(E) The Axiom of Choice
The axiom of choice, AC, has had a rather strange life. It is a principle which is
‘obviously’ true, and one we use every day. Many of the routine parts of Mathematics
depend on this principle. However, it also has some ‘odd’ or even ‘awkward’ consequences.
In these notes we define and look at some simple equivalences of AC. We also look at one
of the awkward consequences.
Contents
1
2
3
4
Introduction . . . . . . . . . . . . . . . . . . .
1.1
Exercises: Introduction . . . . . . . .
The axiom of choice defined . . . . . . . . . .
2.1
Exercises: The axiom of choice defined
2.2
Problems: The axiom of choice defined
Measuring sets of reals . . . . . . . . . . . . .
Conjunctive and disjunctive normal form . .
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1
2
2
7
9
11
15
1 Introduction
Before we begin this analysis let’s list a few consequences of AC, some of which are
equivalent to AC. Most of these are analysed in these notes or a related set of notes.
(1) Each equivalence relation on a non-empty set A has a set of representatives, a subset
B ⊆ A which contains exactly one element from each block (equivalence class).
(2) For each surjective function f : A
f ◦ g = id B .
-
B there is a function g : B
-
A with
- R is -δ continuous at a given point precisely when it is
(3) A function f : R
sequentially continuous at that point.
(4) Each ring (with a 1) has a maximal ideal.
(5) Tychonoff’s theorem, that is the product of a family of compact topological spaces
is itself compact.
(6) There is a subset of R that can not be measured in any sensible way.
Of these assertions the first two, (1, 2), are ‘obviously’ true. However, each is equivalent to AC. We look at this in Theorem 2.6
Condition (3) is very useful. We may formally define the continuity of a function f
using the -δ method. However, if we have to calculate the value of f at a given point a,
then the sequential method is often more useful. We select a sequence x(·) which tends
to a, and such that for each n ∈ N the value f (xn ) is easy to compute. We then have a
sequence
(f xn ) | n ∈ N)
which gives better and better approximations of the value f (a). We will see that a weaker
version of AC is needed to show that the two kinds of continuity are equivalent.
1
As an aside, you might like to find a copy of G.H. Hardy’s ‘A course of Pure Mathematics’ and look at page 186 where he writes down the formal definition of continuity.
You will see that he swaps the use of and δ. I don’t know whether this was standard at
the time, of just an aberration.
The condition (4) is an example of a maximizing principle. We look at two of these,
Zorn’s Lemma and Krull’s Separation Lemma in a separate document. Both of these
depend on a use of AC. However, it is the technique that is important, in several parts of
Mathematics.
Tychonoff’s theorem (5) is another useful tool in point set topology and related topics.
The condition (6) perhaps looks a little odd, and it not what we might expect. We
look at this in some detail in Section 3.
1.1 Exercises: Introduction
1.1 There are times when it is useful to employ a formal language of Predicate Calculus.
- R is not
To see this try to write down the definition that a given function f : R
uniformly continuous. Then do the following. Each is concerned with a arbitrary function
f : R - R.
(a) Write down, in a formal way, the definition that f is continuous, that is f is
continuous at each point a ∈ R.
(b) Write down, in a formal way, the definition that f is uniformly continuous.
(c) Write down, in a formal way, the definition that f is not continuous.
(d) Write down, in a formal way, the definition that f is not uniformly continuous.
(e) Observe the different between the two formal descriptions from (c) and (d).
2 The axiom of choice defined
In this section we give the formal definition of AC, the Axiom of Choice, and we show it
is equivalent to the ‘obviously true’ assertions (1, 2) of Section 1.
The axiom of choice is concerned with getting inside an indexed family
A = Ai | i ∈ I
of non-empty sets. It selects an element from each component Ai . Before we give the
actual definition we need to make sure we know what such a family is.
2.1 DEFINITION. Let I be a non-empty set. An I-indexed family of sets is a function
I
-
i
Sets
-
from I, the index set, to the family of sets.
