Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
thanks Moreland . each a+B is open and arbitrary union of open sets is open – lavy Jul 8 '12 at 6:02
The compact question can be answered by reading – Gerry Myerson Jul 8 '12 at 6:07

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.

share|cite|improve this answer

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.

share|cite|improve this answer
Your proof for the first result is so simple. The best I've seen. – Greg.Paul Oct 21 '15 at 2:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.