Basic Analysis – Gently Done Topological Vector Spaces

56 Basic Analysis
Remark 6.7 The continuity of the map x ↦→ x +a in a topological vector space
X has the following fundamental consequence. If V is a neighbourhood of a, then
b −a+V is a neighbourhood of b, for any a,b ∈ X. Suppose that N a is a local
neighbourhood base at a in X. Let b ∈ X, and let G be any neighbourhood of b.
Then T a−b (G) = a−b+Gisaneighbourhood ofa, sinceT a−b isahomeomorphism,
and so there is an element U in N a such that U ⊆ a−b+G. Hence b−a+U ⊆ G
and b−a+U is a neighbourhood of b. It follows that {b−a+U : U ∈ N a } is a
local neighbourhood base at b. In particular, given any local neighbourhood base
N 0 at 0, {a + U : U ∈ N 0 } is a local neighbourhood base at any point a ∈ X.
Now, in any topological space, if Nx is any open neighbourhood base at x, then
�
x∈X Nx is a base for the topology. It follows that in a topological vector space
the family of translates {x+N 0 : x ∈ X} is a base for the topology, for any open
local neighbourhood base N 0 at 0. Thus, the topological structure of a topological
vector space is determined by any local open neighbourhood base at 0.
Proposition 6.8 For any neighbourhood U of 0 there is a neighbourhood V of
0 such that V + V ⊆ U. Moreover, V can be chosen to be symmetric, i.e., such
that V = −V.
Proof As discussed above, for any neighbourhood U of 0, there are neighbourhoods
V 1 and V 2 of 0 such that V 1 + V 2 ⊆ U. The result follows by setting
V = V 1 ∩V 2 ∩(−V 1 )∩(−V 2 ).
This propositionoffers a substitute for the triangleinequality or ε/2-arguments
which areavailableinanormed space but not inageneral topologicalvector space.
Note that, by induction, for any neighbourhood U of 0 and any n ∈ N, there is
a (symmetric) neighbourhood V of 0 such that V + ···+V ⊆ U, where the left
hand side consists of n terms.
Corollary 6.9 For any topological vector space X and any n ∈ N, the map
((t 1 ,...,t n ),(x 1 ,...,x n )) ↦→ t 1 x 1 +···+t n x n is continuous from K n ×X n to X.
Proof Let((s 1 ,...,s n ),(x 1 ,...,x n )) ∈ K n ×X n , andletW beanyneighbourhood
of z = s 1 x 1 +···+s n x n in X. Then −z +W is a neighbourhood of 0 and so, by
Proposition 6.8, there is a neighbourhood V of 0 such that
V +···+V ⊆ −z +W.
� �� �
n terms
Now, for each 1 ≤ i ≤ n, s i x i +V is a neighbourhood of s i x i and therefore there is
δ i > 0 andaneighbourhoodU i ofx i such thattU i ⊆ s i x i +V whenever |t−s i | < δ i .
Department of Mathematics King’s College, London