If $\mathcal{U}$ is a non-principal ultrafilter on $\mathbb{N}$ then $\mathcal{U}$ does not have the Baire property in $2^\mathbb{N}$ Let $\mathcal{U}$  now be an ultrafilter on $\mathbb{N}$. View $\mathcal{U}$ as a subset of $2^\mathbb{N}$.
Question:

If $\mathcal{U}$ is non-principal then $\mathcal{U}$ does not have the Baire property in $2^\mathbb{N}$.

An ultrafilter is principal if some $x\in X$, $\{x\} \in \mathcal{U}$ or equivalently, $\mathcal{U}=\{A\in2^{\Bbb N}: x\in A \}$ for some $x\in X$.
Thanks.

A free ultrafilter $\mathscr{U}$ is a tail set in $2^{\Bbb N}$, meaning that if $U\in\mathscr{U}$, and $V\in 2^{\Bbb N}$ differs from $U$ in only finitely many coordinates (i.e., $U\mathrel{\triangle}V$ is finite), then $V\in\mathscr{U}$. By Theorem $\mathbf{21.4}$ in Oxtoby’s Measure and Category, if $\mathscr{U}$ had the Baire property, then $\mathscr{U}$ would be either meagre or co-meagre in $2^{\Bbb N}$. However, $\mathscr{U}$ cannot be meagre or co-meagre, because the map $f:2^{\Bbb N}\to 2^{\Bbb N}:A\mapsto\Bbb N\setminus A$ is a homeomorphism that sends $\mathscr{U}$ to its complement.
