Basic Analysis – Gently Done Topological Vector Spaces
Basic Analysis – Gently Done Topological Vector Spaces
Basic Analysis – Gently Done Topological Vector Spaces
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58 <strong>Basic</strong> <strong>Analysis</strong><br />
Proposition 6.13 Any neighbourhood of 0 in a topological vector space X is<br />
absorbing.<br />
Proof For any x ∈ X and neighbourhood V of 0, the continuity of scalar multiplication<br />
(at (0,x)) ensures the existence of δ > 0 such that tx ∈ V whenever<br />
|t| < δ.<br />
In the vector space R 3 , we can readily visualise the proper subspaces (lines or<br />
planes) and we see that these can have no interior points, since any open set must<br />
contain a ball. In particular, no such subspace can be open. This result is quite<br />
general as we now show.<br />
Proposition 6.14 The only open linear subspace in a topological vector space X<br />
is X itself.<br />
Proof Suppose that M is an open linear subspace in a topological vector space X.<br />
Then 0 belongs to M so there is a neighbourhood V of 0 in X such that V ⊆ M.<br />
Let x ∈ X. Since V is absorbing, there is δ > 0 such that tx ∈ V whenever |t| < δ.<br />
In particular, tx ∈ M for some (suitably small) t with t �= 0. But M is linear, so<br />
it follows that x = t −1 tx ∈ M, and we have M = X.<br />
In fact, a proper linear subspace M of a topological vector space X can have<br />
no interior points at all. To see this, suppose that x is an interior point of M.<br />
Then there is a neighbourhood U of x with U ⊆ M. But then −x + U belongs<br />
to M and is a neighbourhood of zero. We now argue as before to conclude that<br />
M = X.<br />
Definition 6.15 A subset non-empty subset B in a vector space X is said to be<br />
balanced if tB ⊆ B for all t ∈ K with |t| ≤ 1.<br />
Note that any balanced set contains 0 and that if B is balanced, then B ⊆ rB<br />
for all r with |r| ≥ 1.<br />
Example 6.16 The balanced subsets of C, considered as a complex vector space,<br />
are all the discs (either with or without their boundary) withcentre at 0—together<br />
with C itself. If C is considered as a vector space over R, then the balanced sets<br />
are given by unions of line segments symmetrical about and passing through 0.<br />
For example, the rectangle {z ∈ C : |Rez| < 1 and |Imz| ≤ 1} is a balanced set<br />
in C when treated as a real vector space.<br />
Department of Mathematics King’s College, London