Ai
Although the I-indexed family is a function (from I as the source) we usually write it
as on the left below
(Indexed) Ai | i ∈ I
Ai | i ∈ I
(Non-indexed)
2
and sometimes give it a name such as A as used above. For each input i ∈ I, of the
function, we write Ai for the output at i, and this is a certain set. It can happen that
Ai = Aj
for distinct indexes i 6= j, and such repetitions do matter. At an extreme case we can
have Ai = Aj for all pairs of indexes. An indexed family is not just a collection of sets as
in (Non-indexed) above.
To illustrate the difference let’s look at a case where I is finite, say {0, 1, . . . , n} for
some n ∈ N. We then have
A0 , A1 , . . . , An
A0 , A1 , . . . , An
for the two cases. On the left the repetitions matter, but on the right they don’t. In an
extreme case we can have A0 = A1 = · · · = An = A, say, and then
A, A, . . . , A
A
are not the same. On the left we have a list of n + 1 sets which happen to be the same,
but on the right we have a set with just one member which happens to be a set.
It is true that in many cases we do not need to be so finicky and distinguish between
(Indexed) and (Non-indexed), but here we have to be more careful.
Notice that in both cases we can take the union of the family to produce
S
S
Ai | i ∈ I =
Ai | i ∈ I
the set of all elements a such that a ∈ Ai for some i. This union forgets the indexed or
non-indexed aspect.
2.2 DEFINITION. Let
A = Ai | i ∈ I
be an indexed family of sets. A choice function for A is a function
S
A such that α(i) ∈ Ai
α:I
for each i ∈ I.
When does a indexed family A have a choice function? Clearly, if there is a choice
function then each component Ai is non-empty, since the choice function produces an
element of Ai . That is the only restriction.
2.3 DEFINITION. The Axiom of Choice, AC, says that for each non-empty index set I
and each I-indexed family of non-empty sets, there is a choice function.
That seems reasonable enough. We are going to look at some of the consequences of
AC. Some of these are what we want, but some seem a bit odd.
Before we start that let’s look at the finite index case.
3
2.4 EXAMPLE. Consider a finite indexed family
A0 , A1 , . . . , An
of non-empty sets. A choice function for this family selects an element from each component, that is it selects some
ai ∈ Ai
for each index i. In other words, a choice function is nothing more that a list
(a0 , a1 , . . . , an )
of elements from the components. Thus the cartesian product
A0 × A1 × · · · × An
is the set of all possible choice functions.
Of course, in this finite indexed case we don’t usually speak of a choice function. We
just use tuples.
This example show us how we can form the product of an arbitrary family of nonempty sets.
2.5 DEFINITION. Let
A = Ai | i ∈ I
be an indexed family of non-empty sets. The product
Q
A
of this family is the set of all choice functions of A.
The axiom of choice, AC, says that such a product is non-empty.
The axiom of choice can be stated in several equivalent ways. Two of these are given
as (1, 2) in Section 1. The proof of the following could be shortened, but looking at the
details given does help to understand what is going on.
Observe that since it is showing something is equivalent to AC, we must be careful
not to use any choice principle to deduce AC. In fact, this is how AC was uncovered
and formalized. Around 1900 various proofs of different facts were offered (such as the
well-ordering of R, the solution of certain differential equations, and other aspects of real
analysis), and it was then seen how a choice principle was being used.
2.6 THEOREM. The following conditions are equivalent.
(0) The axiom of choice, AC.
(1) Each equivalence relation on a non-empty set A has a set of representatives, a subset
B ⊆ A which contains exactly one element from each block (equivalence class).
(2) For each surjective function f : A
f ◦ g = id B .
-
4
B there is a function g : B
-
A with
Proof. (0) ⇒ (1). Consider an equivalence relation ∼ on the set A. For each a ∈ A
let [a] be the block in which a lives, that is
x ∈ [a] ⇐⇒ x ∼ a
for x ∈ A. We use these blocks in two ways.
(A more standard terminology for ‘block’ is ‘equivalence class’, but ‘block’ is neater.)
Firstly, each block is an index, and each index is a block. That is
I = {[a] | a ∈ A}
is the set of indexes. Of course, in this set many indexes will be repeated.
