RIGOROUS PROOF PLS If S Nn and x NotElement S show that S x

RIGOROUS PROOF PLS

If S N_n and x NotElement S, show that S {x} N_n +1.

Solution

Let f be a map from S to N_n. As S is equivalent to N_n. Each member of S has a unique association with an element in N_n. Here, |S| = n. Therefore, the mapping is bijective.

The inclusion of a new element ie. S U {x} has the cardinality n+1, i.e. |S U {x}| = n+1. The element x can not be associated with any element of N_n since x is not in S. Thus, we have a new member to which x is associated. Without loss of generality, let us associate it with n+1 in N_n+1. Now extended map f\' from S U {x} to N_n+1 will be bijective.

Thus, S U {x} ~ N_n+1