`The Schroeder-Bernstein theorem states the following: If A and B are any sets with the property that there is a one-to-one function from B to A, then A and B have the same cardinality. Use this theorem to prove that there are as many functions from Z + to { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } as there are functions from Z + to { 0 , 1 } .

To prove:

There are as many functions from Z+to {0,1,2,3,4,5,6,7,8,9} as there are functions from Z+to {0,1}.

Given information:

A and B are one-to-one function.

Concept used:

A function is said to be one-to-one function if the distinct elements in domain must be mapped with distinct elements in co-domain.

A function is onto function if each element in co-domain is mapped with atleast one element in domain.

Proof:

Consider the Schroeder-Bernstein theorem states that:

Let A and B be any two sets and let f and g be two functions, such that f:A→B and g:B→A. If both f and g are one-to-one functions then the cardinality of A and the cardinality of B are equal.

The object is use the Schroeder-Bernstein theorem to prove that there are many functions from Z+to {0,1,2,3,4,5,6,7,8,9} as there are functions from Z+to {0,1}.

Consider a set A, which is the set of all functions from Z+to {0,1,2,3,4,5,6,7,8,9}

Consider a set B, which is the set of all functions from Z+to {0,1}

Let us show that the cardinality of A is equal to the cardinality of B.

Let f1 be any functions from Z+to {0,1,2,3,4,5,6,7,8,9}. Thus fi∈A.

Let us write a number x1 as follows.

xi=0.fi(1)fi(2)fi(3)......fi(n).....

For example, if the function fi maps 1 to 5,  2 to 8,  3 to 0,  4 to 7, etc. Then

xi=0.5807........

Thus, for any i:0≤xi≤1