Lemma 27.26.6. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume the open sets $X_ s$, where $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ and $n \geq 1$, form a basis for the topology on $X$. Then among those opens, the open sets $X_ s$ which are affine form a basis for the topology on $X$.

Proof. Let $x \in X$. Choose an affine open neighbourhood $\mathop{\mathrm{Spec}}(R) = U \subset X$ of $x$. By assumption, there exists a $n \geq 1$ and a $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that $X_ s \subset U$. By Lemma 27.26.4 above the intersection $X_ s = U \cap X_ s$ is affine. Since $U$ can be chosen arbitrarily small we win. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).