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So I read parts of Rockafellar's "Convex Analysis".

When introducing "locally simplicial" sets, all the examples stated are convex sets, yet he mentions that they do not need to be convex.

I wonder whether all open sets in $\mathbb{R}^n$ are locally simplicial? I think such a fact would have been included in the reference if it were true.

My reasoning goes as follows.

A subset $S \subset \mathbb{R}^n$ is locally simplicial, if for all $x\in S$ there is a finite collection of simplices $\{S_1, \dots, S_m\}$ such that for some neighborhood $U$ of $x$ it holds $$ U \cap S = U \cap (S_1 \cup \dots \cup S_m).$$

If I assume $S$ to be open, then for every $x$ there is a ball with radius $\epsilon$ around $x$ that is still in $S$. If I place a single simplex $S_1$ that contains $x$ in its interior into that ball and take $U=S_1 \setminus \partial S_1$, I have that $U$ is a neighborhood of $x$ and it holds $$U=U\cap S = U \cap S_1.$$

  • You have answered the question in your title by "yes". So what is the purpose of your question? – Paul Frost Jul 21 '20 at 11:30
  • I want to know if there is a flaw in my reasoning. – Lasse H. Jul 21 '20 at 11:37
  • The finite number of simplices has to precede the "for every neighborhood" part. You can't choose a different simplex for each neighborhood of $x$. – Zim Jul 21 '20 at 19:43
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    @LasseH. Of course I knew that ;-) My comment intended to motivate you to formulate future question in a better way. (1) Use the additional tag "proof-verification". (2) Make sure that the text contains a definite question, for example "Is the following proof / argument correct?" As it is, your post asks a question in the title, but the text does not contain any open question, but answers the question in the title. – Paul Frost Jul 21 '20 at 22:20
  • Thank you, Paul. I'll take this into account next time. – Lasse H. Jul 22 '20 at 07:41

1 Answers1

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This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your proof is correct.

Paul Frost
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