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The Axiom of Choice and Zorns Lemma
A short introduction
Jonathan Nilsson
“The axiom of choice“ is an axiom which is usually assumed in most of modern mathematics. It states that given any collection of nonempty sets, we can pick out one element
from each set, or in other words that there exists a function with the union of sets as
domain and f (i) ∈ Xi (that is, the function picks out the object f(i) from the set indexed
i). The axiom may seem to be obviously true, but it can in fact not be proven from the
usual axioms of set theory.
The axiom of choice is equivalent to another statement called ”Zorns Lemma”, which is
sometimes more practical to use. To understand the statement of Zorns lemma we need
some basic theory of partially ordered sets, also called posets.
Definition. Let X be a set with a relation written ≤.
X
1)
2)
3)
is called a partially ordered set if
x ≤ x for all x ∈ X
If x ≤ y and y ≤ z, then x ≤ z
If x ≤ y and y ≤ x, then x = y
the relation ≤ satisfies the following conditions:
(reflexivity)
(transitivity)
(antisymmetry)
The typical example of a poset is the real number with the usual meaning of ≤. This
order is called a total order or a linear order, which means that any two elements are
comparable: For x, y ∈ R either x ≤ y or y ≤ x. This need not hold in general. Totally
ordered sets are also called chains. Chains may of course still have an infinite number of
elements.
For a second example, let S be any set, and let P (S) be the set of subsets of S. P (S) is
ordered by inclusion: We define S1 ≤ S2 iff S1 ⊂ S2 . One can easily check that 1-3 above
holds. Note that this is not a total order.
If S is a subset of a poset P , an upper bound for S is an element b of P (not nessesarily
of S) such that x ≤ b for all x ∈ S. A maximal element of a poset is an element m
beloning to the poset such that m ≤ x implies x = m, in other words, no element is greater
than m. However, unlike the upper bound, a maximal element need not be comparable
to all elements of the poset. Maximal elements does not exist in general, and if it exists
it need not be unique. Lower bound and minimal elements are defined analogously.
Examples:
Let P = {a, b, c, d} where the relation ≤ is defined by d ≤ a and b ≤ a. This poset has
no upper bound since no element is greater than all others. However, both a and c are
maximal elements since no element is greater than any of them. If we consider the subset
{b, d}, both b and d are maximal, and a is an upper bound.
For a second example let P = R with the usual meaning of ≤. P is a totally ordered
set which has no maximal element and no upper bound (and no lower bound/minimal
element). The subset (0, 1) has no maximal element, but any real number ≥ 1 is an upper
bound. The subset (0, 1] has a unique maximal element, namely 1, and every real number
≥ 1 is an upper bound.
Let X be a set and P (X) the set of subsets of X. P (X) is ordered by inclusion as before,
which is not a total order (unless X contains less than 2 elements). X itself is a member
of P (X) and it is an upper bound (and hence a maximal element).
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We are now ready to state Zorns lemma.
Zorns lemma: Let P be a partially ordered set. If every totally ordered subset of P has
an upper bound, then P has at least one maximal element.
We can use Zorns lemma to prove the following useful theorem:
Theorem. Every vector space has a basis.
Proof. Let V be a vectorspace over some field K, and let B be the subset of P (V )
consisting of linearly independent sets of V . B is ordered by inclusion. We shall prove
that every chain in B has an upper bound.
Let S be a chain in B, that is, S consists of a family {Si }i∈I where ... ⊂ Si1 ⊂ Si2 ⊂ Si3 ....
Let M = ∪i∈I Si . This is clearly an upper bound for S as long as it is a member of B.
To show M ∈ B, suppose m1 + ... + mk = 0 where all the mi belongs to M (and are
nonzero). Since S is linearly ordered we can pick r so large that all the mi lies in Sr . But
Sr is linearly independent, a contradiction, so we must in fact have M ∈ B. This shows
that every chain has an upper bound, so by Zorns lemma B has a maximal element Ω.
Finally, we show that the set Ω is our desired basis of V . Being a member of B, we
already know that Ω is a linearly independent subset. Let a ∈ V be an element outside
Ω. Then Ω ( Ω ∪ { a},so in order not to contradict the maximality of Ω, Ω ∪ {a} is not
linearly independent, so a is a combination of elements in Ω. This holds for every such a,
which shows that Ω spans V .
The above proof might be hard to follow, so here is an example of how it would (trivially)
apply to the vector space R4 .
V = R4 , a real vector space of dimension 4. B is then the family of all linearly independent
subsets. Such a subset can of course have at most four elements, but there are infinitly
many such subsets, so B is infinite. The set {(1, 0, 2, 0), (1, 1, 1, 1), (1, 0, 0, 0)} is in B but
{(1, 0, 0, 0), (1, 2, 3, 0), (0, 2, 3, 0)} is not.
Next we let S be a chain in B. For example, we may take
S = {{(1, 0, 0, 0)}, {(1, 1, 0, 0), (1, 0, 0, 0)}, {(1, 2, 3, 0), (1, 0, 0, 0), (1, 1, 0, 0)}}. Note that
S is totally ordered (remember that the order is inclusion). Now the union of the three
sets in S is M = {(1, 2, 3, 0), (1, 0, 0, 0), (1, 1, 0, 0)} which is also in B, and M is an upper
bound for S.
Since this works for any chain in B, Zorns lemma shows that B contains at least one
maximal element Ω. In this finite dimensional case, we can actually find such an Ω
explicitly: just take for example Ω = {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)}. In any
case, Ω will be linearly independent and span V , so it is a basis.
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