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 be a regular uncountable cardinal, and let be stationary. We say that is an -club system if each is a club subset of .

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

**Definition**: If is a club set of ordinals, and , set

.

**Definition**: If and are sets of ordinals with closed in , we define:

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

**Definition**: Let for a regular uncountable, and let be a -club system. For each , define

We suppress mention of the -club system in our notation for because it doesn’t really matter. What’s important is that gives us a way of looking at the gap between and , and filling it in with a club subset of . So we will just think of 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 for a regular uncountable, and let be a -club system. Further, let be a stationary set of limit ordinals. We define by induction on .

- .

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

**Claim 0.14 of Sh413**: Assume

- is inaccessible.
- is a stationary set of limit ordinals such that, if and is stationary in , then .
- is a -indecomposable ideal on extending the non-stationary ideal.
- and .
- and .

Then for some -club system we have that, for any :

.

**Proof**: We begin by fixing a -club system with the property that and such that for any limit (recall that does not reflect outside of itself).

For any club , define if , and otherwise. If there is some club of such that is as required, then we are done. So suppose otherwise. Then for very club , there is a club such that

.

Note that shrinking only makes smaller, so we may replace with in order to assume that . We now inductively pick sets by induction on such that:

- is club;
- ;
- If , then .

This is simple. We first let , and at successor stages we take . At limit stages we let .

Next, let which is itself club in , and so is stationary in . Thus, the set is club in . The following claim will finish the proof.

Claim: For every , there is some such that for every , we have .

To see that this claim suffices, we begin by letting for each . Of course, as by assumption, and . As is -indecomposable, it’s closed under increasing unions of length and in particular . On the other hand, the claim tells us that for every , and so . But by assumption we have that , and as extends the non-stationary ideal we have a contradiction.

**Proof of Claim**: We begin by fixing, for each , an increasing cofinal sequence of elements of . Fix , and note that by assumption , and so . In particular, . Now if we look at the construction of , we see that since , we only add boundedly-many elements of to for each and .

For each , , and , let and note that since , it follows that the sequence is decreasing and hence eventually constant (say it stabilizes at ). Now, since is bounded in , it follows that for each . In fact, we also have that is in for each . Otherwise, we would have that , and so we would have ended up adding a club subset of in the next stage. So far we have shown: For each , and , there is some such that, for every we have

- and
- .

Next, we look at this sequence from the other direction. That is, we consider for a fixed . The best way to visualize this situation is that for each and , we have a matrix where the entry is . 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 is decreasing, and so we have that whenever . Therefore, the sequence is decreasing and eventually stabilizes at some . In particular, this tells us that

Now, since , we know that and so for each , we can find some such that . This is pretty simple, as we just let be the supremum of when running over . Then this supremum will be below as the cofinality of is wrong.

At this point, we would like to show that for every , and , we get that . In particular, we need to show that . Recall that our construction of actually gives us the elements of are of the form . As , it follows that we only need to show that . Recall that stabilization along columns tells us that if :

.

Finally, since is increasing and cofinal in , and , it follows that:

.

Since we already showed that the supremum of the indices involved in the above supremum has supremum equal to . Finally stabilization along the rows told us that eventually ends up in for all sufficiently large . So, we can just pick the correct to achieve a supremum of . That completes the proof of the claim, and consequently the proof of the whole thing.