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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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 $0 \not \in A+B$.
Adayah
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You beat me to it $\ddot{\frown}$! – Rob Arthan Apr 12 '15 at 14:58
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1awesome ! thank you so much. :) – Error 404 Apr 12 '15 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.
Rob Arthan
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In your edit, you have written that "A or B are both closed..." I did not get that. should it be "A and B..." ? – Error 404 Apr 12 '15 at 15:30
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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).
desos
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1If 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. – celtschk Apr 12 '15 at 15:31
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