Problem of whether a given topological space can be realized as a topological product Given a topological space $X$, is it possible to determine whether $X$ is homeomorphic to $\Pi_{i \in I}X_i$, where $I$ is some indexing set and each $X_I$ is a topological space?
Ignoring the trivial solutions of products with one-point sets, of course.
Are there any topological properties that can be imposed on $X$ to yield better results?
 A: A necessary condition for connected spaces $X$: $X$ cannot have a cutpoint (i.e. a point $p$ such that $X\setminus \{p\}$ is disconnected). This follows from the standard theorem that if $X,Y$ are connected, $A$ a proper subset of $X$ and $B$ a proper subset of $Y$, that $(X \times Y) \setminus (A \times B)$ is connected.
So this shows $\Bbb R$ cannot be so written. But subsets of it like $\Bbb Q, \Bbb R\setminus \Bbb Q$ and the Cantor set are homeomorphic to their own square, so can be written as products, as can many natural infinite-dimensional spaces be. 
A: An obvious sufficient condition for a finite non-singleton topological space $X$ to not be homeomorphic to the product of two or more non-singleton spaces is that the cardinality of $X$ is a prime number.
But this is not necessary. For example, the set $\{1,2,3,4\}$ has a composite number of elements, but it together with the following topology is not homeomorphic to the product of two or more non-singleton spaces: $\{\{\},\{1\},\{1,2\},\{1,2,3\},\{1,2,3,4\}\}$.
This only answers the question for finite topological spaces, however. The infinite case is more complicated, because any infinite set $X$ can be placed in bijection with the cartesian product $X \times X$ (assuming the axiom of choice), and so there are no "infinite prime cardinals".
