Uniformity of injectivity for maps associated to linear systems Let $X$ be a compact complex manifold and $L\to X$ a holomorphic line bundle (without any a priori assumption on its positivity). 
Suppose that for each $x,y\in X$, with $x\ne y$, there exists a $k_0\in\mathbb N$ which depends on $x$ and $y$, such that for all $k\in\mathbb N$, $k\ge k_0$ there exists a global holomorphic section $\sigma_k\in H^0(X,L^{\otimes k})$ such that $\sigma_k(x)=0$ and $\sigma_k(y)\ne 0$.
Is it then true that this property is uniform with respect to couples of distinct points of $X$?
In other words, is it then true that there exists a $k_1\in\mathbb N$ such that for all $k\in\mathbb N$, $k\ge k_1$, and for all $x,y\in X$, with $x\ne y$, one can find a global holomorphic section $\tau_k\in H^0(X,L^{\otimes k})$ such that $\tau_k(x)=0$ and $\tau_k(y)\ne 0$?
Thanks in advance!
 A: I think this is true.
The condition implies that for some $n$ the sections of $L^{\otimes n}$ are base point free, so $L^{\otimes n}$ is obtained by pulling back $\mathcal O(1)$ along a map $X \to \mathbb P^N$ for some $N$. This map must be finite, because otherwise there would be a positive dimensional connected subspace of $X$ along which $L^{\otimes n}$ would be trivial, and then its point could not be separated by any power of $L$. By GAGA $X$ is projective, and then $L^{\otimes n}$ is ample (it is a standard result that a pullback of an ample line bundle along a finite map is ample). But this implies that $L$ is ample, so all sufficiently large powers of $L$ must be very ample.
[Addendum]: I guess I am using three facts: that the image $Y$ of $X$ in $\mathbb P^N$ is closed, hence projective, by Chow's theorem; that given a finite map $X \to Y$, if $Y$ is algebraic so is $X$, by GAGA; that the the pullback of an ample line bundle along a finite map is still ample (which implies that $X$ is projective).
A: Maybe I am making a mistake, but you could try the following.  First, for each $k$ the set $U_k$ of points $x$ for which there is a section of $L^k$ not vanishing at $x$ is open.  Moreover, the union of all the $U_k$ is all of $X$, by assumption (case of 0-dimensional $X$ is easy!).  Thus, by compactness, there is a $k_0$ such that the line bundle $L^{k_0}$ is base point free.  Now use $L^{k_0}$ to give a map to projective space and reduce to the projective case.  If you care about the "all greater multiples" business, you can build it in this argument, I believe.
