Let $A$ , $B$ be open subsets of $\mathbb{R}^n$ . Is $A+B$ a open subset of $\mathbb{R}^n$? Same question replacing open sets by compact sets. I know that for closed sets the result is not true.
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thanks Moreland . each a+B is open and arbitrary union of open sets is open – lavy Jul 08 '12 at 06:02
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1The compact question can be answered by reading http://math.stackexchange.com/questions/80974/closed-sum-of-sets – Gerry Myerson Jul 08 '12 at 06:07
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For $A, B$ open, I think the quickest way is to note that $A + B = \bigcup_{a \in A} (a + B)$. If the sets are instead compact, then note that addition is a continuous map $\mathbf R^n \times \mathbf R^n \to \mathbf R^n$, and that the product of two compact spaces is compact.
Dylan Moreland
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For the first question, yes. Suppose $a+b\in A+B$, with $a\in A$ and $b\in B$. we have some $r>0$ such that $B(a,r)\subset A$. If $x\in B(a+b,r)$ then $x-b\in B(a,r)\subset A$, so $x=(x-b)+b\in A+B$. Thus $B(a+b,r)\subset A+B$, so $A+B$ is open.
For the first question, yes as well. Suppose $A,B$ are compact, so any sequence has a convergent sub-sequence. Let $(a_n+b_n)$ be a sequence in $A+B$, and let $(a_{n_k})$ be a convergent sub-sequence of $(a_n)$ in $A$ and $(b_{n_{k_j}})$ ($b$-sub-$n$-sub-$k$-sub-$j$) be a convergent sub-sequence of $(b_{n_k})$ in $B$. Then $$(a_{n_{k_j}}+b_{n_{k_j}})$$ is a convergent sub-sequence of $A+B$. Thus $A+B$ is compact.
Alex Becker
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Your proof for the first result is so simple. The best I've seen. – QuantumEyedea Oct 21 '15 at 02:38