The Relative Consistency of ¬AC
Erik A. Andrejko
October 21, 2004
1

The Axiom in Question
The Axiom of Choice (AC) is the statement:
∀X[∀x ∈ X[x = ∅] =⇒ ∃f∀x ∈ X[f(x) ∈ x]] (AC)
Zermelo added AC to his list of axioms formalizing mathematics in 1904 in order
to prove Cantor’s conjecture that all sets can be well ordered.
2

The Controversy Surrounding AC
The Axiom of Choice is different in character than the other axioms of ZFC since
AC implies the existence of sets without giving their construction.
French mathematicians Borel, Baire and Lebesgue were suspicious of the use of
AC in mathematical proofs, as was Peano who left it out of his Formulario
Mathematico.
The most trivial form of AC:
∀x[x = ∅ =⇒ ∃x ∈ X] (TAC)
Trivial only if one considers constructive and abstract existence to be the same.
For propositions P, the law of excluded middle is
(LEM)
∀P[P ∨ ¬P]
Theorem 1. (LEM) ⇐⇒ (TAC)
The above theorem is intuitionistically valid. See [Age02].
3

Some Paradoxes provable from ZFC
The use of AC allows one to prove existence without construction.
Theorem 2. (Vitali’s Theorem) There exists a non Lebesgue measurable set.
Theorem 3. (Banach-Tarski Paradox) A sphere S can be decomposed into
disjoint sets
S=A∪B∪C∪Q
∼∼ ∼
such that A = B = C, B ∪ C = A, and Q is countable.
4

Some Theorems Equivalent to AC
• Zorn’s Lemma: If every chain in a partial order P has an upper bound in P
then P has a maximal element.
• Tukey’s Lemma: A set X is say to have finite character if
Y ∈ X ⇐⇒ ∀k ∈ Y <ω [k ∈ X]
For X = ∅ with finite character then X has a ⊆-maximal element.
• Basis Theorem: Every vector space has a basis.
Without AC we can still consider the cardinalities of sets. We can define a class
of cardinals p by consider members of the equivalence class given by
x ≡ y ⇐⇒ |x| = |y|
where |x| ≤ |y| if there is a 1-1 function f : x → y and |x| = |y| iff |x| ≤ |y| and
|y| ≤ |x|. Note that |x| ≤ |y| is a partial order.
Then we define p2 = |p × p|.
5

Some Strictly Weaker Implications of AC
• Prime Ideal Theorem: Every Boolean algebra has a prime ideal. Or
equivalently every ideal can be extended to a prime ideal.
• Ultrafilter Theorem: Every filter on X can be extended to an ultrafilter on
X.
• Compactness Theorem: A theory T is satisfiable iff T is finitely satisfiable.
6

Different Notions of Choice
• Principal of Dependent Choice (DC): If R is a relation on A = ∅ such
that ∀x ∈ A∃y ∈ A[xRy] then there is a sequence {xn }n<ω
x0 Rx1 , x1 Rx2 , · · · , xn Rxn+1 , · · ·
• Selection Principal: For every family X of sets such that ∀y ∈ X[|y| ≥ 2]
there exists function f such that
∀y ∈ X[∅ = f(y) X]
7

Ordering Principals
• Well Ordering Principal (WOP): Every set can be well ordered.
• Ordering Principal (OP): Every set can be linearly ordered.
• Order Extension Principal (OEP) Every partial order P can be extended
to a linear order.
8

Restriction of AC to smaller classes
• AC for well ordered sets (ACW): Every set of well ordered sets has a
choice function.
• AC for countable sets (ACℵ0 ): Every set of countable sets has a choice
function
• AC for finite sets (ACF): Every set of finite sets has a choice function.
• AC for pairs (AC2 ): Every set of pairs has a choice function.
• Trivial AC (TAC): Every non empty set contains an element.
9

Relative Strength of Various Choice Principals
AC
KS
Well

Basis
+3 Zorn’s ks +3 Tukey’s ks
p2 = p ks +3 Ordering ks +3
Theorem
Lemma Lemma
Principal
ww
ww
ww
w
www
w
w ww
w 

Selection
+3 Prime Ideal Dependant
Compactness ks
Principal
Theorem Choice
tt
t
ttt
t
tt
t
u} tt
tt
 
Order Extension Ordering
ACℵ0
+3 ACF ks
+3
r 4<
Principal Principal rr
rr
rr
rrr
rr
rr
rrr 
r
r
AC2
ACW

TAC ks +3 LEM
10

The Consistency of the Failure of Various Choice
Principals
The arrows on the previous slide do not reverse. We can prove this by
construction models in which one holds and the negation of the other also holds.
(i.e. Build a model of ZF + OP + ¬OEP to show that OP =⇒ OEP.)
Theorem 4. The Prime Ideal Theorem does not imply AC.
Theorem 5. The Selection Principal does not imply AC.
Theorem 6. The Ordering Principal does not imply the Order Extension
Principal.
11

Relative Consistency Results
G¨del proved in 1930 his famous Incompleteness result that showed we cannot
o
prove the consistency of ZF + ¬AC within ZFC, that is
Con(ZF + ¬AC)
ZFC
So assuming Con(ZFC) we build a model of ZF + ¬AC. Cohen showed such a
construction using forcing in 1963. Fraenkel and Mostowski showed the
unprovability of AC from ZF− .
Theorem 7. Con(ZF− ) =⇒ Con(ZF− + ¬AC).
12

