Question #114844
Prove that the boundary of a subset A of a metric space X is always a closed set
1
Expert's answer
2020-05-11T18:21:03-0400

The boundary A\partial A of AA is the closure of AA minus the interior of AA. We need to prove that every xXx\in X such that every ball around xx intersects A\partial A belongs to A\partial A. Let xXx\in X such that every ball around xx intersects A\partial A.

  • Since A\partial A is included in the closure of AA, every ball around xx intersects the closure of AA. Since every closure is a closed set, xx belongs to the closure of AA.
  • Assume that xx is in the interior of AA. Since the interior of AA is an open set, there is a ball BB around xx that is included in the interior of AA. Hence BB does not intersect A\partial A. This contradicts the condition that every ball around xx intersects A\partial A. Therefore, xx is not in the interior of AA.

Since xx belongs to the closure of AA and is not in the interior of AA, xAx\in \partial A .


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