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

  • 3
    $\begingroup$ This is not Soviet Russia. We are not "Comrades" or "Users". Asaf Karagila, or just Asaf if the context is clear, would suffice. :-) $\endgroup$
    – Asaf Karagila
    Nov 24, 2022 at 23:28
  • $\begingroup$ Apologies - will keep this in mind! $\endgroup$ Nov 25, 2022 at 21:28

1 Answer 1


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.

  • 1
    $\begingroup$ 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 $X(n,i)\in u$. $\endgroup$
    – KP Hart
    Jan 2, 2023 at 22:19

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