Ultrafilters on $\omega$ and lower/upper density 
(Show that)If $\mathscr{U}$ is an ultrafilter on $\omega$, then $\mathscr{U}$ contains a subset $A$ of lower density zero. (But) there is an ultrafilter on $\omega$ such that every $A \in \mathscr{U}$ has positive upper density.

This is an exercise on page 76 of Problems and Theorems in Classical Set Theory, Peter Komjath , Vilmos Totik. I have no clue how to start.
 A: The first question (Nr. $6$ in the book), follows immediately from Nr. $5$:

If $\mathscr{U}$ is an ultrafilter on $\omega$ and $0=n_0<n_1<\dots$ are arbitrary natural numbers, then there exists an $A\in\mathscr{U}$ with $A\cap[n_i,n_{i+1})=\varnothing$ for infinitely many $i<\omega$.

For $k>0$ take $n_k=k!$, say. To prove the result in Nr. $5$, let $\langle n_k:k\in\omega\rangle$ be such that $0=n_0<n_1<\dots~$. Let $$A=\bigcup_{k<\omega}[n_{2k},n_{2k+1})\;;$$ clearly $$\omega\setminus A=\bigcup_{k<\omega}[n_{2k+1},n_{2k+2})\;.$$ Both $A$ and $\omega\setminus A$ have empty intersection with infinitely many of the intervals $[n_k,n_{k+1})$, and since $\mathscr{U}$ is an ultrafilter, one of $A$ and $\omega\setminus A$ is in $\mathscr{U}$.
For the second question, let $\mathscr{I}=\{A\subseteq\omega:\overline{d}(A)=0\}$, where $\overline d$ is the upper density. It’s not hard to verify that $\mathscr{I}$ is an ideal, so $\mathscr{F}=\{\omega\setminus A:A\in\mathscr{I}\}$ is a filter on $\omega$. Now just let $\mathscr{U}$ be any ultrafilter containing $\mathscr{F}$.
