TITLE: Principal Ultrafilters on natural numbers
QUESTION [7 upvotes]: Let $E$ be a countable set of subsets of $\mathbb{N} $. Show that the filter generated by $E$ cannot be a non-principal ultrafilter. 
My idea of solution is:
Let $D$ be the filter generated by $E$. $D$ is not principal $\iff$  for all $x\in\mathbb{N}$   $\{x\}\notin D $  $\iff \mathbb{N} \setminus \{x\}\in D$ if and only if for all $m\in\mathbb{N} $ and for all $X_1,\dots,X_m\in E$  $X_1\bigcap\dots\bigcap X_m\subset \mathbb{N} \setminus \{x\}$. I'd like to conclude that $E$ is not countable. Is that possible?

REPLY [5 votes]: While writing down this answer, I found that @Martin Sleziak gave an answer first with the same idea as in here (with even some details identical). I leave my answer for my personal record, but if people think I should erase it, I will.

Let $\mathcal{D}$ be the filter generated by $E$.
Case 1. If $\cap E \neq \varnothing$, then $\mathcal{D}$ is a principal filter.
Case 2. Assume that $\cap E = \varnothing$. Enumerate $E$ as $E = \{ X_1, X_2, \cdots \}$ and define $(X'_n)$ as the sequence obtained from the list
$$ X_1, \quad X_1 \cap X_2, \quad X_1 \cap X_2 \cap X_3, \quad \cdots $$
by deleting possible repetitions on it. It is straightforward to check that $(X'_n)$ is a decreasing sequence of infinite sets. Finally, define $Y$ as follows:
$$ Y = (X'_1 \setminus X'_2) \cup (X'_3 \setminus X'_4) \cup \cdots. $$
We can check that $Y \notin \mathcal{D}$ and $\Bbb{N}\setminus Y \notin \mathcal{D}$. Indeed,

Assume that $Y \in \mathcal{D}$. Then there exists $n$ such that $X'_{2n} \subseteq Y$. But this is impossible since $X'_{2n}\setminus X'_{2n+1} \subset \Bbb{N}\setminus Y$.
Similarly, assume that $\Bbb{N}\setminus Y \in \mathcal{D}$. Then $X'_{2n-1} \subseteq \Bbb{N}\setminus Y$ for some $n$ but this is again impossible since $X'_{2n-1} \setminus X'_{2n} \subset Y$.