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Consider the following sets in ${\mathbb R}^2$: for each $\alpha \geq 0$, define as $C_\alpha$ the closed segment from the point $(\alpha,0)$ to the point $(0,1)$. Each set is compact and convex, but clearly the convex hull of the union $\bigcup_{\alpha \in {\mathbb R}_{\geq 0}} C_\alpha$ (which in this case is already convex) is not closed (nor bounded).

Question: can something like this occur if the sets are uniformly bounded? That is, assume that for each $\alpha \in {\mathbb R}$ we have compact convex sets $D_\alpha$ such that $D_\alpha \subseteq B$, for some fixed ball $B$ in ${\mathbb R}^2$. Is the convex hull of the union $\bigcup_{\alpha \in {\mathbb R}} D_\alpha$ closed?

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    $\begingroup$ Take the union of $C_\alpha=\{\alpha\}$ for $\alpha\in(0,1)$. $\endgroup$ – logarithm May 14 '19 at 1:34
  • $\begingroup$ @logarithm How silly of me not to have thought of that. Thanks a lot! $\endgroup$ – Ruben May 15 '19 at 0:29
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You can do the same thing with a half circle instead of the $x$-axis. Consider $D_\theta$ the closed segment joining $(0,0)$ to $(\cos\theta,\sin\theta)$, for $-\frac\pi2<\theta<\frac\pi2$. Then $$\bigcup\limits_{\theta\in(-\frac\pi2,\frac\pi2)} D_\theta=\{(x,y)\in\Bbb R^2\,:\, (x^2+y^2\le 1\land x>0)\lor x=y=0\}$$

More generally, consider your favourite bounded and not closed convex set $C$, and take $\{D_{x,y}\,:\, (x,y)\in C\times C\}$ where $D_{x,y}=\{tx+(1-t)y\,:\, t\in[0,1]\}$ is the closed segment joining $x$ to $y$. Then $$\bigcup_{(x,y)\in C\times C} D_{x,y}=C$$

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  • $\begingroup$ Thanks a lot! Those are great examples. $\endgroup$ – Ruben May 15 '19 at 0:31

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