# Convex hull of the union of infinitely many convex compact, uniformly bounded, sets (in ${\mathbb R}^2$).

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?

• Take the union of $C_\alpha=\{\alpha\}$ for $\alpha\in(0,1)$. – logarithm May 14 '19 at 1:34
• @logarithm How silly of me not to have thought of that. Thanks a lot! – Ruben May 15 '19 at 0:29

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 $$\bigcup_{(x,y)\in C\times C} D_{x,y}=C$$