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 $U=\emptyset$.

Title boundary of a closed set is nowhere dense BoundaryOfAClosedSetIsNowhereDense 2013-03-22 18:34:01 2013-03-22 18:34:01 neapol1s (9480) neapol1s (9480) 4 neapol1s (9480) Derivation msc 54A99