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The Osgood's lemma in complex analysis says that if $f \colon \mathbb{C}^n \to \mathbb{C}$ a continuous function such that $f|_{L}$ is holomorphic for every complex line $L \subset \mathbb{C}^n$, then $f$ is holomorphic. I wonder if there is the following geometric analogue of Osgood's lemma:

Let $X$ be a complex manifold and $A \subseteq X$ a subset (maybe not too bad, say a submanifold, or maybe it also has some singularities...).

Question. Assume that for every holomorphic curve $C \subset X$ the intersection $A \cap C$ is analytic in $C$ (the latter just means that either $C \subseteq A$ or $A \cap C$ is a discrete set). Is it true that $A$ is analytic?

One can also ask, is it true that if the intersection of $A \subset \mathbb{C}^n$ with every complex line $L \subset \mathbb{C}^m$ is analytic, then $A$ is analytic. I think this question is stronger, i.e. an affirmative answer to it will also imply a positive answer on the first one by considering a Stein cover and arguing locally.

UPD.: Of course I do not assume the test-curves to be closed, since closed holomorphic curves can simply not exist, as many commentators point out. The question is local, so it can be phrased as follows. Let $U$ be a small neighbourhood of $0$ in $\mathbb{C}^n$ and $A \subset U$ a closed subset. Assume that for every holomorphic embedding of a disk $f \colon \Delta \to U$ with $f(0) \in A$ either $f(\Delta) \subseteq A$ or $f(0)$ is an isolated point in $f(\Delta) \cap A$. Is it true that $A$ is analytic?

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  • $\begingroup$ What exactly do you mean by a holomorphic curve? Do you mean the image of a nonconstant holomorphic map of a Riemann surface? Do you mean a properly embedded complex 1-dimensional submanifold? ... $\endgroup$
    – Moishe Kohan
    Commented May 6 at 15:06
  • $\begingroup$ I guess, an image of a nonconstant map of a Riemann surface. The question is local, so it is enough to check on embedded holomorphic discs. $\endgroup$
    –  V. Rogov
    Commented May 6 at 17:06
  • $\begingroup$ Please correct me if I'm wrong but I will assume that by analytic set you mean complex analytic? If that is the case, what happens if you take X=\mathbb{C}^2 and A=\{ (z, \overline{z}) | z\in \mathbb{C} \} ? Clearly A is not analytic (it is real analytic - even real algebraic), and since holomorphic curves are also analytic (and therefore also real analytic) their intersection will consist of discrete set of points. $\endgroup$
    – Luka Thaler
    Commented May 8 at 13:13
  • $\begingroup$ @LukaThaler That does not work, because the complex curve $\left\{(z,z)\mid z \in \mathbb{C}\right\}$ intersects your $A$ in the real analytic curve $\left\{(x,x)\mid x \in \mathbb{R}\right\}$. $\endgroup$
    – Thomas Kurbach
    Commented May 8 at 14:14
  • $\begingroup$ @ThomasKurbach Indeed, you are totally right! I was to fast with my comment. $\endgroup$
    – Luka Thaler
    Commented May 9 at 13:43

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I think that the answer is no.

Take a complex torus $X$ without any holomorphic curve. If the result you are asking for were true, it would imply (in the empty sense) that every subset $A \subset X$ is analytic, which is false.

For a counterexample with at least one curve, take $X$ as above, let $X'$ be the blow-up of $X$ at one point, and $E \subset X'$ the exceptional divisor. Then $E$ is the unique complex curve inside $X'$.

Now take any non-analytic subset $A$ of $X'$ such that $A$ is disjoint from $E$ and set $$A' := A \cup E.$$

Thus $E \subset A'$ and your condition is satisfied, but $A'$ is not an analytic subset of $X'$.

Note. The OP edited the question after this answer was posted, changing its meaning.

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    $\begingroup$ Of course, I did not assume to test only on the closed curves. The question is local, so you can test on germs of curves (essentially on disks). I added an update to the post. $\endgroup$
    –  V. Rogov
    Commented May 6 at 17:05
  • $\begingroup$ One can also assume that $X$ is algebraic and ask if it is true that $A \cap C$ is an algebraic subset of $C$ for an algebraic curve $C \subset X$ if and only if $A$ is an algebraic subvariety. But I guess this is a completely different question. $\endgroup$
    –  V. Rogov
    Commented May 6 at 17:08
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    $\begingroup$ "Of course" in which sense? Osgood's theorem is about lines, so how could one guess that you were talking about germs? After your update, my answer looks wrong, while it was correct when I wrote it. I think you should have asked another question rather than updating yours after my answer was posted. Anyway, it does not matter... $\endgroup$
    – Francesco Polizzi
    Commented May 6 at 18:49

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