 Monotone threshold activation

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.