Proof] 1-1 and Onto: f(A^c) = [f(A)]^c

[second]

Homework Statement

Show that f : X ! Y is 1-1 and onto if and only if for each set A $\subset$ X, f(A^c) = [f(A)]^c. ...c is a complement ...

Homework Equations

proof by contradiction.
if f is not 1-1,there are x and y with f(x)=f(y)

The Attempt at a Solution

I was thinking if f is not onto ...then i can say A=X...BUT i can't figure it out..i can explain one to one or onto..i am just having hard time here.

[

 

[lippyka]

Solution]Suppose f is 1-1 and onto. Then for any set A \subset X, we have that f(A) = f(A^c)^c. Suppose this is false. Then there exists some set A \subset X such that f(A) \neq f(A^c)^c. Since f is onto, there exists x \in A^c such that f(x) \in f(A). Thus, since f is 1-1, there exists y \in A such that f(y) = f(x). This contradicts the fact that f is 1-1. Conversely, suppose that for all sets A \subset X, f(A) = f(A^c)^c. We will prove that f is 1-1 and onto.First, we will prove that f is 1-1. Suppose not, then there exist x,y \in X such that f(x) = f(y) but x \neq y. Let A = \{ x \}. Then f(A) = \{ f(x) \} while f(A^c) = \{ f(y) \}. This contradicts our assumption that f(A) = f(A^c)^c. Next, we will prove that f is onto. Let y \in Y be arbitrary. Let A = f^{-1}(\{ y \}^c). Then f(A^c) = f(f^{-1}(\{ y \}^c))^c = \{ y \}^c^c = \{ y \}. Thus, by our assumption, f(A) = \{ y \}^c. Since f(A) = \{ y \}^c, there exists x \in A such that f(x) = y. Thus, f is onto.