580 -Cardinal arithmetic (5)

At the end of last lecture we defined club sets and showed that the diagonal intersection of club subsets of a regular cardinal is club.

Definition 10. Let \alpha be a limit ordinal of uncountable cofinality. The set S\subseteq\alpha is stationary in \alpha iff S\cap C\ne\emptyset for all club sets C\subseteq\alpha. 

For example, let \lambda be a regular cardinal strictly smaller than {\rm cf}(\alpha). Then S^\alpha_\lambda:=\{\beta<\alpha : {\rm cf}(\beta)=\lambda\} is a stationary set, since it contains the \lambda-th member of the increasing enumeration of any club in \alpha. This shows that whenever {\rm cf}(\alpha)>\omega_1, there are disjoint stationary subsets of \alpha. Below, we show a stronger result. The notion of stationarity is central to most of set theoretic combinatorics. 

Fact 11. Let S be stationary in \alpha.

  1. S is unbounded in \alpha.
  2. Let C be club in \alpha. Then S\cap C is stationary. 

Proof. 1. S must meet \kappa\setminus\alpha for all \alpha and is therefore unbounded.

2. Given any club sets C and D, (S\cap C)\cap D=S\cap(C\cap D)\ne\emptyset, and it follows that S\cap C is stationary. {\sf QED}

Continuing with the intuition that a club subset of \alpha is large, we should think of stationary sets as being those that are not small. Thinking in terms of measure theory may be helpful: A club set is like a set of full measure, and a stationary set is like a set of positive measure. Clearly, we cannot have disjoint full measure sets, this corresponds to the fact that the intersection of two clubs is club. However, it is possible to have disjoint stationary sets. It is because of this fact that last lecture I made the comment that stationary sets are a manifestation of the axiom of choice: It is consistent with {\sf ZF} (for example, it is a consequence of determinacy) that every stationary subset of \omega_1 contains a club. However, under choice any stationary subset of an ordinal \alpha of cofinality \kappa can be split into \kappa many disjoint stationary subsets. This is a result of Solovay. For \omega_1 or, in general, for any successor cardinal, this was known prior to Solovay’s theorem; it is a consequence of a nice construction due to Ulam. 

Definition 12. Let \kappa be a cardinal. An Ulam matrix on \kappa^+ is defined as follows: For each \beta<\kappa^+ fix an injection f_\beta:\beta\to\kappa. For \iota<\kappa and \alpha<\kappa^+ define A^\iota_\alpha = \{ \beta:\alpha<\beta\mbox{ and }f_\beta(\alpha)=\iota \}.
For \alpha<\kappa^+ define supp(\alpha)=\{ \iota: A^\iota_\alpha \mbox{ is stationary }\}.

The following observations are immediate from the definition:

  1. If \iota\ne\eta then A^\iota_\alpha\cap A^\eta_\alpha=\emptyset for any \alpha<\kappa^+.
  2. If \alpha\ne\gamma then A^\iota_\alpha\cap A^\iota_\gamma=\emptyset for any \iota<\kappa, since the f_\beta are injective.
  3. \bigcup_\iota A^\iota_\alpha=\kappa^+\setminus(\alpha+1) for each \alpha<\kappa^+.

It follows from 3. and Theorem 7 that supp(\alpha)\ne\emptyset for each \alpha, in fact, \bigcup_{\iota\in supp(\alpha)}A^\iota_\alpha contains a club.  It follows from the pigeonhole principle that for some \iota, the set \{\alpha:\iota\in supp(\alpha)\} has size \kappa^+.

This shows that there are \kappa^+ many disjoint stationary subsets of \kappa^+. In fact, more is true: Given S stationary, by replacing each A^\iota_\alpha with S\cap A^\iota_\alpha, the same argument shows:

Theorem 13. (Ulam). For each \kappa, any stationary subset of \kappa^+ can be split into \kappa^+ many disjoint stationary sets. \Box 

A little argument with the pigeonhole principle shows that, for every \eta<\kappa, \{ \alpha<\kappa^+ : supp(\alpha)\subseteq\eta \} has size at most |\eta| and therefore for all but boundedly many \alpha, supp(\alpha) is unbounded in \kappa.

