Given a holomorphic line bundle $L$ on a complex manifold $X$, a point $x\in X$ is called a base point of $L$ if $s(x)=0$ for all $s\in H^0(X,L)$ (the space of global holomorphic sections of $L$). The base locus $\operatorname{Bs}(L)$ is the set of all base points of $L$. Given a basis $s_0,\dots,s_N$ of $H^0(X,L)$, one can define a map \begin{equation} \phi_L:X\backslash\operatorname{Bs}(L)\rightarrow \mathbb{CP}^n \end{equation} sending a point $x$ to $[s_0(x):\dots:s_N(x)]$. This is well-defined, since not all $s_i(x)$ are zero (we exclude $\operatorname{Bs}(L)$ in the domain) and since changing trivialisation scales all components $s_i(x)$ by the same amount.
I am told that the line bundle $L$ is called very ample if for any such basis, the associated map $\phi_L$ is an embedding. $L$ is called ample if there exists a positive integer $m_0$ such that $L^m$ is very ample for all $m\geq m_0$.
In Huybrechts' book Complex Geometry, he says that by definition, a compact complex manifold is projective (i.e. embeds as a complex submanifold of $\mathbb{CP}^n$) if and only if it admits an ample line bundle. But I do not understand how this follows? Do we not only get an embedding of $X\backslash \operatorname{Bs}(L)$ into $\mathbb{CP}^n$, rather than the whole of $X$? I would agree with this statement if $L$ was also assumed to be globally generated, i.e. satisfies $\operatorname{Bs}(L)=\emptyset$, but this is not assumed. Any help would be much appreciated!