.FengandM. e a generic ultrafilter over V\quad periodMagi .. In d V o open r square bracket G closing square bracketLet we $ G can define\subseteq the genericB$ b e a generic ultrafilter over $V . $ \quad In $ V [ultrapower G ] $ Ult we sub can G open define parenthesis the generic V closing parenthesis period Fact 1 period .. The generic ultrapower contains all the reals in the forcing ex hyphen \ begintensio{ a n-line l i g n sub∗} comma i period e period comma P open parenthesis omega closing parenthesis to the power ultrapower Ult { G } (V). of V open square bracket1 80 G Q closing . F e n g square and M bracket . Magi subset d o r equalLet G Ult⊆ subB b G e a open generic parenthesis ultrafilter to over the powerV. In of line-V closing\end{ a parenthesis l i g n ∗} V period[G] we can define the generic To see this comma let a in P open parenthesis omega closing parenthesis cap V open square bracket G closing square\ hspace bracket∗{\ f period i l l } Fact .. Let 1 hatwide-a . \quad bThe e a B generic hyphen name ultrapower for the real contains period .. For all the reals in the forcing ex − n less be a to the power of omega sub maximal toultrapowerUlt the power of commaG(V ). to the power of let A sub n to the power of\noindent prime = opent e n brace s i o X $ in n B− antichainl i n e { sub, } in$ bar i X . turnstileleft e $ . A , sub prime P (n sub\ periodomega n in Let) ˆ{ toV[ the power Fact 1 . The generic ultrapower contains all the reals in the forcing ex - G]of hatwide-a}\ subsubseteq S to the powerUlt of{ orG sub} in( to ˆ the{ lpower i n e −V of} X sub) G . turnstileleft $ n b element-negationslash e tensio n − line i . e ., P (ω)V [G] ⊆ Ult (line−V ). stationary to the power of a-hatwide, closing brace period LetG A sub such to the power of n to the power of = open To see this , let a ∈ P (ω) ∩ V [G]. Let a b e a − name for the real . For brace\ hspace sub∗{\ thatfto i l l the}To power see of this X sub , beta let to the $ a power\ in of n barP( subb open\omegaB parenthesis) sub\ *cap closingV[G parenthesis to] the . power $ \ ofquad betaLet less lambda $ \widehat sub n closing{a} $ brace b holds e a $ B − $ name for the real . \quad For n , 0 a}.LetA X β for the sequence angbracketleftω let AA subn n : n less omega right angbracket periodba∨X .... Let A sub n upharpoonright b n ={ β n < bea = {X ∈ Bantichainin | X ` A0n. n ∈ Let S ∈ G ` nbelement−negationslash such | ∗) < λn}holds S\ [ = n open< brace{ be X sub} betaa ˆ{\maximal to theomega power}ˆ of{ n, bar} beta{ maximal less omega}ˆ sub{ l 2 e closingt A brace ˆ{\ beprime disj oint} { suchn }} = estationary that ( \{ X \ in B { a n t i c h a i n } { in }\mid X \vdash { A } {\prime { n } { . }} Line 1 that B = beta from angbracketleft T sub n to X to the power of nn subset equal : n to the power of S for the sequence hAn : n < ωi. Let An  S = {Xβ | β < ω2} be disj oint such subn less\ in to{ theLet power}ˆ{\ of periodwidehat Let{ omegaa}}ˆ right{\vee angbracket} { S period}ˆ{ X T sub} {\ n =in Then} { unionG }\ of wevdash to the powern{ b of} openelement brace−negationslash X n beta bar have X{ sube beta stationary to the power} ofˆ n{\ turnstileleftwidehat{ na in}\} hatwide-a. closing Let brace A period}ˆ{ Thenn } { such }ˆ{ = n\{} in aˆ Leftrightarrow{ X ˆ{ n } T{\ subbeta n in}} G period{ that Let}\ Linemid 2{ aX = open{ ( brace}ˆ{\ nbeta bar omega} { sub ∗ 2) in} j open< parenthesis\lambda B { n } \}{ holds }\ ] hTn S . [ n closing parenthesisthat subB n= closingβXn ⊆: bracen

\ centerline { Fact 3 . Forcing with $ B $ preserves $ \aleph { 3 } . $ }

\ centerline { If not , let $T \subseteq \omega { 2 }$ be stationary . \quad Let $ f $ b e a name such that }

\ [T \vdash f : \omega { 1 } onto {\rightarrow }\omega { 3 } . \ ]

\noindent For $ \alpha < \omega { 1 } , $ l e t $ A {\alpha }$ be a maximal antichain b elow $ T $ deciding all the possible values of $f ( \alpha ) . $ By $ ( ∗ ) , $ l e t $ S \subseteq T $ b e stationary such that for each $ \alpha < \omega { 1 }$ the s e t $ A {\alpha }\ upharpoonright S $ has cardinality at most $ \aleph { 2 } . $ Then $ S \vdash f $ is bounded , a contradiction .

