Basic Analysis – Gently Done Topological Vector Spaces

7: Locally Convex Topological Vector Spaces 67
If this intersection is empty, there is no more to be done, so suppose that x ∈
A ∩ B. Then x ∈ A and x ∈ B and so there exist U,V ∈ N x such that U ⊆ A
and V ⊆ B. Suppose that U = V(x,p 1 ,...,p m ;r) and V = V(x,q 1 ,...,q n ;s).
Put W = V(x,p 1 ,...,p m ,q 1 ,...,q n ;t) where t = min{r,s}. Then W ∈ N x and
W ⊆ U ∩V ⊆ A∩B. It follows that T is a topology on X.
Let x ∈ X and let U ∈ N x . We shall show that U is open. (The method is the
analogue of showing that for any point in an open disc D in R 2 , there is a (possibly
very small) disc centred on the given point and wholly contained in D.) Suppose
that U = V(x,p 1 ,...,p n ;r) and that z ∈ U. Then p i (z − x) < r for 1 ≤ i ≤ n.
Let δ > 0 be such that δ < r−p i (z−x) for 1 ≤ i ≤ n. For any 1 ≤ i ≤ n and any
y ∈ X with p i (y −z) < δ, we have
p i (y −x) ≤ p i (y −z)+p i (z −x)
< δ +p i (z −x)
< r.
Hence V(z,p 1 ,...,p n ;δ) ⊆ V(x,p 1 ,...,p n ;r) = U and so U ∈ T. Thus Nx is a
neighbourhood base at x, consisting of open sets.
Next we shall show that T is a compatible topology. To this end, suppose that
(xν ,yν ) → (x,y) in X ×X. We want to show that xν +yν → x+y. Let V(x+
y,p1 ,...,p n ;r) be a given basic neighbourhood of x+y. Since (xν ,yν ) → (x,y),
there is ν0 such that (xν ,yν ) ∈ V(x,p 1 ,...,p n ; r
2 ) ×V(y,p 1 ,...,p r
n ; ) whenever 2
ν � ν0 . For any 1 ≤ i ≤ n and ν � ν0 p i (x+y −(x ν +y ν )) ≤ p i (x−x ν )+p i (y −y ν )
< r r
+
2 2
and so x ν +y ν ∈ V(x+y,p 1 ,...,p n ;r). Hence x ν +y ν → x+y, as required.
Now suppose that (t ν ,x ν ) → (t,x) in K×X. Let V(tx,p 1 ,...,p k ;r) be a basic
neighbourhood of tx. For any given ε > 0 and s > 0, there is ν 0 such that
(t ν ,x ν ) ∈ {ζ ∈ K : |ζ −t| < ε}×V(x,p 1 ,...,p k ;s).
Hence, for all 1 ≤ i ≤ k and ν � ν 0 ,
p i (tx−t ν x ν ) ≤ p i (tx−t ν x)+p i (t ν x−t ν x ν )
≤ |t−t ν |p i (x)+|t ν |p i (x−x ν )
< εp i (x)+(|t|+ε)s
< r
, 1 ≤ i ≤ k, and then s > 0 such that
. We conclude
that (t,x) ↦→ tx is continuous and therefore T is a vector space topology on X.
if we choose ε > 0 such that εpi (x) < r
2
(|t|+ε)s < r
2 . Therefore tνxν ∈ V(tx,p 1 ,...,p k ;r) whenever ν � ν0