If $A,B\subset\mathbb{R}^n$ are closed. Let $$M=\{p\in\mathbb{R}^n\,:\,\exists\,t\in[0,1],a\in A,b\in B\;\mathrm{s.t.}\;p=t\cdot a+(1-t)\cdot b\}$$ Is $M$ closed? Why?
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possible duplicate of Is the convex hull of closed set in $R^{n}$ is closed? – Julian Kuelshammer May 08 '13 at 12:52
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@Julian Kuelshammer: But $M$ here is not exactly the convex hall of anything. Moreover, $M$ doesn't have to be convex here – Dennis Gulko May 08 '13 at 12:57
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@DennisGulko True, but the ideas to construct counterexamples are very similar. – Julian Kuelshammer May 08 '13 at 13:00
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Here is a counterexample. Take $A=\{(1,0)\}$ and $B=\{(0,m)\;|\;m\in Z\}$. Note both sets are closed since all points are isolated. On the other hand, $M$ contains all line segments connecting each point in each set. The slope of these lines approach a vertical line, so we can get arbitrarily close to any point on the line $x=1$, but this line is not in $M$.
abnry
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Ok, thanks. Suppose $A,B$ are compact. Can you give a counterexample for that case too? – Anonymous999 May 08 '13 at 13:41
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1Probably not. In the counterexample I gave it is very important that $B$ is unbounded. If it is bounded, then it is likely I don't have enough wiggle room. I would use sequential compactness, or the Heine Borel theorem, to try and show that the set is closed for compact sets. – abnry May 08 '13 at 13:47
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Suppose $A,B$ are compact. Note that $M$ is the image of the compact set $A \times B \times [0,1]$ under the continuous map $f(a,b,t) = ta+(1-t)b$.
GEdgar
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