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Let $F$ be a non-principal ultrafilter of $\mathcal P(\mathbb N)$. My book says that because $F$ is non-principal, it cannot contain a singleton. I don’t see why this is the case. As far as I know, a principal filter is simply a point generated upset, so we have $\{X\in \mathcal P(\mathbb N):A\subset X\}$, for some $A$.

How does $\{x\}\in F$ make $F$ a principal filter? I would say that $F$ in that case contains a principal filter of $\{x\}$. I'm guessing my definition of a principal filter isn't right, but this is what I got from the book and wiki.

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    $\begingroup$ Your definition is right, and your observation is correct. The only thing you have to note is that $F$ cannot contain any set which doesn't contain $x$. Can you see why? $\endgroup$ – Wojowu Mar 16 '18 at 16:12
  • $\begingroup$ O, I see! Otherwise the empty set would be in $F$, and then $F$ would just be $\mathcal P(\mathbb N)$. Thanks! $\endgroup$ – Sha Vuklia Mar 16 '18 at 16:15
  • $\begingroup$ You also cannot have a finite set as well, if you have an ultrafilter, as there we have $A_1 \cup \ldots A_n \in F$ iff one of the $A_i \in F$. $\endgroup$ – Henno Brandsma Mar 16 '18 at 22:59
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If $\{x\} \in F$, then for any $A \in F$, $A \cap \{x\} \neq \emptyset$ by the properties of a filter ($ \emptyset \notin F$ and $F$ closed under intersections), so $x \in A$, so $F \subseteq \{A \subseteq \mathbb{N}: x \in A\}$. The reverse is clear as $x \in A$ implies $\{x\} \subset A$ and so $A \in F$ ($F$ closed under enlargements).

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A useful property of an ultrafilter $F$ is that, for every subset $A$ of the ambient set $X$ (in your case $X=\mathbb{N}$, but it's irrelevant), either $A\in F$ or $A'\in F$ ($A'$ is the complement of $A$).

As soon as $\{x\}\in F$, you have that $\{x\}'\notin F$, so no subset of $\{x\}'$ belongs to $F$.

Suppose $x\in A\subseteq X$. Then $A\in F$ because $A'\subseteq\{x\}'$, so $A'\notin F$. If $x\notin A$, then $A\notin F$.

Therefore $F$ consists of all subsets containing $x$, hence it is a principal ultrafilter.

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