I wonder what are all the topologies on the product of spaces that satisfy certain properties. For example, let $(X_i,\mathcal{T}_i)_{i\in I}$ be a family of topological spaces and let $X:=\prod_{i\in I}X_i$ be the Cartesian product of these sets. Also, let $\pi_i:X\rightarrow X_i,i\in I$ be the canonical projections. Now, the Tychonoff's topology on the set $X$ satisfies these two properties:
- For all $i\in I$, the mapping $\pi_i$ is continuous.
- For all $i\in I$, the mapping $\pi_i$ is open.
It is easy to prove that Tychonoff's topology is the coarsest topology on the set $X$ that satisfies these two properties. The box topology is another topology on the set $X$ that satisfies these two properties. I want to know if the box topology is, as the opposite, the finest topology on the set $X$ that satisfies these two properties. I also want to know (if my previous assumption is false) what is some characterization of all the topologies on the set $X$, beside these two, that satisfy these two properties.
If we take the box product on $\Bbb R^\omega$ for example, in which the "diagonal" $D = e[\Bbb R]$ is not open (where $e: \Bbb R \to \Bbb R^\omega$ maps $x$ to the constant sequence with value $x$.) we can add $D$ to the box topology and generate a strictly larger topology that still obeys 1 and 2. (1 because it's finer than the product topology and 2 because $\pi_n[D]=\Bbb R$ is not going to "spoil" the opennness.). So I think the answer to your question is "no" in general (it's trivially true for a product of discrete spaces, as the box topology is then also discrete).