Secondly, each index i = [a] has to index some subset Ai ⊆ A. And Ai must depend
only on the index i not the particular element a we use to name the block. We let
Ai = [a]
for the index i = [a].
Thus each block is both an index and a subset indicated by an index. Think about
that.
Observe that for this case we have
S
Ai | i ∈ I = A
since each Ai is a subset of A, and each element of A belongs to some Ai .
We now invoke AC to produce a choice function
α:I
-
A
with
α(i) ∈ Ai
that is
α [a] ∈ [a]
for each index i = [a]. In other words we have a function α from blocks to elements such
that
α [a] ∼ a
for each a ∈ A. We show that the range of α
B = {α [a] | a ∈ A}
contains exactly one element from each block. To do that we show that B contains at
least one, and at most one, element from each block.
Consider a typical block. This
has the form [a] for some a ∈ A, and there could be
many such a ∈ A. Let b = α [a] . Then b ∈ B and
b∼a
to show that b ∈ [a]. This shows that B contains an element from each block.
Now suppose that B contains elements b1 , b2 from the same block. We require b1 = b2 .
We have
b1 = α [a1 ]
b2 = α [a2 ]
[a1 ] = [a2 ]
5
since both b1 , b2 are selected by α, and the blocks from which they are selected are the
same. Thus b1 = b2 , as required.
(1) ⇒ (0). Assuming (1), consider any indexed family
A = Ai | i ∈ I
of non-empty sets. Let
A=
S
{Ai × {i} | i ∈ I}
that is A is the set of all ordered pairs (a, i) where i ∈ I and a ∈ Ai . Consider the relation
∼ on A given by
(a, i) ∼ (b, j) ⇐⇒ i = j
which, by an easy argument we see is an equivalence relation. Observe that the blocks of
this relation are the sets
Ai × {i}
for i ∈ I. By the assumption (1) there is a subset
B ⊆A×I
such that for each index i there is a unique a ∈ Ai with (a, i) ∈ B. In other words there
is a function
β A×I
I
such that for each index i ∈ I we have
β(i) = (a, i)
for some a ∈ Ai . Consider the composite
I
β A×I
λ A
where λ simply picks the left component a of each pair (a, i) from A. Then
α=λ◦β
is a choice function for A.
(0) ⇒ (2). Consider any surjective function
f :A
-
B
between two sets. We view B as an index set. For each b ∈ B let Ab be the inverse image
of b across f , that is
a ∈ Ab ⇐⇒ f (a) = b
for each b ∈ B and a ∈ A. This gives an indexed family
A = Ab | b ∈ B
of sets. Furthermore, each Ab is non-empty since f is surjective.
6
By invoking AC we obtain a choice function for this family, that is a function
g:B
A
-
with
g(b) ∈ Ab
for each b ∈ B. By definition of A we have
f (g(b)) = b
for each b ∈ B, that is f ◦ g = id B .
(2) ⇒ (0). Assuming (2), consider any indexed family
A = Ai | i ∈ I
of non-empty sets. Let
A=
S
{Ai × {i} | i ∈ I}
that is A is the set of all ordered pairs (a, i) where i ∈ I and a ∈ Ai . Consider the two
projections
A
(a, i)
f
-
I
A
-
i
(a, i)
h -S
A
-
a
and observe that f is surjective since each Ai is non-empty. By (2) there is some function
g:I
-
A
with f ◦ g = id B . But now, for each index i ∈ I we have
g(i) = (a, i)
for some a ∈ Ai . Thus the composite h ◦ g is a choice function for A.
The three equivalent conditions (0, 1, 2) of this result are not surprising. Most people
will say they are obviously true, (although there one or two nutters out there). In the
next section we go to the other end of the scale. We look at a condition which, when we
first start, we hope can be arranged by some method or other, but then we find that AC
blocks all our attempts.
2.1 Exercises: The axiom of choice defined
2.1 Show, without using any form of choice, that each finite family of non-empty sets,
(A0 , A1 , . . . , An ), has a choice function.
