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 \begin{equation} V_{\alpha}=W_{\alpha}+U_{\alpha} \end{equation} with \begin{equation} \cup W_{\alpha}\supset A, \cup U_{\alpha}\supset B \end{equation} 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.

Any hint would be helpful.


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    $\begingroup$ The sum is a continuous operation. The image $A + B$ of the compact set $A \times B$ is therefore compact. $\endgroup$ – André Caldas Aug 28 '12 at 2:55
  • $\begingroup$ @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? $\endgroup$ – Hui Yu Aug 28 '12 at 2:59
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    $\begingroup$ @HuiYu Let $A$ be the graph of $1/x$, and $B$ the $y$-axis. $\endgroup$ – Alex Becker Aug 28 '12 at 3:08
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    $\begingroup$ 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 $H+K$. $\endgroup$ – Brian M. Scott Aug 28 '12 at 3:08
  • $\begingroup$ @AlexBecker Thanks! $\endgroup$ – Hui Yu Aug 28 '12 at 3:10

Posting André's comment for the sake of having an answer with positive score (to prevent future bumps):

The sum is a continuous operation. The image A+B of the compact set A×B is therefore compact.

  • $\begingroup$ Sorry, Austin... I should have posted a proper answer... $\endgroup$ – André Caldas Aug 31 '12 at 15:27
  • $\begingroup$ @AndréCaldas It's no problem. If you want to go ahead and do so (and receive the rep you deserve), I am happy to delete this one. $\endgroup$ – Austin Mohr Aug 31 '12 at 16:50
  • $\begingroup$ Thanks, Austin. But I guess I will wait for André for a few more days before I accept your answer. $\endgroup$ – Hui Yu Sep 2 '12 at 1:57
  • $\begingroup$ @HuiYu Even if you accept this one and André posts, you can change your accepted answer. $\endgroup$ – Austin Mohr Sep 2 '12 at 2:45
  • $\begingroup$ Austin and @HuiYu, it is fine by me if you accept Austin's answer. I would just suggest that the role played by the product topology be more emphasized and maybe the notation could be enhanced... :-) $\endgroup$ – André Caldas Sep 4 '12 at 0:02

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