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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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    @GEdgar I doubt if the OP knows what he is doing like he mentioned in the question.2012-11-23

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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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    @amWhy: commenting is only possible when reputation is $\geq50$. (and thanks for the dup. link in the closure!)2012-11-23