Tychonoff Theorem in the box topology

Consider the space $2^{\mathbb{N}}$ in the box topology, where $2=\{0,1\}$ is the two-point discrete space. So each factor is a finite space, and clearly compact. In the box topology, however, the product space $2^{\mathbb{N}}$ is discrete, since every individual sequence $x\in 2^{\mathbb{N}}$ is the unique member of the corresponding open set determined by its coordinates. So every point is isolated and the space is discrete. But no infinite discrete space is compact, so Tychonoff fails for the box topology.

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.