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Home / Functional Analysis / Suppose K and C are subsets of a topological vector space X, K is compact, C is closed, and K n C = ∅. Then 0 has a neighborhood of V such that (K + V) n (C + V) = ∅.

Suppose K and C are subsets of a topological vector space X, K is compact, C is closed, and K n C = ∅. Then 0 has a neighborhood of V such that (K + V) n (C + V) = ∅.

Proof: If  K = ∅, then K + V = ∅, and the conclusion of the theorem

is obvious. We therefore assume that K # ∅, and consider a point

x ∈ K. Then x ∉ C ⟹ x ∈ CC. Since C is closed, ∃ a neighborhood W of x such that W ⊂ CC.

Since the topology of X is invariant under translations, – x + v is a neighborhood of  0 (zero). Hence ∃ a symmetric neighborhood vx of such that

vx + vx + vx ⊂ (- x + v)

⟹ x + vx + vx + vx ⊂ W

But W ⊂ CC

∴ x + vx + vx + vx ⊂ CC

⟹ (x + vx + vx + vx) ∩ C = ∅

⟹ (x + vx + vx) ∩ (C – vx ) = ∅

⟹ (x + vx + vx) ∩ (C + vx ) = ∅ [∵ vx is symmetric, ∴ – vx = vx]

Since {x+ vx: x ∈ K} is an open cover of k and K is compact, ∃  x1, x2, …, xn ∈ K.

K ⊂ (x1 + vx1) ∪ (x2 + vx2) ∪ … ∪ (xn + vxn)

Let, V=vx1 ∩  vx2 ∩ … ∩vxn. Then

And the no term in the last union intersect (C + V).

Hence (K + V) ∩ (C + V) = ∅

 

 

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