2.2 This question is concerned with arbitrarary indexed families of sets.
Let
(Ai | i ∈ I)
7
be any indexed family of non-empty sets, and let A be the set of all choice functions
S
α:I
(Ai , i ∈ I)
for this set. For each index i consider the function
A
pi Ai
α
-
α(i)
given by evaluation at i. This is sometimes called the projection function.
Suppose B is an arbitrary set with an indexed family of functions
fi Ai
B
where each one has source B but the target varies with i.
Show there is a unique function
f
B
-
A
such that for each index i the triangle
fi
B
Ai
-
-
pi
-
f
A
commutes, that is fi = pi ◦ f .
2.3 The following four parts illustrate the use of AC in real analysis. For each part show
that the two conditions (i, ii) are equivalent. However, show that (i) ⇒ (ii) does not
require a choice principle, but AC gives (ii) ⇒ (i). Observe also that each use of choice
produces a function of type N - (???) for some set (???).
(a) For arbitrary function f : R
-
R and a ∈ R consider the following conditions.
(i) The function is -δ continuous at a.
(ii) The function is sequentially continuous at a.
(b) For arbitrary subset A ⊆ R consider the following conditions.
(i) The set A is closed (that is its complement R − A is a union of open intervals).
(ii) For each convergent sequence x(·) of elements of A, the limit is also in A.
To sort out this you may like to first describe the interior and the closure of a subset
of R by a couple of -condition.
(c) For arbitrary subset A ⊆ R consider the following conditions.
(i) The set A is bounded above and below.
(ii) Each sequence x(·) of elements of A has a convergent sub-sequence.
The proof of (i) ⇒ (ii) is a bit tricky, but the proof of (ii) ⇒ (i) is much easier.
8
2.2 Problems: The axiom of choice defined
2.4 The Axiom of Countable Choice, ACω , says that ‘Each countably indexed family of
non-empty sets, (An | n ∈ N), has a choice function’. Trivially
AC =⇒ ACω
and it turns out that ACω is strictly weaker then AC (although the proof of this is
complicated).
Show that ACω is sufficient to obtain all the results of Exercise 2.3.
2.5 We have seen AC and ACω , the full axiom of choice and the axiom of countable
choice. In this exercise we look at the axiom of dependent choice which sits between these
two other choice principles.
Consider a set A and a binary relation R on A. Consider the following three conditions
which may or may not hold.
(α) (∀x ∈ A)(∃y ∈ A)[xRy]
(β) (∃f : A
-
A)(∀x ∈ A)[xRf (x)]
(γ) (∃φ : N
-
A)(∀n ∈ N)[φ(n)Rφ(n + 1)]
The Axiom of dependent Choice, DC, says ‘For each set A and binary relation R on A,
we have (α) =⇒ (γ)’.
(a) Show that (β) =⇒ (α)&(γ).
(b) Show that AC is equivalent to ‘For each set A and binary relation R on A, we
have (α) =⇒ (β)’.
(c) Show that AC ⇒ DC ⇒ ACω .
2.6 The following extends Exercise 2.2.
We know that for many kinds of algebraic structures we can take two such structures
A = (A, · · · )
B = (B, · · · )
and form the ‘product’ to obtain a new structure of the same kind. (This is one of the
places where it helps if we distinguish between a structure A and its carrier A. It you
are not very good with Gothic letters then use a different version of the carrier, say an
underlined version.) A standard way to form the product
A×B
is to take the cartesian produce A × B of the carriers, and then furnish this with some
structure to get what we want.
This kind of construction is easily extended to finitely many structures (of the same
kind). But what about an infinite collection of structures?
Let’s see how this is done for monoids.
A monoid is a structure
A = (A, ?, 1)
9
where ? is a binary operation on the set A and 1 is a particular element. These two
gadgets are required to satisfy
(a ? b) ? c = a ? (b ? c)
a?1=a=1?a
for all a, b, c ∈ A. That is ? is associative and 1 is a unit.
Suppose now we have an indexed family
A = (Ai | i ∈ I)
of monoids. Is there any way we can form a ‘product’ of this family?
