Club Guessing At Inaccessibles

Disclaimer: So after having typed this up, I’m a little shaky on a few details here still (mainly with keeping track of indices). But, I think I now have a better idea of how this proof should look if I attempt to rewrite it at some point in the future.

In this post, I want to work through the construction of a club guessing sequence which appears as Claim 0.14 of Sh413. Recall:

Definition: Let \lambda be a regular uncountable cardinal, and let S\subseteq \lambda be stationary. We say that \bar C=\langle C_\delta : \delta \in S\rangle is an S-club system if each C_\delta is a club subset of \delta.

For the construction, we need a few definitions which have become common-place in the literature:

Definition: If C is a club set of ordinals, and \alpha\in nacc(C), set

Gap(\alpha, C)=(\sup(C\cap\alpha), \alpha).

Definition: If C and E are sets of ordinals with E\cap \sup(C) closed in \sup(C), we define:

Drop(C,E)=\{\sup(\alpha\cap E) : \alpha\in C\setminus\min(E)+1\}

Roughly what we’re doing here is dropping ordinals from C above min(E) down to their sup in E in order to create a closed subset of \sup(C).

Definition: Let C, E\subseteq\lambda for a regular uncountable, and let \langle e_\alpha : \alpha<\lambda\rangle be a \lambda-club system. For each \alpha\in nacc(C)\cap acc(E), define

Fill(\alpha, C, E)=Drop(e_\alpha, E)\cap Gap(\alpha, C)

We suppress mention of the \lambda-club system in our notation for Fill because it doesn’t really matter. What’s important is that Fill(\alpha, C, E) gives us a way of looking at the gap between sup (C\cap\alpha) and \sup (E\cap\alpha)=\alpha, and filling it in with a club subset of Gap(\alpha, C). So we will just think of \langle e_\alpha : \alpha\in \lambda\rangle as a fixed collection of clubs which we will draw upon to fill in gaps in an appropriate way.

Now, these describe operations which Shelah refers to as $gl^1$ and $gl^0$ in Cardinal Arithmetic. We’re going be using something called $gl^2$ over the course of our construction, which is the result of filling in multiple gaps and gluing the result together.

Definition: Let C, E\subseteq\lambda for a regular uncountable, and let \langle e_\alpha : \alpha<\lambda\rangle be a \lambda-club system. Further, let A\subseteq\lambda be a stationary set of limit ordinals. We define gl_n(C,E,A) by induction on n<\omega.

  • gl^2_0(C,E,A)=Drop(C,E)
  • gl^2_{n+1}(C,E,A)=gl_n(C,E,A)\cup\bigcup_{\beta\in nacc(gl^2_n(C,E,A))\setminus A}\big(Fill(\beta,E,gl^2_n(C,E,A))\cup\{\sup(\alpha\cap E) : \sup(\alpha\cap E)\geq\sup (\alpha\cap e_\beta)\}\big)
  • gl^2(C,E,A)=\bigcup_{n<\omega}gl^2_n(C,E,A).

This looks horrendous, but the main idea is that at each stage we look at the gaps in the previous stage outside of A, and fill those gaps in. Then at the end we glue it all together (hence gl) and provided that C is club in some \delta\in acc(E), the end result is a club of the same \delta. With this definition in hand, we’re ready to state the theorem and carry out the construction.

Claim 0.14 of Sh413: Assume

  1. \lambda is inaccessible.
  2. A\subseteq \lambda is a stationary set of limit ordinals such that, if \delta<\lambda and A\cap\delta is stationary in \delta, then \delta\in A.
  3. J is a \sigma-indecomposable ideal on \lambda extending the non-stationary ideal.
  4. S\notin J and S\cap A=\emptyset.
  5. \omega<\sigma=cf(\sigma)<\lambda and \delta\in S\implies cf(\delta)\neq\sigma.

Then for some S-club system \bar C=\langle C_\delta : \delta\in S\rangle we have that, for any E\subseteq\lambda:

\{\delta\in S : \delta=\sup(E\cap nacc(C_\delta)\cap A)\}\notin J.

Proof: We begin by fixing a \lambda-club system \bar e=\langle e_\alpha : \alpha<\lambda\rangle with the property that ot(e_\alpha)=cf(\alpha) and such that e_\delta\cap A=\emptyset for any limit \delta\in \lambda \setminus A (recall that A does not reflect outside of itself).

