6
$\begingroup$

Let $\kappa$ be a cardinal (I'm most interested in $\kappa=\aleph_{\omega+1}$ but I suspect a general answer is known). What is the cofinality of $(P(\kappa)/NS,\subseteq)$? By this I mean the least cardinal $\lambda$ such that there exists a subcollection $X\subseteq P(\kappa)/NS$ of size $\lambda$ such that for any $A\in P(\kappa)$ there is $B\in X$ with $A\subseteq B$.

Here we consider $P(\kappa)/NS$ to be without the largest element.

This may be trivial but it's been bugging me all day!

Drake

$\endgroup$
9
  • $\begingroup$ The way you define it, we could take $X=\{\kappa\}$, and have $\lambda=1$, since every $A\subset\kappa$ has $A\subset\kappa\in X$. So I think you haven't said exactly what you mean. That is, since this poset has a largest element, every cofinal collection must include it, and any collection that includes it will be cofinal. $\endgroup$ Aug 18, 2013 at 19:39
  • $\begingroup$ I was thinking of $P(\kappa)$ as just not having the largest element. Is it still trivial? $\endgroup$
    – drizzy
    Aug 18, 2013 at 19:47
  • 3
    $\begingroup$ Turning the Boolean algebra $P(\kappa)/NS$ upside down, we can reformulate the question as asking for the smallest cardinality of a dense subset of the forcing notion $P(\kappa)/NS-\{0\}$. I think a good deal is known about this for $\kappa=\aleph_1$, for example the large-cardinal strength of this cofinality's being $\aleph_1$. But I don't recall seeing similar results for larger $\kappa$ such as $\aleph_{\omega+1}$. (Wait till Andres Caicedo, Philip Welch, and Paul Larson see this question; you'll get plenty of information.) $\endgroup$ Aug 18, 2013 at 21:53
  • 1
    $\begingroup$ @Andreas: Now when you say turning it upside down, do you mean so it would follow the Jerusalem notation or the Universal notation? (And we have weekly arguments on the topic in our students seminar in HUJI... :-)) $\endgroup$
    – Asaf Karagila
    Aug 18, 2013 at 22:13
  • 1
    $\begingroup$ @AsafKaragila Turn it upside down everywhere except Jerusalem; leave it right-side up (and delete the top element instead of $0$) in Jerusalem. $\endgroup$ Aug 19, 2013 at 2:57

1 Answer 1

10
$\begingroup$

I'm going to assume that when you write $\subseteq$, you really mean $\subseteq_{NS}$. We say $A \subseteq_{NS} B$ when $A$ is contained in $B$ except for a nonstationary set, i.e. $A \setminus B \in NS$. If we don't do this, then as Joel said, $\lambda = \kappa$, since the collection of "coatoms" is cofinal.

If we take the set of equivalence classes of stationary sets modulo $NS$, (i.e. $A \sim B$ when $A \triangle B \in NS$) we get an atomless boolean algebra under the set operations modulo $NS$, and this is what is commonly meant by $\mathcal{P}(\kappa)/NS$. Now in general an "upwardly dense" subset $A$ of a boolean algebra $B$ ("cofinal" as you call it) generates a (downwardly) dense set by just taking the complement of every element of $A$.

Now, it is known to be equiconsistent with infinitely many Woodin cardinals that $NS_{\omega_1}$ is $\omega_1$-dense. But by results of Gitik and Shelah, $\mathcal{P}(\kappa)/NS$ never has the $\kappa^+$-chain condition for regular $\kappa > \omega_1$. Hence the density of this algebra is always greater than $\kappa$. Under GCH, it must be equal to $\kappa^+$.

Woodin showed relative to an almost-huge cardinal that it is consistent for a successor cardinal $\kappa > \omega_1$ to have a stationary subset $S$ such that $\mathcal{P}(S)/NS$ has the $\kappa^+$-c.c. (See Foreman's article in Handbook of Set Theory.) I'm pretty sure that the stronger property of being $\kappa$-dense is not known to be consistent from any large cardinal assumption. Under a fragment of GCH, if $\kappa$ is the successor of singular cardinal (for example $\aleph_{\omega+1}$) $NS_\kappa$ is provably not $\kappa$-dense below any stationary set. (This is my result, to appear in my thesis soon.)

$\endgroup$
1
  • 2
    $\begingroup$ Nice result! Please post a link when it is ready. $\endgroup$ Aug 19, 2013 at 2:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.