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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 $\mathbb N\setminus A\in\mathcal F\subseteq\mathcal U$.

For the record, this and this are relevant.

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 $\mathbb N\setminus A\in\mathcal F\subseteq\mathcal U$.

For the record, this and this are relevant.

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 $\mathbb N\setminus A\in\mathcal F\subseteq\mathcal U$.

For the record, this and this are relevant.

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 $\mathbb N\setminus A\in\mathcal F\subseteq\mathcal U$.

For the record, this and this are relevant.

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 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 $\mathbb N\setminus A\in\mathcal F\subseteq\mathcal U$.

For the record, this and this are relevant.