I'm aware that, for an arbitrary topology $\tau\subset \mathscr{P}[U]$ on a set $U$, the collection $\tau^c$ of closed sets is only guaranteed to be closed under finite union.
In the Zariski topology, does this hold? Or does the Nullstellensatz guarantee that $\tau^c$ is closed under arbitrary union (following the observation that the ideal space is closed under arbitrary intersections)?
The context is a problem which parallels this one:
Let $G[1,n]\times \mathbb{P}^n$ be the product of the Grassmannian $G[1,n]$ and projective $n$-space $\mathbb{P}^n$. To every $g\in G[1,n]$ through the origin, associate the unique $\varphi_g\in \mathbb{P}^n$ satisfying $\varphi_g \in g$. Then, for every such $g$ there is a unique element $\alpha_g\;:=\;\big(g,\varphi_g\big)\in G[1,n]\times \mathbb{P}^n$.
Fix $g\in G[1,n]$, then the projection map
$$\big\{g\big\}\times \mathbb{P}^n\overset{p_2}{\longrightarrow} \mathbb{P}^n$$ is a homeomorphism onto its image. For every $g$, $\{\varphi_g\}$ is a closed subset of $\mathbb{P}^n$, so that $p_2^{-1}[\varphi_g]\;=\;\alpha_g$ is closed in the product topology.
Let $L\subset G[1,n]$, then, there is a closed $V\subset G[1,n]\times \mathbb{P}^n$ containing (the preimage of) $L$ which is given by
$$V\;=\;\bigcup_{\ell\in L}\alpha_\ell$$
If the Zariski topology is closed under arbitrary unions, then the argument runs without problem. Otherwise, this is only true when $L$ is countable.
So, is this true?