Let $\mathbb{N}$ be the set of positive integers and for $A\subseteq {\mathbb{N}}$ set $$m(A) = \text{lim sup}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$
Does every ultrafilter ${\cal U}$ on $\mathbb{N}$ have the property that there is $U\in{\cal U}$ with $m(U) = 0$? If not, is it possible that $\inf\{m(U):U\in{\cal U}\} > 0$?
The answer to your second question is no. For any $n$, $\mathcal U$ contains exactly one of the sets $n\mathbb N+k$ for $k=0,\dots,n-1$. These sets have density $\frac{1}{n}$, hence $\inf\{m(U):U\in\mathcal U\}\leq\frac{1}{n}$, hence $\inf\{m(U):U\in\mathcal U\}=0$.
The answer to the first question is no as well: Let $\mathcal F$ be the family of sets with asymptotic density $1$. It's straightforward to see this is a filter. Hence it can be extended to an ultrafilter $\mathcal U$. $\mathcal U$ can't contain any set $A$ of density zero, because $\mathbb N\setminus A\in\mathcal F\subseteq\mathcal U$.