Therefore , \quad forcing with $ B $ \quad p r e s e r v e s \quad both \quad $ \aleph { 1 }$ \quad and \quad $ \aleph { 3 } . $ \quad But \quad ther e i s \quad a condition which forces the cofinality of $ \omega { 2 }$ \quad to be changed to $ \omega , $ \quad i . e $ . , T = $

\ begin { a l i g n ∗} \{\alpha < \omega { 2 }\mid c f ( \alpha ) = \omega \} . \end{ a l i g n ∗}

\ centerline {Now we have a contradiction t o the following lemma due t o Shelah . }

\ hspace ∗{\ f i l l }Lemma ( Shelah ) . If forcing with $ P $ preserves $ \omega { 1 }$ \quad and changes the cofinal −

\noindent i t y o f $ \omega { 2 }$ to $ \omega , $ then $ \aleph { 3 }$ must be collapsed . Reflection of stationary sets .. 1 81 \ hspaceP r o o∗{\ f ..f io lf l } ..Reflection S h e l a h quoteright of stationary s .. l e mm sets a open\quad parenthesis1 81 outline closing parenthesis period .. Assume otherwise period .. In V let P rangbracketleft o o f \quad X subo falpha\quad bar alphaS h e less l omega a h ’ sub s 3\quad right angbracketl emma be ( almost outline disj oint ) . subsets\quad ofAssume omega otherwise . \quad In sub$ V 2 $ period l e t .. In V to the power of P comma let f : omega sub 1 onto right arrow omega sub 2 b e $a one\ langle hyphen toX hyphen{\alpha one function}\mid comma\ andalpha let g : omega< \ rightomega arrow{ omega3 }\ subrangle 2 be cofinal$ be period almost .. disj oint subsets of By$ \ composingomega { f to2 the} power. $ of\quad minusIn 1 and $ V ˆ{ P } , $ l e tReflection $ f ofstationary : \omega sets { 11 81} onto {\rightarrow } \omegag we find{ that2 }$ forP b each r e o alpha o f o less f omega S h e l sub a h 3 ’ theres l e is mm a beta a ( outline less omega ) . sub Assume 1 such otherwise that f to . the In power of prime prime beta iV slet cofinalhXα | α < ω3i be almost disj oint subsets of ω2. In V P, let f : ω1onto→ω2 b e \noindent a one − to − one function , and let $ g : \omega \rightarrow−1 \omega { 2 }$ in X sub alphaa period one - ....to - Since one function omega sub , and 3 is let a cardinalg : ω → largerω2 be than cofinal omega . By sub composing 1 .... in Vf to theand powerg of P be c o f i n a l . \quad By composing $ f ˆ{ − 1 }$ and 00 comma there is alphawe find sub that0 less for omega each subα < 3 ω ....3 there and is a β < ω1 such that f β i s cofinal $ g $ we find that for each $ \alpha < \omega { 3 }$ ther e i s a $ \beta there are alephin subXα 1. .. manySince alphaω3 is quoteright a cardinal s larger which than are smallerω1 thanin V P alpha, there sub is 0α0 .. such< ω that3 forand the same < \omega { 1 }$ such that $ f ˆ{\prime \prime }\beta $ i s c o f i n00 a l beta less omegathere sub are 1 commaℵ1 many f toα the’ s whichpower are of prime smaller prime than α beta0 such is cofinal that for in theX sub same alphaβ < forω1, thosef β alpha quoteright s periodis cofinal in Xα for those α ’ s . \noindent in $ X {\alpha } . $ \ h f i l l Since $ \omega { 3 }$ is a cardinal larger than In V comma for alpha less alphaIn V, subfor 0α comma < α0, pick pickγα gammasuch that alpha{X suchα − thatγα open| α < brace α0} is X a sub pairwise alpha minus $ \omega { 1 }$ \ h f i l l in $ V ˆ{ P } , $ there i s $ \alpha { 0 } < \omega { 3 }$ gamma alpha bar alphadisj oint less family alpha of sub subsets 0 closing of ω brace2 with is each a pairwise| Xα − γα |= ℵ2. But then in V P, \ hdisj f i l l ointand family of subsets of omega sub 2 with each bar X sub alpha minus gamma alpha bar = aleph sub 2 −1 period .... But then in V to the power of P comma{f [Xα − γα] ∩ β | α < α0} \noindentopen bracether f to e the are power $ of\aleph minus 1{ open1 } square$ \quad bracketmany X sub $ alpha\alpha minus$ gamma ’ s whichalpha closing are smaller square than $ \alpha { 0 is}$ an uncountable\quad such family that of disj for oint the nonempty same subsets of β < ω1. This is a bracket cap beta barcontradiction alpha less alpha . sub 0 closing brace $is\ anbeta uncountable< family\omega of disj{ oint1 } nonempty, f ˆsubsets{\prime of beta less\prime omega}\ sub 1beta period$ .... is This cofinal is a in $ X {\alpha }$ f o r those $ \alphaTherefore$ , ’ Claim s . 2 i s proved . contradiction period The contradictory Claim 1 and Claim 2 prove the theorem . Therefore comma Claim 2 i s proved period \ hspace ∗{\ f i l l } In $ V , $ f o r $ \alphaReferences< \alpha { 0 } , $ pick $ \gamma The contradictory[ 1 ] Claim W . 1 F and l e s Claim s n e r 2 , prove private the communication theorem period . \alphaReferences$ such that $ \{ X {\alpha } − \gamma \alpha \mid \alpha < \alpha { 0 }\}[ 2 ]$ M .is F o a r epairwise man , M . M ag i d o r and S . S h e l a h , Ma r − t in ’ s maximum , open square bracketsaturated 1 closing ideals ,square bracket .. W period F l e s s n e r comma private communication period open square bracket 2 closing square bracket .... M period F o r e man comma M period M ag i d o r and S \noindent disj ointand nonregular family ultrafilters of subsets . Pa ofr − t $I \,omega Ann . of Math{ 2 .} 1$ 2 7 with ( 1 988 each ) , 1 – 47 $ .\mid X {\alpha } period S h e l a h comma[ 3 ] Ma T . r-t J e in c h quoteright , Some combinatorial s maximum problems comma concerning saturated uncountable ideals comma cardinals , Ann . −and \gamma nonregular\alpha ultrafilters\ periodmid Pa= r-t I\aleph comma Ann{ 2 period} . of $ Math\ h periodf i l l But 1 2 7 then open in parenthesis $Vˆ{ 1 988P } , $ Math . closing parenthesis comma 1 endash 47 periodLogic 5 ( 1 973 ) , 1 65 – 1 9 8 . open square bracket[4] − 3 − closing − , Set square Theory bracket, Academic .... Press T period , 1 978 J . e c h comma Some combinatorial problems concerning\ [ \{ f uncountable ˆ{ − 1 cardinals} [X comma{\ Annalpha period} Math − period \gamma \alpha ] \cap \beta \mid DEPARTMENT OF MATHEMATICS DEPARTMENT OF MATHEMATICS \alphaLogic 5 open< parenthesis\alpha { 1 9730 }\}\ closing parenthesis] comma 1 65 endash 1 9 8 period NATIONAL UNIVERSITY OF S INGAPORE THE HEBREW UNIVERSITY OF JERUSALEM open square bracket 4 closing square bracket emdash comma Set Theory comma Academic Press comma 1 978 1 0 KENT RIDGE CRESCENT JERUSALEM , ISRAEL SINGAPORE 5 1 1 period Received 8 April 1 991 \noindentDEPARTMENTis an OF uncountable MATHEMATICS family .... DEPARTMENT of disj oint OF nonempty MATHEMATICS subsets of $ \beta < \omegaNATIONAL{ 1 UNIVERSITY} . $ \ h f OF i l l SThis INGAPORE i s a .... THE HEBREW UNIVERSITY OF JERUSALEM 1 0 KENT RIDGE CRESCENT .. JERUSALEM comma ISRAEL \noindentSINGAPOREcontradiction 5 1 1 . Received 8 April 1 991 \ centerline { Therefore , Claim 2 i s proved . }

