Principal Ultrafilters on natural numbers 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? 
 A: You might have a look at older related questions, like:


*

*every non-principal ultrafilter contains a cofinite filter.

*Is an ultrafilter free if and only if it contains the cofinite filter?

*Ultrafilters containing a principal filter at MathOverflow


It is shown there that an ultrafilter is free if and only if it contains the filter consisting of all cofinite sets. (This filter is often called Fréchet filter or cofinite filter.)  
What you ask is a bit different. But thinking about cofinite filter might give you some inspiration for solution of your problem. I will attempt to give a proof below.

I will use $\mathcal D$, $\mathcal E$, $\mathcal F$ for systems of subsets of $\mathbb N$, just to distinguish them notationally from subsets of $\mathbb N$.
Let $\mathcal E=\{E_1,E_2,\dots\}$ be a countable set. We want to describe the filter $\mathcal F$ generated by $\mathcal E$.
To make things a bit easier, let us define 
$$D_k= \bigcap_{i=1}^k E_i$$
and
$$\mathcal D=\{D_1,D_2,\dots\}$$
The systems $\mathcal D$ and $\mathcal E$ generate the same filter, but the situation is a bit simpler now, since we have a non-increasing sequence of sets.
W.l.o.g. we can assume that $D_{i+1} \subsetneq D_i$, i.e., this sequence is strictly decreasing.
The filter generated by this set is
$$\mathcal F=\{A\subseteq \mathbb N; (\exists k\in\mathbb N) A\supseteq D_k\}.$$
If $\bigcap\limits_{k=1}^\infty D_k\ne\emptyset$, then $\bigcap\mathcal F\ne\emptyset$, and the filter $\mathcal F$ is not free.
So it remains to consider the case that $\bigcap\limits_{k=1}^\infty D_k=\emptyset$. We will show that in such case $\mathcal F$ is not an ultrafilter.
For each $k$ let us choose $a_k \in D_k \setminus D_{k+1}$. Let us put
$$A=\{a_{2k}; k\in\mathbb N\}.$$
Neither $A$ nor the complement of the set $A$ belong to $\mathcal F$. Hence  $\mathcal F$ is not an ultrafilter.

In this context, it might be interesting to mention ultrafilter number $\mathfrak u$ which is defined as the smallest cardinality of a base of a free ultrafilter. It can be shown that $\aleph_1\le\mathfrak u=\mathfrak c$. But it is relatively consistent that $\mathfrak u < \mathfrak c$.
A: 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 $X'_{2n-1} \setminus X'_{2n} \subset Y$.

