# 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 $\mathbb{R}$ under the usual order topology, is not open, and its boundary is not nowhere dense, as $\overline{\mathbb{Q}}\cap \overline{{\mathbb{Q}}^{\mathrm{\complement}}}=\mathbb{R}\cap \mathbb{R}=\mathbb{R}$, whose interior is $\mathbb{R}$ 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 |