Let $A$ be a commutative ring with $1_A$. Let $X= \mathrm{Spec}(A)$ be a topological space equipped with the Zariski topology. Let $U \subseteq X$ be an open set and choose $x_1, \ldots, x_n \in U$. Prove that there exists $f \in A$ such that $$\{ x_1, \ldots, x_n \} \subset D(f) \subseteq U.$$
I'm stuck with proving the existence of such $f$. Any help will be appreciated.
Note: For $x \in X=\mathrm{Spec}(A)$, we shall write an ideal with $j_x$. Then $$V(f)= \{x \in X : f \in j_x\} \ \text{and} \ D(f)=X-V(f).$$ Further, we know that $D(f)$ forms a basis for the open sets of $X$.
Best Answer
The key is to use the prime avoidance lemma, which says that if an ideal $I$ is contained in a finite union of prime ideals, then it is contained in one of them.
Let $I$ be an ideal in $A$ such that $U=X\setminus V(I)$. By definition, $x_1,\ldots,x_n$ are primes of $A$ that don't contain $I$, since they are not in $V(I)$. Then we use prime avoidance to choose $$f\in I\setminus \bigcup_{i=1}^n x_i$$
Then $f\in I$, so if a prime contains $I$, it contains $f$. Hence $V(f)\supseteq V(I)$, so $D(f)\subseteq U$. More importantly, $f\not\in x_i$ for any $i$, so $x_i\not\in V(f)$, which means that $x_i\in D(f)$ for all $i$, as desired.