Non-principal filter of $\mathcal P(\mathbb N)$ cannot contain singleton

Question

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.

Answer 1

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).

Answer 2

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.