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!

Answer 1

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.