Basic Analysis – Gently Done Topological Vector Spaces

6: Topological Vector Spaces 55
Proposition 6.5 Let X be a topological vector space. For given a ∈ X and
s ∈ K, with s �= 0, the translation map T a : x ↦→ x+a and the multiplication map
M s : x ↦→ sx, x ∈ X, are homeomorphisms of X onto itself.
Proof Let Φ : X × X → X and Ψ : K × X → X be the mappings introduced
above. Then, by continuity, if ((x α ,y α )) is a net which converges to (x,y) in
X ×X, we have
x α +y α = Φ((x α ,y α )) → Φ((x,y)) = x+y.
Similarly, if (s α ,x α ) → (s,x) in K×X, then
s α x α = Ψ((s α ,x α )) → Ψ((s,x)) = sx.
Hence, for any net (x α ) in X with x α → x, set y α = a for all α and s α = s for
all α, where s �= 0. Then (x α ,y α ) → (x,a) and (s α ,x α ) → (s,x) and we conclude
that
and
T a (x α ) = x α +a → x+a = T a (x)
M s (x) = sx α → sx = M s (x).
Thus both T a and M s are continuous maps from X into X. Furthermore, T a
has inverse T −a and M s has inverse M s −1, which are also continuous by the same
reasoning. Therefore T a and M s are homeomorphisms of X onto itself.
Corollary 6.6 For any open set G in a topological vector space X, the sets
a+G, A+G and sG are open, for any a ∈ X, A ⊆ X, and s ∈ K with s �= 0. In
particular, if U is a neighbourhood of 0, then so is tU for any t ∈ K with t �= 0.
Proof Suppose that G is open in X. For any a ∈ X and s ∈ K, we have
a+G = T a (G) and sG = M s (G). By the proposition, T a and M s , for s �= 0, are
homeomorphisms and so T a (G) and M s (G) are open sets. Now we simply observe
that A+G = �
a∈A (a+G), which is a union of open sets and so is itself open.
If U is a neighbourhood of 0, there is an open set G with 0 ∈ G ⊆ U. Hence
0 ∈ tG ⊆ tU and tG is open so that tU is a neighbourhood of 0.