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[parent] boundary of an open set is nowhere dense (Derivation)

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
$\displaystyle \operatorname{int}(B)=\operatorname{int}(\overline{A})\cap \opera... ...}^\complement \subseteq \overline{A}\cap \overline{A}^\complement =\varnothing.$
The second equality and the inclusion follow from the general properties of the interior operation, the proofs of which can be found here. $ \qedsymbol$

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.



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Cross-references: order topology, reals, subset, rationals, proposition, operation, properties, inclusion, equality, interior axioms, interior, closed, boundary, topological space, open set, nowhere dense

This is version 5 of boundary of an open set is nowhere dense, born on 2008-03-20, modified 2008-03-20.
Object id is 10422, canonical name is BoundaryOfAnOpenSetIsNowhereDense.
Accessed 143 times total.

Classification:
AMS MSC: 54A99 (General topology :: Generalities :: Miscellaneous)
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