# Equipotence and non-empty sets

1. Sep 21, 2007

### jetoso

Hello,
I am trying to prove the following about equipotence:
Let A and B be nonempty sets. We say that A is equipotent with B if there is a bijection between A and B. Then the following hold:
(i) A is equipotent with itself.
(ii) If A is equipotent with B, then B is equipotent with A.
(iii) If A is equipotent with B, and B is equipotent with C, then A is equipotent with C.

Proof:
(i) We can use the identity function Id_A which gives a bijection between A and itself. Shall I need a more formal proof here?

(ii) Let f: A -> B be a 1-1 and onto map. We can use the inverse function f^-1 which will give a bijection between B and A. Same question here, how to give formal proof.

(iii) Let f:A->B, g:B->C be 1-1 and onto. Then the composition h=(g o f) will give a 1-1 and onto map from A onto C. How do you give a formal proof of this?

Hope you guys have some suggestions.

2. Sep 21, 2007

### grossgermany

i. define the trivial function f(A)=A
ii. if f(a)=b then f-1(b)=a
iii. if f(a)=b g(b)=c then f.g(a)=c

3. Oct 10, 2007

### jetoso

Thanks

4. Oct 10, 2007

### matt grime

I do'nt see what you consider to be 'not formal' about the first two. A function is bijective if and only if it is invertible. Thus (i) and (ii) are trivialities - the identity is invertible, and the inverse of an invertible function is invertible.

The third is again simple - if f and g are invertible then so is fg, and to prove so you just write down the inverse and explain why it satisfies the definition of 'inverse'.