$A\subset R^n$ is closed. Proof that there exists $B \subset R^n$ such that $\partial B=A$

Asked Sep 17 '23 at 17:22

Active Sep 18 '23 at 12:52

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Where $\partial B$ is the frontier or boundary of $B$. I tried using some relationships like:

$\bar{A} = int(A) \sqcup \partial A$
$\partial{A} = \partial A^C$
$int(A^C) = \bar{A}^C$

trying to say that $B=A^C$ and maybe reach something but i didn't reach anything. Another way could be using the definition of frontier and closed sets and try to find something but i haven't been succesful.

I would appreciate a hint, solution or guidance on how to proof the title of the question.

Thank you

edited Sep 18 '23 at 12:52

asked Sep 17 '23 at 17:22

Sergio2405

3

Ok rewrite the question in the corpus plz – julio_es_sui_glace Sep 17 '23 at 17:32
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Find some interior empty set that verifies $\bar{B} = A$. You can use that $\mathbb{Q}^n$ is dense – julio_es_sui_glace Sep 17 '23 at 17:35
$B=A$, they are same points. – Bob Dobbs Sep 17 '23 at 17:46
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https://math.stackexchange.com/questions/151265/every-closed-set-in-mathbb-r2-is-boundary-of-some-other-subset – Hgtcl Sep 17 '23 at 17:54
Elements of $\mathbb R^n$ are not the same as subspaces. – M W Sep 17 '23 at 17:57
@BobDobbs that only works when $A$ is nowhere dense. – M W Sep 17 '23 at 17:58
You certainly mean $A \subset \mathbb R^n$ and $B \subset \mathbb R^n$. – Paul Frost Sep 18 '23 at 10:07

$A\subset R^n$ is closed. Proof that there exists $B \subset R^n$ such that $\partial B=A$

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