I need someone to sanity check my understanding: I believe when we define the closed sets of the Zariski topology as:
$$ C \in \mathbb R^n~\text{is closed} \iff \exists S, f_{s \in S}: \mathbb R^n \rightarrow \mathbb R, ~ \left\{ x \in \mathbb R^n : \forall s \in S, f_s(x) = 0 \right\} = C $$
and that the index set $S$ can be infinite. Is this correct?
I believe that it needs to be infinite to prove that the intersection of two closed sets is closed:
$$ \begin{align*} C &\equiv \{ x \in \mathbb R^n: \forall s \in S, f_s(x) = 0 \} \\ D &\equiv \{ x \in \mathbb R^n: \forall t \in T, g_t(x) = 0 \} \\ C \cap D &\equiv \{ x \in \mathbb R^n: \forall s \in S, f_s(x) = 0 \land \forall t \in T, g_t(x) = 0 \} \end{align*} $$
Our index set for $C \cap D $ is now ranging over $S \cup T$. To have infinite intersection, we might potentially need to allow an infinite number of polynomials.
Is there a deeper reason for allowing the closed sets to have potentially infinite sets of polynomials, than just "we need it to satisfy the axioms of a Topology"?