# Convergent sequences in projective varieties

It's very well known that if $$X$$ is an irreducible projective variety (feel free to assume that the base-field is $$\mathbb{C}$$), then any two points $$x,y\in X$$ can be connected by the image of a non-singular curve. An outline of the proof can be found here. Anyway, the stardard reference for the proof is Abelian Varieties-Mumford, Lemma at pg.56. I'm wondering if this result can be adapted in such a way to give a characterization of convergent sequences in $$X$$. More precisely, motivated also by this question, I would prove a result like this:

(RS) Rough Statement: Let $$(x_n)\subset X$$ be a sequence of points in $$X$$. Then $$x_n\to x\in X$$ if and only if there exists a non-singular curve $$C$$ together with a morphism $$f\colon C\longrightarrow X$$ and sequence $$(c_n)\subset C$$ converging to $$c\in C$$ such that $$x_n=f(c_n)$$ for any $$n$$ and $$f(c)=x$$.

Rough Idea of the Proof: Any two points $$x_n$$ and $$x_{n+1}$$ can be connected by the image of a map $$f_n\colon C_n\longrightarrow X$$, where $$C_n$$ is a non-singular curve. Then the first thing coming in mind is taking $$C'_n:=\cup_{i=1}^nC_i$$ and considering the closure (in the Zariski topology) of the limit $$C:=\overline{\lim_n C'_n }.$$

Of course issues could arise in this construction, for example $$C$$ can no lo longer be a curve. Therefore I think that in order to check that $$C$$ is actually a curve (if it is then we get our non-singular curve by taking its normalization) we have to refine the construction of the curves $$C_n$$. It's not clear to me how we can proceed, and especially, if (RS) is true. Any constructive comment or answer is well accepted.

• I don't think this is true. Let $X$ be a projective curve which is smooth except for one double point $s$; let $C$ be its normalization, and let $p,q$ be the two points of $C$ above $s$. Now let $(x_{2n})$ be a sequence in $C$ converging to $p$, and $(x_{2n+1})$ a sequence converging to $q$. The image in $X$ of the sequence $(x_n)$ converges to $s$, but there is no way to lift it to a convergent sequence in $C$. – abx Nov 4 '18 at 16:55
• Another example is to let $X$ be a plane, take a sequence of lines $L_m$ through $x$, and choose a sequence of points $x_{m,n}$ on each $L_m$. Then play Hilbert hotel–reorder the $x_{m,n}$s to get a single sequence $x_n \to x$, and visiting each $L_m$ infinitely often. That makes it impossible for the $x_n$s (or $x_{m,n}$s) to lie on a curve. Really, they're dense in the plane. – Zach Teitler Nov 4 '18 at 18:45
• Very nice example @ZachTeitler! – Vincenzo Zaccaro Nov 4 '18 at 23:34
• I'm wondering what, if any, statement could be salvaged from this. I think it's not even true that there has to exist a curve $C$ passing through infinitely many of the $x_n$'s and through $x$. In the plane with $x=(0,0)$ let $x_n = (1/n, \exp(1/n)-1)$. I guess the condition $x_n \to x$ in the Zariski topology is just too weak to get much out of it, in terms of curves or anything. Perhaps not too surprising, Zariski topology is non-Hausdorff (typically) and so saying $(x_n)$ converges doesn't even determine the limit $x$, let alone any curves... – Zach Teitler Nov 5 '18 at 0:10