Basic Analysis – Gently Done Topological Vector Spaces

Set δ = min{δ1 ,...,δ n }. Then, for any x ′ i ∈ Ui , 1 ≤ i ≤ n,
t 1 x ′ 1 +···+t n x′ n ∈ s 1 x 1 +V +···+s n x n +V
6: Topological Vector Spaces 57
⊆ s 1 x 1 +···+s n x n +(−z +W)
= W
whenever |t i −s i | < δ and the result follows.
Next, we shall show that the continuity of the linear operations implies that
the space is Hausdorff provided every point is closed.
Proposition 6.10 A topological vector space (X,T) is separated if and only if
every one-point set is closed.
Proof Of course, if T is a Hausdorff topology, then all one-point sets are closed.
For the converse, let x and y be distinct points in a topological vector space X,
and let w = x−y, so that w �= 0. Now, if {w} is closed, X \{w} is open. Indeed,
X \{w} is an open neighbourhood of 0. The continuity of addition (at 0) implies
that there are neighbourhoods U and V of 0 such that
U +V ⊆ X \{w}.
Since V is a neighbourhood of 0, so is −V (because the scalar multiplication M −1
is a homeomorphism) and therefore w −V is a neighbourhood of w (because the
translation T w is a homeomorphism). Any z ∈ w−V has the form z = w−v with
v ∈ V, and so w = z +v. Since w /∈ U +V, we must have that z /∈ U. Therefore
U ∩(w −V) = ∅, and so, translating by y, (y +U)∩(y +w−V) = ∅, that is,
(y +U)∩(x−V) = ∅. Hence x−V and y +U are disjoint neighbourhoods of x
and y, respectively, and X is Hausdorff.
Definition 6.11 A subset A in a vector space X is said to be absorbing if for any
x ∈ X there is µ > 0 such that x ∈ tA whenever |t| > µ, t ∈ K. Setting x = 0, it
follows that any absorbing set must contain 0. An equivalent requirement for A
to be absorbing is that for any x ∈ X there is some λ > 0 such that sx ∈ A for all
s ∈ K with |s| < λ.
Example 6.12 It is easy to see that a finite intersection of absorbing sets is absorbing.
However, it may happen that an infinite intersection of absorbing sets fails to
be absorbing. For example, let X be the linear space ℓ∞ of all bounded complex
sequences. For each m ∈ N, let Am be the subset consisting of those (an ) in ℓ∞ such that |am | < 1
m . It is clear that each Am is absorbing but their intersection
�
m A m , however, is not. Indeed, if x = (x n ) is the sequence with x n
n ∈ N, then sx is not in �
m A m
for any s �= 0.
= 1 for each