TITLE: Tychonoff Theorem in the box topology
QUESTION [13 upvotes]: A short question: Why does not the Tychonoff theorem (the arbitrary product of compact spaces is compact) hold in the box topology? 
I don't know how to show that there is no finite sub-cover of any open cover of the product space! Can anyone please give me some hint? Thank you!

REPLY [4 votes]: Here are the cold hard facts. Let $(X_i)_{i \in I}$ be a nonempty family of nonempty topological spaces. Let $X = \prod_{i \in I} X_i$. Let $P$ be $X$ in the product topology. Let $B$ be $X$ in the box topology.
$$ B \text{ is Hausdorff} \Leftrightarrow P  \text{ is Hausdorff } \Leftrightarrow X_i  
\text{ is Hausdorff for each } i \in I$$
$$ B \text{ is compact } \Rightarrow P \text{ is compact } \Leftrightarrow X_i \text{ is compact for each } i \in I$$
The above can be got using Tychonoff's theorem and some elementary arguments. A few remarks: 

the box topology is finer than the product topology (refinements of Hausdorff topologies are Hausdorff, and vice versa for compact topologies) 
the projections are continuous for $B$ and $P$ (and continuous images of compact spaces are compact)
by fixing a base point in X, we can embed the $X_i$ into $B$ or $P$ (note subspaces of Hausdorff spaces are Hausdorff).

The point of the above is that, if $B$ is a compact Hausdorff space, then so is each $X_i$ which means $P$ is a compact Hausdorf space. But any two compact Hausdorff topologies which are comparable must coincide, so this implies $P = B$. 
Moral: if the box topology turns out to be compact Hausdorff, then you're actually looking at the product topology.