Applications
31.1
Statement of Zorn’s Lemma
Zorn’s lemma is a particularly useful form of the axiom of choice, especially for algebraic
applications. Readers who want to learn more about Zorn’s lemma and its applications to
algebra should consult either Lang [65], Appendix 2, §2 (pp. 878-884) and Chapter III, §5
(pp. 139-140), or Artin [3], Appendix §1 (pp. 588-589). For the logical ramifications of
Zorn’s lemma and its equivalence with the axiom of choice, one should consult Schwartz
[89], (Vol. 1), Chapter I, §6, or a text on set theory such as Enderton [32], Suppes [103], or
Kuratowski and Mostowski [64].
Given a set, S, a partial order, ≤, on S is a binary relation on S (i.e., ≤ ⊆ S × S) which
is
(1) reflexive, i.e., x ≤ x, for all x ∈ S,
(2) transitive, i.e, if x ≤ y and y ≤ z, then x ≤ z, for all x, y, z ∈ S, and
(3) antisymmetric, i.e, if x ≤ y and y ≤ x, then x = y, for all x, y ∈ S.
A pair (S, ≤), where ≤ is a partial order on S, is called a partially ordered set or poset.
Given a poset, (S, ≤), a subset, C, of S is totally ordered or a chain if for every pair of
elements x, y ∈ C, either x ≤ y or y ≤ x. The empty set is trivially a chain. A subset, P ,
(empty or not) of S is bounded if there is some b ∈ S so that x ≤ b for all x ∈ P . Observe
that the empty subset of S is bounded if and only if S is nonempty. A maximal element of
P is an element, m ∈ P , so that m ≤ x implies that m = x, for all x ∈ P . Zorn’s lemma
can be stated as follows:
Lemma 31.1. Given a partially ordered set, (S, ≤), if every chain is bounded, then S has a
maximal element.
843
844
CHAPTER 31. APPENDIX: ZORN’S LEMMA; SOME APPLICATIONS
Proof. See any of Schwartz [89], Enderton [32], Suppes [103], or Kuratowski and Mostowski
[64].
Remark: As we noted, the hypothesis of Zorn’s lemma implies that S is nonempty (since
the empty set must be bounded). A partially ordered set such that every chain is bounded
is sometimes called inductive.
We now give some applications of Zorn’s lemma.
31.2
Proof of the Existence of a Basis in a Vector Space
Using Zorn’s lemma, we can prove that Theorem 2.7 holds for arbitrary vector spaces, and
not just for finitely generated vector spaces, as promised in Chapter 2.
Theorem 31.2. Given any family, S = (ui)i∈I, generating a vector space E and any linearly
independent subfamily, L = (uj)j∈J, of S (where J ⊆ I), there is a basis, B, of E such that
L ⊆ B ⊆ S.
Proof. Consider the set L of linearly independent families, B, such that L ⊆ B ⊆ S. Since
L ∈ L, this set is nonempty. We claim that L is inductive. Consider any chain, (Bl)l∈Λ, of
linearly independent families Bl in L, and look at B =
B
l∈Λ
l. The family B is of the form
B = (vh)h∈H, for some index set H, and it must be linearly independent. Indeed, if this was
not true, there would be some family (λh)h∈H of scalars, of finite support, so that
λhvh = 0,
h∈H
where not all λh are zero. Since B =
B
l∈Λ
l and only finitely many λh are nonzero, there
is a finite subset, F , of Λ, so that vh ∈ Bf iff λ
h
h = 0. But (Bl)l∈Λ is a chain, and if we let
f = max{fh | fh ∈ F }, then vh ∈ Bf , for all vh for which λh = 0. Thus,
λhvh = 0
h∈H
would be a nontrivial linear dependency among vectors from Bf , a contradiction. Therefore,
B ∈ L, and since B is obviously an upper bound for the Bl’s, we have proved that L
is inductive. By Zorn’s lemma (Lemma 31.1), the set L has some maximal element, say
B = (uh)h∈H. The rest of the proof is the same as in the proof of Theorem 2.7, but we
repeat it for the reader’s convenience. We claim that B generates E. Indeed, if B does not
generate E, then there is some up ∈ S that is not a linear combination of vectors in B (since
S generates E), with p /
∈ H. Then, by Lemma 2.6, the family B = (uh)h∈H∪{p} is linearly
independent, and since L ⊆ B ⊂ B ⊆ S, this contradicts the maximality of B. Thus, B is
a basis of E such that L ⊆ B ⊆ S.
Another important application of Zorn’s lemma is the existence of maximal ideals.
31.3. EXISTENCE OF MAXIMAL PROPER IDEALS
845
31.3
Existence of Maximal Ideals Containing a Given
Proper Ideal
Let A be a commutative ring with identity element. Recall that an ideal A in A is a proper
ideal if A = A. The following theorem holds:
Theorem 31.3. Given any proper ideal, A ⊆ A, there is a maximal ideal, B, containing A.
Proof. Let I be the set of all proper ideals, B, in A that contain A. The set I is nonempty,
since A ∈ I. We claim that I is inductive. Consider any chain (Ai)i∈I of ideals Ai in A.
One can easily check that B =
A
i∈I
i is an ideal. Furthermore, B is a proper ideal, since
otherwise, the identity element 1 would belong to B = A, and so, we would have 1 ∈ Ai for
some i, which would imply Ai = A, a contradiction. Also, B is obviously an upper bound
for all the Ai’s. By Zorn’s lemma (Lemma 31.1), the set I has a maximal element, say B,
and B is a maximal ideal containing A.
846
CHAPTER 31. APPENDIX: ZORN’S LEMMA; SOME APPLICATIONS
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