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 of or of is closed in or , respectively. Or, if you rather, replace the order type of with . Sorry for the confusion; thanks to Fedor Manin for noticing this.

In exercise 1.(a)iii, “” should be “.” 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 be a given stationary set, and argue by induction on that contains closed copies of with arbitrarily large. (Of course, if the result of the exercise holds, this must be the case: Given any , notice that is stationary, so it must contain a closed copy of , and .)

This strengthened version holds trivially for finite or successor, by induction. So it suffices to show it for limit, assuming it holds for all smaller ordinals. Define a club with increasing enumeration as follows: Let be strictly increasing and cofinal in . Since contains closed copies of for all , with their minima arbitrarily large, by choosing such copies with and taking their union, we see that must contain copies of , *closed in their supremum*, with arbitrarily large minimum element. (I am not claiming that built this way has order type . For example, if and , then would have order type ; but for sure is closed in its supremum and has order type *at least* . So a suitable initial segment of is as wanted.)

Let be the supremum of such a copy of . At limit ordinals , let . Once is defined, find such a copy of inside with minimum larger than , and let be its supremum.

The set so constructed is club, so it meets . If they meet in or in a , this immediately gives us a closed copy of inside . If they meet in a with limit, let be strictly increasing and cofinal in , and consider an appropriate initial segment of , where is a closed copy of in .

For 3, let be regular and be stationary, and set or and is not stationary in . We claim that is stationary. To see this, let be an arbitrary club subset of . Then the set of limit points of is also club (and a subset of ), so it meets , since is stationary. Let . Then either has cofinality , so it is in , or else it has uncountable cofinality. In that case, notice that since , it is a limit of points in , so is club in , so is also club in . Were stationary in , it would meet , and this would contradict the minimality of . It follows that , and therefore is stationary, as wanted.

Let now be an arbitrary point of . If has cofinality , it is the limit of an -sequence of successor ordinals. Let be the increasing enumeration of this sequence, and notice that , since all ordinals in are limit ordinals. Suppose now that has uncountable cofinality, so is not stationary in . Since , it follows that is not stationary either, so there is a club subset of disjoint from , and let be the increasing enumeration of this club set.

With the sequences defined as above for all , we now claim that there is some such that for all , the set and is stationary. The proof is by contradiction, assuming that no is as wanted.

It follows then that for all there is some such that the set as above is non-stationary. Fix a club disjoint from it, and let be the club . Let , where ; here, is the function . Notice that is club, and so is . We claim that . This contradicts that is stationary, and therefore there must be a as claimed.

Suppose then that are points in . Since , then , so (hence, ) for all . We claim that . To see this, let . Then (by definition of ), . Since , then and, since , then . It follows that . Since is cofinal in , we must necessarily have .

Since is continuous and is a limit ordinal (since it is in ), it follows that . But, since is increasing, then also . Hence, . We have finally reached a contradiction, because , but the sequence was chosen so its range is disjoint from . This proves that , which of course is a contradiction since is stationary. It follows that indeed there is some such that all the sets and are stationary for .

Now let be the map Clearly, is regressive. Also, from the definition of , it follows that for all , so is unbounded in , since each is in fact stationary, as we showed above. Given any , since is regressive, there is some (necessarily, ) such that is stationary, by Fodor’s lemma. A simple induction allows us to define a strictly increasing sequence such that is stationary for all . Notice that these many subsets of (hence, of ) so defined are all disjoint. By adding to one of them whatever (if anything) remains of after removing all these sets, we obtain a partition of into many disjoint stationary subsets, as wanted.