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Home / Functional Analysis / If ℬ is a local base for a topological vector space X, then every member of ℬ contains the closure of some member of ℬ.

If ℬ is a local base for a topological vector space X, then every member of ℬ contains the closure of some member of ℬ.

Proof: Let W ∈ ℬ. Set K = { 0 } and C = WC. Then C is closed set and K is compact set and C ∩ K = Ф.

Hence ∃ a neighborhood V of 0 such that (C+ V) ∩ (K+V) = Ф.

Since B is a local base ∃ U∈ ℬ such that U ⊂ V.

(C+ U) ∩ (K+U) = Ф.

Thus (C+ U) ∩ U = Ф.                          [∵K+U = U]

and hence U ∩C+ U = Ф

Since C⊂ C + U,  U ∩C = Ф

U⊂ CC

U⊂ W (Proved)

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