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Let $X=(X,|\cdot|)$ be a Banach space and let $A,B \subset X$. If $A$ is a compact set and $B$ is bounded set, then $$A+B=\{a+b \; ; \; a \in A \: \text{e}\: b \in B\}$$ is a compact set? I tried to prove it as follows: let $(x_n)_{n \in \mathbb{N}} \subset A+B$, then $$x_n=y_n+z_n,\: \forall \; n \in \mathbb{N}, $$ for some, $(y_n)_{n \in \mathbb{N}} \subset A$ and $(z_n)_{n \in \mathbb{N}} \subset B$. Since $A$ is bounded, . Moreover, since $B$ is compact, exist $(z_{n_k})_{k \in \mathbb{N}}$ such that $z_{n_{k}} \longrightarrow z$, for some $z \in X$.

But I do not know what I can conclude with such information.

Is this true or not? If not, is there any additional hypothesis that makes it true?

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    $\begingroup$ Take $A = [0,1]$ and $B = [0,1).$ Is $A+B = [0,2)$ compact? $\endgroup$ – MSDG Aug 30 '18 at 13:25
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    $\begingroup$ How about taking $B$ to be compact as well? $\endgroup$ – MSDG Aug 30 '18 at 13:29
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    $\begingroup$ $A=\{0\}$ is compact. If $B$ is bounded, is $\{0\}+B$ compact? $\endgroup$ – bof Aug 30 '18 at 13:30
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    $\begingroup$ If $A,B$ are both compact, then $A+B$ is a continuous image of the compact set $A\times B$. $\endgroup$ – bof Aug 30 '18 at 13:32
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    $\begingroup$ If $A$ is nonempty, then it's a necessary condition for $A + B$ to be compact that $B$ is relatively compact. So you can't get much beyond "compact + compact is compact". $\endgroup$ – Daniel Fischer Aug 30 '18 at 15:16
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If $I$ and $J$ are bounded intervals in $\mathbb{R}$, then $I+J$ is also an interval and $\sup (I+J)=\sup(I)+\sup(J)$ and $\inf(I+J)=\inf(I)+\inf(J)$. It follows that $I+J$ is compact if and only if both $I$ and $J$ are compact.

In general it is possible for $A+B$ to be compact even if $A$ and/or $B$ is not compact:

Let $A=\{1/n: n\geq 1\} \cup \{0\}$ and $B=\{-1/n: n\geq 1\}$. Then $A$ is compact, $B$ is not compact, but $A+B$ is compact.

Let $A=\{1/n: n\geq 1\}$ and $B=\{-1/n: n\geq 1\}$ then $A$ is not compact, $B$ is not compact, but $A+B$ is compact.

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