boundary of an open set is nowhere dense

This entry provides another example of a nowhere dense set.

Proposition 1.

If A is an open set in a topological spaceMathworldPlanetmath X, then A, the boundary of A is nowhere dense.


Let B=A. Since B=A¯A¯, it is closed, so all we need to show is that B has empty interior int(B)=. First notice that B=A¯A, since A is open. Now, we invoke one of the interior axioms, namely int(UV)=int(U)int(V). So, by direct computation, we have


The second equality and the inclusion follow from the general properties of the interior operationMathworldPlanetmath, the proofs of which can be found here ( ∎

Remark. The fact that A is open is essential. Otherwise, the propositionPlanetmathPlanetmathPlanetmath fails in general. For example, the rationals , as a subset of the reals under the usual order topology, is not open, and its boundary is not nowhere dense, as ¯¯==, whose interior is itself, and thus not empty.

Title boundary of an open set is nowhere dense
Canonical name BoundaryOfAnOpenSetIsNowhereDense
Date of creation 2013-03-22 17:55:41
Last modified on 2013-03-22 17:55:41
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 8
Author CWoo (3771)
Entry type Derivation
Classification msc 54A99