116c- Homework 5

Homework 5

Due Wednesday, May 14 at 2:30 pm.

Update. In both parts of exercise 2, “closed” should actually be “closed in its supremum;” i.e., the subset Xof S or of {mathbb Q} is closed in Scapsup(X) or {mathbb Q}capsup(X), respectively. Or, if you rather, replace the order type alpha of X with alpha+1. Sorry for the confusion; thanks to Fedor Manin for noticing this.

In exercise 1.(a)iii, “m<a” should be “n<b.” Thanks to Michael Conley for pointing this out.

Update. Here are sketches of the solutions of exercises 2.(a) and 3:

For 2.(a), let Ssubseteqomega_1 be a given stationary set, and argue by induction on alpha<omega_1 that S contains closed copies t of alpha+1 with min(t) arbitrarily large. (Of course, if the result of the exercise holds, this must be the case: Given any gamma<omega+1, notice that Ssetminus(gamma+1) is stationary, so it must contain a closed copy t of alpha+1, and min(t)>gamma.)

This strengthened version holds trivially for alpha finite or successor, by induction. So it suffices to show it for alpha limit, assuming it holds for all smaller ordinals. Define a club Csubseteqomega_1 with increasing enumeration {gamma_beta:beta<omega_1} as follows: Let (alpha_n:n<omega) be strictly increasing and cofinal in alpha. Since S contains closed copies A_n of alpha_n+1 for all n, with their minima arbitrarily large, by choosing such copies A_n with min(A_{n+1})>sup(A_n) and taking their union, we see that S must contain copies of alpha, closed in their supremum, with arbitrarily large minimum element. (I am not claiming that A=bigcup_n A_n built this way has order type alpha. For example, if alpha=omega+omega and alpha_n=omega+n, then A would have order type omega^2; but for sure Asubset S is closed in its supremum and has order type at least alpha. So a suitable initial segment of A is as wanted.)

Let gamma_0 be the supremum of such a copy of alpha. At limit ordinals beta, let gamma_beta=sup_{delta<beta}gamma_delta. Once gamma_beta is defined, find such a copy of alpha inside S with minimum larger than gamma_beta, and let gamma_{beta+1} be its supremum.

The set C so constructed is club, so it meets S. If they meet in gamma_0 or in a gamma_{beta+1}, this immediately gives us a closed copy of alpha+1 inside S. If they meet in a gamma_beta with beta limit, let (beta_n:n<omega) be strictly increasing and cofinal in beta, and consider an appropriate initial segment of A=(bigcup_n A_n)cup{gamma_beta}, where A_n is a closed copy of alpha_n+1 in Scap[gamma_{beta_n},gamma_{beta_{n+1}}). {sf QED}

For 3, let kappa be regular and Ssubseteqkappa be stationary, and set T={alphain S:{rm cf}(alpha)=omega or ({rm cf}(alpha)>omega and Scapalpha is not stationary in alpha)}. We claim that T is stationary. To see this, let C be an arbitrary club subset of kappa. Then the set C' of limit points of C is also club (and a subset of C), so it meets S, since S is stationary. Let alpha=min(C'cap S). Then either alpha has cofinality omega, so it is in T, or else it has uncountable cofinality. In that case, notice that since alphain C', it is a limit of points in C, so Ccapalpha is club in alpha, so (Ccapalpha)'=C'capalpha is also club in alpha. Were Scapalpha stationary in alpha, it would meet C'capalpha, and this would contradict the minimality of alpha. It follows that alphain Tcap C, and therefore T is stationary, as wanted.

Let now alpha be an arbitrary point of T. If alpha has cofinality omega, it is the limit of an omega-sequence of successor ordinals. Let f_alpha be the increasing enumeration of this sequence, and notice that {rm ran}(f_alpha)cap T=emptyset, since all ordinals in T are limit ordinals. Suppose now that alpha has uncountable cofinality, so Scapalpha is not stationary in alpha. Since Tsubseteq S, it follows that Tcapalpha is not stationary either, so there is a club subset of alpha disjoint from T, and let f_{alpha} be the increasing enumeration of this club set.

With the sequences f_alpha defined as above for all alphain T, we now claim that there is some xi<kappa such that  for all eta<kappa, the set T_eta={alphain T:xiin{rm dom}(f_alpha) and f_alpha(xi)geeta} is stationary. The proof is by contradiction, assuming that no xi<kappa is as wanted.

It follows then that for all xi<kappa there is some eta=eta(xi)<kappa such that the set T_eta=T_eta^xi as above is non-stationary. Fix a club C_xi disjoint from it, and let C be the club C=bigtriangleup_xi C_xi. Let D=C'cap E, where E={alpha<kappa:eta[alpha]subseteqalpha}; here, eta:kappatokappa is the function ximapstoeta_xi. Notice that E is club, and so is D. We claim that |Dcap T|le 1. This contradicts that T is stationary, and therefore there must be a xi<kappa as claimed.

Suppose then that gamma<alpha are points in Dcap T. Since alphain D, then alphain C, so alphain C_xi (hence, alphanotin T^xi_{eta_xi}) for all xi<alpha. We claim that gammain{rm dom}(f_alpha). To see this, let xiingammacap{rm dom}(f_alpha). Then (by definition of T^xi_{eta_xi}), f_alpha(xi)<eta_xi. Since gammain D, then gammain E and,  since xi<gamma, then eta_xi<gamma. It follows that sup{rm ran}(f_alphaupharpoonrightgamma)legamma<alpha. Since f_alpha is cofinal in alpha, we must necessarily have gammain{rm dom}(f_alpha).

Since f_alpha is continuous and gamma is a limit ordinal (since it is in C'), it follows that f_alpha(gamma)=sup_{xi<gamma}f_alpha(xi)legamma. But, since f_alpha is increasing, then also f_alpha(gamma)gegamma. Hence, f_alpha(gamma)=gamma. We have finally reached a contradiction, because gammain T, but the sequence f_alpha was chosen so its range is disjoint from T. This proves that |Dcap T|le 1, which of course is a contradiction since T is stationary. It follows that indeed there is some xi<kappa such that all the sets T_eta={alphain T:xiin{rm dom}(f_alpha) and f_alpha(xi)geeta} are stationary for eta<kappa.

Now let f:Ttokappa be the map f(alpha)=left{begin{array}{cl}f_alpha(xi)&mbox{ if }xiin{rm dom}(f_alpha),\ 0&mbox{ otherwise.}end{array}right. Clearly, f is regressive. Also, from the definition of f, it follows that {alphain T:f(alpha)geeta}=T_eta for all eta<kappa, so f is unbounded in kappa, since each T_eta is in fact stationary, as we showed above. Given any eta<kappa, since fupharpoonright T_eta is regressive, there is some gamma (necessarily, gammageeta) such that {alphain T_eta:f(alpha)=gamma} is stationary, by Fodor’s lemma. A simple induction allows us to define a strictly increasing sequence (gamma_eta:eta<kappa) such that {alphain T_{gamma_eta+1}:f(alpha)=gamma_{eta+1}} is stationary for all eta. Notice that these kappa many subsets of T (hence, of S) so defined are all disjoint. By adding to one of them whatever (if anything) remains of S after removing all these sets, we obtain a partition of  S into kappa many disjoint stationary subsets, as wanted. {sf QED}

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One Response to 116c- Homework 5

  1. Fedor Manin says:

    There seems to be a problem with 2(a). Say we take S=omega_1 and alpha=omega. Then any subset of S of order type alpha is bounded, but bigcup alpha notin alpha, so it is not closed in S.

    Am I misinterpreting something?

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