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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.

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  • $\begingroup$ 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, 2017 at 13:51
  • $\begingroup$ @drhab thanks, corrected! $\endgroup$
    – Lukas Juhrich
    Dec 30, 2017 at 14:18

1 Answer 1

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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.

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  • $\begingroup$ 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, 2017 at 15:04

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