Question 14. Are there \kappa^+ many values of \alpha such that |supp(\alpha)|=\kappa?

Of course, the answer to Question 14 is yes if \kappa is regular. 

We are now ready to provide a very useful characterization of stationary sets due to Fodor.

Definition 13. Let S be a set of ordinals. A function f : S\to{\sf ORD} is regressive iff f(\alpha)<\alpha for all 0<\alpha\in S.

Theorem 14. (Fodor). Suppose that S is stationary in the regular cardinal \kappa, and f:S\to\kappa is a regressive function. Then there is some \alpha such that f^{-1}\{\alpha\} is stationary.

Proof. Otherwise, for each \alpha<\kappa there is a club C_\alpha disjoint from f^{-1}\{\alpha\}. Let C=\bigtriangleup_\alpha C_\alpha, so C is club by Lemma 9.  Let \alpha\in C\cap S. For all \beta<\alpha, \alpha\in C_\beta, so f(\alpha)\ne\beta. Therefore, f(\alpha)\ge\alpha, and f is not regressive. Contradiction. {\sf QED}

Corollary 15. Let \kappa be an uncountable regular cardinal and let S\subseteq \kappa. The following are equivalent:

  1. S is stationary.
  2. For every regressive function f on S, there is some \alpha such that f^{-1}\{\alpha\} is unbounded.
  3. For every regressive function f on S, there is some \alpha such that f^{-1}\{\alpha\} is stationary.
  4. For every function f on S, there is some stationary subset of S where f is constant, or else there is a stationary subset of S where f is strictly increasing.

Proof. Clearly 3. implies 2., and 1. implies 3. by Theorem 14. Assume that S is not stationary, and let C be a club set disjoint from S. For every \gamma\in S larger than {\rm min}(C), the set C\cap\gamma is nonempty and bounded below \gamma, so it has a largest element. The map that assigns to \gamma this largest element is regressive on S\setminus({\rm min}(C)+1). By mapping every other \gamma\in S (if any) to {}0, the map extends to a regressive function on S. However, the preimage of every point in its range is bounded. This proves that 2. implies 1.

Obviously, 4. implies 1. Suppose that S is stationary in \kappa, that f:S\to\kappa and that T=\{\alpha\in S : f(\alpha)<\alpha\} is not stationary. Let S'=S\setminus T, so S' is stationary and f(\alpha)\ge\alpha for all \alpha\in S'. By induction, define a sequence (a_\alpha:\alpha<\kappa) such that

  1. a_0\in S,
  2. a_{\alpha+1}\in S for all \alpha<\kappa, 
  3. a_{\alpha+1}>a_\alpha for all \alpha<\kappa and in fact
  4. a_{\alpha+1}>f(a_\alpha) whenever a_\alpha\in S, and
  5. a_\alpha=\sup\{a_\beta:\beta<\alpha\} whenever \alpha is a limit ordinal.

(Whenever condition 5 holds in a sequence, we say that the sequence is continuous. If condition 3 holds, this indeed corresponds to the enumeration map being continuous in the order topology.)  

That we can satisfy requirements 1–5 follows from the regularity of \kappa. By construction, C=\{a_\alpha:\alpha<\kappa\} is a club subset of \kappa, and therefore C\cap S' is stationary. By construction, f(a_\alpha)\ge a_\alpha>f(a_\beta) whenever a_\alpha,a_\beta\in S'\cap C and \beta<\alpha. This shows that 4. holds. {\sf QED} 

Remark 16. If \alpha has uncountable cofinality but is not a regular cardinal, it is not necessarily true that every regressive function on every stationary subset of \alpha is constant on a stationary subset. However, a weak version of Fodor’s lemma still holds in this case, namely, for any S stationary in \alpha and any regressive f:S\to\alpha, there is some stationary subset of S in which f is bounded. The proof is an easy extension of the argument for Theorem 14.

Note that the collection of subsets of \kappa that contain a club forms a {\rm cf}(\kappa)-complete filter (the club filter), and the collection of nonstationary sets forms a {\rm cf}(\kappa)-complete ideal (the nonstationary ideal).

