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Let $P$ be certain property. Let $S \subset \mathbb{C}^n\times \mathbb{C}^m$ be a set of closed points such that for any point in $S$, it satisfies the property $P$. I know for any $x \in \mathbb{C}^n$, the set $(x, \mathbb{C}^m) \cap S$ is an open set (in Zariski topology, but may be empty) and for any $y \in \mathbb{C}^m$, the set $(\mathbb{C}^n,y)\cap S$ is an open set. Moreover, I know for any $s \in S$, there is an open set $U_s$ of Euclidean topology such that $s \in U_s \subset S$.

Can I conclude that when $S$ is nonempty, $S$ contains a nonempty open set (of Zariski topology).

This question comes from the intention to do Bertini type result for multipe linear systmes: Suppose for any variety with property $P$, then a general element of any basepoint free linear system also has property $P$. Let $X$ is a variety with property $P$, and $|L_i|, i=1,2$ be finite dimensional basepoint free linear systems, I what to say that $V = V_1 \cap V_2$ with $V_i \in |L_i|$ being the general elements also has property $P$. The difficulty is: when fixed $V_1$, one know that for general $V_2$, $V_1 \cap V_2$ has property $P$, but how to make both of $V_1, V_2$ being general?

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No, you need other assertions. For example, let $f\colon \mathbb{C}\to \mathbb{C}$ be any bijective function. Then, the complement of the graph of $f$ in $\mathbb{C}^2$ satisfies your property: the intersection with each fiber of each projection is just $\mathbb{C}$ minus one point, so is open in the Zariski topology. However, the function can be arbitrary (think for example to a strange field isomorphism) so the graph is certainly not constructible in the Zariski topology. Moreover, the closure of the graph is dense in the Zariski topology if the function is bad enough.

EDIT: Now that the following extra condition was added by the OP: the subset $S$ is open in the Euclidean topology. In this case, the same kind of counterexamples works, it suffices to choose $f$ continuous for the Euclidean topology. For instance, view $\mathbb{C}$ as homeomorphic to $\mathbb{R}^2$ and choose $f$ corresponding to $(x,y)\mapsto (x,y+exp(x)+sin(x))$ or anything of your choice.

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  • $\begingroup$ Thank you for your answer! I added an extra condition that for any point in $S$, there is an Euclidean open set in $S$ containing that point. Do you think that is enough for my purpose. I feel this may rule out the those bad functions $f$ as you pointed -- because I also need some control at nearby points (but only in Euclidean topology). $\endgroup$ – Li Yutong Mar 27 '14 at 12:36

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