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 space $X$, then $\mathrm{\partial }\mathit{}A$, the boundary of $A$ is nowhere dense.

Proof.

Let $B=\partial A$. Since $B=\overline{A}\cap \overline{A^{\mathrm{\complement }}}$, it is closed, so all we need to show is that $B$ has empty interior $\mathrm{int}(B)=\mathrm{\varnothing }$. First notice that $B=\overline{A}\cap A^{\mathrm{\complement }}$, since $A$ is open. Now, we invoke one of the interior axioms, namely $\mathrm{int}(U\cap V)=\mathrm{int}(U)\cap \mathrm{int}(V)$. So, by direct computation, we have

$$\mathrm{int}(B)=\mathrm{int}(\overline{A})\cap \mathrm{int}(A^{\mathrm{\complement }})=\mathrm{int}(\overline{A})\cap {\overline{A}}^{\mathrm{\complement }}\subseteq \overline{A}\cap {\overline{A}}^{\mathrm{\complement }}=\mathrm{\varnothing }.$$

The second equality and the inclusion follow from the general properties of the interior operation, the proofs of which can be found here (http://planetmath.org/DerivationOfPropertiesOnInteriorOperation). ∎

Remark. The fact that $A$ is open is essential. Otherwise, the proposition fails in general. For example, the rationals $\mathbb{Q}$, as a subset of the reals $ℝ$ under the usual order topology, is not open, and its boundary is not nowhere dense, as $\overline{\mathbb{Q}}\cap \overline{{\mathbb{Q}}^{\mathrm{\complement }}}=ℝ\cap ℝ=ℝ$, whose interior is $ℝ$ 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