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Let $X \subseteq \mathbb{P}^n$, $Y \subseteq \mathbb{P}^m$ be quasi-projective varieties (intersections of Zariski-closed and Zariski-open subsets of $\mathbb{P}^n$ and $\mathbb{P}^m$, respectively) over an algebraically closed field. We can view $X \times Y$ as a quasi-projective variety via the Segre embedding. Let $U \subseteq X$ be a nonempty open set and let $V_u \subseteq Y$, be a nonempty open set for each $u \in U$. Is the set $$ \bigcup_{u \in U} \{u\}\times V_u $$ open in $X \times Y$? We can assume X and Y are irreducible if it helps.

I think that if this property holds for $X, Y$ affine, then it will hold in general, because then the set in question would be the union of open sets.

Is there a good reference that describes what the open subsets of $X \times Y$ look like, and finds a base for this (Zariski) topology?

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  • $\begingroup$ $\{u\}\times V_u$ is not open in $X\times Y$. What do you mean exactly with a family of open sets (say in the case $X,Y$ affine) ? Your set is dense. $\endgroup$
    – reuns
    Commented Feb 15, 2020 at 21:43
  • $\begingroup$ @reuns Sorry, that was vague. I have edited the question. Is it clear now? $\endgroup$
    – Ben
    Commented Feb 16, 2020 at 15:49

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A family of open sets may not lead to a open set in the product. Take $k=\mathbb{C}$ and let $X=Y=\mathbb{A}^1_\mathbb{C}$ and let $$V_u = Y\setminus \{e^u\}$$ for each $u\in U=X$. Clearly $V_u$ is open in $Y$.

However $P = \bigcup_{u\in X} \{u\} \times V_u$ is the complement in $\mathbb{A}^2_\mathbb{C}$ of the graph $\Gamma$ of the exponential function. Hence $P$ is open iff $\Gamma$ is closed (in the Zariski topology) and the later is not true because the exponential function is not algebraic.

(Note that it also gives an example of a set that is open in the euclidean topology but not in the Zariski topology.)

The, maybe only, reasonable way to define the Zariski topology of a quasiprojective variety is through an embedding. For instance if $X,Y$ are affine, $X\subset \mathbb{A}^n$, $Y\subset \mathbb{A}^m$, then the Zariski topology of $X\times Y$ is generated by $U\cap X\times Y$ where $U\subset \mathbb{A}^{n+m}$ is open.

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  • $\begingroup$ Thank you! I suspect that $P$ does not even contain a Zariski open dense subset of $X\times Y$, correct? $\endgroup$
    – Ben
    Commented Feb 16, 2020 at 18:44
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    $\begingroup$ Correct. $\Gamma$ is Zariski dense. $\endgroup$
    – Alan Muniz
    Commented Feb 16, 2020 at 20:10

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