Zero Functional Implies Dense Set

The Zero Functional Implies a Subspace is Dense Chris Johnson September 20, 2007 Suppose U is a subspace of a linear space X and assume that if f ∈ X ∗ and f |U = 0 implies f = 0. A corollary of the Hahn-Banach theorem states that if there exists an x ∈ X such that dist(x, U ) = d > 0 then there exists a functional f such that f |U = 0 and f (x) = d. Suppose such an x ∈ X exists and so dist(x, U ) = d and f (x) = d. However, f |U = 0 implies f = 0 (by our original assumption) and so f (x) = 0 so no such x exists. This means for every x ∈ X : dist(x, U ) = inf u∈U ||x − u|| = 0, and so x is arbitrarily close to points in U . Thus we can construct a sequence {un } such that un → x for each x ∈ X. This implies U¯ = X and so U is dense in X.

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