Basic Analysis – Gently Done Topological Vector Spaces

62 Basic Analysis
For each 1 ≤ i ≤ n, let ℓ i : K n → K be the projection map ℓ i ((t 1 ,...,t n )) = t i .
Then ℓ i ◦ ϕ −1 : X → K is a linear functional such that |ℓ i ◦ ϕ −1 (x)| < 1 for all
x ∈ W. By Theorem 6.21, ℓ i ◦ϕ −1 is continuous, for each 1 ≤ i ≤ n, and therefore
ϕ −1 : x ↦→ (ℓ 1 ◦ϕ −1 (x),...,ℓ n ◦ϕ −1 (x))
is continuous. We conclude that ϕ : K n → X is a linear homeomorphism.
Now suppose that | · | is a norm on the finite dimensional vector space X.
From the above, there is a linear homeomorphism ψ : X → K n , where n is the
dimension of X, and so there are positive constants m and M such that, for any
x ∈ X,
�ψ(x)� ≤ M |x| since ψ is continuous
and
|ψ −1 (x)| ≤ m�x� since ψ −1 is continuous
where �·� is the usual Euclidean norm on K n . Hence
1
|x| ≤ �ψ(x)� ≤ M |x|
m
which means that |· | is equivalent to the norm x ↦→ �ψ(x)�, x ∈ X. The result
follows since equivalence of norms is an equivalence relation.
Corollary 6.24 Every linear map from a finite dimensional Hausdorff topological
vector space into a topological vector space is continuous.
Proof Suppose that X and Y are topological vector spaces, with X finite dimensional
and Hausdorff, and let T : X → Y be a linear map. Let {x 1 ,...,x n } be a
basis for X. Then T is the composition of the maps
x = t 1 x 1 +···+t n x n (∈ X) ↦→ (t 1 ,...,t n )(∈ K n ) ↦→ t 1 Tx 1 +···+t n Tx n ∈ Y.
The second of these is continuous, and by Theorem 6.23, so is the first and so,
therefore, is T.
Proposition 6.25 For any non-empty subset A in a topological vector space X
and any x ∈ A, there is a net (a U ) in A, indexed by V 0 , the set of neighbourhoods
of 0 (partially ordered by reverse inclusion), such that a U → x.
Proof ForanyU ∈ V 0 ,x+U isaneighbourhoodofx. Sincex ∈ A,(x+U)∩A �= ∅.
Let a U be any element of (x+U)∩A. Then (a U ) is a net in A such that a U → x.
Indeed, for each U ∈ V 0 , a U −x ∈ U and so a U −x → 0 along V 0 , that is, a U → x.
Department of Mathematics King’s College, London