If two sets $A,B$ in $\mathbb{R^n}$ are closed. Will $A+B$ then be closed or open? In my head it makes sense if it is closed but I can't prove it on paper.
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1I hope this question will be helpful to you. – GNUSupporter 8964民主女神 地下教會 Dec 02 '15 at 12:49
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Just for the record: even if $A+B$ turned out to be closed: "not open" does not mean "closed" (that is: $A+B$ might be neither closed nor open). There is also no reason, why $A+\emptyset$ should be open, if $A$ is closed. – Stefan Perko Dec 02 '15 at 12:50
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Let $A= \cup _{n\in \bf N, n\geq 2} \{n+1/n\}$, and $B= \cup _{n\in \bf N, n\geq 2} \{-n\}$
Then $(A+B ) \cap [0,1]= \{ 1/n, n>2\}$, hence $A+B$ is not closed.
Thomas
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@k.dkhk $A$, $B$, and $A+B$ are each unbounded, and thus not compact. (For example, $100.01\in A$, $-100\in B$, and $(200.005)+(-100)=100.005\in A+B$.) – Akiva Weinberger Dec 02 '15 at 13:37
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If A is $compact$ and B is closed, $A+B$ is closed. Indeed if $a_n+b_n$ converge to $l$, up to a subsequence we may assume that $a_n$ converge to $a$. Then $b_n$ converge to $l-a$; therefore $l-a\in B$ (as $B$ is closed) and $l=a+b\in A+B$ – Thomas Dec 02 '15 at 19:16