Let $((X_k,\tau_k))_{k \in N}$ be topological spaces. Define $X = \prod_{k \in N} X_k$, $\tau$ the box topology on $X$.
If $S_k^c \in \tau_k$ for all $k \in N$, is the set $\prod_{k \in N} S_k$ closed in $(X,\tau)$?
This holds for finite $N$ by induction. It is a key step in proving the closure of product in box topology is equal to the product of closure.
The box topology is finer than the product topology, so every set that is cloesd in the product topology is closed in the box topology.
For either topology, the projection maps $\pi_k \colon X \to X_k$ are continuous, so if $S_k \subseteq X_k$ is closed for all $k$, then $$S = \bigcap_k \pi_k^{-1} (S_k)$$ is closed in $X$.