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I am looking for a 'direct' way to show the following statement:

Problem: Let $V$ be a normed vectorspace, show that if $A$ is compact and $B$ is closed then $A+B:= \lbrace a+b \mid a \in A, b \in B \rbrace$ is closed

My approach: I am trying to show this statement directly i.e. premise $\implies$ conclusion. This was ridiculously easy for $A,B$ compact $\implies A+B$ compact. I don't seem to have much luck with this one though.

Let $(x_n)$ be a convergent sequence in $A+B$ such that $x_n \to x$. If I manage to show that $x \in A+B$ then I am done.

It is true that $x_n = a_n + b_n$ for all $n \in \mathbb{N}$ and $(a_n,b_n) \in A \times B$. Luckily I have that $A$ is compact, that means that there exists a subsequence $(a_{n_k})$ of $(a_n)$ such that $a_{n_k} \to a \in A$ thanks to compact $\equiv $ bound & closed and Bolzano-Weierstrass.

Now I fail to make any statement about the sequence $(b_n) \in B$. I know that if $(b_n)$ converges, then thanks to $B$ being closed it follows that the limiting point would also be a member of $B$. However it feels rather vague to me to say that $(b_n)$ converges because $(x_n) \in A + B$ converges.

Could someone give me some nudges in the right direction? Or do I need to forfeit the direct approach and try to come up with a contradiction if I assume $x \notin A+B$? This seems to be much harder.

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  • $\begingroup$ Correct me if I'm wrong, but I don't think that B-W and Heine-Borel hold in arbitrary normed vector spaces. $\endgroup$ – Mario Carneiro Feb 27 '15 at 15:06
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Hint: You know that both $(x_{n_k})$ and $(a_{n_k})$ converges. Then what can you say about the convergency of $(b_{n_k})=(x_{n_k}-a_{n_k})$? And then think about the limit $b$ of $(b_{n_k})$ along with the $a$ you already got.

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  • $\begingroup$ If I interprete your hint correctly then I would have $b_{n_k} \to x-a=b \in B \implies x = a+b \in (A+B)$. The subsequence $(b_{n_k})$ converges because $(x_{n_k})$ and $(a_{n_k})$ converge. $\endgroup$ – Spaced Feb 27 '15 at 15:15
  • $\begingroup$ @Spaced Yes, you're right. $\endgroup$ – Willard Zhan Feb 27 '15 at 15:18
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I would prove this one directly via the topological definition of closed, i.e. complement is open. Let $x\notin A+B$. Then for every $a\in A$, the set $a+B$ is closed, so $d(x,a+B)>0$ and hence there is a neighborhood $U=B(\frac{d(x,a+B)}2,a)$ of $a$ such that $d(x,U+B)\ge \frac{d(x,a+B)}2>0$. Then since $A$ is compact, we can pick a finite subcover $\{U_k\}$ of this, and then $d(x,A+B)\ge\min_k d(x,U_k+B)$, so there is a neighborhood of $x$ disjoint from $\bigcup_k(U_k+B)\supseteq A+B$.

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