Basic Analysis – Gently Done Topological Vector Spaces

7: Locally Convex Topological Vector Spaces 69
Theorem 7.5 The topology T on a vector space X over K given by a family of
seminorms P is the weakest compatible topology on X such that each member of
P is continuous at 0.
Proof Let T ′ be the weakest vector space topology on X with respect to which
each member of P is continuous at 0. Certainly T ′ ⊆ T. Let p 1 ,...,p n ∈ P and
let r > 0. Then
V(0,p 1 ,...,p n ;r) =
n�
{x ∈ X : pi (x) < r}
i=1
belongs to T ′ since each {x ∈ X : p i (x) < r} = p −1
i ((−∞,r)) is in T ′ , 1 ≤ i ≤ n.
By hypothesis, translations are homeomorphisms and so T ′ contains all sets of the
form V(x,p 1 ,...,p n ;r) for x ∈ X. It follows that T ⊆ T ′ and therefore we have
equality T = T ′ .
Note that we did not use the joint continuity of addition (x,y) ↦→ x +y nor
that of scalar multiplication (t,x) ↦→ tx. This observation leads to the following.
Corollary 7.6 The topology T determined by a given family P of seminorms on a
vector space X is the weakest topology such that each member of P is continuous
at 0 and such that for each fixed x 0 ∈ X the translation x ↦→ x+x 0 is continuous.
Proof The proof is exactly as above—the fact that each translation map T x0 ,
x 0 ∈ X, is a homeomorphism follows from the continuity of the map x ↦→ x+x 0 ,
x ∈ X.
Remark 7.7 It should be noted that the vector space topology on a vector space
X determined by a family P of seminorms is not the same as the σ(X,P)-topology,
the weakest topologyon X making each member of P continuous. For example, let
X be the real vector space R and let P be the family with a single member p given
by p(x) = |x|, x ∈ R. We observe, incidentally, that this family is separating. For
any s,t ∈ R, s < t, p −1 ((s,t)) is equal to the symmetric set I ∪(−I), where I =
[0,∞)∩(s,t). The weakest topology on R making p continuous is that consisting
of all those subsets which are open in the usual sense and symmetrical about the
origin. The sequences (a n ) n∈N = ((−1) n ) n∈N and (b n ) n∈N = ((−1) n+1 ) n∈N both
converge to 1 since every neighbourhood containing the point 1 also contains the
point −1 (so this topology is not Hausdorff.) However, a n +b n = 0 for all n and so
the sequence (a n +b n ) does not converge to 1+1 = 2. We see that this topology
is not compatible with the vector space structure of R. Notice, however, that this
topology and the usual (vector space) topology on R do share a neighbourhood
base at 0; for example, the collection {(−r,r) : r > 0}. The point is that the