Spanned divisors and Base Points Let $X$ be a smooth algebraic variety. 
We say that a line bundle $\xi\in H^1(X,\mathcal{O}^\ast)$ is spanned is for each $x\in X$ there is a global section $s\in H^0(X,\mathcal{O}(\xi))$ with $s(x)\neq 0$.
Let $\xi=[D]$ be the line bundle associated to a divisor $D$. If $\xi$ is spanned, why can we say that the linear system $|D|$ is base-point free?
Basically for each $x$ I want to find $D'\in|D|$ such that $x\notin D'$.

Assume that the line bundle $\mathscr{O}(D)$ is spanned. So given $x \in X$, there exists a section $s \in H^0(X, \mathscr{O}(D))$ such that $s(x) \neq 0$. Let $D'$ denote the divisor of zeroes of the section $s$; then $x$ is not contained in $\mathrm{Supp}(D')$. 
The remaining issue is to show that $D'$ is linearly equivalent to $D$. To do this, we need to show there's a rational function $f$ on $X$ such that $\mathrm{div} f = D'-D$. But there's a correspondence (explained in detail e.g. in Shafarevich Volume 2, VI.1.4) between global sections of the bundle $\mathscr{O}(D)$ and rational functions $f$ on $X$ such that $\mathrm{div} f + D \geq 0$; moreover, if $s \in H^0(X,\mathscr{O}(D))$ has zero-set $D'$, and $s$ corresponds to a rational function $f$, then $D'=\mathrm{div} f + D$. 
So our chosen section $s$ from the first paragraph yields a rational function $f$ such that $D'-D=\mathrm{div} f$, which means the two divisors are linearly equivalent, as we wanted. 
