how can I show that $\overline A + \overline B = \overline{A+B}$

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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.

What does $\bar{A}$ denote? Does it denote closure (or) complement? And I assume $C+D$ stands for ${x+y: x \in C \text{ and }y \in D}$. — , Nov 23 '12 at 00:01
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 00:14
How about using the characterization of closure in terms of sequences? — GEdgar, Nov 23 '12 at 00:20
(Brian's answer in the proposed duplicate answers this question, although suddenly I'm not too sure anymore) — Asaf Karagila, Nov 23 '12 at 00:23
@GEdgar I doubt if the OP knows what he is doing like he mentioned in the question. — , Nov 23 '12 at 00:24

score 0 · Answer 1 · edited Apr 13 '17 at 12:20

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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.

edited Apr 13 '17 at 12:20

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answered Nov 23 '12 at 00:23

anonymous

3,090

Ah, that should be the real duplicate. Not what I had proposed. – Asaf Karagila Nov 23 '12 at 00:28
@amWhy: commenting is only possible when reputation is $\geq50$. (and thanks for the dup. link in the closure!) – Asaf Karagila Nov 23 '12 at 00:31

how can I show that $\overline A + \overline B = \overline{A+B}$

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