Constructing a Permutation Model
By ZF− we denote the axioms of ZF minus the axiom of foundation. In
particular we allow sets such as a = {a} which we will call an atom. Let A be an
infinite set of atoms.
Define Vα (A) by induction of α as follows:
V0 (A) = A
Vα+1 (A) = P(Vα )
Vγ (A) for γ limit
Vα (A) =
γ<α
Finally define V = Vα (A). Then we have
α∈ON
A = V0 (A) ⊆ V1 (A) ⊆ · · · ⊆ Vα (A) · · · ⊆ V
For any x ∈ V we can assign a rank,
rank(x) = least α[x ∈ Vα+1 (A)]
13

Let G be the group of permutations of A. For π ∈ G we extend π to a
permutation of any x ∈ V by induction on ∈ by defining
π(x) = {π(y) : y ∈ x}
and letting π(∅) = ∅. Then G permutes V and fixes the well founded sets
WF ⊆ V.
Lemma 1. For all x, y ∈ V and any π ∈ G.
x ∈ y ⇐⇒ π(x) ∈ π(y)
That is π if an ∈-automorphism of V. From this we can prove that
π({X, Y}) = {π(X), π(Y)} and so
π((X, Y)) = (π(X), π(Y))
π((X, Y, Z)) = (π(X), π(Y), π(Z))
Also by induction on α it is easy to show that
rank(x) = rank(πx)
for all x ∈ V.
14

The Hereditarily Symmetric sets HS
Let a1 , · · · , an ∈ A and define
[a1 , · · · , an ] = {π ∈ G : ∀a1 , · · · , an [π(ai ) = ai ]}
Call a set X ∈ V symmetric if there exists a1 , · · · , an ∈ A such that π(X) = X
for all π ∈ [a1 , · · · , an ]. Define the class HS ⊆ V of hereditarily symmetric
sets
HS = {x ∈ V : x is symmetric and x ⊆ HS}
Call a class N transitive if
∀x ∈ N[x ⊆ N]
and call N almost universal if (for sets S)
∀S ⊆ N[∃Y ∈ N(S ⊆ Y)]
Lemma 2. HS is transitive.
Lemma 3. HS is almost universal.
15

To show that a class N |= ZF− is straight forward for most axioms of ZF− except
for the axiom of Comprehension. To show N is a model of Comprehension it
suffices to show that N is closed under G¨del Operations:
o
G1 (X, Y) = {X, Y}
G2 (X, Y) = X \\ Y
G3 (X, Y) = X × Y
G4 (X) = dom(X)
G5 (X) = ∈ ∩X2
G6 (X) = {(a, b, c) : (b, c, a) ∈ X}
G7 (X) = {(a, b, c) : (c, b, a) ∈ X}
G8 (X) = {(a, b, c) : (a, c, b) ∈ X}
Theorem 8. (ZF) If N is transitive, almost universal and closed under G¨del
o
Operations, then N |= ZF.
Lemma 4. HS is closed under G¨del operations.
o
Lemma 5. HS |= ZF− .
16

HS violates AC
Lemma 6. A ∈ HS.
Lemma 7. Let f : ω → A be 1-1. Then f ∈ HS and so A cannot be well ordered
/
in HS.
Theorem 9. HS |= ZF− + ¬AC.
which completes the proof that Con(ZF− ) =⇒ Con(ZF− + ¬AC). In particular
we have that ZF− AC.
17

Showing the Relative Consistency of ¬AC with ZF
Let B be a complete boolean algebra and let MB be a boolean valued model. Let
G be and M-generic ultrafilter on B. Then
M ⊆ M[G] |= ZFC
Then, as in V the universe of M[G] admits ∈-automorphism even though M does
not. Let π be an automorphism of B, that is
π(x · y) = π(x) · π(y)
π(x + y) = π(x) + π(y)
We can extend π to an automorphism of MB by induction: π(0) = 0, and let
dom(π(x)) = π(dom(x)) Then let
π(x)(π(y)) = π(x(y))
for all π(y) ∈ dom(x). Then π is a permutation of MB and π fixes M. Then as
before we can construct HS ⊆ MB . Finally, using G we can construct N ⊆ M[G]
by using only names for elements of M[G] appearing in HS. Then
M ⊆ N ⊆ M[G]
18

Such an N is called a symmetric extension of M.
Lemma 8. N |= ZF.
Finally let P = Fn(ω × ω, 2), Cohen forcing. Then we define xn to be the Cohen
real
xn = {m ∈ ω : ∃p ∈ G[p(n, m) = 1]}
Let A be set of Cohen reals = {xn : n < ω}. Then A ∈ N and A cannot be well
ordered in N.
To get permutations of π we can extend permutations of ω to P, then to a
Boolean algebra B, then to the boolean valued model MB and finally to M[G].
From this we get that Con(ZFC) =⇒ Con(ZF + ¬AC) and so ZF AC.
19

Some Peculiarities in the Absence of AC
Theorem 10. It is consistent that all uncountable cardinals can have cofinality
ω. (see [Git80])
Theorem 11. There is a model of ZF in which the set of reals has no countable
subset.
Theorem 12. There is a model of ZF with the set of reals a countable union of
countable sets.
Theorem 13. There is a model of M |= ZF + DC in which every set of reals is
Lebesgue Measurable (LM).
Theorem 14. There exists a distance graph G2 in R2 such that under (ZFC)
G2 has chromatic number 2, and under ZF + LM + DC the chromatic number of
G2 cannot equal n ≤ ℵ0 . (see [SS04])
Theorem 15. There is a model with a vector space without a basis.
Theorem 16. The axiom of choice for countable collections of countable sets
does not imply that a countable union of countable sets is countable. (see
[How92])
Theorem 17. (Intuitionistic Logic) The family F = {{a, b}} has no choice
ˇ
function. (see [FS82])
20

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