Explore BrainMass

# Prove that there is no bijection between any set A and its power set P(A) of A.

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

There is no bijection between any set A and its power set P(A) of A.

For finite sets, proof is trivial since |A| = n and |P(A)| = 2^n. For finite sets, this is done by contradiction. Suppose there is a bijection \$ between a set A and its power set P(A). Consider the set B={x|x is a member A where x is not a member \$(x)}For each element x A, since \$ is a function from A to the power set of A, &(x) ia a subset of A. By our earlier assumption on set theory that every element is either in a set or ot in a set, for each x A, w can certainly ask if x is a member of \$(x). Therefore, the set B is well definedB. We know B is a subset of A and \$ is a bijection from A to its power set P(A). For this subset B of A, there exists an element y is a member of A such that \$(y)=B.

Where is y? is y a member of B ? or is y is not a member of B? in both cases we will obtain a contradiction. Therefore, our earlier claim that there is a bijection between A and its power set P(A) is false.

https://brainmass.com/math/basic-algebra/prove-that-there-is-no-bijection-between-any-set-a-and-its-power-set-p-a-of-a-24567

#### Solution Preview

Here I want to clarify some points of the proof.

1. \$ is a map from A to its power set P(A). According to the assumption, \$ is a bijection. This means, for each element B in P(A), we can find an element y in A, such that \$(y)=B. B is a subset of A and B is also an element of P(A). y is an ...

#### Solution Summary

It is proven that there is no bijection between any set A and its power set P(A) of A.

\$2.49