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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.

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Unrelated question: why do I often see "it´s" in place of "is"? Is it just a weird typo? – anon Aug 17 '11 at 3:31
@Daniel: A topological space consists of a set $X$ and a family $\tau$ of subsets of $X$ that contains $\emptyset$, $X$, and is closed under arbitrary unions and finite intersections. What is your set $X$ here, and is $n$ an arbitrary integer, an arbitrary positive integer, an arbitrary natural number, or something else? – Arturo Magidin Aug 17 '11 at 3:34
When you write $n$, do you mean $\{0,\ldots,n-1\}$? What about $2n$ and so on? – Asaf Karagila Aug 17 '11 at 3:35
I don't understand the question for a different reason than nitpicking. As I understand your question you want to check if your sets form the closed sets of a topology, so you should not have to consider arbitrary unions but only finite ones. On the other hand, you must check that arbitrary intersections are of the desired form. – t.b. Aug 17 '11 at 3:39
@Daniel: And what is your $X$? And, as Theo points out: if these are meant to be the closed sets, why are you worried about arbitrary unions? You don't need to consider arbitrary unions to check if a family of subsets are the closed sets in a topology. – Arturo Magidin Aug 17 '11 at 3:50

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.

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Note that if we have infinitely many different sets of the form $(n)$ then the intersection is empty, as any fixed number can be in at most finitely many different ones. – Henno Brandsma Aug 17 '11 at 16:44

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