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I know the proof that If A is compact and B closed then A+B is closed but would like to have an example where both are closed but not A+B.I am not able to figure out.

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marked as duplicate by Marshmallow, G. Sassatelli, Jonas, Robert Soupe, choco_addicted Apr 11 at 1:08

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By $A+B$, do you mean the union, or are you taking a sum in some sort of topological group like $\mathbb{R}^n$? – Brett Frankel Apr 26 '12 at 21:40
What do you mean by '+'? The union? Or... are you working in a topological abelian group, so that '+' means the set of all sums of points from A and B? – Hurkyl Apr 26 '12 at 21:40
I am not saying about union, $A+B=\{x+y:x\in A, y\in B\}$ – Un Chien Andalou Apr 26 '12 at 21:42
Any more information about the space would be helpful. Are you working in $\mathbb{R}^n$ or a more general topological group (in which case it would be nice to know what axioms you're working with)? – Brett Frankel Apr 26 '12 at 21:43
See also this questions:… – Martin Sleziak Apr 27 '12 at 5:19
up vote 14 down vote accepted

Assuming $A+B=\{a+b\mid a\in A, b\in B\}$:

$A=\{\,1,2,3,\ldots\,\}$ and $B=\{ \,-1 +{1\over2}, -2 +{1\over3} ,-3+{1\over4},\ldots\,\}$. The sum contains $\{\,{1\over2},{1\over3},{1\over4},\ldots\,\}$ but not its limit point $0$.

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aha! nice! thank you David – Un Chien Andalou Apr 26 '12 at 21:43

Let $A=\{(x,y):y\ge e^x\}$ and $B=\{(x,0)\}$. Then $A+B=\{(x,y):y>0\}$.

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Is it also enough if $A=\{(x,y):y=e^x\}$? @Julián Aguirre – simon Oct 7 '13 at 13:56
@user47709 Yes, it is. – Julián Aguirre Oct 7 '13 at 15:07

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