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Let $A$ and $B$ be subsets of $\mathbb{R}^n$ (where $\mathbb{R}^n$ is Euclidean n-space). Define $A + B = \{ x + y : x \in A , y \in B \}.$ Now If $A$ and $B$ are closed sets, is $A+B$ also a closed set?

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3 Answers 3

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Not always. Consider in $\mathbb{R}$: $$A = \mathbb{Z}, \quad B = \left\{ n + \frac{1}{n} : n \geqslant 2 \right\}$$ so $\frac{1}{n} \in A + B$ but $0 \not \in A+B$.

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  • $\begingroup$ You beat me to it $\ddot{\frown}$! $\endgroup$
    – Rob Arthan
    Apr 12, 2015 at 14:58
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    $\begingroup$ awesome ! thank you so much. :) $\endgroup$
    – Error 404
    Apr 12, 2015 at 15:15
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Take $n = 2$, take $A$ to be the $y$-axis and take $B$ to be the positive quadrant of the hyperbola $y = \frac{1}x$. Then $A$ and $B$ are both closed, but $A + B$ is the set of $(x, y)$ such that $x > 0$, which is not closed.

If $A$ and $B$ are both closed and one of them is compact, then $A + B$ is closed. See Closed sum of sets for a proof.

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  • $\begingroup$ In your edit, you have written that "A or B are both closed..." I did not get that. should it be "A and B..." ? $\endgroup$
    – Error 404
    Apr 12, 2015 at 15:30
  • $\begingroup$ Yes - thanks for pointing out the typo. Now fixed. $\endgroup$
    – Rob Arthan
    Apr 12, 2015 at 15:37
  • $\begingroup$ Great. now it's fine. $\endgroup$
    – Error 404
    Apr 13, 2015 at 13:22
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\begin{align} A + B &= \{ a + b \mid a \in A, b \in B\} \\ &= \bigcup_{b \in B} \{ a + b \mid a \in A\} \\ &= \bigcup_{b \in B} (A+b)\,. \end{align} If one of the sets $A,B$ is finite, then $A+B$ is closed, because $A+b$ is closed (I think this holds in all normed spaces). But since infinite unions of closed sets are not closed in general, it seems that there should be plenty of counterexamples (two already posted here).

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  • $\begingroup$ thanks for giving your point of view. $\endgroup$
    – Error 404
    Apr 12, 2015 at 15:15
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    $\begingroup$ If one of the sets is finite, it is bounded, and closed bounded sets in $\mathbb R^n$ are compact. So the rule "finite + closed = closed" is a special case of the rule "compact + closed = closed" mentioned by Rob Arthan. $\endgroup$
    – celtschk
    Apr 12, 2015 at 15:31
  • $\begingroup$ Oh thanks I will see that in detail in the link given by him. :) $\endgroup$
    – Error 404
    Apr 13, 2015 at 13:22

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