Jech (3rd edition, page 73) defines a principal filter as {X : Xo ⊆ X} with Xo a non empty subset of S and (page 77) indicates that there is no non-principal σ-complete (i.e. ω1 complete) filter on a countable set S, but provides no proof, so I was hoping that someone would be able to provide a reference to its proof.

I have seen a principal filter defined differently - with Xo being replaced by a single element of S. In this case if S is ω and the filter is ultra, it can't contain any finite subsets of ω. The filter must therefore contain all co-finite subsets of ω and therefore all end segments of ω, whose countable intersection is the empty set. So there is no non-principal σ-complete (ie ω1 complete) ultra-filter on ω.

As Jech strengthens the definition of "principal" from a single element of S to a subset Xo of S, it looks like the ultra-filter requirement (which ensures that any finite subset of S in an ultra-filter is always associated with a single element of S being in the ultra-filter) could be dropped, which Jech statement above suggests is the case.

Using Jech principal filter definition, an ω1 complete non-principal filter on ω must have > |ω| elements, as otherwise it would be principal. But I can't see how this ensures there is always an empty intersection of a countable number of elements in the filter, without getting into Continuum Hypothesis territory.


If a set is countable we can write it as $A=\{a_n\mid n<\omega\}$. If $U$ is a non-principal filter on $A$, then for some $n$'s, $A\setminus\{a_n\}\in U$. Look at $B=\bigcap\{A\setminus\{a_n\}\mid A\setminus\{a_n\}\in U\}$

Now apply $\sigma$-closure and examine what happens when $B\in U$.

  • $\begingroup$ Asaf, the filter in the question above isn't an ultra-filter - I have filled in the result for an ultra-filter in the question. Is there some way your response can be extended to a filter, with Jech definition of principal ? Many thanks. $\endgroup$ – Little Cheese Nov 12 '15 at 13:18
  • $\begingroup$ Ah, you are right. But the idea works without assuming the ultra-ness of the filter. Just takes one more step. $\endgroup$ – Asaf Karagila Nov 12 '15 at 14:17
  • $\begingroup$ Asaf, if the filter isn't an ultra-filter the first step in your response can't be applied, so it can't be guaranteed that all co-finite sets are in the filter (unless you know of a reason why they should be). The filter can't have finite sets, but this doesn't mean that their compliment is in the filter. The filter could only contain non finite, non-cofinite sets, i.e. infinitely long subsets of ω, whose intersection is w long as well. $\endgroup$ – Little Cheese Nov 12 '15 at 14:32
  • $\begingroup$ No, it can't. But it can be applied to some of them. And when you intersect all those you get a set. And can that set be in the filter? $\endgroup$ – Asaf Karagila Nov 12 '15 at 14:33
  • $\begingroup$ Ah, and that's the key question - why should some elements in the filter (that's not an ultra-filter) intersect to lead to a co-finte set and in particular all the co-finite sets that are end segments of ω ? That's what I couldn't work out. $\endgroup$ – Little Cheese Nov 12 '15 at 14:47

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