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Let $X$ be a finite set. Prove that every ultrafilter is a point filter.

My attempt:

Let $\mathcal{U}$ be an ultra filter. Write $X = \bigcup_{i=1}^n \{x_i\} \in \mathcal{U}$. Then there exists $i \in \{1,2, \dots, n\}$ such that $\{x_i\} \in \mathcal{U}$, because it is an ultrafilter. I now claim that $\mathcal{U}$ is the point filter of the point $x_i$, i.e. we have to prove that $\mathcal{U} \subseteq \{F \subseteq X \mid x_i \in F\}$ and equality will follow because it is an ultrafilter.

How to proceed?

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  • $\begingroup$ Well, the claim that $\{x_i\} \in \mathcal{U}$ is already non-trivial, which should follow from the fact that $X$ is finite. For instance, what prevents you from having $X\setminus\{x_i\} \in \mathcal{U}$ for all $i$? $\endgroup$ – Sangchul Lee Mar 10 '18 at 20:11
  • $\begingroup$ I proved that if a union is in an ultrafilter, then one of the sets in the union is in the ultrafilter. $\endgroup$ – user370967 Mar 10 '18 at 20:13
  • $\begingroup$ Oh I see, that is the point where you utilized the finiteness of $X$. $\endgroup$ – Sangchul Lee Mar 10 '18 at 20:14
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Let $\mathcal{F}$ be a filter on a set $X$ such that there is some $x \in X$ with $\{x\} \in \mathcal{F}$. Then

$$\mathcal{F} = \mathcal{F}_x := \{A \subseteq X: x \in A\}$$.

The proof is immediate: if $F \in \mathcal{F}$ then as $\{x\} \in \mathcal{F}$, $\{x\} \cap F \neq \emptyset$ as any two sets in a filter intersect in a member of the filter and all filter members are non-empty (two axioms of filters!). But that just says that $x \in F$ and thus $F \in \mathcal{F}_x$.

On the other hand, if $F \in \mathcal{F}_x$, then $x \in F$ or $\{x\} \subseteq F$ and as $\{x\} \in \mathcal{F}$, $F \in \mathcal{F}$, as filters are closed under supersets (another filter axiom!).

So we have equality of these filters.

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  • $\begingroup$ Thanks. This was very clear. In fact, only one of the two inclusions was necessary, because both things are ultrafilters, right? $\endgroup$ – user370967 Mar 11 '18 at 17:52
  • $\begingroup$ @Math_QED I wanted to show it holds without any more assumptions beyond being a filter. $\endgroup$ – Henno Brandsma Mar 11 '18 at 18:03
  • $\begingroup$ I see . Thanks! $\endgroup$ – user370967 Mar 11 '18 at 18:11
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You've already shown that $\mathcal{U}$ contains a singleton $\{x_i\}$. Now observe that the filter generated by $\{x_i\}$ is an ultrafilter and must be contained in $\mathcal{U}$. But ultrafilters are maximal filters with respect to $\subseteq$, so in fact $\mathcal{U}$ is the filter generated by $\{x_i\}$.

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  • $\begingroup$ Why must the filter generated by $\{x_i\}$ be contained in $\mathcal{U}$? $\endgroup$ – user370967 Mar 10 '18 at 20:59
  • $\begingroup$ The filter $\mathcal{F}$ generated by $\{x_i\}$ is the smallest filter containing $\{x_i\}$. $\mathcal{U}$ is a filter containing $\{x_i\}$, so it must contain $\mathcal{F}$. $\endgroup$ – Hayden Mar 10 '18 at 21:00

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