While working on a research problem (algebraic cycles), I bumped into a question that I want to prove, though I couldn't yet prove. After several days of attempts, I realized that if the following statement on projective geometry holds, then my original question is most likely answered affirmatively.



> 
> **Statement:** Let $k$ be an infinite perfect field (or even suppose algebraically closed, if you wish). Let $X$ be a smooth projective
>  variety over $k$ of dimension $d$, and let $x\_0, x\_1, x\_2, \cdots$ be
>  an infinite sequence of distinct closed points of $X$. Then,
>  there exist an embedding $X\hookrightarrow \mathbb{P}^N$ and a
>  sequence of hyperplanes $H\_1, H\_2, \cdots, H\_d$, each containing
>  $x\_0$, such that $X \cap H\_1 \cap \cdots \cap H\_d$ is a finite set
>  contained in the given infinite countable set $\{ x\_0, x\_1, x\_2, \cdots \}$.
> 
> 
> 


I think that the statement is true when $\dim X = 1$, simply by taking sufficiently many points $x\_0, \cdots, x\_{2g+3}$, say, to form a very ample divisor $D = \sum\_{i=0} ^{2g+3} x\_i$. But, I do not know what to do if $\dim X >1$. Can someone help me in figuring out if this statement is true? I hope I could get some good ideas form mathoverflow. Thank you.



A question on infinitely many closed points on a smooth projective variety and their behavior under embeddings
Let me explain why this statement cannot be true in general. I will give a counterexample where $X$ is a complex K3 surface.


By a result of [Beauville and Voisin, On the Chow ring of a K3 surface], it is possible to find a point $y\in X$ such that whenever $C\_1$, $C\_2$ are curves on $X$, $C\_1\cap C\_2$ is proportional to $y$ in $CH\_0(X)$.


Now, I will construct $x\_0, x\_1,\dots...$ inductively. Suppose that $x\_0,\dots,x\_i$ have been constructed. Then choose $x\_{i+1}$ such that it is independent of $y,x\_0,\dots,x\_i$ in $CH\_0(X)$. To prove that such a point exists necessarily, let me suppose for contradiction that it does not exist : for every $z\in X$, a nonzero multiple of $z$ is in the subgroup of $CH\_0(X)$ generated by $y,x\_0,\dots,x\_i$. Choose a curve $C$ through $y,x\_0,\dots,x\_i$. Our hypothesis is that every $z\in X$ has a nonzero multiple in the image of $Pic(C)\to CH\_0(X)$. Since $Pic^0(C)$ is divisible and $CH\_0^0(X)$ has no torsion by a theorem of Roitman [The torsion of the group of 0-cycles modulo rational equivalence], this implies that $Pic^0(C)\to CH\_0^0(X)$ is surjective. This contradicts a theorem of Mumford [Rational equivalence of $0$-cycles on surfaces] stating that $CH\_0^0(X)$ is infinite-dimensional.


Now, if $C\_1, C\_2$ are two curves on $X$, their intersection is a multiple of $y$ in $CH\_0(X)$, and cannot be supported on the $x\_i$ because $y$ and the $x\_i$ have been chosen independent in $CH\_0(X)$.


