Monotone threshold activation ($v\text{ activates when }|N(v)\cap A|\ge t_v$)

In a finite graph, start with an active vertex set $A_0$ and thresholds $t_v$. An inactive vertex becomes active once sufficiently many neighbours are active, and activated vertices remain active. Add eligible vertices until no more can be added. If $C$ is any threshold-closed set containing $A_0$, induction shows every activated vertex lies in $C$. The terminal active set is itself closed, so it is the least such closure and is independent of the eligible-vertex processing order.

For integer thresholds, compute active-neighbour counts and keep a work queue of vertices newly reaching their thresholds. Each vertex is queued once and each edge is inspected a constant number of times, giving $O(|V|+|E|)$ running time. Strict-majority activation has $t_v=\lfloor \mathrm{deg}(v)/2\rfloor +1$, so an isolated inactive vertex never qualifies. A maximum priority queue of activation margins is an alternative implementation.

Allow individual vertices to switch in either direction according to a fixed neighbour threshold, while the objective is to maximize the number in one designated state. Let $C$ be the monotone threshold activation closure of the initially designated vertices. At every time in any legal switching sequence, the designated set remains a subset of $C$: a loss preserves inclusion, while a gain outside $C$ is impossible because all currently designated neighbours lie in $C$ and the outside vertex has too few neighbours there. This induction proves an upper bound of $|C|$ on every attainable designated population. The monotone sequence attains $C$ itself and is therefore globally optimal. Temporary losses cannot improve this objective.