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I'm trying to show if $X$ is infinite then every uniform filter $F$ on $X$ is contained in a uniform ultra filter $G$ on $X$. A filter is uniform if all the sets in it are of same size.

My thoughts:

  • $F$ is uniform therefore proper
  • Every proper filter is contained in an ultra filter by Zorn's lemma
  • If $Y$ is the set of all $y$ such that neither $y$ nor $X \setminus y$ are in $F$ define $$\text{flt}(F \cup Y) = \{s \subset X: ~\exists k,~~ \exists A_1, \dots ,A_k \in F \cup Y: A_1 \cap \dots A_k \subset s \}$$ This is a filter containing $F$. But I fail to show that it is proper and uniform

Can you help me to prove this? It is not a homework.

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up vote 2 down vote accepted

Let $A$ be any element of $F$, and let $\mathscr{S}=\{S\subseteq A:|A\setminus S|<|A|\}$. For any $B\in F$ and $S\in\mathscr{S}$, $B\cap S\ne\varnothing$, so $F\cup\mathscr{S}$ is a filter subbase. Now extend $F\cup\mathscr{S}$ to an ultrafilter $U$ on $X$. if some $|V|<|A|$ for some $V$ in $U$, then $A\setminus V\in\mathscr{S}\subseteq U$, which is absurd.

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What if $V$ is singleton set? Then $|V| = 1 < |A|$ but $|A \setminus V| = |A|$ does not imply $A\setminus V \in \mathcal S$. Maybe then $V^c$ in $\mathcal S$? – dolan Apr 2 '13 at 15:09
@dolan: Yes, $|V|=1$ does imply that $A\setminus V\in\mathscr{S}$, because $|A\setminus(A\setminus V)|<|A|$. – Brian M. Scott Apr 2 '13 at 20:07
Thank you. And sorry for writing stupid comment, I remember now. (If you have time can you read my proof here, from Henno Brandsma's profile it looks like he is away, thank you) – dolan Apr 3 '13 at 10:10
@dolan: You’re welcome. I took a quick look at the other proof, and I do see a problem: to get your contradiction you assume that $x$ is not in some closed $C\in\mathcal{F}$, which is fine, but then you assume that $C = \prod_i C_i \in \mathcal F$, which is not fine: a closed set in $C$ need not be a product of closed sets. – Brian M. Scott Apr 3 '13 at 18:25
Very many thank you for reading proof! – dolan Apr 4 '13 at 6:54

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