The set $C(X)$ of all continuous, real-value functions no a topological space $X$ will be provided with an algebraic structure and order structure.
zero-set means:
$Z(f) = Z_{X} (f) = \{ x \in X : f(x) = 0 \} \quad ( f \in C(X) )$
$Z(X)= Z[C(X)]=\{ Z(f) : f \in C(X) \}$
I know that $Z(X)$ closed under countable intersection and also $Z(X)$ need not be closed under infinite union.
1: In a general space, Do countable or finite union of zero-sets be a zero-set?Why?
2:Is $Z(X)$ closed under arbitrary intersection?Why?
$Z(f)$ is closed under finite unions : $Z(f) \cup Z(g) = Z(fg)$ plus induction.
In general not closed under countable unions: singletons are in $Z(\mathbb{R})$ but $\mathbb{Q}$ is not.
Also $Z(X)$ will not be closed under arbitary intersections, but you need some non-metrisable counterexample (in metric spaces $Z(X)$ is just the collection of all closed sets which is closed under all intersections). Think (or search) about a counterexample with some closed set that is not a zero-set.