Hello all,

I am currently reading a book on set theory "Joy of Sets" by Keith Delvin. In page 58 he proves a lemma which is stepping stone for proving that axiom of choice(AC) and 'Every set can be well ordered' (WO) are equivalent. The lemma goes something like this

Now, I took an example of A = {x,y,z}, now(AC') Every set of nonempty sets has a choice function

Assume AC'. Let A by any set. Then there is a functionf:P(A) -> A U {A} such that

1.f(A) = A

2.f(X) belongs to A-X, whenever X is subset of A

(P(A) means power set of A, and symbol U means union)

Proof: Let

B = {A-X| X is subset of A}

By AC', let g be a choice function for B. Thusg:B -> U B andg(Y)belongs to Y for all Y belongs to B. Define

f:P(A) -> A U {A}

by

f(A) = A,

f(X) =g(A-X), if X is subset of A.

Clearly, f is as required

AU{A} is {x,y,x,{x,y,z}}

andP(A) is {{},{x},{y},{y},{z},{x,y},{y,z},{z,x},{x,y,z}}

I understandf(X) can be a result from choice functiong(A-X), but I dont understand howf(A) = A.

Also, is the methodology applied to prove the lemma is correct?