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Possible Duplicate:
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 $\{x+y: x \in C \text{ and }y \in D\}$.2012-11-23
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    Not every question about sets has to do with set theory.2012-11-23
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    @Marvis $\overline A$ is the closure2012-11-23
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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 $\overline A + \overline B \supseteq \overline{A+B}$ ??2012-11-23
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    @GEdgar no, i can't2012-11-23
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    How about using the characterization of closure in terms of sequences?2012-11-23
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    (Brian's answer in the proposed duplicate answers this question, although suddenly I'm not too sure anymore)2012-11-23
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    @GEdgar I doubt if the OP knows what he is doing like he mentioned in the question.2012-11-23

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