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 ∈ C^{C}. Since C is closed, ∃ a neighborhood W of x such that W ⊂ C^{C}.

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

v_{x} + v_{x} + v_{x} ⊂ (- x + v)

⟹ x + v_{x} + v_{x} + v_{x} ⊂ W

But W ⊂ C^{C}

∴ x + v_{x} + v_{x} + v_{x} ⊂ C^{C}

⟹ (x + v_{x} + v_{x} + v_{x}) ∩ C = ∅

⟹ (x + v_{x} + v_{x}) ∩ (C – v_{x} ) = ∅

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

Since {x+ v_{x}: x ∈ K} is an open cover of k and K is compact, ∃ x_{1}, x_{2}, …, x_{n} ∈ K.

K ⊂ (x_{1} + v_{x1}) ∪ (x_{2} + v_{x2}) ∪ … ∪ (x_{n} + v_{xn})

Let, V=v_{x1} ∩ v_{x2} ∩ … ∩v_{xn. }Then

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

Hence (K + V) ∩ (C + V) = ∅