# Supremum of infimum of measure of members of a free ultrafilter

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 on $$\omega$$.

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

• This is not Soviet Russia. We are not "Comrades" or "Users". Asaf Karagila, or just Asaf if the context is clear, would suffice. :-) Nov 24, 2022 at 23:28
• Apologies - will keep this in mind! Nov 25, 2022 at 21:28

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.
• Or: every ultrafilter contains, for every $n$, a set of density $\frac1n$: just Take $X(n,i)=\{k:k\equiv i\pmod n\}$ for $n\in\mathbb{N}$, and $i\in\{1,2,\ldots,n\}$. If $u$ is an ultrafilter then for every $n$ there is an $i$ such that $X(n,i)\in u$. Jan 2, 2023 at 22:19