Before stating Silver’s result, we need one additional result. Recall that {\sf SCH} was introduced last lecture in Definition 4, it is the statement that \gimel(\kappa)=\kappa^++2^{{\rm cf}(\kappa)} for all infinite cardinals \kappa. We break the result into two parts, Lemmas 17 and 18, corresponding respectively to Theorems II.1.9 (describing the behavior of the exponential \lambda\mapsto2^\lambda) and II.1.10 (describing the computation of powers \kappa^\lambda) from lecture II.2.

Lemma 17. Assume {\sf SCH}. Let \kappa be singular and let \tau=\sup_{\rho<\kappa}2^{\rho}. Then \displaystyle 2^\kappa=\left\{\begin{array}{cl}\tau&\mbox{if }\rho\mapsto2^\rho\mbox{ is eventually constant below }\kappa,\\ \tau^+&\mbox{otherwise.}\end{array}\right. 

Proof. Assume {\sf SCH}, let \kappa be singular and let \tau be as in the statement of the lemma. By the Bukovský-Hechler theorem, if \rho\mapsto 2^{\rho} is eventually constant below \kappa, then 2^\kappa=\tau, and otherwise, 2^\kappa=\gimel(\tau). By {\sf SCH}, we have that \gimel(\tau)=\tau^+, since we are assuming that 2^{{\rm cf}(\kappa)}<\tau. {\sf QED}

Lemma 18. Assume {\sf SCH}. Let \kappa,\lambda be infinite. Then \displaystyle \kappa^\lambda=\left\{\begin{array}{cl}2^\lambda&\mbox{if }\kappa\le2^\lambda,\\ \kappa&\mbox{if }\lambda<{\rm cf}(\kappa)\mbox{ and }2^\lambda<\kappa,\\ \kappa^+&\mbox{ if }{\rm cf}(\kappa)\le\lambda\mbox{ and }2^\lambda<\kappa.\end{array}\right.

Proof. Now we use Theorem II.1.10. Assume {\sf SCH}. We argue by induction on \kappa that the formula holds for all \lambda. If \kappa\le2^\lambda, then \kappa^\lambda=2^\lambda as before.

Let \tau=\sup_{\rho<\kappa}|\rho|^\lambda. If 2^\lambda<\kappa and \lambda<{\rm cf}(\kappa), then \kappa^\lambda=\kappa\tau=\kappa\sup_{2^\lambda\le\rho<\kappa}|\rho|^\lambda\le\kappa\sup_{2^\lambda\le\rho<\kappa}|\rho|^+=\kappa, where the induction hypothesis is used in the last inequality.

Suppose now that 2^\lambda<\kappa and {\rm cf}(\kappa)\le\lambda. If \rho\mapsto\rho^\lambda is not eventually constant below \kappa, then \kappa^\lambda=\gimel(\kappa)=\kappa^++2^{{\rm cf}(\kappa)}=\kappa^+, by {\sf SCH}. Otherwise, \rho\mapsto\rho^\lambda is eventually constant below \kappa. If \rho<\kappa is sufficiently large, then we have that 2^\lambda<\rho, \tau=\rho^\lambda, and \kappa^\lambda=\tau=\rho^\lambda\le\rho^+\le\lambda, where the last inequality is by the induction hypothesis, and we are done. {\sf QED}

Corollary 19. {\sf SCH} holds iff \kappa^\lambda\le2^\lambda+\kappa^+ for all infinite cardinals \kappa,\lambda. \Box

What the results say is that {\sf SCH} makes all powers as small as possible modulo the size of the exponential function \lambda\mapsto2^\lambda. Notice that the arguments we gave are local, meaning that the computations hold at some \kappa assuming only that {\sf SCH} holds up to \kappa.

Definition 20. (Shelah). Let \kappa be a cardinal. A cardinal \lambda is \kappa-inaccessible iff \mu^\kappa<\lambda for all cardinals \mu<\lambda.

The following is a fairly general statement of Silver’s result (still some additional generality is possible). I’ll present some corollaries and explain how they follow from the theorem. Then I will prove in detail a (very) particular case which, however, contains all the required ingredients for the proof of the general result. The argument I present is due to Baumgartner and Prikry and is combinatorial in nature. Silver’s original proof used the technique of forcing and depended on a generic ultrapower argument. We will see a similar (forceless) argument in Section 4, in the context of large cardinals. For the original approach, see Jack Silver, On the singular cardinals problem, in Proceedings of the International Congress of Mathematicians (Vancouver, B. C., 1974), Vol. 1, . Canad. Math. Congress, Montreal, Que., 1975, 265–268.

