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# 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 $\partial A$, the boundary of $A$ is nowhere dense.

###### Proof.

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

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

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

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 $\overline{\mathbb{Q}}\cap\overline{\mathbb{Q}^{\complement}}=\mathbb{R}\cap%
\mathbb{R}=\mathbb{R}$, whose interior is $\mathbb{R}$ itself, and thus not empty.

## Mathematics Subject Classification

54A99*no label found*

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