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For the other way around, let x ∈ (AUB)'. Use a proof by contradiction to show that x must also be in A'UB'. Assume x∉A' which implies or some r>0, there is Br (x) ⊂ A, where A is equal to the empty set or x. Assume x∉B' which implies or some q>0, there is Bq (x) ⊂ B, where B is equal to the empty set or x. I am not sure where to go ...
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