Theorem 21. (Silver).

  1.  Let \lambda be a cardinal. Assume that {\rm cf}(\lambda)=\kappa>\omega and \lambda is \kappa-inaccessible. Let (\lambda_\alpha:\alpha<\kappa) be a strictly increasing and continuous sequence of cardinals cofinal in \lambda. Suppose that there is some \mu<\kappa such that \{\alpha<\kappa:\prod_{\beta<\alpha}\lambda_\beta\le\lambda_\alpha^{+\mu}\} is stationary in \kappa. Then \lambda^\kappa\le\lambda^{+\mu}. 
  2. In the situation of item 1., if in addition 2^\tau\le\lambda^\kappa for all \tau<\lambda, then 2^\lambda\le\lambda^{+\mu}.
  3. If \aleph_\lambda is a strong limit singular cardinal of uncountable cofinality, and \{\alpha<\lambda:\gimel(\aleph_\alpha)\le\aleph_{\alpha\cdot2}\} is stationary in \lambda, then 2^{\aleph_\lambda}<\aleph_{\lambda\cdot2}. \Box

Corollary 22. (Silver).

  1. Let \lambda be a singular strong limit cardinal of uncountable cofinality \kappa. Let (\lambda_\alpha:\alpha<\kappa) be a strictly increasing and continuous sequence of cardinals cofinal in \lambda. Suppose that \mu<\kappa and \{\alpha<\kappa:\prod_{\beta<\alpha}\lambda_\beta\le\lambda_\alpha^{+\mu}\} is stationary. Then 2^\lambda\le\lambda^{+\mu}.
  2. Let \lambda be a singular cardinal of uncountable cofinality \kappa, and let \mu<\kappa. If \{\delta<\lambda:2^\delta\le\delta^{+\mu}\} is stationary, then 2^\lambda\le\lambda^{+\mu}.
  3. Suppose that \mu<\omega_1, 2^{\aleph_1}<\aleph_{\omega_1}, and \{\alpha<\omega_1:\aleph_\alpha^{\aleph_0}\le\aleph_{\alpha+\mu}\} is stationary. Then \aleph_{\omega_1}^{\aleph_1}\le\aleph_{\omega_1+\mu}.
  4. Suppose that \mu<\omega_1, \aleph_{\omega_1} is strong limit, and \{\alpha<\omega_1:\aleph_\alpha^{\aleph_0}\le\aleph_{\alpha+\mu}\} is stationary. Then 2^{\aleph_{\omega_1}}\le\aleph_{\omega_1+\mu}.
  5. Suppose that \mu<\omega_1 and \{\alpha<\omega_1 : 2^{\aleph_\alpha}\le\aleph_{\alpha+\mu}\} is stationary. Then 2^{\aleph_{\omega_1}}\le\aleph_{\omega_1+\mu}.
  6. The first counterexample to {\sf GCH} is not a singular cardinal of uncountable cofinality
  7. The first counterexample to {\sf SCH} is not a singular cardinal of uncountable cofinality.

Proof. 1. This follows immediately from the theorem, using that 2^\lambda=\lambda^\kappa since \lambda is strong limit.

2. Note that \lambda is strong limit: If \delta<\lambda then \delta^{+\mu}<\lambda, since \mu<\kappa. It follows that 2^\delta<\lambda for unboundedly many, and therefore for all, \delta<\lambda.

Let (\lambda_\alpha:\alpha<\kappa) be a strictly increasing continuous sequence of cardinals cofinal in \lambda. Then \{\alpha:2^{\lambda_\alpha}\le\lambda_\alpha^{+\mu}\} is stationary in \kappa.

But \prod_{\beta<\alpha}\lambda_\beta\le2^{\lambda_\alpha} for all \alpha. Now the result follows from 1.

