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$\;\;\;\;$ Assume $n>1$ and $U$ is a non-empty bounded open convex subset of $\Bbb R^n$. Is it possible that the boundary $\partial U$ of $U$ is also convex or must it be the case that $\partial U$ is not convex ?

$\;\;\;\;$ My intuition immediately says no. The question was asked by Cathy, but with no mention of openness, boundedness or dimension, and answered today in the affirmative by Achille Hui and Paul Frost using the open interval $(0,\infty)$ in the case $n=1$.

Oliver Kayende
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    Look at a half-space. It is the same example, just multiply by the extra factors. $(0,+\infty)\times\mathbb{R}^{n-1}$. Its boundary is the subspace ${0}\times\mathbb{R}^{n-1}$. – conditionalMethod Nov 29 '19 at 20:25
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    Could it also be bounded? Take into account this. – conditionalMethod Nov 29 '19 at 20:31
  • @conditionalMethod. Thank you for the simple generalization. I was imagining a bounded open convex set (non-empty). It seems clear to me in $\Bbb R^1$ that the boundary is not convex and intuitive for me in $\Bbb R^n$ but I want a simple proof. – Oliver Kayende Nov 29 '19 at 20:59
  • @Matthew Daly. Sorry about the confusion. I am indeed asking a different question. – Oliver Kayende Nov 29 '19 at 21:03
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    Well, the proof is not so simple. If the set is bounded, then its closure is convex and compact. By Krein-Milman, its boundary's closed convex hull is the entire set. But if the boundary is convex, then the set is its own boundary. This cannot be for a set with non-empty interior. – conditionalMethod Nov 29 '19 at 21:07
  • @conditionalMethod. Thanks. I think that was pretty straightforward even though you employed the Krein-Milman Theorem. – Oliver Kayende Nov 29 '19 at 21:12

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Choose $x$ be a point on the interior of $U$ (since it was given to be open and nonempty, it must have a non-empty interior). Consider any line $\ell$ that passes through $x$. This line must intersect $\partial U$ on each side of $x$ since $U$ is bounded. Choose $y,z$ to be such points on either side of $x$. Then $y,z\in\partial U$, but $[y,z]\not\subset\partial U$ since $x\in[y,z]$ and $x\not\in\partial U$. Therefore $\partial U$ is not convex.