For a topological space $X$ and $A\subset X$ can topological boundary of $A$ contain an open set?
Boundary in the topological space
2 Answers
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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0Yes. I guess the argument is similar. – 2011-06-06
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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0That's also interesting for me) – 2011-06-08