Basic Analysis – Gently Done Topological Vector Spaces

54 Basic Analysis
subsets of X. We shall use the following notation:
A+B = {x ∈ X : x = a+b, a ∈ A, b ∈ B}
tA = {x ∈ X : x = ta, a ∈ A}, for t ∈ K
A+z = A+{z} = {x ∈ X : x = a+z, a ∈ A}, for z ∈ X.
It should be noted that (s + t)A ⊆ sA + tA but equality need not hold. For
example, ifX = C and A = {1,i}, then 2A = {2,2i}whereas A+A = {2,2i,1+i}.
Remark 6.3 Suppose that X is a topological vector space. Let Φ : X ×X → X
and Ψ : K×X → X denote the mappings Φ((x,y)) = x +y and Ψ((t,x)) = tx,
x,y ∈ X, t ∈ K. By definition, these maps are continuous. In particular, for any
x,y ∈ X, and any neighbourhood W of Φ(x,y), there are neighbourhoods U of x
and V of y such that Φ(U ×V) ⊆ W. That is to say, for any x,y ∈ X, and any
neighbourhood W of x+y, there are neighbourhoods U of x and V of y such that
U +V ⊆ W.
Similarly, the continuity of scalar multiplication implies that for given s ∈ K,
x ∈ X and any neighbourhood W of Ψ(s,x), there is a neighbourhood V of s in
K and a neighbourhood U of x in X such that Ψ(V ×U) ⊆ W. Now, any such V
contains a set of the form {t ∈ K : |t−s| < δ}, for some δ > 0. Hence we may say
that for any neighbourhood W of sx there is a neighbourhood U of x and δ > 0
such that tU ⊆ W for all t ∈ K with |t−s| < δ.
Example 6.4 We know that any non-empty set can be equipped with a Hausdorff
topology(for example, the discrete topology),so onemight ask whether every nontrivialvectorspace(i.e.,notequalto{0})canbegivenaHausdorffvectortopology.
The discrete topology is too fine for this. Indeed, each point is a neighbourhood
of itself in the discrete topology, and so addition is certainly continuous—the
neighbourhood {x} × {y} of (x,y) ∈ X × X is mapped into the neighbourhood
{x+y}ofx+y under addition. (WearetakingU = {x}, V = {y}andW = {x+y}
in the preceding discussion.) However, scalar multiplication is not continuous—for
example, if W = {sx 0 }, a neighbourhood of sx 0 , then tx 0 belongs to W only if
t = s. Thus there is no neighbourhood V of s in K such that tx 0 ∈ W for all
t ∈ V.
However, every vector space X can be normed. To see this, let B be a Hamel
basis for X. Any x ∈ X, with x �= 0, can be written uniquely as
x = t 1 u 1 +···+t n u n
for suitable u 1 ,...,u n in B and non-zero t 1 ,...,t n in K. Set
�x� = |t 1 |+···+|t n |
and �0� = 0. It is clear that �·� is, indeed, a norm on X. The topology induced
by a norm is Hausdorff so that X furnished with this norm (or, indeed, any other)
becomes a topological vector space.
Department of Mathematics King’s College, London