3. For all \alpha<\omega_1, \aleph_\alpha^{\aleph_1}=2^{\aleph_1}\aleph_\alpha^{\aleph_0} by Tarski’s formula, Homework problem 8 from lecture II.3. It follows that \aleph_\alpha^{\aleph_1}\le\aleph_{\alpha+\mu} for stationarily many \alpha. But, for any countable limit ordinal \alpha, we have that \prod_{\beta<\alpha}\aleph_\beta=\aleph_\alpha^{\aleph_0}. The result now follows from the theorem.

4. By 3., since \aleph_{\omega_1} is strong limit.

5. and 6. By 2. 

7. Suppose that {\sf SCH} holds below \lambda, that \lambda is singular, and that \kappa={\rm cf}(\lambda) is uncountable. Suppose first that 2^\kappa\ge\lambda. Then in fact 2^\kappa>\lambda, since {\rm cf}(2^\kappa)>\kappa={\rm cf}(\lambda). We then have that \gimel(\lambda)=2^\kappa=2^\kappa+\lambda^+, and {\sf SCH} also holds at \lambda.

Suppose now that 2^\kappa<\lambda. We claim that in fact \lambda is \kappa-inaccessible. Assume that this is the case, and fix a strictly increasing and continuous sequence of cardinals (\lambda_\alpha:\alpha<\kappa) converging to \lambda. For any limit ordinal \alpha<\kappa, we have that \prod_{\beta<\alpha}\lambda_\beta\le\lambda_\alpha^{|\alpha|}. By {\sf SCH} below \lambda, \lambda_\alpha^{|\alpha|}\le\lambda_\alpha^+ as long as \alpha is sufficiently large to ensure that 2^{|\alpha|}\le2^\kappa<\lambda_\alpha. (We are using Corollary 19 here, more precisely, the fact that Corollary 19 has a local proof as explained above.)

It follows that Theorem 21.1 applies, with \mu=1, and therefore 2^\kappa + \lambda^+ = \lambda^+ \le \gimel(\lambda) \le \lambda^+, and {\sf SCH} also holds at \lambda, as wanted.

All that remains is to argue that \lambda is indeed \kappa-inaccessible. For this, suppose that \delta<\lambda. Using {\sf SCH} below \lambda, by (the local nature of the proof of) Corollary 19 we have that \delta^\kappa\le\delta^++2^\kappa<\lambda, and we are done. {\sf QED}

Now I state and prove the best known particular case of Silver’s result.

Corollary 23. (Silver). \aleph_{\omega_1} is not the first counterexample to {\sf GCH}.

Proof. Suppose that 2^\kappa=\kappa^+ for all \kappa<\aleph_{\omega_1}. In what follows, by \prod_{\alpha<\omega_1}\aleph_{\alpha+1} we mean the set product (the collection of functions) rather than its cardinality, and similarly for other products. Fix injections \pi_\alpha : {\mathcal P}(\aleph_\alpha)\to\aleph_{\alpha+1} for all \alpha<\omega_1. For X\subseteq\aleph_{\omega_1} let f_X\in\prod_\alpha\aleph_{\alpha+1} be the map f_X(\alpha)=\pi_\alpha(X\cap\aleph_\alpha). Set {\mathcal F}=\{f_X : X\subseteq\aleph_{\omega_1}\}.

The following properties of {\mathcal F} are immediate:

  1. Whenever f\ne g\in{\mathcal F}, there is in fact an \alpha such that f(\beta)\ne g(\beta) for all countable \beta\ge\alpha. 
  2. |{\mathcal F}|=2^{\aleph_{\omega_1}}, since the assignment X\mapsto f_X is 1-1.

Whenever a collection {\mathcal G} of functions f:\omega_1\to{\sf ORD} satisfies 1., we will say that {\mathcal G} is almost disjoint.

Lemma 24. Suppose that {\mathcal G}\subseteq\prod_{\alpha<\omega_1}\aleph_{\alpha+1} is almost disjoint, and for all f\in {\mathcal G}, the set \{\alpha : f(\alpha)<\aleph_\alpha\} is stationary in \omega_1. Then |{\mathcal G}|\le\aleph_{\omega_1}.

Proof. Given f\in{\mathcal G} define \hat f:\omega_1\to\omega_1 so that \aleph_0+|f(\alpha)|=\aleph_{\hat f(\alpha)} for all \alpha. Then \hat f is regressive in a stationary set. By Fodor’s lemma (Theorem 14), there is some \gamma_f\in\omega_1 such that S_f=\{\alpha : \hat f(\alpha)=\gamma_f\} is stationary.

