A countable family in $\mathcal P (\mathbb N)$ such that every set in $\mathbb N$ contains some element of the family. Does there exists a countable family of infinite sets $\{A_n:n\in\mathbb N\}\subset\mathcal P(\mathbb N)$ satisfying the following property:
$$\text{For every infinite set }I\in\mathcal P(\mathbb N),\text{ there is }n\in\mathbb N\text{ such that }A_n\subset I\text{ ?}$$
If we require the family to have cardinality $\mathfrak c$, then the question is trivial, but for the countable case I'm stuck.
 A: No. Let $\{A_n\}$ be any sequence of infinite sets. For each $n$ choose an element $x_n$ of $A_n$ that's  different from all the elements $x_k$ chosen so far
and greater than $2n$.   Then the complement of the set of all $x_k$ has none of the $A_n$ as a subset. It is infinite because  it is the complement of an increasing sequence that omits at least half the numbers less than $2n$ for every $n$.
A: Here is a slightly different way to construct a counterexample, one that I find a little simpler. Let $\mathscr{A}=\{A_n:n\in\Bbb N\}$ be a family of infinite subsets of $\Bbb N$. If $n\in\Bbb N$, and distinct $x_k,y_k\in A_k$ have been chosen for each $k<n$, let $X_n=\{x_k:k<n\}$ and $Y_n=\{y_k:k<n\}$. Then let
$$x_n=\min\big(A_n\setminus(X_n\cup Y_n)\big)\,,$$
set $X_{n+1}=X_n\cup\{x_n\}$, let
$$y_n=\min\big(A_n\setminus(X_{n+1}\cup Y_n)\big)\,,$$
and continue. Let $X=\{x_n:n\in\Bbb N\}$ and $Y=\{y_n:n\in\Bbb N\}$; clearly $X$ and $Y$ are disjoint, infinite subsets of $\Bbb N$ such that $x_n\in X\cap A_n$ and $y_n\in Y\cap A_n$ for each $n\in\Bbb N$. Thus, $x_n\in A_n\setminus Y$ and $y_n\in A_n\setminus X$ for each $n\in\Bbb N$, and therefore neither $X$ nor $Y$ contains any member of $\mathscr{A}$.
