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For a topological space $X$ and $A\subset X$ can topological boundary of $A$ contain an open set?

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up vote 5 down vote accepted

Denote by $bd(S)$ the boundary of $S$.

Take $X=\Bbb{R}$ and $A=[0,1] \cap \Bbb{Q}$. Then $bd(A)=[0,1]$ which contains many open sets.

The statement turns true if we are speaking of open sets. Then if $A$ is an open set of $X$, we can write the disjoint decomposition $X=A \cup bd(A) \cup ext(A)$, where $ext(A)$ is the interior of $X\setminus A$. Suppose there exists $U \subset bd(A)$ open. Then pick $x \in U \subset bd(A)$. This makes $U$ an open neighborhood for $x$, which must intersect both $A$ and $ext(A)$, since $x$ is a boundary point. Contradiction with $U\subset bd(A)$.

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And what if $A$ is open itself? – Ilya Jun 6 '11 at 14:32
I have just answered that in my edit of the answer :) – Beni Bogosel Jun 6 '11 at 14:35
Am I right in the following statement: if $B$ is closed then $bd(B)$ does not contain any open sets? I use an argument that $B^c$ is open and $bd(B) = bd(B^c)$ – Ilya Jun 6 '11 at 15:00
Yes. I guess the argument is similar. – Beni Bogosel Jun 6 '11 at 16:29

Yes. Of course in general it contains the boring open set $\emptyset$. For a more exciting example,

$\partial \mathbb{Q}=\mathbb{R}$.

Indeed, the boundary of a dense set with empty interior is the whole space.

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Why do your answer get more upvotes than mine? – Beni Bogosel Jun 6 '11 at 16:38
That's also interesting for me) – Ilya Jun 8 '11 at 10:58

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