2.10. Axiom of choice
Properties of curried composition
Proposition. ∀Set E,F,G,
∀f ∈ FE,
- Im f = F ⇒ Inj(GF ∋ g ↦ g⚬f)
- (Inj(GF ∋ g ↦ g⚬f) ∧ ∃≥2:G)
⇒ Im f = F
- (Inj f ∧ G ≠ ∅)
⇒ {g⚬f | g∈GF} = GE
- ({g⚬f | g∈GF} = GE
∧ ∃≥2:G) ⇒ Inj f
- (E ⊂ F ∧
G ≠ ∅) ⇒ {g|E |
g∈GF} = GE
- (Inj f ∧ E ≠ ∅) ⇒ ∃g∈EF,
g⚬f = IdE
Proofs :
1. done in 2.9 ; simpler proof:
∀g,h∈GF,
(∀x∈E, g(f(x)) = h(f(x)))
⇔ (∀y∈F, g(y) = h(y)) ⇔ g=h
2. ( ) ⇒ ∃z≠z′∈G, (F∋y ↦ z)⚬f =
(F∋y ↦ (y ∈ Im f ? z : z′))⚬f
∴ (∀y∈F, y ∈ Im f ∨ z = z′) ∴
Im f = F
3. Inj f ⇒ ∀z∈G, ∀h∈GE, h =
(F ∋ y ↦ (y ∈ Im f ?
h(f -1(y)) : z))⚬f
4. ∀z≠z′∈G, ∀x∈E, ∀g∈GF,
g⚬f = (E∋y↦
(y = x ? z : z′)) ⇒ ∀y∈E, f(y) = f(x) ⇒ g(f(y))
= g(f(x)) = z ⇒ y = x.
5. and 6. are particular cases of 3.∎
The case G = V2 gives the following properties of h =
(℘(F)∋B↦ f⋆(B)) :
Im f = F ⇔ Inj h
Inj f ⇔ Im h
= ℘(E)
Moreover {f[A]|A ⊂ E} = ℘(Im
f) because ∀B
⊂ Im f, f[f⋆(B)] = B.
Proposition. For any sets E, F and any function g with
domain F,
- Inj g ⇒
Inj(FE
∋ f ↦ g⚬f)
- (Inj(FE
∋ f ↦ g⚬f) ∧ E ≠ ∅) ⇒ Inj g
- ∀SetG, ({g⚬f | f∈FE} =
GE ∧ E ≠ ∅) ⇒ Im g = G.
Proofs:
1. ∀f,f'∈FE, (Inj g ∧
∀x∈E, g(f(x)) = g(f'(x)))
⇒ (∀x∈E, f(x) = f'(x)).
2. ∀y,z∈F, g(y) =
g(z) ⇒ g⚬(E ∋ x ↦ y) =
g⚬(E ∋ x ↦ z) ⇒ (∀x∈E,
y = z) ⇒ (E ≠ ∅ ⇒ y = z).
3. (left as an exercise to the reader).∎
Axiom of choice over a set (ACX)
The axiom of choice (AC) is a statement expressible in any set theory with
powerset, which «feels true» and is convenient to accept as an axiom; but we will
actually not need it in the rest of this work.
It has several equivalent formulations (in clear, it is the
common name of these equivalent statements). The main ones fit the common format
(∀SetX, ACX) where
ACX means any of the following equivalent formulas with
the free variable X in the class of sets:
- (AC1X) ∀SetE, ∀R⊂X×E,
(∀x∈X, ∃y∈E, xRy) ⇒ (∃f∈
EX, ∀x∈X, xRf(x)).
- (AC2X) ∀A : X → Set, (∀x∈X,
Ax ≠ ∅) ⇒ ∏A ≠ ∅
(The product of any family of nonempty sets indexed by X is nonempty)
- (AC3X) ∀Fnc g, X ⊂ Im g
⇒ ∃f ∈ (Dom g)X, g⚬f = IdX
-
(AC4X) ∀grR, X ⊂ Dom R
⇒ ∃f ∈ (Im R)X, Gr f
⊂ R.
