= Solution
Take $\tau$ to be the <ordinary topology on infinite subsets>, whose basic cylinders $[s]$ fix a finite <initial segment>. Let
$$
D=\{Z\in[\mathbb N]^\omega:\mathbb N\setminus Z\text{ is finite}\}.
$$
This is a <dense countable family of cofinite infinite subsets>. It is countable because finite complements form a countable family. Each singleton is $\tau$-<closed>: distinct increasing enumerations differ at some finite coordinate, which separates them by cylinders. No singleton has <interior>, since every cylinder admits more than one infinite continuation. Thus every singleton is $\tau$-<nowhere dense>, and $D$ is $\tau$-<meagre>.
On the other hand, every cylinder $[s]$ contains the <cofinite set> $s\cup\{n:n>\max s\}$, with $\max\varnothing=0$. Hence $D$ is $\tau$-<dense>, its <closure> is all of $X$, and
$$
\boxed{D\text{ is }\tau\text{-meagre but not }\tau\text{-nowhere dense}.}
$$
The paper does not define its symbol $\tau$. If the course instead uses it for the <Ramsey cone topology>, with basic neighborhoods $[A]^\omega$ and no finite stem, use $X$ itself as the example. For $D_j=\{Z:\min Z=j\}$, its cone-<closure> is $\{Z:j\in Z\}$: every cone neighborhood of a set containing $j$ contains an infinite subset of minimum $j$, while a cone based on a set omitting $j$ misses $D_j$. This <closure> has empty <interior>, because every infinite reservoir can be thinned to omit $j$. Thus every $D_j$ is cone-<nowhere dense>, and $X=\bigcup_{j\geq1}D_j$ is cone-<meagre> but is not cone-<nowhere dense>. This supplies the requested example under either convention, without conflating the two topologies.
Back to article page