For each index i ∈ I let
Ai = (Ai , ?i , 1i )
where we now index all the relevant gadgets. This notation is a bit finicky but it is worth
it when you have never seen this construction before.
Let A be the product of the indexed family of sets
(Ai | i ∈ I)
that is the set of all choice function
α:I
-
S
(Ai | i ∈ I)
for the family. We furnish A to produce a monoid (and this is where the indexing of the
operations ?i and the units 1i becomes useful).
Given two members α, β ∈ A let
S
α?β :I
(Ai | i ∈ I)
be the choice function given by
(α ? β)(i) = α(i) ?i β(i)
for each index i ∈ I.
Let
1:I
-
S
(Ai | i ∈ I)
be the choice function given by
1(i) = 1i
for each index i ∈ I.
(a) Show that this structure
A = (A, ?, 1)
is a monoid, that is show that ? is associative and 1 is a unit.
(b) Show that for each index i ∈ I the function
A
pi Ai
α
-
α(i)
is a morphism of monoids. (You will have to decide what such a morphism should be).
10
(c) Suppose for an arbitrary monoid B there is an indexed family of morphisms
fi Ai
B
one to each of the given monoids. Show there is a unique morphism
f
B
-
A
such that for each index i the triangle
fi
B
Ai
-
-
pi
-
f
A
commutes, that is fi = pi ◦ f .
3 Measuring sets of reals
In this section we look at sets of the reals R, and we consider how we might measure such
sets in a sensible way. Before we make the formal definition let’s look at some intuitive
notions to motive the idea of a measure.
Consider the bounded interval of R. There are four kinds depending in whether either
end is open or not. Thus each pair a ≤ b of reals gives four intervals
(a, b)
[a, b)
(a, b]
[a, b]
and it seems reasonable that we take the measure of each of these to be the length
µ(a, b) = µ[a, b) = µ(a, b] = µ[a, b] = b − a
of the interval. In particular, the measure of each singleton [a, a] is 0.
We also allow certain sets to have measure ∞. Thus
µ(a, ∞) = µ[a, ∞) = µ(−∞, b] = µ(−∞, b) = µ(R) = ∞
where here we don’t try to give ‘∞’ a value, but merely use it as a symbol.
Now consider a pair A, B of disjoint sets each of which has a measure. It seems
reasonable to let
µ(A ∪ B) = µ(A) + µ(B)
since no part of A ∪ B is counted twice. Of course we let
∞+r =r+∞=∞+∞=∞
to take care of the symbolic value.
There are also certain collections of subsets which ought to have the attached to them
the same measure.
11
3.1 DEFINITION. Let A ⊆ R and consider any r ∈ R. We set
A + r = {a + r | a ∈ A}
to obtain the r-translate of A.
In this notation it is reasonable to require
µ(A + r) = µ(A)
to hold.
These ideas motivate most of the following notion.
3.2 DEFINITION. A measure on R is a pair (M, µ) where M ⊆ PR and
µ:M
-
[0, ∞]
with the following four properties.
(i) Each interval A is in M and µ(A) is the length of A.
(ii) For disjoint A, B ∈ M we have A ∪ B ∈ M and
µ(A ∪ B) = µ(A) + µ(B)
is the assigned value.
(iii) The family M is closed under translations and
µ(A + r) = µ(A)
for each A ∈ M and r ∈ R.
(iv) If
A = {An | n ∈ N}
S
is an ascending chain of members of M, then A ∈ M and
S
µ( A) = sup{µ(An ) | n ∈ N}
where this value may be ∞.
The target [0, ∞] of µ is the positive reals together with a symbolic infinite value ∞. Only condition (iv) is not discussed above. Let’s see what that means and why we
need it.
