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.
[Math] Open and Closed Set in Zariski Topology
algebraic-geometry
Best Answer
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.