For a set $A\subseteq \omega$ we let the *upper density* of $A$ be defined as $d^+(A) := \lim\sup\_{n\to\infty}\frac{|A\cap(n+1)|}{n+1}$. Let $\text{FrU}(\omega)$ be the collection of [free ultrafilters](https://en.wikipedia.org/wiki/Ultrafilter) on $\omega$.


[Asaf Karagila](https://mathoverflow.net/users/7206/asaf-karagila) provided a convincing argument that there is a free ultrafilter ${\cal U}$ on $\omega$ such that $d^+(U) > 0$ for all $U\in {\cal U}$.


**Question.** What is $$\sup\big\{\inf \{d^+(U): U \in {\cal U}\}: {\cal U} \in \text{FrU}(\omega)\big\}\;?$$



Supremum of infimum of measure of members of a free ultrafilter
The answer is: zero.


The reason is that every ultrafilter has zero as the infimum of the upper density of its members. To see this, observe that if a set $U$ is in the ultrafilter $\mathcal{U}$, with some positive upper density, then we can split $U$ in half $U=A\sqcup B$ each with half the upper density (just take every other element of $U$ into $A$, the others into $B$). One of these sets will be in the ultrafilter, and so we will have a set in $\mathcal{U}$ with half the upper density of $U$. By iterating this, we can make the upper density of the sets in $\mathcal{U}$ as low as desired, so the infimum over the members is zero.