The given ascending chain is
A0 ⊆ A1 ⊆ · · · ⊆ An ⊆ · · ·
(n ∈ N)
an N-indexed family of subsets of R that increases with n. Suppose each such An can be
measured, that is An ∈ M. Then surely we want
µ(A0 ) ≤ µ(A1 ) ≤ · · · ≤ µ(An ) ≤ · · · (n ∈ N)
to hold. S
Also, we when start to use µ for other more delicate purposes we find that we
do need A ∈ M with the indicated value. For what we do here we need not go into
the details of this second part. However, condition (iv) does have a couple of important
consequences, one of which we have already mentioned as an informal justification.
12
3.3 LEMMA. Let (M, µ) be any measure.
(a) Consider any pair A ⊆ B of members of M. Then µ(A) ≤ µ(B).
(b) Let B be any countable sub-family of M which is pairwise disjoint. Suppose also
that each member of B has the Ssame measure, that is there is some b ∈ [0, ∞] with
µ(B) = b for each B ∈ B. Then B ∈ M and
S
∞ if b 6= 0
µ( B) =
0 if b = 0
holds.
Proof. (a) Consider the N-chain A given by
A0 = A
An = B
for all n 6= 0. This is ascending and constant from A1 onwards. We have
that property (iv) gives
S
µ(A) ≤ sup{µ(A), µ(B)} = µ( A) = µ(B)
S
A = B, so
for the required comparison.
(b) We are given that B is a countable, and hence we have
B = {Bn | n ∈ N}
for some enumeration of B. (This does not require a use of AC, for it is simply what
being countable means). For each n ∈ N let
An = B0 ∪ · · · ∪ Bn
to obtain an N-indexed ascending chain. By multiple uses of property (ii) we have An ∈ M
with
µ(An ) = µ(B0 ) + · · · + µ(Bn ) = (n + 1)b
where b is the common value of the µ(B• ). We also have
S
S
A= B
and hence by property (iv) we have
S
S
µ( B) = µ( A) = sup{(n + 1)b | n ∈ N}
which gives the required result.
The grand scheme of setting up a measure is to attach a value to as many sets as
possible. We want to make M as big as possible, and it would be nice if we could get
M = PR. However, AC ensures that there is one particular set V which will always be
a nuisance.
13
3.4 DEFINITION. We make use of AC in the form of condition (1) of Theorem 2.6.
Consider the interval [0, 1) of R and the equivalence relation on this interval given by
x ∼ y ⇐⇒ x − y ∈ Q
for 0 ≤ x, y < 1. By the indicated equivalent of AC there is some set V ⊆ [0, 1) which
meets each block of ∼ at precisely one real number.
Let’s see what this set V does. Consider any real x ∈ R. By taking a suitable, and
unique, natural number m ∈ N we have 0 ≤ x−m < 1. Consider the block of this number
x − m. By the selection of V we have
x−m∼v
for some unique v ∈ V . The difference
x−m−v
is rational, so we have
x=v+q
for some unique q ∈ Q. This leads to the following awkward conclusion.
3.5 THEOREM. Consider the set V described above, and consider any measure (M, µ).
Then AC ensures that V ∈
/ M.
Proof. Let
I = Q ∩ (−1, 1)
and use this as an index set. Let
B = V +q |q ∈I
and observe that the members of B are pairwise disjoint. We also have
S
[0, 1) ⊆ B ⊆ (−1, 2)
by the positions of V and I.
Now, by way of contradiction, suppose that V ∈ M. Then, by property (iii), we have
µ(V + q) = µ(V )
for each index q ∈ I By Lemma 3.3(b) we have
S
µ( B) = 0 or ∞
depending on the value µ(V ). By Lemma 3.3(a) and the property (i) we have
S
1 = µ[0, 1) ≤ µ( B) ≤ µ(−1, 2) = 3
which gives the contradiction.
There are some things in life that can never be achieved. Oh dear!
14
4 Conjunctive and disjunctive normal form
Before we look at the main idea let’s look at a motivating example.
