Twelve matchsticks are laid out to form one large square that is divided into four small squares (a $2\times 2$ arrangement). What is the least number of matchsticks that must be taken away so that no square of any size is left in the figure?

Two removals are not enough. The large outer square can only be broken by taking one of the 8 boundary matchsticks, and each boundary stick belongs to just one small square; that leaves three small squares to be broken by the single remaining removal, but one matchstick lies on at most two small squares. Three removals do work: take the upper and lower arms of the inner cross (these two between them touch all four small squares) and any one boundary stick. The trap is 2 — it destroys all four small squares but leaves the big square standing.