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.

  • $\begingroup$ I suspect that the difference is mostly visible with infinite products, just like the difference between box product and categorical product. $\endgroup$ – Arthur Dec 30 '17 at 13:51
  • $\begingroup$ @drhab thanks, corrected! $\endgroup$ – Lukas Juhrich Dec 30 '17 at 14:18

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.

  • $\begingroup$ Ah, That example seems pretty clear. However, the only proof I could imagine would say that every set in our generator consists only of interior points, and finite intersections and arbitrary unions preserve this property, which is however violated in $(0,0)$ of our set. Is there a more general, “elegant” way of proving this, which makes no use of the specific nature of the topologies of $X,Y$ as being induced from the metric on $\mathbb R$? $\endgroup$ – Lukas Juhrich Dec 30 '17 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.