\ centerline {The contradictory Claim 1 and Claim 2 prove the theorem . }

\ centerline { References }

\noindent [ 1 ] \quad W. F l e s s n e r , private communication .

\noindent [ 2 ] \ h f i l l M.Foreman ,M.Mag idorandS.She l ah ,Ma $ r−t $ in ’ s maximum , saturated ideals ,

\ centerline {and nonregular ultrafilters . Pa $ r−t$ I ,Ann. ofMath. 127(1988 ) ,1 −− 47 . }

\noindent [ 3 ] \ h f i l l T . J e c h , Some combinatorial problems concerning uncountable cardinals , Ann . Math .

\ centerline { Logic 5 ( 1 973 ) , 1 65 −− 1 9 8 . }

\noindent $ [ 4 ] −−−{ , }$ Set Theory , Academic Press , 1 978 .

\noindent DEPARTMENT OF MATHEMATICS \ h f i l l DEPARTMENT OF MATHEMATICS

\noindent NATIONAL UNIVERSITY OF S INGAPORE \ h f i l l THE HEBREW UNIVERSITY OF JERUSALEM

\noindent 1 0 KENT RIDGE CRESCENT \quad JERUSALEM , ISRAEL SINGAPORE 5 1 1

\ centerline { Received 8 April 1 991 }