Жиын тең өлшемді екенін дәлелдеу әдісі: Биективті дәлелдеме

Technique for proving sets have equal size
In combinatorics, bijective proof is a proof technique for proving that two sets have equally many elements, or that the sets in two combinatorial classes have equal size, by finding a bijective function that maps one set one to one onto the other. This technique can be useful as a way of finding a formula for the number of elements of certain sets, by corresponding them with other sets that are easier to count. Additionally, the nature of the bijection itself often provides powerful insights into each or both of the sets.

The key idea of the proof may be understood from a simple example: selecting k children to be rewarded with ice cream cones, out of a group of n children, has exactly the same effect as choosing instead the n − k children to be denied ice cream cones. More abstractly and generally, the two quantities asserted to be equal count the subsets of size k and n − k, respectively, of any n element set S. Let A be the set of all k element subsets of S, the set A has size Let B be the set of all n−k subsets of S, the set B has size There is a simple bijection between the two sets A and B: it associates every k element subset (that is, a member of A) with its complement, which contains precisely the remaining n − k elements of S, and hence is a member of B. More formally, this can be written using functional notation as, f : A → B defined by 1=f(X) = X^(c) for X any k element subset of S and the complement taken in S. To show that f is a bijection, first assume that 1=f(X1) = f(X2), that is to say, 1=X1^(c) = X2^(c). Take the complements of each side (in S), using the fact that the complement of a complement of a set is the original set, to obtain 1=X1 = X2. This shows that f is one to one. Now take any n−k element subset of S in B, say Y. Its complement in S, Y^(c), is a k element subset, and so, an element of A. Since 1=f(Y^(c)) = (Y^(c))^(c) = Y, f is also onto and thus a bijection. The result now follows since the existence of a bijection between these finite sets shows that they have the same size, that is, .