(AC3X) and (AC4X)
are more often used in cases of equality (respectively X = Im g and X = Dom R)
which are equivalent to the general cases, as the proofs of equivalence work the same :
- (AC1X)⇒(AC2X) by R =
∐ A ∧ E = Im R ;
- (AC2X)⇒(AC3X) by
A = (g•(x))x∈X ;
- (AC3X)⇒(AC4X) :
X ⊂ π0[R] ⇒ ∃(h×f) ∈ RX,
h = IdX ∴ Gr f = Im (h×f) ⊂ R ;
- (AC4X)⇒(AC1X) obvious.∎
Other easy deductions are - (AC2X)⇒(AC1X)
by Ax = R⃗(x) ;
- (AC4X)⇒(AC3X) by
R = tGr g.
Dependencies between diverse ACX
A big achievement of mathematical logic was to prove the independence of AC
from the rest of set theory: even the very strong ZF system,
if consistent, can neither prove nor refute it, as each universe where AC is true includes
another one where it is false, and vice versa. This will be further commented in 5.5
(without the details of the proofs which are much too difficult).
The diverse ways in which AC may fail in different universes, can be partly sorted out by
specifying for which X does ACX hold.
The different ACX depend on each other as follows:
- ∀SetX,Y, (ACX ∧
∃f, f : Y ↪ X) ⇒ ACY
- ∀SetX,Y, (ACX ∧ ACY) ⇒
(ACX∪Y ∧ ACX×Y)
- (ACI ∧ ∀i∈I, ACAi)
⇒ AC∐i∈I Ai
The proofs of these implications are easy and left as an exercise to the reader.
Finite choice theorem. If X is finite then ACX.
This will be easily deducible from the case of singletons AC{x} (which is trivial)
and the preservation by binary union (ACX ∧ ACY) ⇒
ACX∪Y once finiteness will be formalized (4.6).
Now let us just suggest a schema of proofs, namely one for each number of
elements of X. For example the case of X with 3 elements comes down to
(AC2V3) whose proof can be written
∀SetX,Y,Z,
(X ≠ ∅ ∧ Y ≠ ∅ ∧ Z ≠ ∅) ⇒ (∃x∈X, ∃y∈Y,
∃z∈Z, (x,y,z) ∈ X×Y×Z)
⇒ X×Y×Z ≠ ∅.∎
More statements simply equivalent to AC
Theorem. The following statements are equivalent
to the axiom of choice:
(AC5) Any set E of nonempty sets has a choice function: ∅ ∉ E ⊂ Set
⇒ ∏ IdE ≠ ∅.
(AC6) For any partition P of a set E,
∃K⊂E, ∀A∈P, ∃!: K∩A.
(AC7) For any sets E,F,G and any
g: F ↠ G, {g⚬f | f
∈ FE} = GE.
Proofs:
(AC2)⇒(AC5) is obvious ;
(AC5)⇒(AC6) : (x∈ ∏ IdP ∧ K
= Im x) ⇒ (K⊂E ∧ ∀A∈P,
xA ∈ K∩A ∧ ∀B∈P,
xB ∈ A ⇒ A∩B ≠ ∅ ⇒ A = B)
(AC6)⇒(AC3) : ∀Fnc g, X ⊂ Im g ⇒ ∃K⊂Dom g,
(∀A∈ Im(g•), ∃! : K∩A)
∴ ∃f ∈ (Dom g)X, (∀x∈X, {f(x)}
= K∩g•(x)) ∴ g⚬f = IdX.
(AC1E)⇒(AC7) : ∀h∈GE,
(∀x∈E, ∃y∈F, g(y) = h(x)) ⇒
(∃f∈FE, ∀x∈E, g(f(x)) = h(x))
(AC3G)⇒(AC7) : ∃j∈FG,
g⚬j = IdG ∴ ∀h∈GE,
j⚬h ∈ FE ∧ g⚬j⚬h = h.
(AC7)⇒(AC3) : ∀SetX, ∀Fncg,
X = Im g ⇒ IdX
∈ XX = {g⚬f | f ∈ (Dom g)X}. ∎
Note: (AC5)⇒(AC2) is also easy : ∅ ∉ Im A ⊂ Set ⇒ ∃f ∈ ∏ IdIm A,
f⚬A ∈ ∏ A.
There are many more statements equivalent to AC, which will not be of any use in this work,
and for which the proofs of equivalence are much harder. Interested readers can find about
them elsewhere.
Other languages:
FR : 2.10. Axiome du choix