I have some problems with the arbitrary union and finite intersections of sets.
Let C be the set of finite unions of subsets of $\mathbb{N}$ of the form $$
\left( n \right) = \left\{ {n,2n,3n,…} \right\}
$$ and the empty set.
The problem is to show that the sets of this form are the closed sets of a topology on the set of natural numbers.
Best Answer
Since the sets of the form $C$ are supposed to be the closed sets of a topology, you need to show that the collection of all such sets is closed under finite union and arbitrary intersection.
That it is closed under finite union is clear. If you have a finite number of sets each themselves a finite union of sets like $(n)$, then their union will again be a finite collection of sets like $(n)$.
For arbitrary intersections, think about a smaller example first. The set $(2)$ is the set of all multiples of 2. The set $(3)$ is the set of all multiples of 3. Thus, the set $(2) \cap (3)$ is the set of natural numbers that are multiples of both 2 and 3. In other words, $(2) \cap (3) = (6)$. In general, you can show $$ \bigcap_{i \in I} (n_i) = (\operatorname{lcm}(n_1, n_2, \dots)), $$ where $I$ is some indexing set and "lcm" denotes the least common multiple.
As an aside, you also need to show $\mathbb{N}$ is closed. Since $\mathbb{N} = (1)$, it is indeed closed.