I'm trying to show that the following statement is false:

Let $x$ be a non-empty set. Let $F$ be a proper countably complete filter on $X$. Then there is a countably complete proper ultra filter $H$ on $X$ s.t. $F \subset H$.

Countably complete means the following: $$ \forall H \subset F (H \text{ is countable } \rightarrow \bigcap H \in F)$$ $F$ is called proper if $\varnothing \notin F$.

My idea was to pick an uncountable set $x$ and let $F = \{ y \subset x \mid y^c \text{ is countable } \}$ be the cocountable filter. Then pick any $y \in F$ so that $y^c$ is countable. Let's denote it $y^c = \{ y_n \mid n \in \mathbb N_{>0} \}$. Let $(P, \le)$ be the set of all proper countably complete filters $G$ containing $F$. I wanted to make a chain extending $F$. But I can't seem to make it work. Something like $C_0 = F$ and either $C_n = flt(C_{n-1}, \{y_1, \dots, y_n \})$ or $C_n = flt(C_{n-1}, y^c \setminus \{y_1, \dots, y_n \})$ where $flt(F,s) = \{ w \subset x \mid \exists z \in F ( z \cap s \subset w \}$ doesn't work. How can I construct a chain such that $\bigcup \mathcal C$ is not closed with respect to countable intersection?

Thanks for your help.


Background information: I've done parts (a) and (b) of the following exercise and this is part (c):

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  • $\begingroup$ Does "proper" mean nonprincipal? $\endgroup$ – Asaf Karagila Nov 29 '12 at 14:41
  • $\begingroup$ @AsafKaragila No it means that $\varnothing \notin F$. $\endgroup$ – Rudy the Reindeer Nov 29 '12 at 14:43
  • $\begingroup$ what is that book you were using? $\endgroup$ – user292542 Nov 22 '15 at 20:37
  • 1
    $\begingroup$ I added "on X" , and " proper" to your conditions on H. Otherwise we could let H be the set of all subsets of X. $\endgroup$ – DanielWainfleet Nov 22 '15 at 21:31

Your choice of $F$ is a good one, and it provides a counterexample as long as the uncountable set $x$ that it lives on is not of measurable or greater cardinality. The reason is that on such an $x$, the only countably complete ultrafilters are the principal ones, and those don't extend your $F$.

For a more specific and explicit example, let's take $x$ to be the set of infinite sequences of 0's and 1's, and suppose, toward a contradiction, that $U$ were a countably complete ultrafilter extending your filter $F$ of co-countable sets. For each natural number $n$, we can partition $x$ into the set of those sequences that have 0 in their $n$-th component and those that have 1 there. As $U$ is an ultrafilter, it contains one of these, say the set $A(n,i_n)$ of sequences with $i_n$ (= $0$ or $1$) in the $n$-th component. Being countably complete, $U$ would contain the intersection of those sets $A(n,i_n)$. But that intersection is a singleton, containing only the one sequence $(i_0,i_1,i_2,\dots)$. So $U$ would be principal.

  • $\begingroup$ Thank you very much for this example! $\endgroup$ – Rudy the Reindeer Nov 30 '12 at 9:29

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