The boundary ∂A of A is the closure of A minus the interior of A. We need to prove that every x∈X such that every ball around x intersects ∂A belongs to ∂A. Let x∈X such that every ball around x intersects ∂A.
- Since ∂A is included in the closure of A, every ball around x intersects the closure of A. Since every closure is a closed set, x belongs to the closure of A.
- Assume that x is in the interior of A. Since the interior of A is an open set, there is a ball B around x that is included in the interior of A. Hence B does not intersect ∂A. This contradicts the condition that every ball around x intersects ∂A. Therefore, x is not in the interior of A.
Since x belongs to the closure of A and is not in the interior of A, x∈∂A .
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