For \gamma\in\omega_1 and S\subseteq\omega_1, let {\mathcal A}_{\gamma,S}=\{f\in{\mathcal G}:\gamma_f=\gamma\mbox{ and }S_f=S\}. Then either {\mathcal A}_{\gamma,S} is empty, or else the map that assigns to f\in {\mathcal A}_{\gamma,S} its restriction f\upharpoonright S is injective, since {\mathcal G} is almost disjoint and S is stationary in \omega_1 and therefore unbounded.

By definition of \gamma_f, if f\in{\mathcal A}_{\gamma,S} then f\upharpoonright S:\omega_1\to\aleph_{\gamma+1}. But then it follows from {\sf GCH} that |{\mathcal A}_{\gamma,S}|\le\aleph_{\gamma+2}.

The result follows, since {\mathcal G}=\bigcup_{\gamma,S}{\mathcal A}_{\gamma,S}, and there are only \aleph_1 many possible values of \gamma and \aleph_2 many possible values of S. {\sf QED}

Corollary 25. Let g\in\prod_{\alpha<\omega_1}\aleph_{\alpha+1} and let {\mathcal G} be almost disjoint. Suppose that for all f\in{\mathcal G}, \{\alpha<\omega_1 : f(\alpha)<g(\alpha)\} is stationary. Then |{\mathcal G}|\le\aleph_{\omega_1}.

Proof. This is immediate from the lemma since \prod_\alpha g(\alpha) injects into \prod_\alpha\aleph_\alpha. {\sf QED}

To complete the proof of the theorem, we now argue that {\mathcal F} can be split into \aleph_{\omega_1 +1} many sets to which the corollary applies.

Definition 26. Let {\mathcal C}_{\omega_1} be the club filter on \omega_1. Given functions f,g:\omega_1\to{\sf ORD}, we say that f<_* g iff \{\alpha:f(\alpha)<g(\alpha)\} contains a club.

Suppose that g\in{\mathcal F}. Then, by Corollary 22, at most \aleph_{\omega_1} many functions in {\mathcal F} are below g on a stationary set, i.e., g<_*f for all but at most \aleph_{\omega_1} many functions f\in{\mathcal F}.

By induction, we can therefore define a <_*-increasing sequence (f_\eta : \eta<\aleph_{\omega+1}) of members of {\mathcal F}. But notice that for any f\in{\mathcal F}, it is the case that f<_* f_\eta for some \eta. Otherwise, f contradicts the previous paragraph. It follows that {\mathcal F}=\bigcup_\eta{\mathcal F}_\eta, where {\mathcal F}_\eta=\{f\in{\mathcal F} : f<_* f_\eta\}, and we are done by Corollary 22. {\sf QED}

The arguments above (see the proof of Corollary 22.7, for example) suggest that {\sf SCH} guarantees that Tarski’s conjecture (see lecture II.3) holds. This easily follows from Lemma 18. We can in fact say a bit more:

Homework problem 9. (Jech-Shelah). Assume that Tarski’s conjecture fails, so for some limit ordinal \beta there is some strictly increasing sequence of cardinals (\kappa_\alpha:\alpha<\beta) with limit \kappa such that \prod_{\alpha<\beta}\kappa_\alpha<\kappa^{|\beta|}.

  1. Show that {\rm cf}(\beta)<|\beta|<\beta and there is some cardinal \lambda<\kappa such that \lambda^{|\beta|}>\kappa.
  2. Suppose that \beta is least such that there is a counterexample as above. Then (\aleph_0<|\beta| and) \beta=|\beta|+\omega.
  3. If Tarski’s conjecture fails, then there is a cardinal \aleph_\gamma of uncountable cofinality \tau such that \gamma>\tau, \aleph_\nu^\tau<\aleph_\gamma for all \nu<\gamma, and \gimel(\aleph_{\gamma+\omega})<\gimel(\aleph_\gamma). 

This entry was posted on Friday, February 13th, 2009 at 2:47 pm and is filed under 580: Topics in set theory. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

6 Responses to 580 -Cardinal arithmetic (5)

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