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Example where closure of $A+B$ is different from sum of closures of $A$ and $B$
need one counter example for sum of two closed set need not be closed

Given $A$ and $B$ two non empty set in $\mathbb R$ with $A$ bounded how can I show that $$\overline A + \overline B = \overline{A+B}$$

I have no idea how to approach this question.

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What does $\bar{A}$ denote? Does it denote closure (or) complement? And I assume $C+D$ stands for $\{x+y: x \in C \text{ and }y \in D\}$. –  user17762 Nov 23 '12 at 0:01
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Not every question about sets has to do with set theory. –  Asaf Karagila Nov 23 '12 at 0:03
    
@Marvis $\overline A$ is the closure –  Prince Nov 23 '12 at 0:10
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Equality of sets can be viewed as two inclusions. Can you do one of the inclusions: $\overline A + \overline B \subseteq \overline{A+B}$ or $\overline A + \overline B \supseteq \overline{A+B}$ ?? –  GEdgar Nov 23 '12 at 0:14
    
@GEdgar no, i can't –  Prince Nov 23 '12 at 0:16
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marked as duplicate by Asaf Karagila, amWhy, Marvis, Brian M. Scott, martini Nov 23 '12 at 6:28

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1 Answer

It's not true. need one counter example for sum of two closed set need not be closed

The above link provides an example of sets A and B which are closed such that A+B is not closed. Using the above, we'd have that $\bar{A} = A$, and $\bar{B} = B$, but since $A +B$ is not closed, the proposition fails.

Whoops. Beaten to the punch.

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Ah, that should be the real duplicate. Not what I had proposed. –  Asaf Karagila Nov 23 '12 at 0:28
    
@amWhy: commenting is only possible when reputation is $\geq50$. (and thanks for the dup. link in the closure!) –  Asaf Karagila Nov 23 '12 at 0:31
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