Answer to Question #132140 in Real Analysis for VIRENDRA BALKI

Question

State & Prove cantor's

theorem.

Answer 1

CANTOR'S THEOREM-Definition:

The cardinality of the set of all subsets of any set is strictly greater than the cardinality of the set; i.e., for any set A,|P(A)| > |(A)|Proof:

To prove the theorem we must show that there is a one-to-one correspondence between A and a subset of P(A) but not vice versa.

Let the function f : A→P(A) defined by f(a)={a} is one-to-one into P(A).

Thus, $|P(A)|\ge|(A)|$

To prove that the |P(A)| $\ne$ |(A)| let us assume there were a one-to-one onto mapping between A and P(A), say f : A→P(A).

There are some elements of A which map into subsets of A of which they are a member and there are some which map into a subset which they are not a member of.

Let N be the set of elements of A that do not map into a subset they are member of; i.e., N = {x∈A and x∉f(x)}

Since N is a subset of A and f is one-to-one onto there must be an element z such that N=f(z), which sets up a contradiction.

If z belongs to N it cannot belong to N. If it does not belong to N then it must belong to N.

Therefore the assumption of the existence of a one-to-one onto function between A and P(A) leads to a contradiction and therefore must be false.

Thus, for any set A, |P(A)| > |(A)|

Thus for any cardinality there is another cardinal of a higher order.