Compact topology between box and product topology

Question

Suppose we have a family $(X_i)_i$ of compact $T_1$, but not necessarily Hausdorff spaces.

Suppose $\tau$ is a topology on $\prod_i X_i$ which is compact, refines the product topology, and is refined by the box topology.

Can $\tau$ be distinct from the product topology?

Note that if $X_i$ are all Hausdorff, then the answer is no, even if we don't assume that the box topology is finer.

Answer 1

Sure. For instance, let us consider the product $X=Y^\mathbb{N}$, where $Y$ denotes $\mathbb{N}$ with the cofinite topology. Let $A\subset X$ be the set of all constant sequences. Note that in the product topology, $A$ has the cofinite topology, and in particular is compact. However, $A$ is not closed in the product topology. On the other hand, $A$ is closed in the box topology: if $s\in X$ is a nonconstant sequence, then by a simple diagonalization you can construct a box-open neighborhood of $s$ which contains no constant sequences.

So now, let $\tau$ be the topology on $X$ generated by the product topology and the set $X\setminus A$. Since $A$ is compact in the product topology, $\tau$ will still be compact. Since $A$ is closed in the box topology, $\tau$ is contained in the box topology.

Answer 2

One way to construct such topologies is described in this preprint: we take a (potentially trivial) filter $\mathcal{F}$ on the index set $I$, (we have a family $X_i, i \in I$), and define a base for $\prod_{i \in I} X_i$ by all sets $\prod_{i \in I} U_i$, where $U_i$ is open in $X_i$ for all $i \in I$, and the set $\{i \in I: U_i = X_i\} \in \mathcal{F}$.

It's easy to check (and the paper does) that this is indeed a base for a topology $\mathcal{T}(X_i, I, \mathcal{F})$ on $\prod_i X_i$.

Clearly if $\mathcal{G}$ is another filter on $I$ such that $\mathcal{F} \subseteq \mathcal{G}$, then $\mathcal{T}(X_i, I, \mathcal{F}) \subseteq \mathcal{T}(X_i, I, \mathcal{G})$, so finer filters give finer topologies.

The product topology is exactly the case where $\mathcal{F}$ is the cofinite filter on $I$, and the box-topology is where $\mathcal{F}$ is the powerset of $I$ (the trivial filter). So taking any filter inbetween those gives a topology that's inbetween as well. We then also get topologies that make the projections open and continuous etc., and such that products of Hausdorff spaces are still Hausdorff,etc. Nice properties (like preserving compactness, connectedness etc.) are probably out, in most cases.