This question is from Mumford's The Red Book of Varieties and Schemes (Section I.4).
Let $X\subseteq k^n$ be an irreducible algebraic set, $R$ its affine coordinate ring. Since $X$ is irreducible, $I(X)$ is prime and $R$ is an integral domain. Let $K$ be its field of fractions.
Let $\underline{o}_x=\{f/g\mid f, g\in R, g(x)\neq 0\}\subseteq K$. Now for $U$ open in $X$, let $$\underline{o}_X(U)=\bigcap_{x\in U}\underline{o}_x.$$
Let $X_f=\{x\in X\mid f(x)\neq 0\}$.
The author said: Since the sets $X_f$ are a basis of the Zaiski topology of $X$, we have $$\varinjlim_{x\in U}\underline{o}_X(U)=\varinjlim_{x\in X_f}\underline{o}_X(X_f).$$
My Question: Why? I think the first part I have to prove is that $$\bigcup_{x\in U}\underline{o}_X(U)=\bigcup_{x\in X_f}\underline{o}_X(X_f).$$ But I cannot prove it.
The sets $X_f$ are cofinal in the poset of open sets of $X$ containing $x$ (this follows from the $X_f$ being a basis for the topology of $X$). Since stalks are calculated as colimits, it suffices to calculate along any cofinal collection - which is exactly the claim the author makes.