Basic Analysis – Gently Done Topological Vector Spaces

42 Basic Analysis
Zorn’s lemma can be shown to be equivalent to Hausdorff’s maximality principle
and the axiom of choice. These are the following.
Hausdorff’s maximality principle Any non-empty partially ordered set contains a
maximal chain, i.e., a totally ordered subset maximal with respect to being totally
ordered.
Axiom of Choice Let {Aα : α ∈ J} be a family of (pairwise disjoint) non-empty
sets, indexed by the non-empty set J. Then there is a mapping ϕ : J → �
αAα such that ϕ(α) ∈ Aα for each α ∈ J.
Thus, the axiom says that we can ‘choose’ a family {a α } with a α ∈ A α , for
each α ∈ J, namely, the range of ϕ. (The requirement that the {A α } be pairwise
disjoint is not essential and can easily be removed—by replacing A α by B α =
{(α,a) : a ∈ A α }.) As a consequence, this axiom gives substance to the cartesian
product �
α A α .
We are now in a position to attack the existence problem of a Hamel basis.
Theorem 5.7 Every vector space X (�= {0}) possesses a Hamel basis.
Proof Let S denote the collection of linearly independent subsets of X, partially
ordered by inclusion. Let {Sα : α ∈ J} be a totally ordered subset of S. Put
S = �
αSα . We claim that S is linearly independent. To see this, suppose that
x 1 ,...,x m are distinct elements of S and suppose that
λ 1 x 1 +···+λ m x m = 0
for non-zero λ1 ,...,λ m ∈ K. Then x1 ∈ Sα1 ,...,x m ∈ Sαm for some α1 ,...,α m ∈
J. Since {Sα } is totally ordered, there is some α ′ ∈ J such that Sα1 ⊆ Sα ′, ... ,
Sαm ⊆ Sα ′. Hence x1 ,...,x m ∈ Sα ′. But Sα ′ is linearly independent and so we
must have that λ1 = ··· = λm = 0. We conclude that S is linearly independent,
as claimed.
It follows that S is an upper bound for {Sα } in S. Thus every totally ordered
subset in S has an upper bound and so, by Zorn’s lemma, S possesses a maximal
element, M, say. We claim that M is a Hamel basis.
To see this, let x ∈ X, x �= 0, and suppose that an equality of the form
x = λ 1 u 1 +···+λ k u k
is impossible for any k ∈ N, distinct elements u 1 ,...,u k ∈ M and non-zero
λ 1 ,...,λ k ∈ K. Then, for any distinct u 1 ,...,u k ∈ M, an equality of the form
αx+λ 1 u 1 +···+λ k u k = 0
Department of Mathematics King’s College, London