Basic Analysis – Gently Done Topological Vector Spaces

5: Vector Spaces 43
must entail α = 0. This means that λ 1 = ··· = λ k = 0, by independence,
and so x,u 1 ,...,u k are linearly independent. It follows that M ∪{x} is linearly
independent, which contradicts the maximality of M. We conclude that x can be
written as
x = λ 1 u 1 +···+λ m u m
for suitable m ∈ N, u 1 ,...,u m ∈ M, and non-zero λ 1 ,...,λ m ∈ K; that is, M is
a Hamel basis of X.
The next result is a corollary of the preceding method of proof.
Theorem 5.8 Let A be a linearly independent subset of a vector space X. Then
there is a Hamel basis of X containing A; that is, any linearly independent subset
of a vector space can be extended to a Hamel basis.
Proof Let S denote the collection of linearly independent subsets of X which
contain A. Then S is partially ordered by set-theoretic inclusion. As above, we
apply Zorn’s lemma to obtain a maximal element of S, which is a Hamel basis of
X and contains A.
Now we turn to the relationship between subspaces of a vector space and linear
functionals. First we need a definition.
Definition 5.9 A subspace V in a vector space X is said to be a maximal proper
subspace if it is not contained in any other proper subspace.
Proposition 5.10 Let V be a linear subspace of a vector space X. The following
statements are equivalent.
(i) V is a maximal proper subspace of X.
(ii) There is some z ∈ X, with z �= 0, such that X = {v +tz : v ∈ V, t ∈ K}.
(iii) V has codimension one, i.e., X/V has dimension one.
Proof Suppose that V is a maximal proper subspace of X. Then there is some
z ∈ X with z /∈ V. The set {v + tz : v ∈ V, t ∈ K} is a linear subspace of X
containing V as a proper subset. By the maximality of V, this set must be the
whole of X, thus, (ii) follows from (i).
Suppose that (ii) holds and let [x] be any member of X/V, i.e., [x] is the
equivalence class in X/V which contains the element x. Then, writing x as x =
v + tz, for some v ∈ V and t ∈ K, we see that x is equivalent to tz, i.e., [x] =
[tz] = t[z]. It follows that X/V is one-dimensional (with basis element given by
[z]), which proves (iii).