Prove that the Zariski space $\text{Zar} \space (K,A)$ is compact. I posted part of the proof from Matsumura's Commutative Ring Theory. I got stuck in the last sentence where it says "Hence the intersection of all the elements of $\mathcal{A}$ is the same thing as the intersection of the sets of the form $U(x)^c$ belonging to $\mathcal{A}$. I am not sure why this is true...
Note: $U(x_1, ... , x_n) := \text{Zar} \space (K, A[x_1, ... , x_n])$, where $A$ is a subring of the field $K$.


I'm not sure why the last sentence is true and would appreciate any kind of help...
 A: It seems the following.
I don’t perfectly understand the algebraic constructions from the proof, but it is not necessary, because
its topological constructions are clear. :-) In fact, by Alexander subbase theorem, a space $X$ is compact iff it has a subbase $\mathcal B$ that that each cover of $X$ by elements of $\mathcal B$ has a finite subcover.
So, let $F\in\mathcal A$. Then $F^c=\bigcup_\lambda U_\lambda$ for some family $U_\lambda$ of (canonical?) open sets, and hence $F=\bigcap_\lambda U^c_\lambda$. By Property $\gamma$, $ U^c_\lambda\in\mathcal A$ for each $\lambda$. Moreover, $U^c_\lambda=U(x_1,\dots,x_n)^c=\bigcup_{i=1}^n U(x_i)^c$ for some $x_1,\dots, x_n$. By Property $\beta$ (which follows from the maximality of the family $\mathcal A$), one of the $U(x_i)^c$ must belong to $\mathcal A$.  Denote this set $U(x_i)^c$ as $G_\lambda$. Then $F=\bigcap_\lambda U^c_\lambda\supset \bigcap_\lambda G_\lambda$. Therefore each set $F\in\mathcal A$ contains an intersection ($\bigcap_\lambda G_\lambda$) of the sets of the form $U(x)^c$  belonging to $\mathcal A$.
