Open and Closed Set in Zariski Topology I'm confused about the definition closed and open set in Zariski Topology, it is said that the set $$V(I)=\{P \in \operatorname{Spec}(R)\mid I \subseteq P\}$$ are the closed set in Zariski Topology. But it is said in James Munkres's Topology that a subset $U$ of $X$ is an open set of $X$ if $U$ belong to the collection $\tau$. So assuming that $V(I)$ is the closed set of Zariski Topology on $\operatorname{Spec}(R)$, shouldn't the collection of $D(r)$ in which $$D(r)=\{P \in \operatorname{Spec}(R)\mid r \notin P\}$$ are the topology $\tau$ in $\operatorname{Spec}(R)$?
But, in a lecture notes about Zariski Topology www.math.kth.se/~laksov/courses/algebradr01/notes/rings5.pdf‎ , proposition 5.3 to be precise, the one that is proved to be the topology $\tau$ is the collection $V(I)$ instead. Could somebody explain this to me? Thank you.

What's proved is that arbitrary intersections and finite unions of such sets are again sets of that form. This is exactly dual to the requirements of a topology - and so the complements of sets $V(I)$ are the open sets in the Zariski topology.
Complements of $V(I)$ are not generally of the form $D(r)$, but they are unions of such sets - so we call the collection of $D(r)$ a basis of the topology.
