Question

Show that if A andBare sets|A| = |B|then P (A) = P (B) .

Short answer

If g is one-to-one and onto, g is one-to-one correspondence from P (A) to P (B) is P (A) = P (B) .

Step-by-step solution

Definition 

The power set of S is the set of all subsets of S .

Notation: P (s)

The function is onto if and only if for every element $b\in B$there exist an element $a\in A\text{such that}f\left(a\right)=b.\mid$

The function f is one-to-one if and only if (a) = f (b) implies that a = b for all a and b in the domain.

f is a one-to-one correspondence if and only if is one-to-one and onto.

Definition (1): |A| = |B| if and only if there is a one-to-one correspondence from A to B.

Proof

Given: A and B are sets with |A| = |B|

To proof: |P (A)| = |P (B)|

By definition : There exists a one-to-one correspondence f from A to B.

We can then define the function g as $C\in P\left(A\right)$and f (C) is the image of the subset with f (C)$\in$P (B) :

$g:P\left(A\right)\to P\left(B\right),g\left(C\right)=f\left(C\right)$

g is one-to-one: Let g (C) = g (D) , by definition of g : f (C) = g (D) which implies C = D since F is a one-to-one correspondence and thu g is one-to-one .

g is onto: Let $D\in P$(B) . Since f is a one-to-one correspondence, for each element c $\in$A there exists an element $c\in A$ such that f (C) = D and thus f(C) = D with C

the set contain all element c for which f (c) = d We have thus then shown that is onto.

Since g is one-to-one and onto, g is one-to-one correspondence from P (A) to P (B) .

By definition (1) : P (A) = P (B)