Stupid Product Topology Question Please forgive me for asking such an easy question. I'm still getting used to the concept of product topologies. Problem: Find a non-empty topological space $X$ such that $X \cong X \times X$ (with product topology). So my thought for a finite topological space is just the set {$a$} with discrete topology. It seems like it works to me, but it's so stupid I thought something had to be wrong. Also, is this the only case for finite topological spaces? Because if #$X >1$, then #$X \times X >$ #$X$, so we can't establish a bijection. So if I wanted to find such a topological space without the discrete topology, can anyone think of such an example?
 A: You are correct: taking a one-point space (which carries exactly one topology!) works.
It is a little strange though that you speak of "finite topological spaces", since for a finite space your example works only if it has a single element (or, even more trivially, no elements at all).  In general if $X \times X \cong X$ then we need $\# X \cdot \# X  = \# X$.  For finite sets this is only true if the cardinality is $0$ or $1$, but assuming the Axiom of Choice it holds for all infinite sets.
Added: I read your question more carefully and what you are saying no longer seems strange: yes, as you say, your example and the empty set are the only ones among finite topological spaces.
Any infinite set $X$ endowed with the discrete topology gives an example.  For another example....take any infinite set endowed with the indiscrete topology $\tau_X = \{ \varnothing, X \}$.  
There are more interesting examples.  For instance, if you start with any topological space $Y$ and put $X = \prod_{i=1}^{\infty} Y$, then $X$ is such an example.  If you take $Y$ to be for instance the unit circle then you get an example with a compact Hausdorff space which is not discrete.
