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Let $X\subseteq \mathbb{R}^n$ be an open set and let $Y\subseteq X$ be convex. Can I always construct an open and convex set $Z$ such that $Y\subseteq Z\subseteq X$?

Edit: As Ilya Bogdanov pointed out, the claim is wrong in general. But, what if $X$ is the interior of a finite union of polytopes? Any hint will be greatly appreciated.

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    $\begingroup$ Consider a line $y=0$ inside the set $x^2y^2<1$. There also exist bounded examples. $\endgroup$
    – Ilya Bogdanov
    Commented Nov 29, 2019 at 9:17
  • $\begingroup$ @Ilya Bogdanov Thanks! You mentioned that there exist bounded examples. May I have a counterexample when $X$ and $Y$ are both bounded? $\endgroup$
    – Lemma1
    Commented Nov 29, 2019 at 9:23

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A bounded example: $$ X=\left\{(x,y)\colon \sqrt{|x|}+\sqrt{|y|}<1\right\}, \quad Y=\{(x,0)\colon |x|<1\} $$

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  • $\begingroup$ Thanks! I guess I will need more restrictions. Currently I may restrict $X$ to be the interior of a finite union of polytopes. Understanding that I may already have asked too much, it would be great if you can give me some hint on that... $\endgroup$
    – Lemma1
    Commented Nov 29, 2019 at 9:39
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On the positive side: if $Y$ is a compact convex subset of an open set $X$ in a locally convex TVS, then there is an open convex neighborhood $U$ of the origin so small that the open convex set $Z:=Y+U$ is included in $X$.

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Yet a bit more on the positive side: If $Y$ is a convex subset of an open convex subset $X$ of a normed space, then the set $Z:=\bigcup_{y\in Y}B_y^X$ will do, where $B_y^X$ is the largest open ball centered at $y$ that is contained in $X$.

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