When does an "infinite polynomial" make sense? Suppose I pick a collection $A \subset \mathbb{C}$ of points in the complex plane and attempt to construct a "polynomial" with those roots via,
$$f(z):=\Pi_{\alpha \in A} (z-\alpha).$$
If $A$ is finite, we get a polynomial. 
If $A=\{n\pi:n \in \mathbb{Z}\}$, according to Euler we get $f(x)=\sin(x)$. Edit: this example is not right as Qiaochu has pointed out; see his answer for more details.
What about other subsets of the complex plane? Other countable subsets without accumulation points? Countable sub with accumulation points like $A=\{1/n:n \in \mathbb{Z}\}$? Uncountable subsets?? When does the product converge, and if it does how does the spatial distribution of $A$ effect the properties of $f$?
This question was motivated by the question here: Determining the density of roots to an infinite polynomial
 A: That is not what Euler's product expansion of the sine looks like. It is very much supposed to be in the form
$$\frac{\sin z}{z} = \prod_{n \ge 1} \left( 1 - \frac{z^2}{\pi^2 n^2} \right).$$
The product you've written down does not converge for $A = \{ n \pi \}$ unless $z \in A$. Indeed, its factors don't go to $1$, which is a necessary condition exactly analogous to the condition for infinite series that the terms need to go to $0$. In fact one can switch between infinite sums and infinite products using the logarithm, which can be used to prove the following.
(First I need to mention that the theorem below requires the convention wherein a product which tends to $0$ is said to diverge. This is because the logarithm of such a product diverges to $-\infty$.) 
Theorem: Let $a_n \in \mathbb{C}$ be a sequence such that $\sum |a_n|^2$ converges. Then $\prod (1 + a_n)$ converges if and only if $\sum a_n$ converges. 
Sketch. Use the fact that $\log (1 + a_n) = a_n + O(|a_n|^2)$. 
So we can make sense of the "infinite polynomial"
$$\prod_{\alpha \in A} \left( 1 - \frac{z}{\alpha} \right)$$
for countable $A$ such that $\sum_{\alpha \in A} \frac{1}{|\alpha|^2}$ and $\sum_{\alpha \in A} \frac{1}{\alpha}$ both converge. See also the Weierstrass factorization theorem.
Note that by the identity theorem, the zeroes of a holomorphic function are isolated, so if you want your product to be holomorphic with $A$ as its zero set, $A$ needs to be discrete. 
Infinite sums and products do not behave well for uncountably many terms, the basic reason being the following.
Theorem: Let $S$ be an uncountable set of positive real numbers. Then for any positive real $r$, there is a finite subset of $S$ whose sum is greater than $r$. 
(In other words, no sum with uncountably many terms can converge absolutely.)
Proof. The sets $S_{\epsilon} = \{ s : s \in S, s > \epsilon \}$ for $\epsilon$ a positive rational are a countable collection of sets whose union is $S$. Since a countable union of countable sets is countable, it follows that there exists $\epsilon$ such that $S_{\epsilon}$ is uncountable. Then the result is clear. 
