There is a recurrent infinite/finite duality in topology, with one appearing in the opposite direction of the other; union/intersection, directed sets work their way upwards finitely while never caring about whatever infinite lies downwards, FIP vs arbitrary open covers, etc.

This appears too in the most natural ways to define topology on arbitrary Cartesian products: For both the product and box topology a base are products of open sets, the first one having only finite ones different from the whole space, while the other one employs products of arbitrary open sets. Also, projections are continuous and open maps in both, being the product topology the coarsest topology by which projections are continuous. This is because the space is created using finite intersections of cylinders.

This strongly suggests me that the box topology might be the finest topology by which projections are open maps, giving a bit of symmetry to the matter, being boxes the "arbitrary" version of cylinders, and open maps being the inverse concept to continuous maps.

I just couldn't find neither a prove nor a counter-example, nor a fruitful internet search.

  • $\begingroup$ Or maybe there's an even finer topology I'm missing $\endgroup$ – Emilio Martinez Sep 30 '18 at 14:46

Let $\mathcal{U}$ be the collection of all subsets $U$ of $\prod_i X_i$ such that $\pi_i[U]$ is open in $X_i$ for all $i$. This is much bigger than the set of open boxes (e.g. consider diagonals and antidiagionals in the plane etc.). There could be a lot of topologies between the box topology and this collection, and all of them would have open projections.

  • $\begingroup$ You are right. The topology with base given by balls lacking a finite number of straight lines through the centre is strictly finer and projections are still open. Thanks. $\endgroup$ – Emilio Martinez Sep 30 '18 at 15:51

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