Given an infinite cardinal , Shelah defines an ideal as follows.
Definition (Shelah, implicit in here). A set is in iff and there exists a collection , and some club , so that for every , there exists a cofinal subset such that:
- (in particular, is singular);
- for every , there exists some such that .
In other words, contains all the initial segments of some small cofinal subset, , of .
Easy observations. 1) ;
2) if for all , then ;
3) if is nonstationary, then ;
4) if holds, then . (the definition of may be found in here.)
Proof hints. 1) Let for all .
2) Let for all .
3) Take a club which is disjoint from .
4) If is a -sequence, then simply let for all , and restrict your attention to .
So, is rather trivial in the presence of . On the other hand, it is a very sophisticated result of Mitchell, that may behave as the other extreme:
Theorem (Mitchell, 2009). Starting with a cardinal which is -Mahlo, in some forcing extension, has the property that is nonstationary for every .
In other words, restricted to cofinality may coincide with nonstationary ideal . Next, let us ask what about restricted to cofinality ? it turns out that we are back on track here, and this follows from the trivial fact that is a successor of a regular cardinal, together with the following less-trivial proposition:
Proposition (Shelah, 1991). for every regular cardinal .
Proof. Of course, we may assume that . For every ordinal , fix an injection . Notice that for every :
It follows that for all , the following set is a club in :
Next, for all , let Clearly, . We now fix an arbitrary limit ordinal , and show that contains all the initial segments of some small cofinal subset of .
Let be a cofinal subset of of minimal order-type. In particular, , and hence is a club in . Put , and . Then:
- ;
- , and hence ;
- , and hence ;
- for every , as , we have ;
- for every , letting , we get that
This completes the proof.
We now arrive to the main result of today’s post:
Theorem (Shelah, 1993). If is a regular cardinal, and , then contains a stationary subset of .
Proof. We already know that . We also know that in the case . Thus, we shall assume that .
In an earlier post, we proved that carries a club guessing sequence, and since here is regular and uncountable, the same argument shows that carries a club-guessing sequence. Thus, let us fix such a club-guessing sequence . Next, fix a large enough regular cardinal , and let be an increasing sequence of elementary submodels of such that for all :
We shall now make a somewhat naive move, and simply let for all . Since for all , the following set is obviously in ,
On the other hand, the next statement is not entirely obvious.
Subclaim. is stationary.
Proof. Given a club , we shall seek some . Wlog, .
Let be a sequence of elementary submodels of such that for all :
- , with ;
- ;
- ;
- whenever is a limit ordinal.
For all , as and , we get by elementarity that . Recalling that is closed, we infer that for all . Thus, it make sense to define a function by letting:
By the second and fourth defining properties of , we get that is increasing and continuous, and that for all . Put . Since , we get (by elementarity) the existence of some strictly-increasing and cofinal function that lies in .
Since and are both continuous and cofinal in , a standard back-and-fourth argument yields the existence of a club for which . Since is a club guessing sequence, we may find some such that . Put and . Then , and is a cofinal subset of of order-type . Thus, we are left with showing that for all .
Fix , and let . Then:
- ;
- , and the latter belongs to , since ;
- , and the latter belongs to since .
Denote . Then , and . Consequently: It follows that there exists some such that . In particular, .
So, is an example of a stationary subset of that belongs to .
1. Notice that in the above proof, we could have replaced with any regular cardinal for which carries a club-guessing sequence. In particular, the above proof shows:
Theorem (Shelah). If are all regular cardinals, then there exists a subset such that:
- , and reflects in the following sense:
- is stationary in .
2. The proof may be adapted to show that if are regular, then there exists a sequence , and an enumeration , such that for every club , there exists with:
- ;
- is a club subset of ;
- for every , there exists such that .
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I have a question. Assuming that the ideal is not trivial, is the forcing notion interesting? What does forcing with it add?