Math Is Fun Forum

ziabing

Member

Registered: 2021-09-18

Posts: 6

Proving that a function is bijective + finding its inverse function.

Hello!

I'm trying to prove that the following function is bijective and find its inverse function.

Let X be a set and A a subset of X.

f : P(X) → P(X)
    A →  the complement of set A

It's the first time I'm asked to do this with power sets and sets, so I have no idea of what I'm supposed to do.

All help is much appreciated!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,631

Re: Proving that a function is bijective + finding its inverse function.

hi ziabing

Sorry this didn't get answered ages ago.  I found the post again when looking at your recent post.

A power set is a set of sets.  It's members are all the subsets (including the empty set and the whole set) .

For every subset there is a complement set. 

eg.  If X = {w,x,y,z} and if  A = {x,y} is a subset and A' = {w,z} is it's complement.

The mapping maps every such subset to its complement.  So if you take any member of the complement set (A') let's say {x,y,z} ,there exists a member of A that maps onto it; in this example {w}

So the function is surjective.

Similarly if A =  {w} is a set in X then it maps onto {x,y,z} in the complement set so it is injective.

The inverse function is the same function again as (A')' = A

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob