boundary of a closed set is nowhere dense

Let $A$ be closed. In general, the boundary of a set is closed. So it suffices to show that $\\partial A$ has empty interior.

Let $U\\subset\\partial A$ be open. Since $\\partial A\\subset\\overline{A}=A$ , this implies that $U\\subset A$ . Since $\\operatorname{int}(A)$ is the largest open subset of $A$ , we must have $U\\subset\\operatorname{int}(A)$ . Therefore $U\\subset\\partial A\\cap\\operatorname{int}(A)$ . But $\\partial A\\cap\\operatorname{int}(A)=(\\overline{A}-\\operatorname{int}(A))\\cap%\\operatorname{int}(A)=\\emptyset$ , so $U=\\emptyset$ .

单词	BoundaryOfAClosedSetIsNowhereDense
释义	boundary of a closed set is nowhere dense Let $A$ be closed. In general, the boundary of a set is closed. So it suffices to show that $\\partial A$ has empty interior. Let $U\\subset\\partial A$ be open. Since $\\partial A\\subset\\overline{A}=A$ , this implies that $U\\subset A$ . Since $\\operatorname{int}(A)$ is the largest open subset of $A$ , we must have $U\\subset\\operatorname{int}(A)$ . Therefore $U\\subset\\partial A\\cap\\operatorname{int}(A)$ . But $\\partial A\\cap\\operatorname{int}(A)=(\\overline{A}-\\operatorname{int}(A))\\cap%\\operatorname{int}(A)=\\emptyset$ , so $U=\\emptyset$ .
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