For any club E\subseteq\lambda, define C_{\delta,E}:=gl^2(e_\delta,E,A) if \delta\in acc(E), and C_{\delta,E}=e_\delta otherwise. If there is some club E of \lambda such that \bar C_E=\langle C_{\delta, E}: \delta\in S\rangle is as required, then we are done. So suppose otherwise. Then for very club E\subseteq\lambda, there is a club D(E)\subseteq\lambda such that

Y_E:=\{\delta\in S : \sup(D(E)\cap A\cap nacc (C_{\delta, E}))=\delta\in J.

Note that shrinking D(E) only makes Y_E smaller, so we may replace D(E) with D(E)\cap E in order to assume that D(E)\subseteq E. We now inductively pick sets E_\epsilon by induction on \epsilon<\sigma such that:

  1. E_\epsilon\subseteq\lambda is club;
  2. \xi<\epsilon\implies E_\epsilon\subseteq\ E_\xi;
  3. If \epsilon=\xi+1, then E_\epsilon\subseteq D(E_\epsilon).

This is simple. We first let E_0=\lambda, and at successor stages we take E_\epsilon=D_\xi. At limit stages we let E_\epsilon=\bigcap_{\xi<\epsilon}E_\xi.

Next, let E=\bigcap_{\epsilon<\sigma}E_\epsilon which is itself club in \lambda, and so E\cap A is stationary in \lambda. Thus, the set E'=\{\delta\in E : \delta=\sup(E\cap A\cap \delta)\} is club in \lambda. The following claim will finish the proof.

Claim: For every \delta\in S\cap E', there is some \epsilon_\delta<\sigma such that for every \epsilon_\delta\leq \epsilon<\sigma, we have \delta\in Y_{E_\epsilon}.

To see that this claim suffices, we begin by letting Y_{\epsilon}=\bigcap_{\epsilon\leq\xi<\sigma}Y_{E_\xi} for each \epsilon<\sigma. Of course, Y_\epsilon\in J as Y_\epsilon\subseteq Y_{E_\epsilon}\in J by assumption, and \epsilon_1<\epsilon_2\implies Y_{\epsilon_1}\subseteq Y_{\epsilon_2}. As J is \sigma-indecomposable, it’s closed under increasing unions of length \sigma and in particular \bigcup_{\epsilon<\sigma} Y_{\epsilon}\in J. On the other hand, the claim tells us that \delta\in Y_{\epsilon_\delta} for every \delta\in S\cap E', and so S\cap E'\in J. But by assumption we have that S\notin J, and as J extends the non-stationary ideal we have a contradiction.

Proof of Claim: We begin by fixing, for each \delta\in S\cap E', an increasing cofinal sequence \langle \beta^i_\delta : i<cf(\delta)\rangle of elements of A\cap E\cap S. Fix \delta\in S\cap E', and note that by assumption \delta\notin A, and so e_\delta\cap A=\emptyset. In particular, \{\beta^i_\delta : i<cf(\delta)\}\cap e_\delta=\emptyset. Now if we look at the construction of gl^2(e_\delta, E_\epsilon, A), we see that since \beta^i_\delta\in A\setminus e_\delta, we only add boundedly-many elements of \beta^i_\delta to gl^2_n(e_\delta, E_\epsilon, A) for each \epsilon<\sigma and i< cf(\delta).

For each i<cf(\delta), \epsilon<\sigma, and n<\omega, let \beta^i_\delta(n,\epsilon)=\min(gl^2_n(e_\delta, E_\epsilon, A)\setminus\beta^i_\delta) and note that since gl^2_n\subseteq gl^2_{n+1}, it follows that the sequence \langle \beta^i_\delta(n,\epsilon) : n<\omega\rangle is decreasing and hence eventually constant (say it stabilizes at n(i,\delta,\epsilon). Now, since \beta^i_\delta\cap gl^2_n(c_\delta, E_\epsilon, A) is bounded in \beta^i_\delta, it follows that \beta^i_\delta(n,\epsilon)\in nacc(C_{\delta, E_\epsilon}) for each n>n(\i,\delta,\epsilon). In fact, we also have that \beta^i_\delta(n,\epsilon) is in A for each n\geq n(i, \delta, \epsilon). Otherwise, we would have that \beta^i_\delta(n,\epsilon)\in nacc (gl^2_n(e_\delta, E_\epsilon, A)\setminus A, and so we would have ended up adding a club subset of \beta^i_\delta(n,\epsilon) in the next stage. So far we have shown: For each i<cf(\delta), and \epsilon<\sigma, there is some n(i,\delta,\epsilon)<\omega such that, for every \beta^i_\delta(n,\epsilon) we have

