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

Question

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.

Thanks!

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 02:55
@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 02:59
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 03:08
@AlexBecker: I always thought that $y$-axis was an open set. Is it clopen? — Libertron, Sep 02 '13 at 23:54

score 8 · Accepted Answer · answered Aug 31 '12 at 14:06

8

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.

answered Aug 31 '12 at 14:06

Austin Mohr

25,662

Sorry, Austin... I should have posted a proper answer... – André Caldas Aug 31 '12 at 15:27
@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. – Austin Mohr Aug 31 '12 at 16:50
Thanks, Austin. But I guess I will wait for André for a few more days before I accept your answer. – Hui Yu Sep 02 '12 at 01:57
@HuiYu Even if you accept this one and André posts, you can change your accepted answer. – Austin Mohr Sep 02 '12 at 02:45
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... :-) – André Caldas Sep 04 '12 at 00:02

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

1 Answers1