Finding Bijection Proof: X to p(X)

[bedi]

Homework Statement

Let X be a set. Suppose that f is a bijection from p(X) to p(X) such that $f(A)\subseteq f(B)$ iff $A\subseteq B$ for all subsets A,B of X.
Show that there is a bijection g from X to X such that for all $A\subseteq X$ one has f(A)=g(A).

Homework Equations

p(X) is the power set of X.

The Attempt at a Solution

This seems too elementary and I doubt that there is something to prove. Can't I just take f=g?

 

[micromass]

Homework Statement

Let X be a set. Suppose that f is a bijection from p(X) to p(X) such that $f(A)\subseteq f(B)$ iff $A\subseteq B$ for all subsets A,B of X.
Show that there is a bijection g from X to X such that for all $A\subseteq X$ one has f(A)=g(A).

Homework Equations

p(X) is the power set of X.

The Attempt at a Solution

This seems too elementary and I doubt that there is something to prove. Can't I just take f=g?

But f is a function from P(X) to P(X). And g is a function from X to X. So taking f=g makes no sense. The arguments of f should be subsets of X. The arguments of g should be elements of X.

 

[bedi]

I can't believe I didn't see it:D thanks

 

[hedipaldi]

Define g(x) as the unique element satisfying {g(x)}=f{x}.show that g(A)=U{g(a)}=Uf{a}=
f(A)