For the [extra credit](, I _want_ to say that the poset on \\(\mathbb{Z}^+\\) ordered by divisibility is isomorphic to the product of a countably infinite number copies of \\(\mathbb{N}\\). It's really the lattice of [multisets]( ("bags") over the set of primes, which falls out neatly from the [prime factorization theorem]( (A [lattice]( is a poset with all finite joins and meets.) I'm not sure this really meets the bar of a "characterization", but I'll write it out anyway.

(**EDIT:** This argument fails to account for the fact that the multiset of prime factors for a positive integer _is finite_. We're essentially throwing out any multiset that's "too large". So my final statement is a bit of a lie, because we have to restrict our attention to a certain kind of multiset. I _think_ this is similar to the difference between the [product topology]( and the [box topology]( The product topology requires all but finitely many components to be trivial, whereas the box topology relaxes this constraint entirely. Essentially, I'm assuming a box product below, where I need to use the categorical product. I think this is solvable by sprinkling in the phrase "of finite support" liberally.)

(**EDIT 2:** I suspect this might be a kind of [restricted wreath product](, which is based on the idea of a [direct sum](, which makes the same finite support distinction as the box/product topology above. It seems like we're actually looking at a categorical _coproduct_, even though in the case of topological spaces it's the _product_ that requires finite support. This is fascinating, and I hope I'm reading all of this properly.)

Recall that the [indicator function]( of a set \\(X \subseteq U\\) is a map \\(\mathbf{1}\_X : U \to \mathbb{B}\\) such that \\(\mathbf{1}\_X(x) = \mathrm{true}\\) iff \\(x \in X\\). We can generalize this to multisets by considering maps \\(\mathbb{N}\_X : U \to \mathbb{N}\\), mapping each element of \\(U\\) to the number of times it appears in \\(X\\). The subset relation for multisets is given by \\(X \subseteq Y\\) iff \\(\mathbb{N}\_X(u) \le \mathbb{N}\_Y(u)\\) for all \\(u \in U\\). (Notice how regular sets are just multisets that range only over \\(\\{0, 1\\} \subseteq \mathbb{N}\\)!)

If we take \\(U\\) to be \\(\mathbb{P}\\), the set of prime numbers, we can assign every positive integer \\(n \in \mathbb{Z}^+\\) a unique map \\(\mathbb{P}\_x : \mathbb{P} \to \mathbb{N}\\) such that $$x = \prod_{p \in \mathbb{P}} p^{\mathbb{P}_x(p)}.$$ The prime factorization theorem gives us a one-to-one correspondence between positive integers and multisets of [finite support]( (so that the product above is well-defined). In other words, the poset of positive integers ordered by division \\(\langle \mathbb{Z}^+ , \mid \rangle\\) is isomorphic to the poset of multisets over \\(\mathbb{P}\\). Of course, \\(\mathbb{P}\\) is only used as an index set at this level of abstraction; all we require is that \\(\mathbb{P}\\) be a countably infinite set. In other words, this poset is also isomorphic to the poset of maps \\(\mathbb{N} \to \mathbb{N}\\), which we can also denote as \\(\mathbb{N}^\mathbb{N}\\), ordered as above. (Exponential notation is nice because it makes it clear why I called this the product of a countably infinite number copies of \\(\mathbb{N}\\) above: \\(\mathbb{N}^\mathbb{N} \cong \mathbb{N} \times \mathbb{N} \times \mathbb{N} \times \cdots\\).)

So I guess I want to say that this is the unique poset (up to isomorphism) that factors into a countably infinite number of copies of the totally ordered chain \\(\mathbb{N}\\).