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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 $\omega$.

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

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 $\omega$.

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

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

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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 $\omega$.

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