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Suppose $n \ge 2$ and $X$ is a path-connected subset of $\Bbb{CP}^n$ under the manifold topology. If for every complex hyperplane $H \subset \Bbb{CP}^n$, $H \cap X$ is a degree $k$ algebraic hypersurface in $H$ (could be singular or reducible), is it true that $X$ itself is a degree $k$ algebraic hypersurface in $\Bbb{CP}^n$? This question arose when I was constructing an algebra-free definition of quardric surface (where $n=3$ and $k=2$ respectively), and I have no idea how to prove or disprove this.
Edit$1$: Since a "hypersurface" in $\Bbb{CP}^1$ is a set of finite points (thus no enough rigidity appears), the answer should be carefully discussed when $n=2$.
Edit$2$: According to the comment a counterexample can be easily found for $n=2$, so I'm wondering if the conjecture is also false when $n>2$.

Suppose $n \ge 3$ and $X$ is a path-connected subset of $\Bbb{CP}^n$ under the manifold topology. If for every complex hyperplane $H \subset \Bbb{CP}^n$, $H \cap X$ is a degree $k$ algebraic hypersurface in $H$ (could be singular or reducible), is it true that $X$ itself is a degree $k$ algebraic hypersurface in $\Bbb{CP}^n$? This question arose when I was constructing an algebra-free definition of quardric surface (where $n=3$ and $k=2$ respectively), and I have no idea how to prove or disprove this.
Appendix: The $n=2$ case can be easily proved to be false as stated in the comment. However the reason might be that a "hypersurface" in $\Bbb{CP}^1$ is just a set of finite points, thus no enough rigidity occurs. I wonder whether things change in higher dimensions.

Suppose $n \ge 2$ and $X$ is a path-connected subset of $\Bbb{CP}^n$ under the manifold topology. If for every complex hyperplane $H \subset \Bbb{CP}^n$, $H \cap X$ is a degree $k$ algebraic hypersurface in $H$ (could be singular or reducible), is it true that $X$ itself is a degree $k$ algebraic hypersurface in $\Bbb{CP}^n$? This question arose when I was constructing an algebra-free definition of quardric surface (where $n=3$ and $k=2$ respectively), and I have no idea how to prove this.
Edit$1$: Since a "hypersurface" in $\Bbb{CP}^1$ is a set of finite points (thus no enough rigidity appears), the answer should be carefully discussed when $n=2$.
Edit$2$: According to the comment a counterexample can be easily found for $n=2$, so I'm wondering whether we can get the result with the following additional condition: Not only does every $H \cap X$ is a degree $k$ algebraic hypersurface in $H$, but also for almost all $H$, $H \cap X$ is non-singular. Here "almost all" means that if we consider $H$ as points in the dual space, then the set $\{ H | H \cap X \; \text{is singular in} \; H \}$ is of Hausdorff dimension $0$ with respect to the round metric on $\Bbb{CP}^n$.

Suppose $n \ge 2$ and $X$ is a path-connected subset of $\Bbb{CP}^n$ under the manifold topology. If for every complex hyperplane $H \subset \Bbb{CP}^n$, $H \cap X$ is a degree $k$ algebraic hypersurface in $H$ (could be singular or reducible), is it true that $X$ itself is a degree $k$ algebraic hypersurface in $\Bbb{CP}^n$? This question arose when I was constructing an algebra-free definition of quardric surface (where $n=3$ and $k=2$ respectively), and I have no idea how to prove or disprove this.
Edit$1$: Since a "hypersurface" in $\Bbb{CP}^1$ is a set of finite points (thus no enough rigidity appears), the answer should be carefully discussed when $n=2$.
Edit$2$: According to the comment a counterexample can be easily found for $n=2$, so I'm wondering if the conjecture is also false when $n>2$.

Suppose $n \ge 2$ and $X$ is a path-connected subset of $\Bbb{CP}^n$ under the manifold topology. If for every complex hyperplane $H \subset \Bbb{CP}^n$, $H \cap X$ is a degree $k$ algebraic hypersurface in $H$ (could be singular or reducible), is it true that $X$ itself is a degree $k$ algebraic hypersurface in $\Bbb{CP}^n$? This question arose when I was constructing an algebra-free definition of quardric surface (where $n=3$ and $k=2$ respectively), and I have no idea how to prove this.
Edit: Since a "hypersurface" in $\Bbb{CP}^1$ is a set of finite points (thus no enough rigidity appears), the answer should be carefully discussed when $n=2$.

Suppose $n \ge 2$ and $X$ is a path-connected subset of $\Bbb{CP}^n$ under the manifold topology. If for every complex hyperplane $H \subset \Bbb{CP}^n$, $H \cap X$ is a degree $k$ algebraic hypersurface in $H$ (could be singular or reducible), is it true that $X$ itself is a degree $k$ algebraic hypersurface in $\Bbb{CP}^n$? This question arose when I was constructing an algebra-free definition of quardric surface (where $n=3$ and $k=2$ respectively), and I have no idea how to prove this.
Edit$1$: Since a "hypersurface" in $\Bbb{CP}^1$ is a set of finite points (thus no enough rigidity appears), the answer should be carefully discussed when $n=2$.
Edit$2$: According to the comment a counterexample can be easily found for $n=2$, so I'm wondering whether we can get the result with the following additional condition: Not only does every $H \cap X$ is a degree $k$ algebraic hypersurface in $H$, but also for almost all $H$, $H \cap X$ is non-singular. Here "almost all" means that if we consider $H$ as points in the dual space, then the set $\{ H | H \cap X \; \text{is singular in} \; H \}$ is of Hausdorff dimension $0$ with respect to the round metric on $\Bbb{CP}^n$.