When I learned about the product topology, I did not really take the definition from the textbook, but rather tried to construct a “senseful topology” on my own as follows: If we think about $\mathbb R^2$, then every open set I could imagine had the property that for every point, every horizontal and vertical “fiber” had to be open in $X, Y$, respectively.
To be rigorous: A subset $S \subset X\times Y$ is called open, if for every $(x, y)\in S$, $\pi_X^{-1}(x) \cap S = \{x\}\times U$ with $U$ open in $Y$, and similarly for $y$. It is easily checked that this indeed forms a topology on $X\times Y$.
When I revisited some parts about topology, I figured out what a categorical product is and what the actual definition of the Tychonoff product is.
I later even figured out that my guessed topology has a name, it's the “cross topology”. However, Just like the box topology, which is generated by the product of open sets $U\times V$, it is said to be strictly finer (see e.g. [1]) than the categorical, or “Tychonoff” product topology, which is the coarsest topology making each of the projections continuous. That came as quite a shock to me, as I took it for granted that cross and tychonoff are just different characterizations!
Since However, I fail to see a counterexample, i.e. two topological spaces $X, Y$, and a set $S\subset X\times Y$, which is open with respect to the cross topology, but not the tychonoff topology.
I sat down with my professor and we tried to figure out some obvious examples (I only remember constructing things with the countable-complement topology), but we didn't end up with a satisfying answer.
As the Tychonoff topology is defined implicitly by being the coarsest topology generated by $\pi_X^{-1}(\mathcal T_X),\, \pi_Y^{-1}(\mathcal T_Y)$, I'm not even sure anymore how we show that some candidate set has to be contained (or not) in there, i.e. when exactly a set can be generated by finite intersections and arbitrary unions of the projection's “stripes”.0
What am I missing / what would be a useful approach here?
[1] Example 1.2.6 in “Topological Groups and Related Structures”, Atlantis press, 2008.
Let $U:=\{\langle x,y\rangle\in\mathbb R^2\mid |y|<\frac13|x|\}$.
Let $V:=\{\langle x,y\rangle\in\mathbb R^2\mid |x|<\frac13|y|\}$.
Then $U\cup V\cup\{\langle 0,0\rangle\}$ is open in the cross product topology, but not in the usual product topology.