# If $A$ and $B$ are compact, then so is $A+B$.

This is an exercise in Chapter 1 from Rudin's Functional Analysis.

Prove the following:

Let $X$ be a topological vector space. If $A$ and $B$ are compact subsets of $X$, so is $A+B$.

My guess: Let $\cup V_{\alpha}$ be an open covering of $A+B$, if we can somehow split each $V_{\alpha}$ into two parts $$V_{\alpha}=W_{\alpha}+U_{\alpha}$$ with $$\cup W_{\alpha}\supset A, \cup U_{\alpha}\supset B$$ then we can easily pass the compactness of $A$ and $B$ to $A+B$.

However, I cannot find such a way to split $V_{\alpha}$. I admit this is the only nontrivial part of this problem.

Thanks!

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The sum is a continuous operation. The image $A + B$ of the compact set $A \times B$ is therefore compact. – André Caldas Aug 28 '12 at 2:55
@AndréCaldas Thanks! Here is a related problem. Find an example to show that sum of closed sets may fail to be closed. Could you have a look at this? – Hui Yu Aug 28 '12 at 2:59
@HuiYu Let $A$ be the graph of $1/x$, and $B$ the $y$-axis. – Alex Becker Aug 28 '12 at 3:08
In $\Bbb R^2$ let $$H=\left\{\left\langle x,\frac1x\right\rangle:x>0\right\}$$ and $$K=\left\{\left\langle x,-\frac1x\right\rangle:x>0\right\}\;.$$ Then $H+K\supseteq\{\langle x,0\rangle:x>0\}$, so $\langle 0,0\rangle$ is a limit point of $H+K$ that is not in $H+K$. – Brian M. Scott Aug 28 '12 at 3:08
@AlexBecker Thanks! – Hui Yu Aug 28 '12 at 3:10