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I'm wondering if the following exercise from Just and Weese's Discovering Modern Set Theory II is correct:

${\bf Exercise 13.20}$ Let $\kappa$ is an infinite cardinal. Suppose $\langle P, \leq\rangle$ is a p.o. with the $\kappa$-c.c. and $F$ is a $\kappa$-complete ultrafilter in $\langle P, \leq\rangle$. [Then] there exists $p\in F$ with no incompatible elements below $p$... (p. 7)

A ${\it filter}$ on $\langle P, \leq\rangle$ is a subset of $P$ which is closed upwards and contains lower bounds for any two of its elements. An ${\it ultrafilter}$ is a filter which is not a proper subset of any other filter.

I ask because two exercises leading up to it are, seemingly, incorrect; and the only way I can think of proving Exercise 13.20 is by using them. In particular, Exercise 13.18(a) is:

... if $C$ is a finite predense subset of a p.o. $\langle P, \leq\rangle$ and $F$ is an ultrafilter in $\langle P, \leq\rangle$, then $F\cap C \not = 0$. (p.7)

This seems to fall foul of the following counterexample:

We define the following ordering $<^*$ on $\omega\cup\{\langle n, 0\rangle:n\in\omega\}\cup \{\langle 0, 1\rangle\}$:

$m <^* n$ iff $n< m$

$\langle n, 0\rangle <^* m$, for $m\leq n$

$\langle n, 0\rangle <^* \langle 0, 1\rangle$

Then $\{\langle 0, 1\rangle\}$ is a finte predense subset in the new ordering, but $\omega$ is an ultrafilter.

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  • $\begingroup$ Using a $\kappa$-complete atomless filter over $P$, you can build a binary tree of height $\kappa$ where each node is a condition in $P$ with incompatible children. Then you can take a fishbone in this tree consisting of $\kappa$ many pairwise incompatible conditions contradicting the $\kappa$-cc-ness of $P.$ $\endgroup$
    – hot_queen
    Nov 2 '13 at 21:53
  • $\begingroup$ Ok, good. So $\langle 2n, 0\rangle$ has nothing below it. So if it's compatible with $2n+1$, $\langle 2n, 0\rangle<^* 2n+1$. But it isn't - it's only below $m\leq 2n$. Does that sound right? $\endgroup$
    – GME
    Nov 2 '13 at 22:00
  • $\begingroup$ Oops, you're right. I am going to delete my bogus comments, go drink some coffee, and then look at the question again, hopefully making fewer silly mistakes. $\endgroup$ Nov 2 '13 at 22:04
  • $\begingroup$ :) - thanks for looking at it! And sorry again for writing it down wrong! $\endgroup$
    – GME
    Nov 2 '13 at 22:05
  • $\begingroup$ Perhaps there is some unstated assumption on the partial ordering, such as separativity? Otherwise, can't we get an even simpler counterexample just by considering the two-element linear ordering $\{0,1\}$ with the usual ordering? $\{1\}$ is an ultrafilter that is disjoint from the finite pre-dense set $\{0\}$. $\endgroup$ Nov 2 '13 at 22:19
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Suppose for every $p \in F$, there are incompatible conditions in $P$ below $p$. Inductively construct a fishbone $\{(p_i, q_i) : i < \kappa \}$ such that:

(1) For each $i < \kappa$, $p_i \in F$ and $q_i$ is incompatible with $p_i$.

(2) For each $j < i < \kappa$, both $p_i, q_i \leq p_j$.

At stage $i$, pick a condition $p \in F$ below all $p_j$'s, $j < i$. Then take a maximal antichain $A$ (of size at least 2) in $P$ below $p$ and notice that your ultrafilter $F$ contains exactly one of these conditions (since $F$ is $\kappa$-complete, $p \in F$ and $|A| < \kappa$) say $q$. Put $p_i = q$ and let $q_i$ be any other member of this antichain.

But now $\{q_i : i < \kappa\}$ is an antichain in $P$ which is impossible.

Addendum 1: List $A = \{a_i : i < \lambda\}$, where $\lambda < \kappa$. Suppose $F$ is disjoint with $A$. Then by maximality of $F$, for each $i < \lambda$, there is some $b_i \in F$ such that $a_i$ is incompatible with $b_i$. Using $\kappa$-completeness of $F$, pick some $b \in F$ below all $b_i$'s and also below $p$. But then, $A \cup \{b\}$ is a larger antichain in $P$ below $p$.

Addendum 2: Suppose $F$ is a $\kappa$-complete ultrafilter over a $\kappa$-cc separative poset $P$ and $p \in P$ is compatible with every member of $F$. Then $p \in F$.

Proof: Suppose first that there is some $q \leq p$ such that every extension of $q$ is compatible with every member of $F$. Then every member of $F$ is above $q$ as $P$ is separative. Hence the principal filter above $q$ is proper extension of $F$ which is impossible. To obtain such a $q$, consider a maximal antichain $A$ of conditions below $p$ which are incompatible with some member of $F$. Using $\kappa$-cc-ness of $P$ and $\kappa$-completeness of $F$, we can assume that there is some $r \in F$ which is incompatible with every member of $A$. Let $q$ be a condition extending both $r$ and $p$.

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  • $\begingroup$ Thanks. Could you just say why exactly F will have to contain an element of A? Sorry for being slow! $\endgroup$
    – GME
    Nov 3 '13 at 4:29
  • $\begingroup$ No problem. I added a little more explanation. $\endgroup$
    – hot_queen
    Nov 3 '13 at 5:02
  • $\begingroup$ So, I had a proof which would have worked if whenever $p$ is compatible with everything in $F$, then $p\in F$. It almost looks like you're using that in the second line - ``Then by maximality...". Why is that true here? (I think it's that move I'm getting stuck on). $\endgroup$
    – GME
    Nov 3 '13 at 5:27
  • $\begingroup$ If $p$ is compatible with everything in $F$, then you can add $p$ to $F$ and take the generated filter which is at least as big as $F$. $\endgroup$
    – hot_queen
    Nov 3 '13 at 5:29
  • $\begingroup$ Generated as in: add lower bounds and then close upwards? $\endgroup$
    – GME
    Nov 3 '13 at 5:30

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