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 A has empty interior.

Let UA be open. Since AA¯=A, this implies that UA. Since int(A) is the largest open subset of A, we must have Uint(A). Therefore UAint(A). But Aint(A)=(A¯-int(A))int(A)=, so U=.

Title boundary of a closed setPlanetmathPlanetmath 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