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Prove/Disprove: for all closed $S\subseteq\mathbb{R}^n$, there exists some $A\subseteq\mathbb{R}^n$ s.t $$\partial A=S$$ I've tried defining $$A=\partial S\uplus\left(\mathbb{Q}^{n}\cap\text{int}(S)\right)$$ because intuitively it feels to me like it would satisfy the equality, but i did not succeed in proving that $\partial A=S$.

MJD
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Ziv
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  • Consider $S =$ a closed ball. – I H Dec 30 '21 at 14:22
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    @IH How about $S\cap\mathbb Q^n$? –  Dec 30 '21 at 14:22
  • @xpaul How about $S\cap\mathbb Q^n$? –  Dec 30 '21 at 14:25
  • @Stinking Bishop Ah, I see. Now this question seems indeed interesting. – I H Dec 30 '21 at 14:26
  • I think the OP is well aware of this approach with intersecting with $\mathbb Q^n$. It seems to work. Does it work in all cases, or are there counterexamples? That is the question here. –  Dec 30 '21 at 14:27
  • Here's a different approach to try: Let $B$ be a countable base for $\mathbb{R}^n$. For each $U_n\in B$ such that $U_n\cap S \neq \emptyset$, pick an element $x_n\in U_n\cap S$. The set of all these $x_n$ should be a good candidate for $A$.

    Although I see Florian has given a proof that your answer works.

     – Brandon du Preez Dec 30 '21 at 15:05
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    I'm skeptical of this $\Bbb Q^n$ approach. What if $S$ is a segment of a curve that contains no rational points? – Karl Dec 30 '21 at 15:17
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    @BrandonduPreez That your approach works can even be proved much quicker! – Florian R Dec 30 '21 at 15:21
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    @Karl That is why it's important to include $\partial S$ in the set $A$. That takes care of it. – Florian R Dec 30 '21 at 15:22
  • Oh, I see. In my example you get $A=S$, which works. – Karl Dec 30 '21 at 15:29

1 Answers1

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Can you verify the following two facts?

(i) The set $A$ as you defined has empty interior.

(ii) The closure of the set $\mathbb{Q}^n \cap \mathrm{int}(S)$ is the closure of $\mathrm{int}(S)$.

Then you can conclude as follows: $$ \partial A = \bar{A} \backslash \mathrm{int}(A) = \bar{A} = \overline{\partial S \cup (\mathbb{Q}^n \cap \mathrm{int}(S))} = \overline{\partial S} \cup \overline{\mathbb{Q}^n \cap \mathrm{int}(S)} = \partial S \cup \overline{\mathrm{int}(S)} = S. $$

If you'd like to see a proof of the two facts, I'll gladly add them.

Florian R
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