8/10/2019 On Aoc First Forms

1/2

09/21/14: Some tidbits on the forms of the axioms of choice that we have proven so far...

AC0. If f is a function from A to B and fis surjective, then there exists g a function s.t. f g = IB.

ACI. For any R a relation, there exists g a function s.t. g R and domg = dom R.

ACII. For any set I, if H is a function s.t. dom H= I, and H(i) =/ for each i I,Q

iIH(i) =/ , whereQiI

H(i) = {f:fa function with dom F= I and f(i) H(i) for each i I}.

AC0ACI

()Suppose R is a relation. Define f: R dom R by ha, bi 7 a.

Claim 1. fis a function.

Proof. Suppose hha, bi, a0i, hha, bi, a00i f. Then note that f(ha, bi) = a = a0, and that f(ha, bi) = a = a00, sothat a 0= a00. X

Claim 2. f is surjective.

Proof. Suppose thata dom A. Then by definition, there exists t s.t. ha, ti R. Further, f(ha, ti) = a, so fis onto. X

By Axiom of Choice 0, there exists a function g with f g = Idom R.

Now, define h : dom R ran R by a 7 b g(a) = ha, bi.

Claim 3. h is a function.

Proof. Suppose ha, bi and ha, b0i are in h. Then g(a) = ha, bi and g(a) = ha, b0i, but g is a function. Hence,ha, bi = ha, b0i, which implies that b = b0. X

Claim 4. h R and dom h = dom R.

Proof. h dom R ran R implies that h R. Further, by construction, dom h = dom R. X

his the witness to the existence claim we wanted.

() Suppose f is an onto function from A to B. Then note that f1 is a relation, so that by AC1, thereexists a function g f1 with domg = domf1 = ranf=since f is onto B.

Claim 1. f g = IB.

()Supposehx, y i f g. Then note that there existstwith hx, ti gand ht, y i f. Sinceg f1, ht, xi f.However, f is a function, so that x =y. Further, f g domg ran f= B B. Therefore,hx, y i IB.

() Suppose hb, b i IB. Since b domg, there exists t with hb, t i g. However, since g f1, ht, b i f.Then by definition,hb, bi f g. X

g is the witness to the existence claim that we wanted. This completes the proof.

//

ACI AC2

() Suppose Iis a set and that H is a function with dom H= I, and that for each i I, H(i) =/ . Definea relation R with dom R = I, so that hi, y i R i I and y H(i).

By ACI, there exists a function f R with domf= dom R.

Claim 1. domf= I.

Proof. By construction, domf= dom R = I. X

1

8/10/2019 On Aoc First Forms

2/2

Claim 2. f(i) H(i)for each i I.

Proof. By construction, hi, y i f i I and y = f(i) H(i). X

Hence, fQ

iIH(i), so that

QiI

H(i) =/ .

() Suppose R is a relation. Call domR= I. Define H with domH=I and ranH= (ranR), so thati 7 {y : hi, y i R}.

Claim 1. His a function.

Proof. Suppose hi, ti and hi, t i are in H.

This implies that t= {a: hi, ai R } and that t ={b: hi, bi R }. However, these sets are both equal to

R[[{i}]] = R {i} = {t ran R : x {i}s.t. hx, ti R}, so that t = t . X

Claim 2. H(i) =/ for each i I.

Proof. Supposei I. Then note that sinceI= dom R, there existst withhi, ti R, so that H(i) = {t} =/ . X

By ACII, there exists a function f with domf= I= dom R, and that f(i) H(i)for each i I.

Claim 3. f R.

Proof. Supposehi, y i f. Then note thati I= dom R, and y = f(i) H(i) (ran R), so that H(i) ran R,which shows that f(i) ran R. It follows that f R.

Claim 4. domf= dom R.

Proof. By construction, domf= dom R = I.

Then fis a witness to our existence claim. This completes the proof.

2