  1. \beta^i_\delta(\epsilon, n(i,\delta,\epsilon))=\beta^i_\delta(\epsilon,n) and
  2. \beta^i_\delta(\epsilon, n)\in nacc(C_{\delta, E_\epsilon})\cap A.

Next, we look at this sequence from the other direction. That is, we consider \langle \beta^i_\delta(\epsilon, n) : \epsilon<\delta\rangle for a fixed n. The best way to visualize this situation is that for each \delta\in S\cap E' and i<cf(\delta), we have a \sigma\times\omega matrix M_{i,\delta} where the \epsilon, n entry is \beta^i_\delta(\epsilon, n). So far, we’ve shown that the values are decreasing and eventually constant along the rows, and at this point we’d like to show the same for the columns. We begin by noting that the sequence of the sets E_\epsilonis decreasing, and so we have that \sup(\alpha\cap E_{\epsilon_1})<\sup (\alpha\cap E_{\epsilon_2}) whenever \epsilon_2>\epsilon_1. Therefore, the sequence \langle \beta^i_\delta(\epsilon, n) : \epsilon<\delta\rangle is decreasing and eventually stabilizes at some \epsilon(i,\delta, n)<\sigma. In particular, this tells us that

(\forall \xi>\epsilon(i,\delta, n))(\min (C_{\delta, E_\xi}\setminus \beta^i_\delta)=\min (C_{\delta, E_{\epsilon(i,\delta, n)}}\setminus \beta^i_\delta)=\beta^i_\delta(\epsilon(i,\delta, n), n))

Now, since \delta\in S, we know that cf(\delta)\neq\sigma and so for each \delta, we can find some \epsilon_\delta such that \sup\{i: \epsilon(i,\delta, n)\leq\epsilon_\delta \}=cf(\delta). This is pretty simple, as we just let \epsilon_\delta be the supremum of \epsilon(i,\delta, n) when running over i< cf(\delta). Then this supremum will be below \sigma as the cofinality of \delta is wrong.

At this point, we would like to show that for every \delta\in S\cap E', and \epsilon\geq\epsilon_\delta, we get that \delta\in Y_{E_\epsilon}. In particular, we need to show that \delta=\sup(D(E_\epsilon)\cap A\cap nacc(C_{\delta,E_\epsilon}). Recall that our construction of C_{\delta,E_\epsilon} actually gives us the elements of C_{\delta,E_\epsilon} are of the form \sup(\alpha\cap E_{\epsilon}. As D(E_\epsilon)\subseteq E_\epsilon, it follows that we only need to show that \delta=\sup(A\cap nacc(C_{\delta, E_\epsilon})). Recall that stabilization along columns tells us that if \epsilon_\delta\leq\epsilon<\sigma:

\min (C_{\delta, E_\epsilon}\setminus \beta^i_\delta)=\min (C_{\delta, E_{\epsilon(i,\delta, n)}}\setminus \beta^i_\delta)=\beta^i_\delta(\epsilon(i,\delta, n), n).

Finally, since \langle \beta^i_\delta : i<cf(\delta)\rangle is increasing and cofinal in \delta, and \beta^i_\delta<\beta^i_\delta(\epsilon,n)<\delta, it follows that:

\delta=\sup\{min(C_{\delta, E_\epsilon}\setminus \beta^i_\delta: \epsilon(i,\delta, n)\leq\epsilon_\delta\}.

Since we already showed that the supremum of the indices i involved in the above supremum has supremum equal to cf(\delta). Finally stabilization along the rows told us that eventually \beta^i_\delta(\epsilon, n) ends up in A\cap nacc C_{\delta, E_\epsilon} for all sufficiently large n<\omega. So, we can just pick the correct \beta^i_\delta(\epsilon, n) to achieve a supremum of \delta. That completes the proof of the claim, and consequently the proof of the whole thing.

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