Basic Analysis – Gently Done Topological Vector Spaces

6: Topological Vector Spaces 63
Proposition 6.26 For any non-empty sets A and B in a topological vector space
X,
(i) A+B ⊆ A+B, and
(ii) A = �
U∈N0 (A+U), where N0 is any neighbourhood base at 0.
Proof Let x ∈ A and y ∈ B. Then there are nets (a ν ) and (b ν ), indexed by
V 0 , such that a ν → x, and b ν → y. By continuity of addition, it follows that
a ν +b ν → x+y. Hence x+y ∈ A+B, which proves (i).
To prove (ii), suppose that x ∈ A and let N 0 be any neighbourhood base at 0.
For any U ∈ N 0 , −U is a neighbourhood of 0 and so x − U is a neighbourhood
of x. Since x ∈ A it follows that (x − U) ∩ A �= ∅, that is, x ∈ A + U, and so
A ⊆ �
�
(A + U). On the other hand, suppose that x ∈ (A + U) and
U∈N0 U∈N0
that V is any neighbourhood of x. Then −x+V is a neighbourhood of 0 and so
also is −(−x+V) = x−V. Therefore there is some U ∈ N0 such that U ⊆ x−V.
Now, x ∈ A + U, by hypothesis, and so x ∈ A + U ⊆ A + x − V. Hence there
is a ∈ A and v ∈ V such that x = a + x − v, i.e., a = v. It follows that every
neighbourhood of x contains some element of A and therefore x ∈ A.
Corollary 6.27 The closed balanced neighbourhoods of 0 form a local neighbourhood
base at 0.
Proof Let U be any neighbourhood of 0. By Proposition 6.8, there is a neighbourhood
V of 0 such that V + V ⊆ U. Also, by Proposition 6.18, there is a
balanced neighbourhood of 0, W, say, with W ⊆ V, so that W + W ⊆ U. By
Proposition 6.26, W ⊆ W +W ⊆ U, and W is a neighbourhood of 0. We claim
that W is balanced. To see this, let x ∈ W and let t ∈ K with |t| ≤ 1. There is a
net (w ν ) in W such that w ν → x. Then tw ν → tx, since scalar multiplication is
continuous. But tw ν ∈ W, since W is balanced, and so tx ∈ W. It follows that
W is balanced, as claimed.
Proposition 6.28 If M is a linear subspace of a topological vector space X, then
so is its closure M. In particular, any maximal proper subspace is either dense or
closed.
Proof Let M be a linear subspace of the topological vector space X. We must
show that if x,y ∈ M and t ∈ K, then tx+y ∈ M. There are nets (x ν ) and (y ν ) in
M, indexed by the base of all neighbourhoods of 0, such that x ν → x and y ν → y.
It follows that tx ν → tx and tx ν +y ν → tx+y and we conclude that tx+y ∈ M,
as required.
If M is a maximal proper subspace, the inclusion M ⊆ M implies that either
M = M, in which case M is closed, or M = X, in which case M is dense in X.