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 $\mathrm{int}(A)$ is the largest open subset of $A$, we must have $U\subset \mathrm{int}(A)$. Therefore $U\subset \partial A\cap \mathrm{int}(A)$. But $\partial A\cap \mathrm{int}(A)=(\overline{A}-\mathrm{int}(A))\cap \mathrm{int}(A)=\mathrm{\varnothing }$, so $U=\mathrm{\varnothing }$.

Title	boundary of a closed set is nowhere dense
Canonical name	BoundaryOfAClosedSetIsNowhereDense
Date of creation	2013-03-22 18:34:01
Last modified on	2013-03-22 18:34:01
Owner	neapol1s (9480)
Last modified by	neapol1s (9480)
Numerical id	4
Author	neapol1s (9480)
Entry type	Derivation
Classification	msc 54A99