# Finding a bijection

1. Nov 16, 2012

### bedi

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

p(X) is the power set of X.

3. 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?

2. Nov 16, 2012

### micromass

Staff Emeritus
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.

3. Nov 16, 2012

### bedi

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

4. Nov 16, 2012

### hedipaldi

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