Basic Analysis – Gently Done Topological Vector Spaces

6: Topological Vector Spaces 65
Proposition 6.34 Anyconvergent sequence intopologicalvector spaceisbounded.
Proof Suppose that (x n ) is a sequence in a topological vector space (X,T) such
that x n → z. For each n ∈ N, set y n = x n − z, so that y n → 0. Let U be
any neighbourhood of 0. Let V be any balanced neighbourhood of 0 such that
V ⊆ U. Then V ⊆ sV for all s with |s| ≥ 1. Since y n → 0, there is N ∈ N
such y n ∈ V whenever n > N. Hence y n ∈ V ⊆ tV ⊆ tU whenever n > N and
t ≥ 1. Set A = {y 1 ,...,y N } and B = {y n : n > N}. Then A is a finite set so
is bounded and therefore A ⊆ tU for all sufficiently large t. But then it follows
that A∪B ⊆ tU for sufficiently large t, that is, {y n : n ∈ N} is bounded and so is
{x n : n ∈ N} = z +(A∪B).
Remark 6.35 A convergent net in a topologicalvector space need not be bounded.
For example, let I be R equipped with its usual order and let x α ∈ R be given by
x α = e −α . Then (x α ) α∈I is an unbounded but convergent net (with limit 0) in
the real normed space R.