Suppose we have a finite collection of sets
. . . , Ar , . . . , Bs , . . . , Ct , . . .
which have been separated into several parts. We can combine these using union and
intersection. Thus we can first take the union of each part
· · · ∪ Ar ∪ · · ·
· · · ∪ Bs ∪ · · ·
· · · ∪ Ct ∪ · · ·
and then we can take the intersection of these bits
TS
· · · ∩ (· · · ∪ Ar ∪ · · · ) ∩ (· · · ∪ Bs ∪ · · · ) ∩ (· · · ∪ Ct ∪ · · · ) ∩ · · ·
TS
to form a
combination. We can also form the intersections
· · · ∩ Ar ∩ Bs ∩ Ct ∩ · · ·
P (r, s, t)
for all possible lists . . . , r, s, t, . . . of indexes. We can then take the union of these parts
ST
S
(
)
{P (r, s, t) | all possible r, s, t}
ST
to form a
combination.
The two resulting sets are the same. One is in conjunctive normal form and the other
is in disjunctive normal form.
We will look at an infinitary version of this construction.
4.1 DEFINITION. Let I be any non-empty set, and let
I = {I(j) | j ∈ J}
be a partition of I indexed by some set J.
In other words we have an equivalence relation on I and we have separated I into the
associated blocks (equivalence classes). We use I and J as indexing sets. We also use
certain choice functions.
4.2 DEFINITION. Let I be the set of all functions
α:J
-
I
such that
for each j ∈ J.
α(j) ∈ I(j)
In other words, such a function selects an index i from each block of the partitioning
of I. Observe that AC ensures that I is non-empty.
We now work with an I-indexed family
A = (Ai | i ∈ I)
15
of non-empty sets. We let
A=
S
A
and everything we do take place in A.
Using the partition on I we let
Bj =
S
{Ai | i ∈ I(j)}
for each j ∈ J. This gives a family
B = (Bj | j ∈ J)
and we form
T
B
S
C
TS
to obtain a
combination.
Going the other way we let
Cα =
T
{Aα(j) | j ∈ J}
for each α ∈ I. This gives a family
C = (Cα | α ∈ I)
and we form
ST
to obtain a
combination.
We need a more explicit description of these two subsets of A.
4.3 LEMMA. We have
T
x ∈ B ⇐⇒ (∀j ∈ J)[x ∈ Bj ] ⇐⇒ (∀j ∈ J)(∃i ∈ I(j))[x ∈ Ai ]
S
x ∈ C ⇐⇒ (∃α ∈ I)[x ∈ Cα ] ⇐⇒ (∃α ∈ I)(∀j ∈ J)[x ∈ Aα(j) ]
for each x ∈ A.
Proof. This is nothing more than an unravelling of the two constructions.
We can now obtain a more general version of the conjunctive-disjunctive normal form
result.
4.4 LEMMA. We always have
S
Proof. Consider first any x ∈
C⊆
S
T
B, and AC ensures that these two sets are equal.
C. Then we have
(∀j ∈ J)[x ∈ Aα(j) ]
for some α ∈ I. For this α we have i = α(j) ∈ I(j) so that
(∀j ∈ J)(∃i ∈ I(j))[x ∈ Ai ]
T
to give x ∈ B.
To obtain
T the required equality it suffices to show the converse inclusion. Thus consider
any x ∈ B. This gives
(∀j ∈ J)(∃i ∈ I(j))[x ∈ Ai ]
16
and so a use of AC provides a function
i = α(j) ∈ I(j) and x ∈ Ai
S
for each j ∈ J. This α is in I, and so x ∈ C, as required.
α:J
-
I
with
This show that a use of AC gives an infinitary conjunctive-disjunctive equality, a kind
of normal form result. What is T
a bit surprising
is that such an equality requires AC. We
S
show that a particular equality B = C implies AC.
As above we fix I with the J-indexed partition I. We also let I be the choice functions
for this partition. Our problem is to show that I 6= ∅.
4.5 THEOREM. The conjunctive-disjunctive equality given above ensures AC.
Proof. As remarked, it suffices to show that I is non-empty. To this end let
Ai = {?}
for each i ∈ I.
For each
T j ∈ J we knowSthere is at least one i ∈ I(j), and then ? ∈ Ai . This show
that ? ∈ B. But now ? ∈ C, so that
(∃α ∈ I)[· · · ]
to show that I is non-empty, as required.
17