Proof that an infinite product of discrete spaces may not be discrete I am trying to prove that an infinite product of discrete spaces may not be discrete.  I tried taking the simplest nontrivial discrete space, $X:=\{0,1\}$ with the discrete topology, and tried to find a sequence of $0$s and $1$s in $\displaystyle\prod_{i=1}^\infty X_i$ that can't be produced by taking intersections of preimages of open sets in the factor spaces under the projection maps, but couldn't come up with anything.
Is this a good approach, or is this actually too simple of an example?
How would you go about solving this problem?  I'd be interested in hearing the thought process behind the proof as well.  Thanks.
 A: The product you describe is compact by Tychonoff's theorem, but the only compact discrete spaces are the finite ones. 
The thought process behind this proof is strongly informed by the material in this blog post. A product of finite discrete spaces is an example of a profinite set or Stone space, and these behave in very particular ways. 
Alternatively, the product you describe is second-countable, which an uncountable discrete space isn't. 
A: More elementarily: The open sets in $\prod_{\mathbb N}\{0,1\}$ are precisely the (possibly infinite) unions of cylinder sets. Because even a single cylinder set is infinite, so is every nonempty open set. In particular, a singleton cannot be open, so the space is not discrete.
A: Here is an example derived directly from this answer at mathoverflow. Let $(0)$ be the constant sequence of $0$'s. For $i\in\omega$, let $Z_i$ be the set of all those sequences that have $0$ at the i-th cordinate. By the definition of the product topology, for every open set $V$ containing $(0)$ there can only be finitely many $Z_i$'s such that $V\subseteq Z_i$. But $\{(0)\}=\bigcap_{i\in\omega}Z_i$, so $\{(0)\}$ is not open.  
Of course Henning's answer is more general, but this gives some additional insight at what is going on in the product.
