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Unformatted text preview: TOPOLOGICAL VECTOR SPACES 1. Topological vector spaces and local base Definition 1.1 . A topological vector space is a vector space over R or C with a topology τ such that • every point is closed; • the vector space operations are continuous. This means that X \ { x } is open and the operations (1.1) X × X ( x, y ) → x + y ∈ X, F × X ( k, x ) → kx ∈ X are continuous, where F = R or F = C and X × X and F × X have the product topology. Since the operations (1.2) T a x = x + a M λ x = λx are homeomorphism of X , then the topology is translation invariant . This means that a set E is open if and only if E + a is open. As a consequence, the topology τ is completely determined by the neighborhoods of 0. We define a local base B of 0 by requiring that B ∈ B are open and every neighborhood of 0 contains a set in B . Clearly the topology τ is generated by the sets E = [ α,β ( x α + B β ) , x α ∈ X, B β ∈ B , i.e. by unions of translated sets belonging to the local base at 0. In the following by local base B we denote the local base at 0. We say that a set E in a topological vector space is bounded if for every neighborhood V of 0 there corresponds a s ∈ F such that E ⊂ tV for all t > s . 2. Separation properties Aim of this section is to prove that the compatibility of the topology with the vector operation implies some good properties on the structure of the local base B . We say: • X is locally convex if there is a local base B whose members are convex, i.e. (2.1) aB + (1 a ) B ⊂ B ∀ B ∈ B ; • X is locally bounded if 0 has a bounded neighborhood, i.e. there is A ∈ B such that (2.2) ∀ B ∈ B ∃ s ∈ F : ∀ t > s, A ⊂ tB. One can check that this implies that the local base B can be chosen to contain only bounded neighborhood of 0; • X is locally compact if 0 has a neighborhood with compact closure. As before, this implies that B can be generated by relatively compact open neighborhood of 0. The first result is that the topological vector spaces are “normal” topological spaces. Proposition 2.1. If K is compact, C closed, K ∩ C = ∅ then there is V neighborhood of such that ( K + V ) ∩ ( C + V ) = ∅ . Proof. If K = ∅ , there nothing to prove. Otherwise, let x ∈ K : by the invariance with translation, we can assume x = 0. Then X \ C is an open set of 0. Since addition is continuous, by 0 = 0 + 0 + 0, there is a neighborhood U such that 3 U ⊂ X \ C. 1 2 TOPOLOGICAL VECTOR SPACES By defining ˜ U = U ∩ ( U ) ⊂ U, we have that ˜ U is open, symmetric and 3 ˜ U ⊂ X \ C . This means that ∅ = n 3 x, x ∈ ˜ U o ∩ C = n 2 x o ∩ n y x, y ∈ C, x ∈ ˜ U o ⊃ ˜ U ∩ ( C + ˜ U ) ....
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 Spring '10
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 Logic, Topology, Vector Space

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