Sections $s$ such that "${s(x) \neq 0 }$ is affine" is a vector subspace Let $X$ be a scheme, $\mathcal{L}$ be a line bundle on $X$.
For section $s \in H^0 (X, \mathcal{L})$ denote by $X_s$ an open subscheme of $X$ defined by $s \neq 0$.
Finally denote by $A =  \{ s \in H^0 ( X, \mathcal{L} )$ such that $X_s$ is affine}.
Question Why $A$ is a vector subspace of $H^0 (X, \mathcal{L})$ ?
 A: For a vector space structure, you need a field, so I guess $X$ is a scheme over some field. But this does not really touch us, since $A$ is obviously closed under scalar multiplication. The interesting part is closedness under addition.
Use Exercise 2.17 in Hartshorne: 

A scheme $S$ is affine if and only if there is a finite set of global
  sections $f_1, \dotsc, f_r \in H^0(S,\mathcal O_S)$, that generate the
  unit ideal and each $S_{f_i}$ is affine.

Let $s,t \in H^0(X,\mathcal L)$ be global sections, such that $X_s$ and $X_t$ are affine.
Note that $$\mathcal O_{X|X_{s+t}} \xrightarrow{\cong} \mathcal L_{|X_{s+t}}, 1 \mapsto s+t$$, hence the ideal generated by $s_{|X_{s+t}}$ and $t_{|X_{s+t}}$ is the unit ideal in $H^0(X_{s+t},\mathcal O_{X|X_{s+t}})$.
We are left to show that $(X_{s+t})_s = X_s \cap X_{s+t}$ is affine ($(X_{s+t})_t$ is of course dealt with the same way). But this is clear, since we have $(X_{s+t})_s = X_s \cap X_{s+t}=(X_{s})_{s+t}$ and the latter is a principal open subscheme of an affine scheme, hence affine.
